TS Inter 2nd Year Botany Study Material Chapter 2 Mineral Nutrition

Telangana TSBIE TS Inter 2nd Year Botany Study Material 2nd Lesson Mineral Nutrition Textbook Questions and Answers.

TS Inter 2nd Year Botany Study Material 2nd Lesson Mineral Nutrition

Very Short Answer Type Questions

Question 1.
Define hydroponics.
Answer:

  1. The technique of growing plants in a specified nutrient solution is known as hydroponics.
  2. Julius Von Sachs (1860) demonstrated this technique for the first time.

Question 2.
How do you categorize a particular essential element as a macro or micronutrient?
Answer:
1. Macronutients :
Elements needed in high quantities and are present is large amounts (in excess of lOmmole Kg-1 of dry matter) in plant tissues.

2. Micronutrients :
Trace elements that are needed in very small amounts and present is less than lOmmole Kg-1 of dry matter in plants.

Question 3.
Give two examples of essential elements that act as activators for enzymes.
Answer:
1. Mg 2+ is an activator of RuBp carboxylase oxygenase.
2. Zn 2+ is an activator of alcohol dehydrogenase.

Question 4.
Name the essential mineral elements that play an important role in photolysis of water.
Answer:
Calcium, Manganese and chlorine are the mineral elements that help is splitting of water molecules to liberate O2 during photosynthesis.

Question 5.
Out of the 17 essential elements which elements are called non-mineral essential elements?
Answer:

  1. Carbon, Hydrogen and Oxygen obtained from CO2 and H2O are called non – mineral essential elements.
  2. These frame work elements are not absorbed from the soil as mineral nutrients. Name two amino acids in which sulphur is present.

TS Inter 2nd Year Botany Study Material Chapter 1 Transport in Plants

Question 6.
Name two amino acids in which sulphur is present.
Answer:

  1. Cysteine and methionine are sulphur containing amino acids.
  2. They help in formation of disulphide bridges and stabilizing the protein structure.

Question 7.
When is an essential element said to be deficient?
Answer:

  1. The concentration of an essential element below which plant growth is retarded is called critical concentration.
  2. The element is said to be deficient when present below the critical concentration.

Question 8.
Name two elements whose symptoms of deficiency first appear in younger leaves.
Answer:

  1. Sulphur and Iron.
  2. Deficiency symptoms of immobile elements first appear in young leaves.

Question 9.
Explain the role of the pink colour pigment in the root nodule of legume plants. What is it called?
Answer:

  1. The pink colour pigment in root nodule of legume plant is Leg-haemoglobin.
  2. Enzyme nitrogenase is highly sensitive to the molecular oxygen and requires anaerobic condition. Leg haemoglobin which is an oxygen scavenger protects enzyme nitrogenase from oxygen.

Question 10.
Excess Mn in soils leads to deficiency of Ca, Mg and Fe. Justify.
Answer:

  1. Manganese competes with iron and Mg for uptake and with magnesium for binding with enzymes. It also inhibits calcium translocation in the shoot apex.
  2. Therefore, excess of manganese induce deficiencies of iron, magnesium and calcium.

Question 11.
What acts as a reservoir of essential elements for plants? By what process is it formed?
Answer:

  1. Soil acts as a reservoir of essential elements for plants.
  2. It is formed due to weathering and breakdown of rocks.

TS Inter 2nd Year Botany Study Material Chapter 1 Transport in Plants

Question 12.
Which element is regarded as the 17th essential element? Name a disease caused by its deficiency.
Answer:

  1. Nickel is regarded as the 17th essential element.
  2. Its deficiency causes mouse ear in pecan.

Question 13.
Nitrogen fixation is shown by prokaryotes only. Why not by eukaryotes?
Answer:

  1. Nitrogenase enzyme, which is capable of nitrogen reduction is present exclusively in prokaryotes.
  2. In eukaryotes, nitrogenase enzyme is absent and so they cannot fix Nitrogen.

Question 14.
Give an example for each of the aerobic and anaerobic nitrogen fixing prokaryotes.
Answer:

  1. Aerobic nitrogen fixing prokaryotes. Eg: Azotobacter, Beijernickia.
  2. Anaerobic nitrogen fixing prokaryotes. Eg: Rhodospirillum.

Question 15.
Non-legume plants also form root nodules. Justify.
Answer:

  1. The microbe Frankia, produces nitrogen fixing nodules on the roots of non-leguminous plants Eg: Alnus.
  2. Frankia is free living in soil, but as a symbiont can fix atmospheric nitrogen.

Question 16.
Name the essential elements present in nitrogenase enzyme. What type of essential elements are they?
Answer:

  1. Nitrogenase enzyme consists of 2 essential elements Molybdenum and Iron.
  2. They are micronutrients.

Question 17.
Write the balanced equation of nitrogen fixation.
Answer:
N2 + 8H+ + 8e + 16 ATP → 2NH3 + H2 + 16 ADP + 16 Pi

Question 18.
How many ATPs of energy is required to fix one molecule of atmospheric nitrogen by biological mode? What is the source of that energy?
Answer:

  1. 16 ATP molecules, are required to fix one N2 molecule and to produce 2 NH3 molecules.
  2. Source of that energy is obtained from the respiration of the host cells.

Question 19.
Why are amides transported through xylem?
Answer:

  1. Amides contain more nitrogen than amino acids and are highly toxic to living cells.
  2. Hence, they are transported to other parts of the plant through non living xylem vessels. Name any two essential elements and the deficiency diseases caused by them.

Question 20.
Name any two essential elements and the deficiency diseases caused by them.
Answer:

  1. Zinc deficiency – mottled leaf in citrus.
  2. Molybdenum deficiency – whip tail in cauliflower.

Short Answer Type Questions

Question 1.
“All elements that are present in a plant need not be essential for its survival.” Justify.
Answer:
Most of the elements present in the soil enter plants through roots. All the elements that are present in a plant need not be essential for its survival. The criteria for essentiality of elements are given below :

  1. The element must be absolutely necessary for supporting normal growth and reproduction. In the absence of the element the plant do not complete their life cycle or set the seeds.
  2. The requirement of the element must be specific and not replaceable by another element. In other words, deficiency of any one element cannot be met by supplying some other element.
  3. The element must be directly involved in the metabolism of the plant.

Question 2.
Name at least five different deficiency symptoms in plants. Describe them and correlate them with the concerned mineral deficiency.
Answer:
Deficiency symptoms in plants are chlorosis, necrosis, inhibition of cell division, delay flowering.

Chlorosis :
Chlorosis is the loss of chlorophyll leading to yellowing in leaves. This symptom is caused by the deficiency of elements N, K, Mg, S, Fe, Mn, Zn and Mo.

Necrosis :
It is the death of tissue, particularly leaf tissue. It is due to the deficiency of Ca, Mg, Cu, K.

Inhibition of cell division :
Cell division stops. It is due to the deficiency of lack or low level of N, K, S, Mo.

Delay flowering :
Flowering is delayed due to low concentration of N, S, Mo.

Mottled leaf in citrus :
It is due to deficiency of Zn,

Question 3.
Explain the steps involved in the formation of root nodule. [Mar. 2019, 18, 17, ’14]
Answer:
Steps involved in the formation of root nodule :

  1. Rhizobia attracted by the sugars, amino acids etc., released by the host legume, multiply and colonise the surroundings of roots and get attached to epidermal and root hair cells.
  2. The root hairs curl and the bacteria invade the root hair.
  3. An infection thread is produced, carrying the bacteria into the cortex of the root.
  4. Bacteria initiate nodule formation in the cortex of the root.
  5. Then the bacteria released from the thread into the cortical cells of the host stimulate the host cells to divide. Thus leads to the differentiation of specialised nitrogen fixing cells.
  6. The nodule thus formed establishes a direct vascular connection with the host for the exchange of nutrients.

TS Inter 2nd Year Botany Study Material Chapter 2 Mineral Nutrition 1
Development of root nodules in soyabean :
(a) Rhizobium bacteria contact a susceptible root hair, divide near it. (b) Successful infection of the root hair causes it to curl, (c) Infected thread carries the bacteria to the inner cortex. The bacteria get modified into rod-shaped bacteroids and cause inner cortical and pericycle cells to divide. Division and growth of cortical and pericycle cells lead to nodule formation, (d) A mature nodule is complete with vascular tissues continuous with those of the root.

TS Inter 2nd Year Botany Study Material Chapter 1 Transport in Plants

Question 4.
Some angiospermic plants have adapted to absorb molecular nitrogen from atmosphere. Explain, citing two examples.
Answer:

  1. Plants cannot use atmospheric nitrogen directly. But some of the plants in association with N2 – fixing bacteria, especially roots of legumes, can fix atmospheric nitrogen.
  2. Leguminous plants (Eg : alfalfa, sweet clover, sweet pea etc.,) are associated with Rhizobium bacteria in nodular roots. Non-leguminous plants (Eg: Alnus) are associated with Frankia in nodula roots.
  3. Both Rhizobium and Frankia are free – living in soil but as symbionts can fix atmospheric nitrogen.
  4. Molecular nitrogen (N2) is available abundant in air. Only prokaryotic species are capable of nitrogen fixation.
    Reduction of nitrogen to ammonia by living organisms is called biological nitrogen fixation.
  5. The enzyme nitrogenase is capable of nitrogen reduction. It is present only in prokaryotes. Such microbes are called N2 – fixers.
    TS Inter 2nd Year Botany Study Material Chapter 2 Mineral Nutrition 2

Question 5.
Write in brief how plants synthesize amino acids.
Answer:
NH4+ is used to synthesize amino acids in plants. There are two main ways in which this can take place :
1) Reductive amination
2) Transamination.

1) Reductive amination :
In these processes, ammonia reacts with α-ketoglutaric acid and forms glutamic acid as indicated in the equation given below
TS Inter 2nd Year Botany Study Material Chapter 2 Mineral Nutrition 3

2) Transamination :
It involves the transfer of an amino group from an amino acid to the keto group of a keto acid. Glutamic acid is the main amino acid from which the transfer of NH2, the amino group, takes place and other amino acids are formed through transamination. The enzyme transaminase catalyses all such reactions. For example
TS Inter 2nd Year Botany Study Material Chapter 2 Mineral Nutrition 4

Question 6.
What will happen if a healthy plant is supplied with excess essential elements? Explain.
Answer:

  1. Excess essential element may inhibit the uptake of another element.
  2. For example, the prominent symptom of manganese toxicity is the appearance of brown spots surrounded by chlorotic veins.
  3. Manganese competes with iron and magnesium for uptake and with magnesium for binding with enzymes.
  4. Manganese also inhibits calcium translocation in the shoot apex.
  5. Therefore, excess of manganese induce deficiencies of iron, magnesium and calcium.

Question 7.
Explain in brief how plants absorb essential elements.
Answer:
1. The process of absorption of elements can be demarcated into two main phases. In the first phase, there is an initial rapid up take of ions into the free space or outer space of cells – the apoplast. It is a passive process. In the second phase of uptake, the ions are taken in slowly into the ‘inner space’ – the symplast of the cells.

2. The movement of ions into the apoplast along the concentration gradient is passive process. The movement of ions to and from the symplast against the concentration gradient requires the expenditure of metabolic energy. It is an active process.

TS Inter 2nd Year Botany Study Material Chapter 1 Transport in Plants

Question 8.
Nitrogen is fixed into the soil not only by biological processes. Elaborate. [May 2014]
Answer:
Nitrogen is fixed into the soil not only by biological processes, it is also done by abiological processes.

a) In nature, due to thunders and lightening dinitrogen is converted into nitrogen oxides.
N2 + O2 → 2NO (nitric oxide)
2NO + O2 → 2NO2 (nitrogen dioxide)
2NO2 + H2O → HNO2 / HNO3 (nitrous/nitric acid)
HNO3 + Ca/K salts → Ca/K nitrates
When nitrous/nitric acid reach the soil reacts with the alkali radical to form nitrates.

b) Industrial combustions, forest fires, automobile exhausts and power-generating stations are also the sources of atmospheric nitrogen oxides.

Long Answer Type Questions

Question 1.
Explain the nitrogen cycle, giving relevant examples. [Mar. 2020]
Answer:
The cyclic movement of nitrogen from the atmosphere to soil and from soil back into the atmosphere through plants, animals and micro-organisms is termed as nitrogen cycle. Nitrogen cycle involves five steps :

  1. Nitrogen fixation
  2. Nitrogen assimilation
  3. Ammonification
  4. Nitrification
  5. Denitrification

1) Nitrogen fixation :
The process of conversion of molecular nitrogen (N2) to ammonia or nitrogen oxides, nitrites and nitrates is termed as nitrogen – fixation. It occurs both by biological and physical method.

Biological method :
Conversion of molecular nitrogen into ammonia by prokaryotes is called biological methed.
TS Inter 2nd Year Botany Study Material Chapter 2 Mineral Nutrition 5
Eg : Free – living nitrogen – fixing aerobic microbes – Azotobacter – Beijernickia.
Free living nitrogen – fixing anaerobic microbes – Rhodospirillum
Cyanobacteria (blue green algae) – Nostoc & Anabaena
Symbiotic bacteria – Rhizobium (roots of leguminous plant)
Symbiotic bacteria – Frankia (roots of non – leguminous plant)

Physical or abiological method :
In nature lightning and ultraviolet radiation provide enough energy to convert nitrogen to nitrogen oxides (NO, NO2, NO3). Industrial combustions, forest fires, automobile exhausts and power – generating stations are also sources of atmospheric nitrogen oxides.
N2 + O2 → 2NO
2NO + O2 → 2NO2
2NO2 + H2O → HNO2 + HNO3
HNO3 + Ca/K salts → Ca or K nitrates

2) Nitrogen assimilation :

  1. The process of absorbing nitrates, ammonia to produce organic nitrogen constitutes is called nitrogen assimilation.
  2. Nitrates and ammonia formed in 1st step are absorbed by plants and converted into organic nitrogen constitute like proteins, enzymes, nucleic acid etc.
  3. When plants are eaten by animats, this organic nitrogen is passed into animal body.

3) Ammonification :

  1. Decomposition of organic nitrogen of dead plants and animals into ammonia is called ammonification.
  2. Bacteria responsible for this are called ammonifying bacteria.

4) Nitrification :
The conversion of ammonia into nitrites and nitrates by bacteria is called nitrification. Such bacteria are called nitrifying bacteria (chemo auto trophs). It occurs in two steps.

  1. Ammonia is first oxidised to nitrite by the bacteria Nitrosomonas and Nitrococcus.
    2NH3 + 3O2 → 2NO2 + 2H+ + 2H2O
  2. The nitrite is further oxidised to nitrate with the help of the bacterium Nitrobacter.
    2NO2 + O2 → 2NO3

The nitrate thus formed is absorbed by plants and is transported to the leaves. In leaves, it is reduced to form ammonia that finally forms the amine group of amino acids.

5) Denitrification :
Conversion of nitrates from soil into molecular nitrogen is called denitrification. Denitrification is done by bacteria like Pseudomonas and Thiobacillus.
TS Inter 2nd Year Botany Study Material Chapter 2 Mineral Nutrition 6

Question 2.
Trace the events starting from the coming in contact of Rhizobiurri with a leguminous root till nodule formation. Add a note on the importance of leg haemoglobin.
Answer:
Various stages of nodule formation :

  1. Roots of legume plant secrete sugars, amino acids etc.
  2. Attracted by this, Rhizobium bacteria move to the root. It multiply and colonise the surroundings of roots and get attached to epidermal and root hair cells.
  3. The root – hairs curl and the bacteria invade the root – hair.
  4. An infection thread is produced, carrying the bacteria into the cortex of the root.
  5. Bacteria initiate nodule formation in the cortex of the root.
  6. Then the bacteria released from the thread into the cortical cells of the host stimulate the host cells to divide. Thus leads to the differentiation of specialised nitrogen fixing cells.
  7. The nodule thus formed establishes a direct vascular connection with the host for exchange of nutrients.

TS Inter 2nd Year Botany Study Material Chapter 2 Mineral Nutrition 1
Development of root nodules in soyabean :
(a) Rhizobium bacteria contact a susceptible root hair, divide near it. (b) Successful infection of the root hair causes it to curl, (c) Infected thread carries the bacteria to the inner cortex. The bacteria get modified into rod-shaped bacteroids and cause inner cortical and pericycle cells to divide. Division and growth of cortical and pericycle cells lead to nodule formation, (d) A mature nodule is complete with vascular tissues continuous with those of the root.

The nodule contains all the necessary biochemical components, such as the enzyme nitrogenase and leg-haemoglobin. The enzyme nitrogenase is a Mo-Fe protein and catalyses the conversion of atmospheric nitrogen to ammonia, the first stable product of nitrogen fixation.
N2 + 8H+ + 8e + 16 ATP → 2NH3 + H2 + 16 ADP + 16 Pi

The enzyme nitrogenase is highly sensitive to the molecular oxygen, it requires anaerobic conditions. To protect these enzymes, the nodule contains an oxygen scavenger called leg -haemoglobin. These microbes live as aerobes under free living conditions but during nitrogen – fixing events, they adapt to anaerobic conditions, thus protecting the nitrogenase enzyme.

Intext Question Answers

Question 1.
Who should be credited for initiation of hydroponics?
Answer:
Julius von Sachs

Question 2.
Are all the essential elements required by plants mineral elements? Explain.
Answer:
Yes. Whenever the supply of an essential element becomes limited plant growth is retarded.

Question 3.
Which essential element is needed to activate the enzymes required for CO2 fixation?
Answer:
Magnesium and Manganese.

TS Inter 2nd Year Botany Study Material Chapter 1 Transport in Plants

Question 4.
Name a cation and an anion that maintain osmotic balance in cells.
Answer:
Potassium K+ and Chlorine Cl

Question 5.
Which element is required for the formation of mitotic spindle?
Answer:
Calcium

Question 6.
What is the role of sulphur in plant life?
Answer:
Sulphur is obtained in the form of sulphate SO42- ions. Sulphur is present in two amino acids – cysteine and methionine and is the main constituent of several coenzymes, vitamins (thiamine, biotin, coenzyme A) and Ferredoxin. Sulphur forms disulphide bridges which help in stabilizing the protein structure.

Question 7.
Which microelement is required in more quantity than the other micronutrients ?
Answer:
Iron

Question 8.
Which element is necessary for the synthesis of the chief photosynthetic pigment without being its structural component?
Answer:
Iron and Mg

Question 9.
Which micronutrient necessary for photolysis of water is absorbed by plants in anionic form?
Answer:
Cl

Question 10.
Which enzyme is activated by the 17th essential element?
Answer:
Nickel is the 17th essential element which acts as an activator for Urease.

Question 11.
When is an element considered to be toxic?
Answer:
Any mineral ion concentration in tissues that reduces the dry weight of tissues by about 10 percent is considered to toxic.

Question 12.
Which element when supplied in excess leads to appearance of brown spots surrounded by chlorotic veins?
Answer:
Magnesium

Question 13.
Name an anaerobic, free living, photo-heterotrophic nitrogen fixing bacterium.
Answer:
Rhodospirillum

Question 14.
Which microorganism produces nitrogen-fixing nodules in Alnus?
Answer:
Frankia

TS Inter 2nd Year Botany Study Material Chapter 1 Transport in Plants

Question 15.
When the cross section of root nodules of ground – nut plants are observed under microscope, they appear pinkish. Why?
Answer:
Due to presence of leg haemoglobin or leguminous haemoglobin.

Question 16.
Apart from the cortical cells, which other cells are stimulated to divide by the bacteroids inside the root nodules?
Answer:
Pericycle cells.

Question 17.
What is the ratio of electrons and protons required for the fixation of atmospheric mplecular nitrogen through biological mode?
Answer:
8 protons and 8 electrons i.e., 1:1 ratio.

Question 18.
What acts as oxygen scavenger in the legume-root nodule combination?
Answer:
Leg-haemoglobin

Question 19.
In what way does asparagine differ from aspartic acid?
Answer:
Asparagine is an amide found in plants as a structural part of protein. It is formed from aspartic acid by additional of another amino group.

Question 20.
Through which tissue the amino acids are transported inside the plant body?
Answer:
Amino acids are transported to other parts of the plant body via xylem vessels.

TS Inter 2nd Year Botany Study Material Chapter 1 Transport in Plants

Question 21.
Plants like the Pitcher and Venus fly trap have special nutritional adaptations. Name the essential element and its source for which they show such adaptations.
Answer:

  1. N2 (Nitrogen)
  2. Nitrogen is absorbed from the insect body.

TS Inter 2nd Year Botany Study Material Chapter 1 Transport in Plants

Telangana TSBIE TS Inter 2nd Year Botany Study Material 1st Lesson Transport in Plants Textbook Questions and Answers.

TS Inter 2nd Year Botany Study Material 1st Lesson Transport in Plants

Very Short Answer Type Questions

Question 1.
What are porins? What role do they play in diffusion? [March 2018]
Answer:

  1. Porins are proteins that form huge pores in the outer membranes of the plastids, mitochondria and some bacteria.
  2. Molecules upto the size of small proteins to pass through the channels formed by porins due to facilitated diffusion.

Question 2.
Define water potential. What is the value of water potential of pure water?
Answer:

  1. Water potential is a relative term, which refers to the chemical potential of pure water to that of chemical potential of water in a solution.
  2. Water potential of pure water not under pressure is zero (0) at standard temperature.

Question 3.
Differentiate osmosis from diffusion. [March 2017]
Answer:
1. Osmosis :
Movement of water (solvent) from a region of its higher concentration to a region of lower concentration through a selectively permeable membrane, due to pressure and concentration gradient.

2. Diffusion :
Movement of gases and liquids from a region of thejr higher concentration to the region of lower concentration due to concentration gradient.

Question 4.
Compare transpiration and evaporation.
Answer:

  1. Transpiration is a physiological process, in which water is lost as vapour from the plant surface.
  2. Evaporation is a physical process, in which water is lost from any free surface in the form of vapour.

Question 5.
What are apoplast and symplast? [March 2019]
Answer:

  1. Apoplast is the continuous system (network) of cell walls and intercellular spaces in a plant body. This allows movement of water molecules without crossing the cell membranes,
  2. Symplast is the system of interconnected protoplasts by means of plasmodesmata. This allows movement of water from one cell to another through their cytoplasm.

TS Inter 2nd Year Botany Study Material Chapter 1 Transport in Plants

Question 6.
How does guttation differ from transpiration?
Answer:
1. Guttations :
The loss of water from the leaves of grasses and many herbaceous plants In liquid form.

2. Transpirations :
The loss of water from the plant surface in the form of vapour.

Question 7.
What happens when a pressure greater than the atmospheric pressure is applied to pure water or a solution?
Answer:

  1. Water potential increases, if a pressure greater than atmospheric pressure is applied to pure water or a solution.
  2. It is equivalent to pumping water from one place to another.

Question 8.
Explain what will happen to a plant cell if it is kept in a solution having higher water potential?
Answer:

  1. Endosmosis occurs when a plant cell is placed in a solution having higher water potential (hypotonic solution)
  2. Hence, the cytoplasm build up a pressure against the cell wall, which is called turgor pressure.

Question 9.
What are the physical factors responsible for the ascent Of sap through xylem in plants?
Answer:
1. Cohesion :
Mutual attraction between water molecules.

2. Adhesion :
Attraction of water molecules to polar surfaces (such as the surface of tracheary elements)

3. Transpiration pull :
Driving force for upward movement of water.

Question 10.
Explain why xylem transport is unidirectional while that in phloem is bidirectional.
Answer:

  1. Xylem transport is unidirectional due to one way flow of water in transpiration.
  2. Food in phloem sap can be transported bidirectionally so long as there is a source of sugar and a sink which is able to use, store or remove the sugar.

Question 11.
With reference to transportation within plant cells, what are source and sink?
Answer:

  1. Source is understood to be that part of the plant which prepare food i.e., the leaf.
  2. Sink is the part that needs or store of food i.e., root, stem or fruits.

TS Inter 2nd Year Botany Study Material Chapter 1 Transport in Plants

Question 12.
Does transpiration occur at night? Give an example.
Answer:

  1. Yes, transpiration occurs at night is succulent plants (E.g.: Cacti, Bryophyllum)
  2. Guard cells is those plants become turgid and open at night due to accumulation of organic acids.

Question 13.
Compare the pH of guard cells during the opening and closing of stomata.
Answer:

  1. The pH of guard cells increases during the opening of stomata due to influx of K+ ions and efflux of H+ ions.
  2. The pH of guard cells decreases during closing of stomata due to effux of K+ and influx of H+ ions.

Question 14.
In the wake of transpirational loss, why do the C4 plants are more efficient than C3 plants?
Answer:

  1. C4 plant loses only half as much water as a C3 plant for the same amount of CO2 fixed.
  2. C4 plants are more efficient than C3 plants in minimizing water loss.

Question 15.
What is meant by transport saturation? How does it influence facilitated diffusion?
Answer:

  1. Transport saturation : A protein transporters are being used (saturation) and so transport rate reaches maximum.
  2. During transport saturation all the active sites of carrier proteins are occupied with transporting molecules.

Question 16.
Pressure potential in plant systems can be negative. Elaborate.
Answer:

  1. Pressure potential in xylem is negative and plays a major role is upward water transport in stem.
  2. The tension in water column in the xylem is negative (- Ψp) dute to transpirational pull.

TS Inter 2nd Year Botany Study Material Chapter 1 Transport in Plants

Question 17.
How does ABA bring about the closure of stomata under water stress conditions? [Mar. 2020]
Answer:

  1. Abscisic acid (ABA) is a natural anti-transpirant.
  2. This drives the K+ ions out of guard cells under water stress conditions making them close.

Short Answer Type Questions

Question 1.
Define and explain water potential. [Mar. 2019, ’18]
Answer:
Water potential (Ψw) is a relative term to understand water movement. It represents the difference between chemical potential of water in a system and that of a pure water. Water molecules possess kinetic energy. The greater the concentration of water in a system, the greater is its kinetic energy or water potential. Hence, pure water will have greatest water potential. The water potential of pure water at standard temperatures which are not under any pressure, is taken to be zero. Solute potential (Ψs ) and pressure potential (Ψp) are the two main components that determine water potential.

Solute potential :
All solutions have a lower water potential than pure water. The magnitude of this lowering due to dissolution of a solute is called solute potential denoted by Ψs. It is always negative. The more the solute molecules, the lower is the Ψs. For a solution at atmospheric pressure water potential Ψw = solute potential Ψs.

Pressure potential :
The pressure which developed due to entry of water in a cell is called pressure potential. It is always positive, it is denoted as Ψp.

Water potential Ψw of a cell is affected by both solute potential and pressure potential.
Ψw = Ψs + Ψp

Question 2.
Write short notes on facilitated diffusion.
Answer:

  1. Membrane protein provides sites at which certain molecules can cross the membrane. This process is called facilitated diffusion.
  2. Porins are proteins that form huge pores in the outer membranes of the plastids, mitochondria and some bacteria, allowing molecules up to the size of small proteins to pass through.
  3. Concentration gradient must already be present for molecules to diffuse even if facilitated by the proteins.
  4. In’facilitated diffusion, special proteins help to move substance across the membrane without expenditure of ATP energy.
  5. Facilitated diffusion cannot cause net transport of molecules from a low to a high concentration this would require input of energy.
  6. Transport rate reaches a maximum when all of the protein transporters are being used.
  7. Facilitated diffusion is very specific, it allows cell to select substance for uptake.
  8. It is sensitive to inhibitors which react with protein side chain.

Question 3.
What is meant by plasmolysis? How is it practically useful to us?
Answer:
Shrinkage of protoplast when the cell is placed in hypertonic solution is called plasmolysis. It occurs when water moves out of the cell and the cell membrane of the cell shrinks away from the cell wall.

The salting of pickles and preserving of fish and meat in salt are good examples of practical application of this phenomenon.

TS Inter 2nd Year Botany Study Material Chapter 1 Transport in Plants

Question 4.
How does ascent of sap occur in tall trees?
Answer:
In tall trees ascent of sap occurs due to transpiration. It depends mainly on the following physical properties of water.

  1. Cohesion – mutual attraction between water molecules.
  2. Adhesion – attraction of water molecules to polar surfaces (such as the surface of tracheary elements)
  3. Transpiration pull – driving force for upward movement of water.

These properties give water high tensile strength i.e., an ability to resist a pulling force and high capillarity i.e-., the ability to rise in thin tubes.

Photosynthesis needs water and it is supplied by xylem vessels from roots to leaf veins. Transpiration pull is created due to evaporation of water through the stomata.

Question 5.
Stomata are turgor operated valves. Explain.
Answer:
TS Inter 2nd Year Botany Study Material Chapter 1 Transport in Plants 1

  1. The cause of opening or closing of the stomata is a change in the turgidity of the guard cells.
  2. Guard cells are thick towards inner side and thin towards outer side.
  3. When turgidity increases within the two guard cells, the thin outer walls bulge out and force the inner walls into a crescent shape, stoma opens.
  4. When the guard cells lose turgor due to water loss (or water stress) the elastic inner walls regain their original shape, the guard cells becomes flaccid and the stoma closes.

Question 6.
Explain pressure flow hypothesis of translocation of sugars in plants.
Answer:
Translocation of sugars from source to sink can be explained by pressure flow or mass flow hypothesis. As glucose is prepared at the source (by photosynthesis) it is converted to sucrose.

The sugar is then moved in the form of sucrose into the companion cells and then into the living phloem sieve tube cells by active transport. This process of loading at the source produces a hypertonic condition in the phloem. Water in the adjacent xylem moves into the phloem by osmosis. it into the energy starch or cellulose. As sugars are removed, the osmotic pressure decreases and water moves out of the phloem and reaches xytem.

This can be shown by following illustration.
TS Inter 2nd Year Botany Study Material Chapter 1 Transport in Plants 2

Question 7.
“Transpiration is a necessary evil” – Explain. [Mar. 2020, May, Mar. 2017, ’14]
Answer:
Beneficial and harmful effects of transpiration are below :

Beneficial effects of transpiration :

  1. It creates transpiration pull for absorption and transportation in plants.
  2. It supplies water for photosynthesis.
  3. It transports minerals from the soil to all parts of the plant.
  4. It cools leaf surfaces, sometimes 10 to 15 degrees, by evaporative cooling.
  5. It maintains the shape and structure of the plants by keeping cells turgid.

Harmful effects of transpiration :

  1. Photosynthesis is limited by available water which is depleted by transpiration.
  2. The humidity of rain forests is largely due to cycling of water from the root to the leaf to the atmosphere and back to the soil.

Even though large amount of water absorbed is lost during transpiration, it is beneficial to plant in many ways. Hence it is considered as “necessary evil”.

Question 8.
A gardener forgot to water a potted plant for a day in summer. What will happen to the plant? Do you think it is reversible? Explain.
Answer:
Plants show wilting.

Yes, it is reversible by watering.
As the potted plant was not water for a day, it will show incipient wilting or temporary wilting. During incipient wilting there is no external symptoms of wilting but the mesophyll cells have lost sufficient water due to transpiration being higher“than the availability of water.

Temporary wilting is the temporary drooping down of leaves and young shoots due to loss of turgidity during noon. It can be corrected only after rate of transpiration decreases accompanied by replenishment of water around root hairs.

Question 9.
Explain the type of molecular movement which is highly selective and requires special membrane proteins, but does not require any metabolic energy.
Answer:
Facilitated diffusion is the type of molecular movement membrane which is highly selective and requires special membrane protein but does not require any metabolic energy.

The random movement of molecules takes place from high concentration to low concentration till equilibrium reaches is called simple diffusion. Membrane provides special proteins to move substances across the membrane by diffusion. This process is called Facilitated diffusion. When all these protein transporters are used transport rate reaches maximum.

Facilitated diffusion is very specific, it allows the cell to select substances for uptake. It is sensitive to inhibitors which react with protein side chain. These special proteins form channels which may be always open or controlled.

Porins are proteins that form huge pores in the outer membranes of the plastids, mitochondria, and some bacteria allowing molecules.

Extra cellular molecules are bound to the transport protein. The transport protein then rotates and releases the molecules inside the cell.
TS Inter 2nd Year Botany Study Material Chapter 1 Transport in Plants 3

Question 10.
How does most of the water move within a healthy plant body and by which path? [May 2014]
Answer:
Water is absorbed along with mineral solutes by the root hairs purely by diffusion. Once water is absorbed by the root hair, it moves by two distinct path ways.
a) Apoplast pathway
b) Symplast pathway

a) Apoplast pathway :
Apoplast is the continuous system of cell walls and intercellular spaces in plant tissues. In apoplast pathway, water movement is through the cell wall without crossing membranes.

b) Symplast pathway :
Symplast comprises of network of cytoplasm of all cells inter-connected by plasmodesmata. The movement of water occurs from one cell to another through plasmodesmata.

Most of the water flows in the cortex occurs via the apoplast. Since the cortical cells are loosely packed and hence offer no resistance. Endodermis radial walls are thickened with casparian strip. Passage cells allows water by symplast, through pericycle it reaches xylem.
TS Inter 2nd Year Botany Study Material Chapter 1 Transport in Plants 4

Question 11.
Transpiration and photosynthesis – a compromise. Explain.
Answer:

  1. Transpiration has more than one purpose : it
    a) creates transpiration pull for absorption and transportation in plants.
    b) supplies water for photosynthesis.
    c) transports minerals from the soil to ail parts of the plant.
    d) cools leaf surfaces, sometimes 10 to 15 degrees, by evaporative cooling.
    e) maintains the shape and structure of the plants by keeping cells turgid.
  2. An actively photosynthesising plant has an insatiable need for water.
  3. Photosynthesis is limited by available water which can be swifty depleted by transpiration.

TS Inter 2nd Year Botany Study Material Chapter 1 Transport in Plants

Question 12.
Do different species of plants growing in the same area show the same rate of transpiration at a particular time? Justify your answer.
Answer:
No. Different species of plant growing in the same area does not show the same rate of transpiration at a particular time. Plants take up huge amount of water daily but most of it is lost to the air through evaporation from the leaves he., transpiration.

A mature corn plant absorbs almost three litres of water in a day while a mustard plant absorbs water equal to its own weight in about 5 hours.

Long Answer Type Questions

Question 1.
Explain how plants absorb water.
Answer:
Water is absorbed along with mineral solutes by the root hairs purely by diffusion. Once water is absorbed by the root hair, it moves by two distinct pathways.
a) Apoplast pathway
b) Symplast pathway

a) Apoplast pathway :
it is “the system of adjacent cell walls that is continuous throughout plant, except at the casparian strips of endodermis in roots.” Apoplastic movement of water takes place through intercellular spaces and cell walls. Movement through apoplast does not include crossing cell membrane. It is dependent on the gradient. It does not provide any barrier to water movement. The water movement is through mass flow. When the water evaporates into intercellular spaces or atmosphere, a tension develops in continuous stream of water in apoplast. Therefore mass flow of water takes place as a result of adhesive and cohesive properties of water.

b) Symplast system :
It is “the system of interconnected protoplasts.” The adjacent cells are connected through cytoplasmic strands that extend through plasmodesmata. In it, water travels through the cells – their cytoplasm; intercellular movement is through plasmodesmata. Water enters the cells only through the cell membrane, so the movement is relatively slower. The movement is again down a potential gradient. It may cytoplasmic streaming.

Most of the water flow in roots takes place by apoplast as cortical cells are loosely packed. They offer no resistance to water movement. Endodermis is impervious to water due to casparian strip. Water moves through symplast and again crosses a membrane to reach the cells of the xylem. In the endodermis, the movement of water is symplastic. This is the only way water and other solutes can enter the vascular cylinder.
TS Inter 2nd Year Botany Study Material Chapter 1 Transport in Plants 5

Question 2.
Define transpiration. Explain the structure and mechanism of opening and closing of stomata.
Answer:
Transpiration is defined as the loss of water in the form of vapour from the living tissues of aerial parts of the plants.

TS Inter 2nd Year Botany Study Material Chapter 1 Transport in Plants 6
Structure of Stomata :
Each stoma is composed of two bean shaped cells known as guard cells. In grasses, the guard cells are dumb-bell shaped. The outer walls of guard cells are thin and inner walls are thickened. The guard cells possess chloroplasts and regulate the opening and closing of stomata. Sometimes, a few epidermal cells, in the vicinity of the guard cells become specialised in their shape and size and are known as subsidiary cells. The stomatal aperture, guard cells and the surrounding subsidiary cells are together called stomatal apparatus.

Opening and closing of stomata :
The stoma is the turgor operated valve in the epidermis.

  1. The immediate cause of opening or closing of the stomata is a change in the turgidity of the guard cells.
  2. The inner wall of each guard cell towards the pore or stomatal aperture is thick and elastic.
  3. When turgidity increases within the two guard cells flanking each stomatal aperture or pore, the thin outer walls bulge out and force the inner walls into a crescent shape,
  4. The opening of the stoma is also aided by the orientation of the microfibrils in the cell walls of the guard cells.
  5. Cellulose microfibrils are oriented radially rather than longitudinally, making it easier for the stoma to open.
  6. When the guard cells lose turgor due to water loss the elastic inner walls regain their original shape, the guard cells becomes flaccid and the stoma closes.

Levitt proposed K+ pump theory to explain the mechanism of opening and closing of photoactive stomata.

Opening of stomata :

  1. According to this theory accumulation of K+ ions into the guard cells from the subsidiary cells occurs in the presence of light.
  2. This coupled with efflux of protons leads to increase in pH of the guard cells.
  3. Accumulation of K+ ions into the guard cells is associated with passive influx of Cl ions thereby decreasing the water potential of the guard cells.
  4. Water thereby enters the guard cells making them turgid.
  5. As the outer walls are thin and elastic, the guard cells expand outwardly, leaving a minute pore at the centre open.

Closing of stomata :

  1. At night, in the absence of light, the K+ and Clions move out of the guard cells due to which the water potential of guard cells increases and water starts moving out of them leading to closure of stomata.
  2. Under water stress conditions, abscisic acid (ABA) a natural anti-transpirant drives the K+ ions out of guard cells making them close.
  3. In Succulent plants, the water potential gradient established due to accumulation of organic acids at night makes the guard cells become turgid, hence stomata opens at night.

TS Inter 2nd Year Botany Study Material Chapter 1 Transport in Plants 7

Intext Question Answers

Question 1.
Differentiate uphill and downhill transport.
Answer:
Active transport uses energy to pump molecules against a concentration gradient. That means from a low concentration to a high concentration. It is called uphill.

Passive transport does not use energy. It transport from high concentration to low concentration. It is called downhill.

TS Inter 2nd Year Botany Study Material Chapter 1 Transport in Plants

Question 2.
Compare facilitated diffusion and simple diffusion.
Answer:
The random movement of individual molecules from a region of higher concentration to a region of lower concentration is called simple diffusion. Facilitated diffusion is a type of molecular movement which is highly selective and required special membrane proteins. These proteins form channels in the membrane for molecules to pass through. These channels may be open always or controlled.

Question 3.
What happens when two solutions of different concentrations are separated by egg membrane? State the reason.
Answer:
When two solutions of different concentrations are separated by egg membrane osmosis takes place. Osmosis occurs spontaneously in response to a driving force. The net direction and rate of osmosis depends on both the pressure gradient and concentration gradient.

Question 4.
Compare imbibing capacities of pea and wheat seeds.
Answer:
Proteins have a very high imbibing capacities followed by starch and cellulose the least. That is why proteinaceous pea seeds swell move on imbibition than starchy wheat seeds.

Question 5.
In general in a plant which path of water movement is more and why?
Answer:
Major proportion of water flow in the root cortex occurs via the apoplast since the cortical cells are loosely packed and hence offer no resistance to water movement. However, the inner boundary of the cortex, the endodermis, is impervious to water because of a band of suberised matrix called the casparian strip. Water flow is by symplast, through pericycle and ultimately reaches xylem.

Question 6.
Why do Pinus seeds fail to germinate in the absence of mycorrhiza?
Answer:
A mycorrhizae is a symbiotic association of a fungus with a root system. The fungus provides minerals and water to the roots, in turn the roots provide sugars and N – containing compounds to the mycorrhizae. Some plants have an obligate association with the mycorrhizae. So pinus seeds fail to germinate in the absence of mycorrhizae.

Question 7.
Which structures do you think the Pinus plant does not possess due to which its seeds fail to germinate?
Answer:
Mycorrhizae.

Question 8.
What do you think is the driving force for ascent of sap?
Answer:
Transpiration pull.

Question 9.
Why do stomata close under water stress conditions?
Answer:
Under water stress conditions, abscisic acid (ABA) a natural anti-transpirant drives the K+ ions out of guard cells making them close.

Question 10.
How are stomata distributed in a typical monocot plant?
Answer:
Stomata are distributed .equally, both in upper epidermis and lower epidermis in typical monocot plant.

Question 11.
In what form the sugars are transported through phloem?
Answer:
Sucrose

Question 12.
The inward movement of water into a plant begins either as symplast or apoplast. How does it conclude before entering into xylem?
Answer:
Symplast

Question 13.
Why does the root endodermis transports ions in one direction only?
Answer:
Endodermis have casparian strips on their radial walls. Hence the movement of water through the root layers is symplastic in endodermis. This is the only way water and other solutes can enter vascular cylinder. So in endodermis transportation of ions occurs in one direction.

Question 14.
If a ring of bark is removed from an actively growing plant, what will happen and why?
Answer:
In the absence of downward movement of food the portion of the bark above the ring on the stem becomes swollen after a few weeks. This shows that phloem is the tissue responsible for translocation of food and that transport takes place in one direction i.e., towards the roots.

TS Inter 2nd Year Botany Study Material Chapter 1 Transport in Plants

Question 15.
A flowering plant is planted in an earthen pot and watered. Urea is added to make the plant grow faster, but after some time the plant dies. Why?
Answer:
As urea is added the concentration of water in the pot increases. Instead of root hairs absorbing water, the water in the plant comes out due to exosmosis. The cells get plasmolyzed. After some time the plant dies.

TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Telangana TSBIE TS Inter 1st Year Zoology Study Material 8th Lesson Ecology and Environment Textbook Questions and Answers.

TS Inter 1st Year Zoology Study Material 8th Lesson Ecology and Environment

Very Short Answer Type Questions

Question 1.
Define the term ecology. (T.Q.)
Answer:
Ecology was defined as “the study of the relationship of organisms with their environment”.

Question 2.
What is autecology?
Answer:
Autecology is the ecology of a single species/population in relation to its environment. It is also known as species (population) ecology.

Question 3.
What do you call the study of interactions of organisms of a community?
Answer:
Synecology is a branch of ecology that deals with the structure, development and distribution of ecological communities.

Question 4.
What is an ecological population? (T.Q.)
Answer:
Population is a group of organisms of the same species, living in a specific area at a specific time.

Question 5.
Define a community. (T.Q.)
Answer:
Community is an association of the interacting members of populations of different autotrophic and heterotrophic species in a particular area.

Question 6.
What is an ecosystem? (T.Q.)
Answer:
Ecosystem is the next level of organisation above the level of biological community. An ecosystem is a functional unit of the biosphere in which members of the community interact among themselves and with surrounding environment.

TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 7.
Distinguish between ecosystem and biome.
Answer:

  1. Ecosystem is a functional unit of biosphere in which members of the community interact among themselves and with surrounding environment.
  2. A biome is a large community of plants and animals that occupies a vast region.

Question 8.
What is a biome? Name any two biomes you studied. (T.Q.)
Answer:
A biome is a large community of plants and animals that occupies a vast region.
eg : Tropical rain forest, desert, coniferous forest, tundra etc.

Question 9.
What is meant by ecosphere? (T.Q.)
Answer:
All the habitable zones on the Earth constitute the ecosphere or biosphere. It is the part of the Earth that supports life.

Question 10.
Define the term habitat.
Answer:
Habitat is the place in which an organism lives. It is comparable to the address of an organism.

Question 11.
Explain the difference between the ‘niche* of an organism and its ‘habitat’. (T.Q.)
Answer:

  1. Niche : With in a community, each organism occupies a particular biological role or Niche. Niche is the functional role of an organism in an ecosystem.
  2. Habitat is the place in which an organism lives.

Question 12.
A population has more genetically similar organisms than those on biotic community. Justify the statement. (T.Q.)
Answer:

  1. Population is a group of organisms of the same species, living in a specific area at a specific time.
  2. Community is an association of the interacting members of populations of different autotrophic and heterotrophic species in a particular area.

TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 13.
Among the red, green and brown algae that inhabit the sea, which is likely to be found in the deepest waters? Why?
Answer:
There are microbes like archaebacteria that flourish in hotsprings and in some parts of deep seas, where temperatures far exceed 100°C. Brown algae is likely to be found in the deepest waters as it is an alchea bacterium.

Question 14.
What is the source of energy for deep sea inhabitants?
Answer:
The source of energy for deep sea inhabitants is by the action of brown algae releasing energy during synthesis of their food. The spectral quality of solar radiation is also important in life. Also detritus food of sea bottom.

Question 15.
How do the fish living in Antarctic waters manage to keep their body fluids from freezing? (T.Q.)
Answer:
Many fish thrive in Antarctic waters where the temperature is always below zero. Many species have evolved a relatively constant internal environment. It permits all biochemical reactions and physiological functions to proceed with maximal efficiency and thus, enhance the overall fitness of the species.

Question 16.
How does your body solve the problem of altitude sickness, when you ascend tall mountains? (T.Q.)
Answer:
Altitude also causes variations in temperature. For instance, the temperature decreases gradually as we move to the top of the mountains we experience altitude sickness if we ever been to any high altitude place. Its symptoms include nausea, fatigue and heart palpitations. We can solve the problem because the body compensates low oxygen availability by increasing red blood cell production and increasing the rate of breathing.

Question 17.
Name the structural components of an ecosystem.
Answer:
The structural components of ecosystem are of two types: Abiotic and biotic factors. Abiotic factors are two types : physical and chemical.
Physical are light, temperature, soil, pressure etc.
Chemical are oxygen, carbondioxide, minerals of soil / water.

Question 18.
What is the effect of light on body pigmentation? (T.Q.)
Answer:
Light influences the colour of the skin. The animals which live in regions of low intensity of light have less pigmentation than that of the animals exposed to light.

TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 19.
Distinguish the terms phototaxis and photokinesis. (T.Q.)
Answer:

  1. Phototaxis is oriented locomotor movement of an organism towards or away from the direction of light.
  2. Photokinesis is the influence of light on non-directional movement of organisms as seen in the larvae of Pinnotheres macculatus-the mussel crab. Intensity of light influences the velocity of the movement of organism.

Question 20.
What is the primardial source of energy for all living organisms?
Answer:
The primardial source of energy for all living organisms is Sunlight.

Question 21.
What are biological rhythms?
Answer:
In the bodies of organisms, many behavioural activities are repeated at regular intervals and these are called biological rhythms.

Question 22.
What are circadian rhythms?
Answer:
Biological rhythms that occur in a time period of 24 hours are circadian rhythms.

Question 23.
What is photoperiodism? (T.Q.)
Answer:
The duration of the light hours / exposure to light in a day is known as photoperiod. The response of organisms for the photoperiod is called Photoperiodism.

Question 24.
Distinguish between photoperiod and critical photoperiod. (T.Q.)
Answer:
The duration of the light hours / exposure to light in a day is known as Photoperiod. The specific day length which is essential for the initiation of seasonal events is called ‘Critical Photoperiod’.

Question 25.
Explain Bioluminescence.
Answer:
Production of light by certain living organisms is called Bioluminestence. The light emitted by living organisms is devoid of infrared rays and so it is called cold light.

TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 26.
Mention the advantages of some UV rays to us. [March 2014] (T.Q.)
Answer:
UV radiation helps in the conversion of sterols present in the skin into vitamin D in mammals.

Question 27.
Distinguish between the terms heat and temperature.
Answer:
Temperature is a measure of the intensity of heat.

Question 28.
Distinguish between minimum effective temperature and maximum effective temperature.
Answer:

  1. The lowest temperature at which an organism can live indefinitely is called minimum effective temperature.
  2. The maximum temperature at which a species can live indefinitely in an active state is called maximum effective temperature.

Question 29.
What is optimum temperature?
Answer:
The temperature at which the metabolic activities occur at the climax level is called “optimum temperature”.

Question 30.
What is cyclomorphosis? Explain its importance in Daphnia. [March 2019] (T.Q.)
Answer:
The cyclic seasonal morphological variations among certain organisms is called cyclomorphosis. In Daphnia (water flea) the body may become elongated with hood or short with out hood depending upon seasons. Cyclomorphosis is a seasonal adaptation to changing densities of water in lakes based on seasons.

Question 31.
What are “regulators”? (T.Q.)
Answer:
Some organisms are able to maintain homeostasis by physiological (sometimes behavioural) means which ensures constant body temperature constant osmotic concentration. These are called regulators.

There are no perfect regulators. (Eg : Birds and Mammals)

Question 32.
What are ‘conformers’? (T.Q.)
Answer:
In aquatic animals, the osmotic concentration of the body fluids changes along with that of the surrounding water. Such animals are described as conformers. There are no perfect conformers.

TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 33.
Define commensalism. Give one example. [May 2017 – A.P.] (T. Q.)
Answer:
The interaction where one species is benefited and the other is neither benefited nor harmed is called “commensalism”.
Example : Barnacles growing on the back of a whale benefit while the whale derives no noticeable benefit.

Question 34.
Define mutualism. Give one example. [March 2015 – A.P.] (T. Q.)
Answer:
This type of interaction benefits both the interacting species.
Example: Lichens represent an intimate mutualistic relationship between a fungus and photosynthetic algae.

Question 35.
Define parasitism. Give one example.
Answer:
In parasitism only one species benefits and the interaction is detrimental to the other species. Parasite generally harms ho’st. Malarial parasite causes harm to the host man by causing malaria.

Question 36.
Define amensalism. Give one example. (T.Q.)
Answer:
In amensalism one species is harmed whereas the other is unaffected.

Question 37.
What is predation? Give one example.
Answer:
Predation :
Only one species benefits and the interaction is detrimental to the other species. Tiger predates upon a deer.

Question 38.
What is meant by interspecific competition? Give one example. (T.Q.)
Answer:
Interspecific competition is a potent force in the process of organic evolution, involving natural selection.
Example : In intertidal communities of the American Pacific Coast, the Starfish Pisaster is an important predator. In a field experiment, when all the starfish were removed from an enclosed intertidal area, more than 10 species of invertebrates became extinct within a year, because of increased inter-specific competition.

TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 39.
Distinguish between predation and parasitism.
Answer:
In both parasitism and predation only one species benefits (parasite and predator, respectively) and the interaction is detrimental to the other species (Host and Prey respectively). Sometimes predator controls over population of prey. Majority of parasites harm the host.

Question 40.
Distinguish between the interactions commensalism and amensalism.
Answer:
The interaction where one species is benefited and the other is neither benefited nor harmed is called “commensalism”.
Example: Barnacles growing on the back of a whale benefit while the whale derives no noticeable benefit.

In amensalism on the other hand one species is harmed whereas the other is unaffected.

Question 41.
In an ecological food chain, what types of interactions exist between trophic levels?
Answer:
Types of interactions exist between trophic levels are mutualism, commensalism, parasitism, amensalism, parasitism and predation.

Question 42.
What is camouflage? Give its significance. (T.Q.)
Answer:
Some species of insects and frogs are cryptically coloured (camouflaged) to avoid being detected easily by the predator.
Example : Stick insect.

Question 43.
What is Gause’s principle ? When is it applicable? (T.O.)
Answer:
Gause’s principle of competitive exclusion states that two closely related species competing for the same resources cannot co-exist indefinitely and the competitively inferior one will be eliminated in due course of time.

TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 44.
Name the association that exist in micorrhiza. (T.O.)
Answer:
Mycorrhizae are associations between fungi and the roots of higher plants. The fungi help the plant in the absorption of essential nutrients from the soil while the plant in turn provides the fungi with energy yielding carbohydrates. It is mutualism.

Question 45.
What are lichens?
Answer:
Lichens represent an intimate mutualistic relationship between a fungus and photosynthesising algae or cyanobacteria.

Question 46.
Name the major types of ecosystems.
Answer:
Major types of ecosystems are basically two types : Natural and Artificial. Natural ecosystems are aquatic and terrestrial ecosystems. Artificial ecosystems are agro-ecosystems like cropland ecosystems, aquaculture ponds and aquaria.

Question 47.
Distinguish between natural ecosystem and an artificial ecosystem.
Answer:
Natural Ecosystems :
These are naturally occurring ecosystems and these is no role of humans in the formation of such types of ecosystems.

Artificial ecosystems :
These are man made ecosystems such as agricultural or agro ecosystems. They include cropland ecosystems, aquaculture ponds and aquaria.

Question 48.
What is an estuary?
Answer:
Estuary is the zone where river joins the sea. Sea water ascends up into the river twice a day.

Question 49.
How does an estuarine ecosysterri differ from freshwater ecosystem?
Answer:
The Freshwater Ecosystem :
Is the smallest aquatic ecosystem. It includes rivers, lakes, ponds etc. Fresh water ecosystem is studied under Limnology.

Estuarine Ecosystem :
Estuary is the zone where river joins the sea. Sea water ascends up into the river twice a day. Estuarine organisms are capable of with standing the fluctuations in salinity.

TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 50.
Distinguish between lotic and lentic habitats. [March 2018 – A.P.] (T.Q.)
Answer:

  1. The still water bodies like ponds, lakes, reservoirs, etc., fall under the category of lentic ecosystems.
  2. Streams, rivers and flowing water bodies are called lotic ecosystems.

Question 51.
What is limnology?
Answer:
The study of freshwater ecosystem is called as Limnology.

Question 52.
What is euphotic zone?
Answer:
It is the shallow part of the lake closer to the shore. Light penetrates up to the bottom. It is Euphotic zone. It has rich vegetation and higher rate of photosynthesis, hence rich in oxygen.

Question 53.
What is zone of compensation in an aquatic ecosystem? (T.Q.)
Answer:
The imaginary line that separates the limnetic zone from the profundal zone is known as zone of compensation / compensation point / light compensation level.

Question 54.
Distinguish between phytoplankton and zooplankton. (T.Q.)
Answer:

  1. Chlorophyll bearing floating micro organisms present in the water form the phytoplankton, eg : diatoms, volvox.
  2. Small microscopic floating animals present in the surface waters form the zooplankton, eg: rotifers, copepods, Daphnia.

Question 55.
Distinguish between neuston and nekton. (T.Q.)
Answer:

  1. The animals living at the air – water interface constitute the neuston.
  2. The animals such as fishes, amphibians, water snakes which are capable of swimming constitute the nekton.

Question 56.
What is periphyton? (T.Q.)
Answer:
The animals that are attached to / creeping on the aquatic plants, such as the water snails, nymphs of insects, bryozoans, hydras constitute the periphyton.

TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 57.
What is benthos?
Answer:
The animals that rest on or move on the bottom of the lake constitute the ‘benthos’ eg: red annelids, chironomid larvae, crayfishes, some isopods, amphipods, clams, etc.

Question 58.
Write three examples for man-made ecosystems. (T.Q.)
Answer:
Examples for man-made ecosystems are 1) Cropland ecosystems 2) Aquaculture ponds 3) Aquaria.

Question 59.
What is meant by osmotrophic nutrition?
Answer:
Intake of pre-digested food material through the body surface is called osmotrophic nutrition.

Question 60.
Explain the process of ‘leaching’.
Answer:
By the process of leaching, water soluble inorganic nutrients go down into the soil and get precipitated as unavailable salts.

Question 61.
What is catabolism?
Answer:
Bacterial and fungal enzymes degrade detritus into simpler inorganic substances. This process is called as catabolism.

Question 62.
What is humification? Name the organisms which act on it.
Answer:
Humification and mineralization occur during decomposition in the soil. Humification leads to accumulation of a dark coloured amorphous substance called humus. Microbes work on humus and make it further degraded.

TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 63.
What is PAR? (T.Q.)
Answer:
Sun is the only source of energy for all ecosystems on earth. Of the incident solar radiation less than 50 percent of it is photosynthetically active radiation(PAR). Plants capture only 2-10 percent of the PAR and this small amount of energy sustains the entire living world.

Question 64.
What is the percentage of PAR, in the incident solar radiation? (T.Q.)
Answer:
Of the incident solar radiation less than 50 percent of it is photosynthetically active radiation (PAR).

Question 65.
Define entropy. (T.Q.)
Answer:
Energy lost or not available for work in a system is called ENTROPY.

Question 66.
What is standing crop? (T.Q.)
Answer:
Each trophic level has a certain mass of living material at a particular time, and it is called the standing crop. The standing crop is measured as the mass of living organisms or the number of organisms per unit area.

Question 67.
If you were to count the number of insects feeding on a big tree, and asked to graphically represent the relation of structure and function between the tree and insects, what kind of pyramid does it form ?
Answer:
It forms an inverted pyramid of Numbers.

Question 68.
Explain the terms GPP, NPP. (T.Q.)
Answer:
1) Gross primary productivity (GPP) of an ecosystem is the rate of production of organic matter during photosynthesis.

2) Net primary productivity (NPP) :
Gross primary productivity minus respiratory loss (R) is the net primary productivity (NPP). On average about 20 – 25 percent of GPP is used for the catabolic (respiratory) activity.

TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 69.
Distinguish between GFC and DFC.
Answer:
a) GFC :
Grazing Food Chain also known as predatory food chain.

b) DFC :
Detritus Food Chain which begins with dead organic matter.
Some of the organisms of DFC may form the prey of GFC animals.

Question 70.
Name the trophic level(s) at which ‘decomposers’ feed.
Answer:
Decomposers break down complex organic matter into simple inorganic substances like CO2, water and nutrients and the process is called decomposition. It occupies first trophic level.

Question 71.
Distinguish between production and decomposition.
Answer:
a) The green paints in the ecosystem terminology are called producers. They synthesize the food by using solar energy. They form main source of food for the organisms.

b) Breaking down of complex organic matter into simple inorganic substances like carbondioxide, water and nutrients by decomposess (microbes) and this process is called decomposition.

Question 72.
Distinguish between upright and inverted ecological pyramids. (T.Q.)
Answer:

  1. In most ecosystems, all the pyramids of numbers, energy and biomass are upright i.e., produces are more in number and biomass than the herbivores, herbivores more in number and biomass than the carnivores.
  2. In case of a parasitic food chain, the pyramid of numbers is inverted. Here in each trophic level from the bottom to the top, th’e numbers of organisms increase and form an inverted pyramid of numbers.

Question 73.
Distinguish between food chain and food web.
Answer:

  1. When the path of food energy is ‘linear’ the components resemble the links of a chain and it is called food chain.
  2. The natural interconnections of food chains form a network called food web.

Question 74.
Distinguish between litter and detritus. (T.Q.)
Answer:

  1. Plant parts such as leaves, bark, flowers and dead remains of animals, including their faecal matter, constitute the detritus.
  2. Litter contains only dead decayed plant material mixed with mud and soil.

TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 75.
Distinguish between primary and secondary productivity. (T.Q.)
Answer:

  1. Primary productivity is defined as the amount of biomass or organic matter produced per unit area over a period of time by plants, during photosynthesis.
  2. Secondary productivity is defined as the rate of formation of new organic matter by consumers.

Question 76.
What is the effect of Carbon monoxide pollution on human beings?
Answer:
CO interferes with O2 transport. CO causes symptoms such as headache and blurred vision at lower concentrations. In higher concentrations. It leads to coma and death.

Question 77.
What is ‘Green House Effect’?
Answer:
The green house effect is a naturally occurring phenomenon that is responsible for heating of the Earth’s surface and atmosphere.

Question 78.
Which air pollutants are chiefly responsible for acid rains? (T.Q.)
Answer:
Sulphur dioxide and nitrogen oxides are the major causes of acid rains.

Question 79.
What is BOD? (T.Q.)
Answer:
BOD (biological oxygen demand) is a measure of the content of biologically degradable substances in sewage. Micro organisms involved in biodegradation of organic matter in water bodies consume a lot of oxygen, and as a result there is a sharp decline in dissolved oxygen causing death offish and other aquatic animals.

Question 80.
What is biological magnification? [March 2020, ’13] (T.Q.)
Answer:
Increase in the concentration of the pollutant or toxicant at successive trophic levels in an aquatic food chain is called Biological Magnification or Bio-magnification.

Question 81.
Distinguish between ‘Global warming’ and ‘Thermal pollution’?
Answer:

  1. Increase in the level of green house gases has led to considerable heating of the earth leads to Global warming. Global warming is causing climatic changes.
  2. Water pollution caused by hot water coming out from industries, thermal power plants which is harmful to aquatic organisms is called thermal pollution.

TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 82.
Why are incinerators used in hospitals? [May/June 2014] (T. Q.)
Answer:
Hospitals generate hazardous wastes that contain disinfectants, harmful chemicals and also pathogenic micro- organisms. Such wastes also require careful treatment and disposal. The use of incinerators (to burn wastes) is essential for disposal of hospital waste.

Question 83.
Why are catalytic converters used in automobiles?
Answer:
Fitting catalytic converters to the automobiles having expensive metals namely platinum, palladium and rhodium as catalysts which reduce emission of poisonous gases.

Short Answer Type Questions

Question 1.
Write about ecological hierarchy.
Answer:
Ecological Hierarchy :
Hierarchy means arrangement into a ‘graded series’. Ecological organization consists of eleven integrative levels, ranging from Cell to Ecosphere – cell, tissue, organ, organ – system, organism, population, community, ecosystem, landscape, biome and ecosphere (also called Biosphere).

Population :
Population is a group of organisms of the same species, living in a specific area at a specific time.

Community :
It is an association of the interacting members of populations of different autotrophic and heterotrophic species in a particular area. In a community, generally one or a few species dominate with reference to their numbers or size.

Ecosystem :
It is the next level of organization above the level of biological community. An ecosystem is a functional unit of the biosphere in which members of the community interact among themselves and with the surrounding environment, involving ‘flow of energy’ forming a well defined trophic structure.

Landscape :
It is the unit of land containing different ecosystems (mosaic of ecosystems) surrounded by natural boundaries. It is the level of organization higher than ‘ecosystem’.

Biome :
A ‘biome’ is a large community of plants and animals that occupies a vast region. There are ‘terrestrial biomes’ and ‘aquatic biomes’.

Ecosphere (Biosphere) :
All the habitable zones on the Earth constitute the ecosphere or biosphere. It is the part of the Earth that supports ‘life’.

Question 2.
Write a note on habitat and medium.
Answer:
Habitat and Medium: Ecologically, habitat is the place in which an organism lives. It is comparable to the ‘address’ of a person (as mentioned in the introduction page to ecology). For instance, the habitat of fish is a pond, lake, sea etc., the habitat of a lion is forest, the habitat of Ascaris is the ‘small intestine’ of man, and so on. The water surrounding the body of a fish is called the medium and the medium of a lion is the air around its body.

TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 3.
Considering the benefits of a constant internal environment to the organism, we tend to ask ourselves ‘why the conformers had not evolved to become regulators’? (T.Q.)
Answer:
Majority (99 percent) of animals cannot maintain a constant internal environment. Their body temperature changes with the ambient (surrounding) temperature. In aquatic animals the osmotic concentration of the body fluids changes along with that of the surrounding water. Such animals are described as conformers. Conformer means adapt from one condition to a new or different conditions.

Thermoregulation is energetically expensive for many organisms. (Animals such as camels can be conformers upto a particular range of temperature and regulator afterwards). Heat loss or heat gain is a function of the surface area.

During the course of evolution, the costs and benefits of maintaining a constant internal environment are taken into consideration.

Question 4.
The individuals who have fallen through the ice and been submerged under cold water for long periods can sometimes be revived – explain. (T.Q.)
Answer:
Many fish thrive in Antarctic waters where the temperature is always below zero. Having realized that the abiotic conditions of many habitats may vary over a time period. One would expect that during the course of millions of years of their existence, many species would have evolved a relatively constant internal environment. It permits all biochemical reactions and physiological functions to proceed with maximal efficiency and thus, enhance the overall fitness of the species. The familiar case of polar bears going into hibernation during winter is an example of escape in time. Some snails and fish go into aestivation to avoid summer related problems- heat and desiccation.

Question 5.
What is summer stratification? Explain. [March 2020] (T.Q.)
Answer:
Summer stratification :
During summer in temperate lakes, the density of the surface water decreases because of increase in its temperature (21-25°C). This ‘upper more warm layer’ of a lake is called epilimnion. Below the epilimnion there is a zone in which the temperature decreases at the rate of 1° C per meter in depth, and it is called thermocline or metalimnion. The bottom layer is the hypolimnion, where water is relatively cool, stagnant and with low oxygen content (due to absence of photosynthetic activity).
TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment 1

During autumn (also called Fall), the epilimnion cools down, and the surface water becomes heavy when the temperature is 4°C, and sinks to the bottom of the lake. Overturns bring about ‘uniform temperature’ in lakes during that period. This circulation during the autumn is known as the fall or autumn overturn. The upper oxygen rich water reaches the hypoliminion and the nutrient rich bottom water comes to the surface. Thus there is uniform distribution of nutrients and oxygen in the lake.

Question 6.
What is the significance of stratification in lakes? (T.Q.)
Answer:
Temperature variations occur with seasonal changes in the temperate regions. These differences in the temperature form ‘thermal layers’ in water. These phenomena are called thermal stratifications.
TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment 2

Water shows maximum density at 4°C. Rise or faII of temperatures above or below 4°C decreases its density. This anomalous property of water and the seasonal variations in temperature are responsible for the thermal stratification in temperate lakes.

Due to overturns, there is uniform distribution of nutrients and oxygen in the lake.

Question 7.
Explain van’t Hoff rule. (T.Q.)
Answer:
van’t Hoff’s rule :
van’t Hoff, a Nobel Laureate in thermochemistry, proposed that, with the increase of every 10°C, the rate of metabolic activities doubles. This rule is referred to as the van’t Hoff’s rule, van’t Hoff’s rule can also be stated in reverse saying that the reaction rate is halved with the decrease of every 10°C. The effect of temperature on the rate of a reaction is expressed in terms of temperature coefficient or Q10 value. Q10 values are estimated taking the ratio between the rate of a reaction at X°C and rate of reaction at (X – 10°C). In the ‘living systems’ the Q10 value is about 2.0. If the Q10 value is 2.0, it means, for every 10°C increase, the rate of metabolism doubles.

Question 8.
Unlike mammals the reptiles cannot tolerate environmental fluctuations in temperature. How do they adapt to survive in desert conditions? (T.Q.)
Answer:
Some organisms show behavioural responses to cope with variations in their environment. Desert lizards manage to keep their body temperature fairly constant by behavioural means. They ‘bask’ (staying in the warmth of sunlight) in the sun and absorb heat when their body temperature drops below the comfort zone, but move into shade when the temperature starts increasing. Some species are capable of burrowing into the soil to escape from the excessive heat above the ground level.

TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 9.
Write a short note on soil as an ecological abiotic factor.
Answer:
Soil :
The nature and properties of soil in different places vary depending on the climate, and the ‘weathering’ processes involved. Various characteristics of the soil such as soil composition, grain size and aggregation determine the percolation and water-holding capacity of the soils. These characteristics, along with the parameters such as pH, mineral composition etc., determine to a large extent the vegetation in any area. This in turn dictates the type of aninmals that can be supported. Similarly, in the aquatic environment, the sediment-characteristics often determine the type of benthic animals that can live there. ‘

Question 10.
How do terrestrial animals protect themselves from, the danger of dehydration of bodies? (T.Q.)
Answer:
In the absence of an external source of water, the kangaroo rat of the North American deserts is capable of meeting all its water requirements through oxidation of its internal fat (in which water is a by product – metabolic water). It also has the ability to concentrate its urine, so that minimal volume of water is lost in the process of removal of their excretory products.

Question 11.
How do marine animals adapt to hypertonic seawater? (T.Q.)
Answer:
Seawater is high in salt content compared to that of the body fluids. So, the marine animals continuously tend to lose water from their bodies by exosmosis and face the problem of dehydration. To overcome the problem of water loss, marine fishes have aglomerular kidneys with less numbers of nephrons. Such kidneys minimize the loss of water through urine. To compensate water loss the marine fish drink more water, and along with this water, salts are added to the body fluids and disturb the internal equilibrium.

To maintain salt balance (salt homeostasis) in the body, they have salt secreting chloride cells in their gills. Marine birds like sea gulls and penguins eliminate salts in the form of salty fluid that drips through their nostrils. In turtles the ducts of chloride secreting glands open near the eyes. Some cartilaginous fishes retain urea and trimethylamine oxide (TMO) in their blood to keep the body fluid isotonic to the sea water and avoid dehydration of the body due to exosmosis.

Question 12.
Discuss the various types of adaptations in freshwater animals. (T.Q.)
Answer:
Adaptations in freshwater habitat :
Animals living in freshwaters have to tackle the problem of endosmosis. The osmotic pressure of freshwater is very low and that of the body fluids of freshwater organisms is much higher. So water tends to enter into bodies by endosmosis. To maintain the balance of water in the bodies, the freshwater organisms acquired several adaptations such as, contractile vacuoles in the freshwater protozoans, large glomerular kidneys in fishes, etc. They send out large quantities of urine, along which some salts are also lost.

To compensate the ‘salt loss’ through urine, freshwater fishes have salt absorbing ‘chloride cells’ in their gills. The major problem in freshwater ponds is – in summer most of the ponds dry up. To overcome this problem most of the freshwater protists undergo encystment. The freshwater sponges produce asexual reproductive bodies, called gemmules, to tide over the unfavourable conditions of the summer. The ‘African lungfish’, Protopterus, burrows into the mud and forms a ‘gelatinous cocoon’ around it, to survive, in summer.

Question 13.
Compare the adaptations of animals with freshwater and seawater mode of life. (T.Q.)
Answer:
Seawater is high in salt content compared to that of the body fluids. So, the marine animals continuously tend to lose water from their bodies by exosmosis and face the problem of dehydration. To overcome the problem of water loss, marine fishes have aglomerular kidneys with less numbers of nephrons. Such kidneys minimize the loss of water through urine. To compensate water loss the marine fish drink more water, and along with this water, salts are added to the body fluids and disturb the internal equilibrium. To maintain salt balance (salt homeostasis) in the body, they have salt secreting chloride cells in their gills.

Marine birds like sea gulls and penguins eliminate salts in the form of salty fluid that drips through their nostrils. In turtles the ducts of chloride secreting glands open near the eyes. Some cartilaginous fishes retain urea and trimethylamine oxide (TMO) in their blood to keep the body fluid isotonic to the sea water and avoid dehydration of the body due to exosmosis.

Adaptations in freshwater habitat :
Animals living in freshwaters have to tackle the problem of endosmosis. The osmotic pressure of freshwater is very low and that of the body fluids of freshwater organisms is much higher. So water tends to enter into bodies by endosmosis. To maintain the balance of water in the bodies, the freshwater organisms acquired several adaptations such as, contractile vacuoles in the freshwater protozoans, large glomerular kidneys in fishes, etc. They send out large quantities of urine, along which some salts are also lost. To compensate the ‘salt loss’ through urine, freshwater fishes have salt absorbing ‘chloride cells’ in their gills.

The major problem in freshwater ponds is – in summer most of the ponds dry up. To overcome this problem most of the freshwater protists undergo encystment. The freshwater sponges produce asexual reproductive bodies, called gemmules, to tide over the unfavourable conditions of the summer. The ‘African lungfish’, Protopterus, burrows into the mud and forms a ‘gelatinous cocoon’ around it, to survive, in summer.

TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 14.
Distinguish between euryhaline and stenohaline animals. (T.Q.)
Answer:
For aquatic organisms the quality (chemical composition, pHr etc.,) of water becomes important. The salt concentration is less than 5 percent in inland, waters, and 30 – 35 percent in the sea water. Some organisms are tolerant to a wide range of salinities (euryhaline), but others are restricted to a narrow range (stenohaline). Many freshwater animals cannot live for long in sea water and vice versa because of the osmotic problems, they would face.

The animals of brackish water are adapted to withstand wide fluctuations in salinity. Such organisms are called euryhaline animals and those that can’t withstand are known as stenohaline.

Question 15.
How do the non migratory animals overcome the unfavourable climatic conditions?
Answer:
In bacteria, fungi and lower plants, various kinds of thick-walled spores are formed which help them survive unfavourable conditions. They germinate (come out of the spore wall and produce a normal active organism) on the return of suitable environmental conditions.

Some animals can avoid the stress by escaping in ‘time’ (migration is escaping in ‘space’). The familiar case of ‘polar bears’ going into hibernation during winter is an example of escape in time. Some snails and fish go into aestivation to avoid summer-related problems – heat and desiccation.

Diapause :
Certain organisms show delay in development, during periods of unfavourable environmental conditions and spend some period in a state of ‘inactiveness’ called ‘diapause’. This dormant period in animals is a mechanism to survive extremes of temperature, drought, etc. It is seen mostly in insects and embryos of some fish. Under unfavourable conditions many zooplankton species in lakes and ponds are known to enter diapause.

Question 16.
Many tribes living in high altitude of Himalayas normally have higher red blood cell count (or) total haemoglobin that the people living in the plains. Explain. (T.Q.)
Answer:
Pressure is another factor that changes dramatically with depth in the ocean. Organisms on land face less than one ‘atmosphere’ of pressure at the sea level. Since water is much heavier than air, marine organisms are under much more pressure than those pn land. The pressure in water increases at the rate of 1 atmosphere per 10m depth. The organisms living in such extreme environments show a wide range of biochemical adaptations. Some organisms possess adaptations that are physiological and allow them to respond quickly to a stressful situation. If you had ever been to any high altitude place (e.g. > 3,500m Rohtang Pass near Manali and Manasarovar, in Tibet) you must have experienced what is called altitude sickness.

Its symptoms include nausea (vomiting sense), fatigue (tiredness) and heart palpitations (abnormality in heart beat). This is because in the low atmospheric pressure of high altitudes, the body does not get enough oxygen. But, you gradually get acclimatized and overcome the altitude sickness. How did your body solve this problem ? The body compensates low oxygen availability by increasing red blood cell production and increasing the rate of breathing. (Note : decreasing the binding capacity of haemoglobin).

Question 17.
An orchid plant is growing on the branch of mango tree. How do you describe this interaction between the orchid and mango tree? (T.Q.)
Answer:
This is an interaction called commensalism in which one species benefits and the other is neither harmed nor benefited. An orchid growing as an epiphyte on mango branch, gets the benefit of exposure to hight, while the mango tree does not derive any noticeable benefit.

Question 18.
Do you believe that an ideal parasite should be able to thrive within the host without harming it? Then why didn’t natural selection lead to the evolution of such totally harmless parasites?
Answer:
No. I don’t believe that an ideal parasite should be able to thrive within the host without harming it. It is not possible.

Considering that the parasitic mode of life ensures free ‘lodging’ and ‘meals’, it is not surprising that parasitism has evolved in so many taxonomic groups from plants to higher vertebrates. Many parasites have evolved to be host – specific (they can parasitize only a specific species of host) in such a way that both host and the parasite tend to co – evolve; that is, if the host evolves special mechanisms for rejecting or resisting the parasite, the parasite has to evolve mechanisms to ‘counteract’ and ‘neutralize1 them, in order to continue successful parasitic relationship with the same host species. In order to lead successful parasitic life, parasites evolved special adaptations, such as.

a) loss of sense organs b) loss of digestive system and presence of high reproductive capacity c) presence of adhesive organs such as suckers and hooks d) complex life cycle.

TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 19.
The female mosquito is not considered completely a parasite, although it needs our blood for reproduction. Can you explain why?
Answer:
Considering that the parasitic made of life ensures free lodging and meals, it is not surprising that parasitism has evolved in so many taxonomic groups from plants to higher vertebrates. Many parasites have evolved to be host specific (they can parasitize only a specific species of host) in such a way that both host and the parasite tend to co-evolve, that is, if the host evolves special mechanism for rejecting or resisting the parasite, the parasite has to evolve mechanism to counteract and neutralise them, in order to continue successful parasitic relationship with the same host species.

The female mosquito is not considered completely a parasite although it needs our blood for reproduction as warm conditions are necessary to stimulate the production of eggs, because it otherwise had a semi independent and free life as it is a temporary ecto parasite. It needs no shelter, no anaerobic respiration.

Question 20.
Predation is not an association. Support the statement. (T.Q.)
Answer:
Predation :
We think of predation as nature’s way of transferring the energy fixed by plants to higher trophic levels. When we think of predatorand prey, most probably it is the tiger and the deer that readily come to our mind, but a sparrow eating any seed is also a type of predator (a seed predator also called granivore). Although animals eating plants are categorized separately as herbivores, they are, in a broad ecological context, not very different from predators.

Besides acting as ‘conduits’ / ‘pipelines’ for energy transfer across trophic levels, predators play other important roles. They keep the prey populations under control. In the absence of predators, the prey species could achieve very high population densities and cause instability in the ecosystem. Predators have different types of functions to play in nature. They include :

a) Predator as ‘a biological control b) Predators maintain species diversity c) Predators are prudent pertaining to preys.

Question 21.
Assigning the sign ‘+’ for beneficial, for detrimental, and ‘O’ for neutral interactions. Explain the different types of interspecific interactions in an ecosystem.
Answer:
Inter-specific interactions arise from the interaction of populations of two different species. They could be beneficial, detrimental or neutral (neither harmful nor beneficial) to one of the species or both. Assigning a ‘+’ sign for beneficial interaction,’-‘ sign for detrimental and ‘O’ for neutral interaction, let us look at all the possible outcomes of inter-specific interactions.

Population Interactions – Types

Name of InteractionSpecies ASpecies B
Mutualism++
Competition
Predation+
Parasitism+
Commensalism +0
Amensalism_0

TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 22.
Predation has a significant role in maintaining of species diversity – discuss.
Answer:
Predators maintain ‘species diversity’: Predators also help in maintaining species diversity in a community, by reducing the intensity of competition among competing prey species. In the rocky intertidal communities of the American Pacific Coast, the starfish Pisaster is an important predator. In a field experiment, when all the starfish were removed from an enclosed intertidal area, more than 10 species of invertebrates became extinct within a year, because of increased inter-specific competition.

Question 23.
What is the biological principle behind the biologiqal control method of managing pest insects? (T.Q.)
Answer:
Predator as a biological control: The prickly pear cactus introduced intcTAustralia in the early 1920s caused havoc by spreading rapidly into millions of hectares of rangeland (vast natural grass lands). Finally, the invasive cactus was brought under control only after a cactus feeding predator (a moth) was introduced into the country. Biological control methods adopted in agricultural pest control are based on the ability of the predators to regulate prey populations.

Question 24.
Name important defence mechanisms in plants against herbivory.
Answer:
For plants, herbivores are the predators. Nearly 25 percent of all insects are known to be phytophagous (feeding on plant sap and other parts of plants). The problem is particularly severe for plants because, unlike animals, they cannot escape from their predators. Plants therefore have evolved a variety of morphological and chemical defences against herbivores.

  1. Thoms (Acacia, Cactus, etc.,) are the most common morphological means of defense. Many plants produce and store chemicals that make the herbivore sick when they are eaten, inhibit feeding or digestion, disrupt its reproduction or even kill it.
  2. You must have seen the weed Calotropis growing in abandoned fields. The plant produces highly poisonous cardiac glycosides and that is why you never see any cattle or goats browsing on this plant.
  3. A wide variety of chemical substances that we extract from plants on a commercial scale (nicotine, caffeine, quinine, strychnine, opium, etc.,) are produced by them actually as defences against grazers and browsers.

Question 25.
Discuss competitive release. (T.Q.)
Answer:
Competitive release :
Another evidence for the occurrence of competition in nature comes from what is called ‘competitive release’. Competitive release occurs when one of the two competing species is removed from an area, thereby releasing the remaining species from one of the factors that limited its population size. A species, whose distribution is restricted to a small geographical area because of the presence of a competitively superior species, is found to expand its distributional range dramatically when the competing species is experimentally removed. This is due to the phenomenon called ‘competitive release’.

Connell’s ‘field experiments’ showed that, on the rocky sea coasts of Scotland, the larger and competitively superior barnacle Balanus dominates the intertidal area, and excludes the smaller barnacle Chathamalus from that zone. When the dominant one is experimentally removed, the populations of the smaller ones increased. In general, herbivores and plants appear to be more adversely affected by competition than the carnivores.

Question 26.
Write a short note on the parasitic adaptations. (T.Q.)
Answer:
In order to lead successful parasitic life, parasites evolved special adaptations such as

  1. Loss of sense organs (which are not necessary for most parasites).
  2. Presence of adhesive organs such as suckers, hooks to cling on to the host’s body parts.
  3. Loss of digestive system and presence of high reproductive capacity.
  4. The life cycles of parasites are often complex, involving one or two intermediate hosts or vectors to facilitate parasitisation of their primary hosts.
    eg : 1 : The human liver fluke depends on two intermediate (secondary ) hosts (a snail and fish) to complete its life cycle.
    e.g. – 2 : The malaria parasite needs a vector (mosquito) to spread to other hosts.

TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 27.
Explain brood parasitism with a suitable example. (T.Q.)
Answer:
Brood parasitism :
Certain birds are fascinating examples of a special type of parasitism, in which the parasitic bird lays its eggs in the nest of its host and lets / allows the host incubates them. During the coruse of evolution, the eggs of the parasitic bird have evolved to resemble the host’s egg in size and colour to reduce the chances of the host bird detecting the foreign eggs and ejecting them from the nest, eg: Cuckoo (Koel) laying its eggs in crow’s nest.

Question 28.
How do predators act as biological control? (T.Q.)
Answer:
Predator as a biological control: The prickly pear cactus introduced into Australia in the early 1920s caused havoc by spreading rapdily into millions of hectares of rangeland (vast natural grass lands). Finally, the invasive cactus was brought under control only after a cactus feeding predator (a moth) was introduced into the country. Biological controlmethods adopted in agricultural pest control are based on the ability of the predators to regulate prey populations.

Question 29.
Explain the interaction mechanism between fig trees and wasps.
Answer:
In many species of fig trees, there is a one-to-one relationship with the pollinator species of wasp. It means that a given fig species can be pollinated only by its ‘partner’ wasp species and no other species. The female wasp uses the fruit not only as a site for oviposition (egg-laying site), but also uses the developing seeds within the fruit for nourishing its larvae. The wasp pollinates the flowers of the fig plant while searching for suitable egg-laying sites. In return for the favour of pollination the fig offers the wasp some of its developing seeds, as food for the developing wasp larvae.

Question 30.
Write notes on the structure and functioning of an ecosystem. (T.Q.)
Answer:
An ‘ecosystem’ is a functional unit of nature, where living organisms interact among themselves and also with the surrounding physical environment. Ecosystem varies greatly in size from a small pond to a large forest or a sea. Many ecologists regard the entire biosphere as a ‘global ecosystem’, as a composite of all local ecosystems on Earth. Since this system is too big and complex to be studied at one time, it is convenient to divide it into two basic categories, namely natural and artificial. The natural ecosystems include aquatic ecosystems of water and terrestrial ecosystems of the land. Both types of natural and artificial ecosystems have several subdivisions.

Question 31.
Explain the different types of aquatic ecosystems.
Answer:
Aquatic Ecosystems :
Based on the salinity of water, three types of aquatic ecosystems are identified marine, freshwater, and estuarine.

i) The Marine Ecosystem :
It is the largest of all the aquatic ecosystems. It is the most stable ecosystem.

ii) Estuarine Ecosystem :
Estuary is the zone where river joins the sea. Sea water ascends up into the river twice a day (effect of high tides and low tides). The salinity of water in an estuary also depends on the seasons. During the rainy season out flow of river water makes the estuary less saline and the opposite occurs during the summer. Estuarine organisms are capable of withstanding the ‘fluctuations’ in salinity.

iii) The Freshwater Ecosystem :
The freshwater ecosystem is the smallest aquatic ecosystem. It includes rivers, lakes, ponds, etc. It is divided into two groups – the lentic and lotic. The still water bodies like ponds, lakes, reservoirs, etc., fall under the category of lentic ecosystems. The communities of the above two types are called lentic and lotic communities respectively. The study of freshwater ecosystem is called as limnology.

Question 32.
Explain the different types of terrestrial ecosystems. (T.Q.)
Answer:
The Terrestrial Ecosystems: The ecosystems of land are known as terrestrial ecosystems.

Some examples of terrestrial ecosystems are the forest, grassland and desert.
i) The Forest Ecosystems :
The two important types of forests seen in India are i) tropical rain forest and ii) tropical deciduous forests.

ii) The Grassland Ecosystems :
These are present in the Himalayan region in India. They occupy large areas of sandy and saline soils in Western Rajasthan.

iii) Desert Ecosystems :
The areas having less than 25 cm rainfall per year are called deserts. They have characteristics flora and fauna. The deserts can be divided into two types – hot type and cold type deserts. Thar Desert in Rajasthan is the example for hot type of desert. Cold type desert is seen in Ladakh.

TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 33.
Draw a diagram of the lake ecosystem and its physical or ecological divisions.
Answer:
TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment 3

Question 34.
Write about the producers of the littoral zone with suitable examples.
Answer:
Producers of the littoral zone :
Littoral zone is rich with pedonic flora (especially up to the depth of the effective light penetration). At the shore proper emergent vegetation is abundant with firmly fixed roots in the bottom of the lake and shoots and leaves are exposed above the level of water. These are amphibious plants. Certain emergent rooted plants of littoral zone are the cattails (Typha), bulrushes (Scirpus), arrowheads (Sagittaria). Slightly deeper are the rooted plants with floating leaves, such as the water lilies (Nymphaea), Nelumbo, Trapa, etc. Still deeper are the submerged plants such as Hydrilla, Chara, Potamogeton, etc. The free floating vegetation includes Pistia, Wolffia, Lemna (duckweed), Azolla, Eichhornia, etc.

The phytoplankton of the littoral zone composed of diatoms (Coscinodiscus, Nitzschia, etc.), green algae (Volvox, Spirogyra, etc.), euglenoids (Euglena, Phacus, etc.), and dinoflagellates (Gymnodinium, Cystodinium, etc.).

Question 35.
Write a short note on the limnetic zone of a lake ecosystem.
Answer:
Limnetic zone :
It is the open water zone away from the shore. It extends up to the effective light penetration level, vertically. The imaginary line that separates the limnetic zone from the profundal zone is known as zone of compensation/ compensation point / light compensation level. It is the zone of effective light penetration. Here the rate of photosynthesis is equal to the rate of respiration. Limnetic zone has no contact with the bottom of the lake.

Biota of the limnetic zone :
Limnetic zone is the largest zone of a lake. It is the region of rapid variations of the level of the water, temperature, oxygen availability, etc., from time to time. The chief autotrophs of this region are the phytoplankton such as the euglenoids, diatoms, cyanobacteria, dinoflagellates and green algae. The consumers of the limnetic zone are the zooplanktonic organisms such as the copepods. Fishes, frogs, water snakes, etc., form the limnetic nekton.

Question 36.
Write a short note on the profundal zone of a lake ecosystem.
Answer:
Profundal zone :
It is the deep water area present below the limnetic zone and beyond the depth of effective light penetration. Light is absent. Photosynthetic organisms are absent and so the water is poor in oxygen content. It includes mostly the anaerobic organisms which feed on detritus.

The organisms living in lentic habitat are classified into pedonic forms, which live at the bottom of the lake and those living in the open waters of lakes, away from the shore vegetation are known as limnetic forms.

Biota of the profundal zone :
It includes the organisms such as decomposers (bacteria), chironomid larvae, Chaoborus (phantom larva), red annelids, clams, etc., that are capable of living in low oxygen levels. The decomposers of this zone decompose the dead plants and animals and release nutrients which are used by the biotic communities of both littoral and limnetic zones.

Question 37.
Give a brief account of a lake ecosystem.
Answer:
The lake ecosystem performs all the functions of any ecosystem and of the biosphere as a whole, i.e., conversion of inorganic substances into organic material, with the help of the radiant solar.energy by the autotrophs; consumption of the autotrophs by the heterotrophs; decomposition and mineralization of the dead matter to release them back for reuse by the autotrohs (recycling of minerals).

Question 38.
How is a lake ecosystem described as a ‘micro-model’ for the entire biosphere?
Answer:
Lake Ecosystem :
To understand the fundamentals of an aquatic ecosystem, let us take a ‘lake’ as an example. This is fairly a self-sustainable unit and rather a simple, example that explains even the complex interactions that exist in an aquatic ecosystem.

Lakes are large inland water bodies containing standing/still water (Recall: Lentic community). They are deeper than ponds (pond is not an ideal example as it is very shallow). Most lakes contain water throughout the year. In deep lakes, light cannot penetrate more than 200 meters, in depth. They are vertically stratified in relation to light intensity, temperature, pressure, etc., Deep water lakes contain three distinct zones namely, i) littoral zone, ii) limnetic zone, and iii) profundal zone. Hence lake ecosystem is described as a micro model for the entire biosphere.
TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment 4

Question 39.
In GFCs the number of trophic levels is restricted. Give reason.
Answer:
I. Grazing Food Chain (GFC) :
It is also known as predatory food chain. It begins with the green plants (producers) and the second, third and fourth trophic levels are occupied by the herbivores, primary carnivores and secondary carnivores respectively. In some food chains there is yet another trophic level – the climax carnivores. The number of trophic levels in food chains varies from 3 to 5 generally. Some examples for grazing food chain (GFC) are given below.
TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment 5 TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment 6

II. Parasitic food chain :
Some authors included the ‘Parasitic Food Chains’ as a part of the GFC. As in the case of GFCs, it also begins with the producers, the plants (directly or indirectly). However, the food energy passes from large organisms to small organisms in the parasitic chains. For instance, a tree which occupies the 1st trophic level provides shelter and food for many birds. These birds host many ectoparasites and endo-parasites. Thus, unlike in the predator food chain, the path of the flow of energy includes fewer, l^rge sized organisms in the lower trophic levels, and numerous, small sized organisms in the successive higher trophic levels.

Question 40.
What are the ecological limitations for ecological pyramids?
Answer:
Limitations of Ecological Pyramids :
There are certain limitations of ecological pyramids, such as-

  1. It does not take into account the same species belonging to two or more trophic levels,
  2. It assumes a simple food chain, something that almost never exists in nature,
  3. It does not accommodate a food web,
  4. moreover, saprophytes are not given any place in ecological pyramids even though they play a vital role in the ecosystem.

Question 41.
How is the second law of thermodynamics applicable to the functional part of an ecosystem?
Answer:
The ecosystems are not exempted from the Second Law of thermodynamics. It states that no process involving energy transformation will spontaneously occur unless there is degradation of energy. As per the second law of thermodynamics – the energy dispersed is in the form of unavailable heat energy, and constitutes the entropy (energy lost or not available for work in a system). The organisms need a constant supply of energy to synthesize the molecules they require. The transfer of energy through a food chain is known as energy flow.

A constant input of mostly solar energy is the basic requirement for any ecosystem to function. The important point to note is that the amount of energy available decreases at successive trophic levels. When an organism dies, it is converted to detritus or dead biomas that serves as a source of energy for the decomposers. Organisms at each trophic level depend on those at the lower trophic level, for their energy demands.

TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 42.
Discuss the main reason for the low productivity of ocean. (T.Q.)
Answer:
The element carbon constitutes 49 percent of the dry weight of organisms and is next only to water. Among the total carbon quantity present on the Earth, 71 percent is found dissolved in oceans. This ‘oceanic reservoir’ regulates the amount of carbon dioxide in the atmosphere. It will be interesting to know that the atmosphere only contains about 1 percent of the total global carbon. This is the main reason for the low productivity of ocean.

Question 43.
Explain the terms saprotrophs, detritivores, and mineralizers. (T.Q.)
Answer:
a) Saprotrophs :
The detritus food chain (DFC) begins with dead organic matter (such as leaf litter, bodies of dead organisms). It is made up of decomposers which are heterotrophic organisms, mainly the fungi and bacteria. They meet their energy and nutrient requirements by degrading dead organic matter or detritus. These are also known as saprotrophs.

b) Detritivores :
Plant parts such as leaves, bark, flowers and dead remains of animals, including their faecal matter, constitute the detritus. Detritus is the raw material for the decomposition.

Detritivores (eg : earthworm) break down detritus into smaller particles and this process is called fragmentation. Bacterial and fungal enzymes degrade detritus into simpler inorganic substances. This process is called as catabolism.

c) Mineralizers :
It is important to note that all the steps in decomposition operate simultaneously on the detritus. Humification and mineralization occur during decompositon in the soil. Humification leads to accumulation of a dark coloured amorphous substance called humus that is highly resistant to microbial action and undergoes decompositon at an extremely slow rate. Being colloidal in nature it serves as a reservoir of nutrients. The humus is further degraded by some microbes and release of inorganic nutrients occurs by the process known as mineralization. Hence the microbes are called mineralizers.

Question 44.
Discuss the factors that influence the process of decomposition.
(OR)
Question 45.
Define decomposition and describe the process and products of decomposition. (T.Q.)
Answer:
Decomposition :
Decomposition is largely an oxygen-requiring process. The rate of decomposition is controlled by chemical composition of the detritus and climatic factors. In a particular climatic condition, decomposition rate is slower if detritus is rich in lignin and chitin, and quicker, if detritus is rich in nitrogen and water-soluble substances like sugars. Temperature and soil moisture are the most important climatic factors that regulate decomposition through their effects on the activities of soil microbes.

Warm and moist environment favours decomposition whereas low temperature and anaerobic environment ‘inhibit’ the decomposition resulting in build up of organic materials. As most of the decomposers are very small microscopic forms, they are also called ‘micro-consumers’.

Question 46.
Explain GFC.
Answer:

Question 47.
Write a short notes on the various trophic levels in a typical ecosystem. [March 2013]
Answer:
Energy flows into biological systems (ecosystems) from the Sun. The biological systems of environment include several food levels called trophic levels. A trophic level is composed of those organisms which have the same source of energy and having the same number of steps away from the sun. Thus a plant’s trophic level is one, while that of a herbivore – two, and that of the first level carnivore – three.

The second and third levels of the carnivores occupy fourth and fifth trophic levels respectively.
TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment 7

A given organism may occupy more than one trophic level simultaneously. One must remember that the trophic level represents a functional level. A given species may occupy more than one trophic level in the same ecosystem at the same time; for example, a sparrow is a primary consumer when it eats seeds, fruits, and a secondary consumer when it eats insects and worms.
Trophic levels in an ecosystem

Question 48.
Give examples for GFCs with three, four and five trophic levels from your locality. Explain the parasitic food chain. How does it differ from GFC?
Answer:
Grazing food chain :
TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment 8

Question 49.
Write a note on DFC. Give its significance in a terrestrial ecosystem. (T.Q.) [March 2018 – A.P.]
Answer:
Detritus Food Chain (DFC): Dead organisms of the grazing food chain form a source of energy for other organisms. Similarly, waste materials passed from the bodies of living organisms are also a source of energy for some organisms called decomposers like saprotrophs, detritivores and mineralizers. Saprotrophs are microorganisms such as fungi and bacteria, which live on dead organic matter. Detritivores ingest small fragments of decomposing organic materials, termed detritus. Mineralisers effect the mineralisation of humus.

Detritivores may be eaten by the carnivores, which may then be consumed by other carnivores, thus building up a food chain based on detritus. This is termed detritus food chain. Those feeding directly on the dead bodies or detritus are primary detritus feeders and those preying on these organisms are secondary detritus feeders.

Two typical detritus food chains of the woodland ecosystem are :
TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment 9

Question 50.
What is primary productivity? Give a brief description of the factors that affect primary productivity. (T.Q.)
Answer:
The rate of production of biomass is called productivity. It can be divided into primary and secondary productivities.

I. Primary productivity is defined as the amount of biomass or organic matter produced per unit area over a period of time by plants, during photosynthesis. It can be divided into gross primary productivity (GPP) and net primary productivity (NPP).

a) Gross primary productivity of an ecosystem is the rate of production of organic matter during photosynthesis. A considerable amount of GPP is utilized by plants for their catabolic process (respiration).

b) Net primary productivity Gross primary productivity minus respiratory loss (R), is the net primary productivity (NPP). On average about 20 – 25 percent of GPP is used for the catabolic (respiratory) activity.
GPP-R = NPP

The net primary productivity is the biomass available for the consumption of the heterotrophs (herbivores and decomposers).

II. Secondary productivity is defined as the rate of formation of new organic matter by consumers.

TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 51.
Define ecological pyramids and describe with examples, pyramids of numbers and biomass. (T.Q.)
Answer:
Ecological Pyramids (Eltonian pyramids) :
You must be familiar with the shape of a pyramid. The base of a pyramid is broad and it narrows down towards the apex. The trophic relationship is expressed in terms of numbers; biomass or energy arranged one on the top of its lower trophic level, resulting in a pyramidal shape. It is a graphical representation of the trophic structure and function of an ecosystem.

The base of each pyramid represents the producers or the first trophic level, while the apex represents the tertiary or top level / top order consumer. The three types of ecological pyramids that are usually studied are (a) pyramid of number, (b) pyramid of biomass and (c) pyramid of energy. These pyramids were first represented by Elton, hence the name ELTONIAN pyramids / Ecological pyramids.

Any calculations of energy content, biomass, or numbers has to include all organisms at that trophical evel. No generalizations we make will be true if we take only a few individuals of any trophic level into account. In most ecosystems, all the pyramids – of numbers, energy and biomass are upright i.e., producers are more in number and biomass than the herbivores, and herbivores are more in number and biomass than the carnivores. Also energy (available) at a lower trophic level is always more than that at a higher level.

There are exceptions to this generalization. In the case of a parasitic food chain, the pyramid of numbers is inverted.

Question 52.
Can you work out the number of trophic levels at which human beings can function in a food chain?
Answer:
Man occupies 3rd trophic level being represented by secondary consumer. Some times man also occupies 2nd tropic level as he becomes primary consumer feeding only on producers.

If he feeds on animals like fish, goat, prawns etc., he becomes secondary consumer in a food chain. But never goes to 4th trophic level of tertiary consumer.

So total number of trophic levels at which human beings can function in a food chain are two or three levels.

Question 53.
Measurement of biomass in terms of dry weight is more accurate. Justify.
Answer:
Each trophic level has a certain mass of living material at a particular time, and it is called the standing crop. The standing crop is measured as the “mass” of living organisms (biomass) or the number of organisms per unit area. The biomass of a species is expressed in terms of fresh or dry weight (dry weight) is more accurate because water contains no usable energy.

Question 54.
What are the deleterious effects of depletion of ozone in the stratosphere? [March 2020, ’19, May/June 2014] (T.Q.)
Answer:
The depletion of ozone is particularly marked over the Antarctic region. This has resulted in the formation of a large area of thinned ozone layer, commonly called as the ‘ozone hole’.

UV radiation with wavelengths shorter than that of UV-B, are almost completely absorbed by Earth’s atmosphere, provided that the ozone layer is intact. But UV-B damages DNA and may induce mutations. It causes ageing of skin, damage to skin cells and various types of skin cancers. In human eye, cornea absorbs UV-B radiation, and a high dose of UV-B causes inflammation of cornea, called snow blindness, cataract, etc. Such exposure may permanently damage the cornea.

Question 55.
Write a note on ‘algal blooms’.
Answer:
Algal blooms: Presence of large amounts of nutrients in waters also causes excessive growth of planktonic algae and the phenomenon is commonly called ‘algal blooms’. Algal blooms impart distinct colour to the water bodies and deteriorate the quality of water. It also causes mortality of fish. Some algae which are involved in algal blooms are toxic to human beings and animals.

Excessive growth of aquatic plants such as the common water hyacinth (Eichhornia crassipes), the world’s most problematic aquatic weed which is also called ‘Terror of Bengal’ causes blocks in our water ways. They grow faster than our ability to remove them. They grow abundantly in eutrophic water bodies (water bodies rich in nutrients) and lead to imbalance in the ecosystem dynamics of the water body.

Question 56.
Describe ‘Greenhouse Effect’. [May 17; Mar. 2020, 17, 14] (T.Q)
Answer:
The greenhouse effect is a naturally occurring phenomenon that is responsible for heating of the Earth’s surface and atmosphere. It would be surprising to know that without greenhouse effect the average temperature of the Earth’s surface would have been a chilly – 18° C rather than the present average of 15° C.

When sunlight reaches the outermost layer of the atmosphere, clouds and gases reflect about one-fourth of the incoming solar radiation, and absorb some of it. Almost half of the incoming solar radiation falls on the Earth’s surface and heats it up, while a small proportion is reflected back. The Earth’s surface re-emits heat in the form of infrared radiation but part of this does not escape into space as atmospheric gases (e.g., carbondioxide, methane, etc.,) absorb a major fraction of it. The molecules of these gases radiate heat energy, and a major part of which again comes back to the Earth’s surface, thus heating it up once again.The above – mentioned gases – carbon dioxide and methane – are commonly known as greenhouse gases, because they are responsible for the greenhouse effect.

Increase in the level of greenhouse gases has led to considerable heating of the Earth leading to global warming.

Question 57.
Write notes on the following.
a) Radioactive waste disposal
b) e-wastes management
Answer:
Radioactive wastes :
Initially, nuclear energy was hailed as a non-polluting way for generating electricity. Later on, it was realised that the use of nuclear energy has two very serious inherent problems. The first is accidental leakages, as occurred in the Three Mile Island (USA) and Chernobyl (Russia) and the second is the safe disposal of radioactive wastes.

Radiation, that is released from nuclear waste is extremely dangerous to biological organisms, because it induces mutations. Exposure to high doses of nuclear radiation is lethal as it can lead to cancers (e.g. leukemia). Therefore, nuclear waste is an extremely potent pollutant and has to be dealt with utmost caution. Storage of nuclear wastes should be done in suitably shielded containers and buried deep in the soil or oceans (about 500 meters). Even when done so, geological upheavals can bring them up, some day and cause radiation.

Electronic wastes (e-wastes) :
Irreparable computers and other electronic goods constitute the modern day pollutants called electronic wastes (e-wastes), wases). E-wastes are buried in landfills or incinerated. Over half of the e-wastes generated in the developed world are exported to developing countries, mainly to China, India and Pakistan, where metals like copper, iron, silicon, nickel and gold are recovered during recycling process.

Unlike developed countries, which have specifically built facilities for recycling of e-wastes, recycling in developing countries often invovles manual participation thus exposing workers to toxic substances present in e-wastes. Eventually recycling is the only solution for the treatment of e-wastes provided it is carried out in an environmental friendly manner.

Question 58.
Discuss the role of women and communities in protection and conservation of forests in India.
Answer:
People’s participation in ‘protecting forests’ has a long history in India. In 1731, the king of Jodhpur in Rajasthan asked one of his ministers to arrange wood for constructing a new palace. The minister and workers went to a forest near a village, inhabited by Bishnois, to cut down trees. The Bishnoi community is known for its peaceful co-existence with nature. The effort to cut down trees by the king was thwarted by the Bishnois. A Bishnoi woman Amrita Devi showed exemplary courage lost their lives in their effort to save trees.

Nowhere in history do we find a commitment of this magnitude when human beings sacrificed their lives for the cause of the environment. The Government of India has recently instituted the “Amrita Devi Bishnoi Wildlife Protection Award” for individuals or communities from rural areas that have shown extraordinary courage and dedication in protecting wildlife.

The Chipko Movement of Garhwal Himalayas :
In 1974, local women showed enormous bravery in protecting trees from the axe of contractors by hugging the trees. People all over the world have acclaimed the Chipko movement. Realising the significance of participation by local communities the Governemnt of India in 1980s has introduced the concept of Joint Forest Management (JFM) so as to work closely with the local communities for protecting and managing forests. In return for their services to the forest, the communities get benefit of various forest products (e.g. frutis, gum, rubber, medicine, etc.).

Question 59.
Discuss briefly the following : (T.Q.)
a) Greenhouse gases
b) Noise pollution
c) Organic farming
d) Municipal solid wastes
Answer:
a) Greenhouse gases :
Carbon dioxide and methane are commonly known as greenhouse gases, because they are responsible for the greenhouse effect.

b) Noise pollution :
In India, Air (Prevention and control of pollution) Act came into force in 1981. In 1987 it was amended to include noise also as an air pollutant. Undesirably high sound constitute NOISE POLLUTION. Sound is measured in units called decibels (dB). The human ear is sensitive to sounds ranging from 0 to 180 dB. 0 dB is threshold limit of hearing and 120 dB is threshold limit for sensation of pain in the ear. Any noise above 120 dB is considered to be a noise pollution.

A brief exposure to extremely high sound level, 150 dB or more generated by jet planes while taking-off may damage eardrums causing permanent hearing impairment. Even long term exposure to a relatively lower level of noise of cities may also cause hearing impariment. Noise also causes auditory fatigue, anxiety, sleeplessness (insomnia), increased heart beat, altered breathing pattern thus causing considerable stress to humans.

c) Organic farming :
Integrated organic farming is a zero-waste procedure, where recycling of waste products is efficiently carried out. Wastes originated from one process are used as nutrients for other processes. This allows the maximum utilisation of resource and increases the efficiency of production. A method practised by Ramesh Chandra Dagar, a farmer in Sonipat, Haryana, is a very good example for this. He integrated bee-keeping, dairy management, water harvesting, composting and agriculture in a chain of processes.

All these processes support one another and allow an extremely economical and substainable venture. Crop waste and cattle excreta (dung) are used to create compost, which can be used as a natural fertilizer. Natural bio-gas generated in the process can be used for meeting the energy needs of the farm. Enthusiastic about spreading information and helping in the practice of integrated organic farming, Dagar has created by Haryana Kisan Welfare Club.

d) Municipal solid wastes :
Any thing (substance/ material / articles / goods) that is thrown out as waste in solid form is referred to as solid waste. Municipal solid wastes are wastes from homes, offices, institutions, shops, hotels, restaurants etc., in towns and cities.

The municipal solid wastes generally consist of paper, food wastes, plastics, glass, metals, rubber, leather, textile, etc. The wastes are burnt to reduce the volume of the wastes. But generally wastes are not completely burnt and left as open dumps which often serve as the breeding grounds for rats and flies. As the substitute for open-burning dumps, sanitary landfills are adopted. In a sanitary landfill, wastes are dumped in a depression or trench after compaction, and covered with dirt everyday. These is a danger of seepage of chemicals and pollutants from these landfills, which may contaminate the underground water resources.

Question 60.
Discuss the causes and effects of global warming. What measures need to be taken to control ‘Global Warming’? [March 2015 – T.S. & A.P.] (T.Q.)
Answer:
Increase in the level of greenhouse gases has led to considerable heating of the earth leading to global warming. During the past century, the temperature of the earth has increased by 0.6°C. Most of it during the last three decades. Scientists believe, that this rise in temperature is leading to severe changes in the environment. Global warming is causing climate changes and is also responsible for the melting of polar ice. caps and other snow caps of mountains such as the Himalayas. Over many years, this will result in a rise in sea levels all over the world that can sub-merge many coastal areas. The total spectrum of changes that gldbal warming can bring about is a subject that is still under active research.

Global warming – Control measures :

  1. The measures include cutting down use of fossil fuels
  2. Improving efficiency of energy usage
  3. Planting of trees, and avoiding deforestation
  4. Slowing down the growth of human population.

TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 61.
Write critical notes on the following : (T.Q.)
a) Eutrophication
b) Biological magnification
c) Ground water depletion and ways for its replenishment.
Answer:
a) Eutrophication :
Natural ageing of a lake by nutrient enrichment of its water is known as eutrophication. In a young lake, the water is cold and clear, supporting little life. Gradually nutrients, such as nitrates and phosphates are carried into the lake via streams, in course of time. This encourages the growth of aquatic algae and other plants. Consequently, the animal life proliferates, and organic matter gets deposited on the bottom of the lake. Over centuries, as silt and organic debris piles up, the lake grows shallower and warmer. As a result, the aquatic organisms thriving in the cold environment are gradually replaced by warm-water organisms.

Marsh plants appear by taking root in the shallow regions of the lake. Eventually, the lake gives way to large masses of floating plants (bog) and finally converted into land.
TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment 10

b) Biomagnification :
Increase in the concentration of the pollutant or toxicant at successive trophic levels in an aquatic food chain is called Biological Magnification or Biomagnification. This happens in the instances where a toxic substance accumulated by an organism is not metabolized or excreted and thus passes on to the next higher trophic level. This phenomenon is well known regarding DDT and mercury pollution.

As shown in the above example, the concentration of DDT is increased at successive trophic levels starting at a very low concentration of 0.003 ppb (ppb = parts per billion) in water, which ultimately reached an alarmingly high concentration of 25 ppm (ppm = parts per million) in fish-eating birds, through biomagnification. High concentrations of DDT disturb calcium metabolism in birds, which causes thinning of egg shell and their premature breaking, eventually causing decline in bird populations.

c) Ground water depletion and ways for its replenishment :
Sewage arising from homes and hospitals may contain undesirable pathogenic microorganisms. If it is released untreated into water courses, there is a likelihood of outbreak of serious diseases, such as dysentery, typhoid, jaundice, cholera etc.

Untreated industrial effluents released into water bodies pollute most of the rivers, fresh water streams, etc. Effluents contain a wide variety of both inorganic and organic pollutants such as oils, greases, plastics, metallic wastes, suspended solids and toxins. Most of them are non-degradable. Arsenic, Cadmium, Copper, Chromium, Mercury, Zinc, and Nickel are the common heavy metals discharged from industries.

Effects :
Orgnic substanes present in the water deplete the dissolved oxygen content in water by increasing the BOD (Biological Oxygen Demand) and COD (Chemical Oxygen Demand). Most of the inorganic substances render the water unfit for drinking.

Removal of dissolved salts such as nitrates, phosphates and other nutrients and toxic metal ions and organic compounds is much more difficult. Domestic sewage primarily contains biodegradable organic matter, which will be readily decomposed by the action of bacteria and other microorganisms.

Water-logging and soil salinity :
Irrigation without proper drainage of water leads to water-logging in the soil. Besides affecting the crops, water-logging draws salt to the surface of the soil (salinisation of the top soil). The salt then is deposited as a thin crust on the land surface or starts collecting at the roots of the plants. This increased salt content is inimical (unfavourable) to the growth of crops and is extremely damaging to agriculture. Water-logging and soil salinity are some of the problems that have come in the wake of the Green Revolution.

Essay Answer Type Questions

Question 1.
Write an essay on temperature as an ecological factor. (T.Q.)
Answer:
Temperature :
Temperature is a measure of the intensity of heat. The temperature on land or in water is not uniform. On land the temperature variations are more pronounced when compared to the aquatic medium, because land absorbs or loses heat much quicker than water. The temperature on land depends on seasons and the geographical area on this planet. Temperature decreases progressively when we move from the equator to the poles. Altitude also causes variations in temperature. For instance, the temperature decreases gradually as we move to the top of the mountains.

The Effects of Temperature in Lakes :
Thermal Stratification :
Temperature variations occur with seasonal changes in the temperature regions. These differences in the temperature form ‘thermal layers’ in water. These phenomena are called thermal stratifications.

Water shows maximum density at 4°C. Rise or fall of temperatures above or below 4°C decreases its density. This anomalous property of water and the seasonal variations in temperature are responsible for the thermal stratification in temperate lakes.
TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment 2

Summer stratification :
During summer in temperate lakes, the density of the surface water decreases because of increase in its temperature (21-25° C). This ‘upper more warm layer’ of a lake is called epilimnion. Below the epilimnion there is a zone in which the temperature decreases at the rate rate of 1°C per meter in depth, and it is called thermocline or metalimnion. The bottom layer is the hypolimnion, where water is relatively cool, stagnant and with low oxygen content (due to absence of photosynthetic activity).

TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment 1
During autumn (also called fall), the epilimnion cools down, and the surface water becomes heavy when the temperature is 4° C, and sinks to the bottom of the lake. Overturns bring about ‘uniform temperature’ in lakes during that period. This circulation during the autumn is known as the fall or autumn overturn. The upper oxygen rich water reaches the hypoliminion and the nutrient rich bottom water comes to the surface. Thus there is uniform distribution of nutrients and oxygen in the lake.

Winter stratification / stagnation :
The ‘Fall’ is followed by ‘Winter’. In this season the surface water cools down. The upper water freezes when the temperature reaches 0°C. Below the upper icy layer, the cool (4° C) water occupies the lake. The aquatic animals continue their life below the icy layer. At lower temperatures the activity of bacteria and the rate of oxygen consumption by aquatic animals decrease. Hence, organisms can survive below the frozen upper water without being subjected to ‘hypoxia’ (low oxygen availability).

In the ‘Spring season’ the temperatures start rising. When it reaches 4°C, the water becomes more dense and heavy and sinks to the bottom, taking ‘oxygen rich water’ to the bottom. The upper oxygen rich water sinks down and the bottom ‘nutrient rich water’ reaches the surface. It is called ‘spring overturn’. The lakes which show overturns twice a year are called ‘dfmictic lakes’. Thus ‘stratifications’ and ‘overturns’ help survival of organisms at all levels in deep lakes.
TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment 11

Biological effects of Temperature :
Temperature Tolerance :
A few organisms can tolerate and thrive in a wide range of temperatures they are called eurythermal, but, a vast majority of organisms are restricted to a narrow range of temperatures (such organisms are called stenotherma. The levels of thermal tolerance of different species determine their geographical distribution.

Temperature and Metabolism :
Temperature affects the working of enzymes and through it, the basal metabolism, and other physiological functions of organism. The temperature at which the metabolic activities occur at the climax level is called the ‘optimum temperature. The lowest temperature at which an organism can live indefinitely is called minimum effective temperature. If an animal or plant is subjected to a temperature below the minimum effective limit, it enters into a condition of inactiveness called chilLcoma. The metabolic rate increases with the rise of temperature from the minimum effective temperature to optimum temperature.

The maximum temperature at which a species can live indefinitely in an active state is called maximum effective temperature. If the temperature is raised above the maximum effective temperature, the animals enter into ‘heat coma’. The maximum temperature varies much in different animals.

van’t Hoff’s rule :
van’t Hoff, a Nobel Laureate in thermochemistry, proposed that, with the increase of every 10°C, the rate of metabolic activities doubles. This rule is referred to as the van’t Hoff’s rule, van’t Hoffs rule can also be stated in reverse saying that the reaction rate is halved with the decrease of every 10°C. The effect of temperature on the rate of a reaction is expressed in terms of temperature coefficient or Q10 value. Q10 values are estimated taking the ratio between the rate of a reaction at X°C and rate of reaction at (X – 10°C). In the ‘living systems’ the Q10 value is about 2.0. If the Q10 value is 2.0, it means, for every 10° C increase, the rate of metabolism doubles.

Cyclomorphosis :
The cyclic seasonal morphological variations among certain organisms is called cyclomorphosis. This phenomenon has been demonstrated in the cladoceran (a sub group of Crustacea) Daphnia (water flea). In the winter season the head of Daphnia is ’round’ in shape (typical or non-helmet morph). With the onset of the spring season, a small ‘helmet’/ ‘hood’ starts developing on it. The helmet attains the maximum size in summer. In ‘autumn’ the helmet starts receding. By the winter season, the head becomes round. Some scientists are of the opinion that cylomorphosis is a seasonal adaptation to changing densities of the water in
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lakes – in summer as the water is less dense Daphnia requires a larger body surface to keep floating easily. During winter the water is more dense, and so it does not require a larger surface area of the body to keep floating. Others believe that these cyclic changes are adaptations to ‘stabilize the movement’ in water. Compared to the ’typical morphs’, the ‘helmeted morphs’ can resist the water currents better to stay in the water rich in food materials.

Temperature adaptations :
Temperature adaptations in animals can be dealt under three heads a) Behavioural adaptations, b) Morphological and Anatomical adaptations and c) Physiological adaptations.

a) Behavioural adaptations :
Some organisms show behavioural responses to cope with variations in their environment. Desert lizards manage to keep their body temperature fairly constant by behavioural means. They ‘bask’ (staying in the warmth of sunlight) in the sun and absorb heat when their body temperature drops below the comfort zone^ but move into shade when the temperature starts increasing. Some species are capable of burrowing into the soil to escape from the excessive heat above the ground level.

b) Morphological and anatomical adaptations :
In the polar seas, aquatic mammals such as the seals have a thick layer of fat (blubber) that acts as an insulatpr and reduces the loss of body heat, underneath their skin.

The animals which inhabit the colder regions have larger body size with greater mass. The body mass is useful to generate more heat. As per Bergmann’s rule mammals and other warm blooded animals living in colder regions have less surface area to body volume ratio’, than their counterparts living in the tropical regions.

The small surface area helps to conserve heat. For instance, the body size of American moose/Eurasian elk (Alces alces), increases with the latitudes in which they live. Moose of northern part of Sweden shows 15-20% more body mass than the same species (counterparts) living in the southern Sweden.

Mammals from colder climates generally have shorter earlobes and limbs (extremities of the body) to minimize heat loss. Large earlobes and long limbs increase the surface.area without changing the body volume. This is known as Allen’s rule. For instance, the polar fox, Vulpes lagopus (formerly called Alopex lagopus), has short extremities to minimize the heat loss from the body. In contrast, the desert fox, Vulpes zerda, has large earlobes and limbs to facilitate better ‘heat loss’ from the body.

c) Physiological adaptations :
In most animals, all the physiological functions proceed ‘optimally’ in a narrow temperature range (in humans, it is 37° C). But there are microbes (archaebacteria) that flourish in hot springs and in some parts of deep seas, where temperatures far exceed 100° C. Many fish thrive in Antarctic waters where the temperature is always below zero. Having realized that the abiotic conditions of many habitats may vary over a time period, we now ask – How do the organisms living in such habitats manage with stressful conditions? One would expect that during the course of millions of years of their existence, many species would have evolved a relatively constant internal (within the body) environment.

It permits all biochemical reactions and physiological functions to proceed with maximal efficiency and thus, enhance the over all’ fitness’ of the species. This constancy, could be chiefly in terms of optimal temperature and osmotic concentration of body fluids. So the organism should try to maintain the constancy of its internal environment (homeostasis) despite varying external environmental conditions that tend to upset its homeostasis. This is achieved by the processes described below.

(i) Regulate :
Some organisms are able to maintain homeostasis by physiological (sometimes behavioural also ) means which ensures constant body temperature, constant osmotic concentration, etc. All birds and mammals, and a very few lower vertebrate and invertebrate species are indeed capable of such regulation (thermoregulation and osmoregulation). Evolutionary biologists believe that the ‘success’ of mammals is largely due to their ability to maintain a constant body temperature and thrive whether they live in Antarctica or in the Sahara desert.

The mechanisms used by most mammals to regulate their body temperature are similar to the one that we, the humans use. We maintain a constant body temperature of 37°C. In summer, when outside temperature is more than our body temperature, v.e sweat profusely. The resulting ‘evaporative cooling’ brings down the body temperature. In winter when the temperature is much lower than 37°C, we start to shiver (a kind of exercise which produces heat and raises the body temperature – a type of body’s own defence mechanism against low temperature). Plants, on the other hand, do rvot have such mechanisms to maintain internal temperatures.
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ii) Conform :
Majority (99 percent) of animals cannot maintain a constant internal environment. Their body temperature changes with the ambient (surrounding) temperature. In aquatic animals, the osmotic concentration of the body fluids changes along with that of the surrounding water. Such animals are described as ‘conformers’.

(iii) Partially regulate :
Animals such as ‘camels’ can be ‘conformers’ up to a particular range of temperature and ‘regulator’ afterwards. So, they are described as ‘partial regulators’ or ‘partial conformers’.

Thermoregulation is energetically ‘expensive’ for many organisms. This is particularly true in small animals like shrews and humming birds. Heat loss or heat gain is a function of the surface area. Since small animals have a larger surface area relative to their volume, they tend to lose body heat very fast when it is cold outside; then they have to spend much energy to generate Body heat through metabolism. This is the main reason why very small animals are rarely found in the ‘polar regions’. During the course of evolution, the costs and benefits of maintaining a constant internal environment are taken into consideration. Some species have evolved the ability to regulate, but only over a limited range of environmental conditions, beyond which they simply conform.

If the stressful external conditions are localized or remain only for as short duration, the organism has two other alternatives.

(iv) Migrate :
The organism can move away temporarily from the ‘stressful habitat’ to a more ‘hospitable’ (comfortable) area and return when the stressful period is over. In human analogy (comparison), this strategy is comparable a person moving from Delhi to Shimla for the duration of summer. Many animals, particularly birds, during winter undertake long-distance migrations to more hospitable areas. Every winter, many places in India including the famous Keoladeo or Keoladeo Ghana National park (Formerly – Bharatpur bird sanctuary) in Rajasthan and Pulicat Lake in Andhra Pradesh host thousands of ‘migratory birds’ coming from Siberia and other extremely cold northern regions.

(v) Suspend life activities :
In bacteria, fungi and lower plants, various kinds of thick-walled spores are formed which help them survive unfavoruable conditions. They germinate (come out of the spore wall and produce a normal active organisms) on the return of suitable environmental conditions.

Some animals can avoid the stress by escaping in ‘time’ (migration is – escaping in space’). The familiar case of ‘Polar bears’ going into hibernation during winter is an example of escape in time. Some snails and fish go into aestivation to avoid summer-related problems – heat and desiccation.

Diapause :
Certain organisms show delay in development, during periods of unfavourable environmental conditions and spend some period in a state of ‘inactiveness’ called ‘diapause’. This dormant period in animals is a mechanism to survive extremes of temperature, drought, etc. It is seen mostly in insects and embryos of some fish. Under unfavourable conditions many zooplankton species in lakes and ponds are known to enter diapause.

TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 2.
Write an essay on water as an ecological factor. (T.Q.)
Answer:
Water :
Water is another important factor influencing the life of organisms. Life is unsustainable without water. Its availability is so limited in deserts that only certain special adaptations make it possible for them to live there. You might think that organisms living in oceans, lakes and rivers should not face any water-related problems, but it is not true. For aquatic organisms the quality (chemical composition, pH, etc.,) of water becomes important.

The salt concentration is less than 5 percent in inland waters, and 30 – 35 percent in the seawater. Some organisms are tolerant to a wide range of salinities (euryhaline), but others arerestricted to a narrow range (stenohaline). Many freshwater animals cannot live for long in sea water and vice versa because of the osmotic problems, they would face.

Adaptations in freshwater habitat: Animals living in freshwaters have to tackle the problem of endosmosis. The osmotic pressure of freshwater is very low and that of the body fluids of freshwater organisms is much higher. So water tends to enter into bodies by endosmosis. To maintain the balance of water in the bodies, the freshwater organisms acquired several adaptation such as, contractile vacuoles in the freshwater protozoans, large glomerular kidneys in fishes, etc. They send out large quantities of urine, along which some salts are also lost.

To compensate the ‘salt loss’ through urine, freshwater fishes have ‘salt absorbing’ ‘chloride cells’ in their gills. The major problem in freshwater ponds is – in summer most of the ponds dry up. To overcome this problem most of the freshwater protists undergo encystment. The freshwater sponges produce asexual reproductive bodies, called gemmules, to tide over the unfavourable conditions of the summer. The ‘African lungfish’, Protopterus, burrows into the mud and forms a ‘gelatinous cocoon’ around it, to survive, in summer.

Adaptations in marine habitat :
Seawater is high in salt content compared to that of the body fluids. So, the marine animals continuously tend to lose water from their bodies by exosmosis and face the problem of dehydration. To overcome the problem of water loss, marine fishes have aglomerular kidneys with less number of nephrons. Such kidneys minimize the loss of water through urine. To compensate water loss the marine fish drink more water, and along with this water, salts are added to the body fluids and disturb the internal equilibrium.

To maintain salt balance (salt homeostasis) in the body, they have salt secreting chloride cells in their gills. Marine birds like sea gulls and penguins eliminate salts in the form of salty fluid that drips through their nostrils. In turtles the ducts of chloride secreting glands open near the eyes. Some cartilaginous fishes retain urea and trimethylamine oxide (TMO) in their blood to keep the body fluid isotonic to the sea water and avoid dehydration of the body due to exosmosis.

Water related adaptations in brackish water animals :
The animals of brackish water are adapted to withstand wide fluctuations in salinity. Such organisms are called euryhaline animals and those that can’t withstand are known as stenohaline. The migratory fishes such as salmon and Hilsa are anadromous fishes i.e., they migrate from the sea to freshwater, for breeding; Anguilla bengalensis is a catadromous fish i.e., it migrates from the river to sea, for breeding. In these fishes their glomerular kidneys are adjusted to changing salinities.

The chloride cells are adapted to excrete or absorb salts depending on the situation. On entering the river salmon drinks more freshwater to maintain the concentration of body fluids equal to that of the surround water.

Water related adaptations for terrestrial life :
In the absence of an external source of water, the kangaroo rat of the North American deserts is capable of meeting all its water requirements through oxidation of its internal fat (in which water is a by product – metabolic water). It also has the ability to concentrate its urine, so that minimal volume of water is lost in the process of removal of their excretory products.

Question 3.
Give an account of various types of interactions among the animal species of an ecosystem.
Answer:
Inter – specific Interactions :
Inter – specific interactions arise from the interaction of populations of two different species. They could be beneficial, detrimental or neutral (neither harmful nor beneficial) to one of the species or both. Assigning a V sign for beneficial interaction, sign for detrimental and ‘O’ for neutral interaction, let us look at all the possible outcomes of inter-specific interactions.

The interactions between species are grouped into four types. They are mutualism, commensalism, parasitism and amensalism. Both the species benefit in mutualism and both lose in competition in their interactions with each other. The interaction where one species is benefited and the other is neither benefited nor harmed is called commensalism. In amensalism on the other hand one species is harmed whereas the other is unaffected. In both parasitism and ‘predation’only one species benefits (parasite and predator, respectively) and the interaction is detrimental to the other species (host and prey, respectively). Predation, parasitism and commensalisms share a common characteristic – the interacting species live closely together.

Population Interactions – Types

Name of InteractionSpecies ASpecies B
Mutualism++
Competition
Predation+
Parasitism+
Commensalism+0
Amensalism_0

Predation :
What would happen to all the energy fixed by autotrophic organisms if the community has no animals to eat the plants? We can think Of predation as nature’s way of transferring the energy fixed by plants to higher trophic levels. When we think of predator and prey, most probably it is the tiger and the deer that readily come to our mind, but a sparrow eating any seed is also a type of predator (a seed predator also called granivore). Although animals eating plants are categorized separately as herbivores, they are, in a broad ecological context, not very different from predators.

Besides acting as ‘conduits’ / ‘pipelines’ for energy transfer across trophic levels, predators play other important roles. They keep the prey populations under control. In the absence of predators, the prey species could achieve very high population densities and cause instability in the ecosystem. Predators have different types of functions to play in nature. They include :

A. Predator as a biological control :
The prickly pear cactus introduced into Australia in the early 1920s caused havoc by spreading rapidly into millions of hectares of rangeland (vast natural grass lands). Finally, the invasive cactus was brought under control only after a cactus feeding predator (a moth) was introduced into the country. Biological control methods adopted in agricultural pest control are based on the ability of the predators to regulate prey populations.

B. Predators maintain ‘species diversity’ :
Predators also help in maintaining species diversity in a community, by reducing the intensity of competition among competing prey species. In the rocky intertidal communities of the American Pacific Coast, the starfish Pisaster is an important predator. In a field experiemnt, when all the starfish were removed from an enclosed intertidal area, more than 10 species of invertebrates became extinct within a year, because of increased inter-specific competition.

C. Predators are prudent (practical) pertaining to preys :
If a predator is too efficient and overexploits its prey, then the prey might become extinct and following it, the predator will also become extinct due to lack of food. This is the reason why predators in nature are ‘prudent’.

Prey species have evolved various defenses to lessen the impact of predation they include :

a) Preys fool (deceive) or avoid their predators :
Some species of insects and frogs are cryptically – coloured (camouflaged) to avoid being detected easily by the predator. Some are poisonous and therefore avoided by the predators.

b) Preys defend by becoming distasteful to predators :
The Monarch butterfly is highly distasteful to its predator (bird) because of a special chemical present in its body. Interestingly, the butterfly acquires this chemical during its caterpillar stage by feeding on a poisonous weed.

c) Plants too have their defensive mechanisms :
For plants, herbivores are the predators. Nearly 25 percent of all insects are known to be phytophagous (feeding on plant sap and other parts of plants). The problem is particularly severe for plants because, unlike animals, they cannot escape from their predators. Plants therefore have evolved a variety of morphological and chemical defences against herbivores.

i) Thorns (Acacia, Cactus, etc.,) are the most common morphological means of defense. Many plants produce and store chemicals that make the herbivore sick when they are eaten, inhibit feeding or digestion, disrupt its reproduction or even kill it.

ii) We must have seen the weed Calotropis growing in abandoned fields. The plant produces highly poisonous cardiac glycosides and that is why we never see any cattle or goats browsing on this plant.

iii) A wide variety of chemical substances that we extract from plants on a commercial scale (nicotine, caffeine, quinine, strychnine, opium, etc.,) are produced by them actually as defences against grazers and browsers. Competition: When Darwin spoke of the struggle for existence and survival of the fittest in nature, he was convinced that interspecific competition is a ’potent force’ in the process of organic evolution, involving Nature Selection. It is generally believed that competition occurs when closely related species compete for the same resources that are limited, but this is not entirely true.

Parasitism :
Considering that the parasitic mode of life ensures free ‘lodging’ and ‘meals’, it is not surprising that parasitism has evolved in so many taxonomic groups from plants to higher vertebrates. Many parasites have evolved to be host-specific (they can parasitize only a specific species of host) in such a way that both host and the parasite tend to co-evolve; that is, if the host evolves special mechanisms for rejecting or resisting the parasite, the parasite has to evolve mechanisms to ‘counteract’ and ‘neutralize’ them, in order to continue successful parasitic relationship with the same host species. In order to leacl successful parasitic life, parasites evolved special adaptations, such as :
In order to lead successful parasitic life, parasites evolved special adaptations such as.

  1. Loss of sense organs (which are not necessary for most parasites).
  2. Presence of adhesive organs such as suckers, hooks tq cling on to the host’s body parts.
  3. Loss of digestive system and presence of high reproductive capacity.
  4. The life cycles of parasites are often complex, involving one or two intermediate hosts or vectors to facilitate parasitisation of their primary hosts.

e.g : 1: The human liver fluke depends on two intermediate (secondary ) hosts (a snail anda fish)to complete its life cycle.

e.g : 2 : The malaria parasite needs a vector (mosquito) to spread to otehr hosts. Majority of the parasites harm the host. They may reduce the survival, growth and reproduction of the host and reduce its population density. They might render the host more vulnerable to predation by making it physically weak.
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Commmensalism :
This is the interaction in which one species benefits and the other is neither harmed nor benefited. Barnacles growing on the back of a whale benefit while the whale derives no noticeable benefit.

Mutualism :
This type of interaction benefits both the interacting species.

The most common examples of mutualism are found in plant-animal relationships. Plants need the help of animals for pollinating their flowers and dispersing their seeds. Animals obviously have to be paid ‘fees’ for the services that plants derive from them. Plants offer rewards in the form of pollen and nectar for pollinators and juicy and nutritious fruits for seed dispersing animals.

TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 4.
Describe lake as an ecosystem giving examples for the various zones and the biotic components in it. [March 2015 – T.S ; March 2013] (T.Q.)
Answer:
Lake Ecosystem :
To understand the fundamentals of an aquatic ecosystem, let us take a ‘lake’ as an example. This is fairly a self-sustainable unit and rather a simple example that explains even the complex interactions that exist in an aquatic ecosystem.

Lakes are large inland water bodies containing standing/still water (Recall: Lentic community). They are deeper than ponds (pond is not an ideal example as it is very shallow). Most lakes contain water throughout the year. In deep lakes, light cannot penetrate more than 200 meters, in depth. They are vertically stratified in relation to light intensity, temperature, pressure, etc. Deep water lakes contain three distinct zones namely, i) littoral zone, ii) limnetic zone, and iii) profundal zone.
TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment 4

Littoral zone :
It is the shallow part of the lake closer to the shore. Light penetrates up to the bottom. It is ‘euphoric’ (having good light), has rich vegetation and higher rate of photosynthesis, hence rich in oxygen.

Limnetic zone :
It is the open water zone away from the shore. It extends up to the effective light penetration level, vertically. The imaginary line that separates the limnetic zone from the profundal zone is known as zone of compensation/ compensation point / light compensation level. It is the zone of effective light penetration. Here the rate of photosynthesis is equal to the rate of respiration. Limnetic zone has no contact with the bottom of the lake.

Profundal zone :
It is the deep water area present below the limnetic zone and beyond the depth of effective light penetration. Light is absent. Photosynthetic organisms are absent and so the water is poor in oxygen content. It includes mostly the anaerobic organisms which feed on detritus.

The organisms living in lentic habitat are classified into pedonic forms, which live at the bottom of the lake and those living in the open waters of lakes, away from the shore vegetation are known as limnetic forms.

Biota (animal and plant life of a particular region) of the littoral zone :
Littoral zone is rich with pedonic flora (especially up to the depth of the effective light penetration.) At the shore proper emergent vegetation is a abundant with firmly fixed roots in the bottom of the lake and shoots and leaves are exposed above the level of water. These are amphibious plants. Certain emergent rooted plants of littoral zone are the cattails (Typha), bulrushes (Scirpus), arrowheads (Sagittaria). Slightly deeper are the rooted plants with floating leaves, such as the water lilies (Nymphaea), Nelumbo, Trapa, etc. Still deeper are the submerged plants such as Hydrilla, Chara, Potamogeton, etc. The free floating vegetation includes Pistia, Wolffia, Lemna (duckweed), Azolla, Eichhornia, etc.

The phytoplankton of the littoral zone composed of diatoms (Coscinodiscus, Nitzschia, etc.), green algae (Volvox, Spirogyra, etc.), euglenoids (Euglena, Phacus, etc.), and dinoflagellaes (Gymnodinium, Cystodinium, etc.)

Animals, the consumers of the littoral zone, are abundant in this zone of the lake. These are categorized into zooplankton, neuston, nekton, periphyton, and benthos. The zooplankton of the littoral zone consists of ‘water fleas’ such as Daphnia rotifers and ostracods.

The animals living at the air – water interface constitute the ‘neuston’. They are of two types, the epineuston and hyponeuston. Water striders (Gerris), beetles, water bugs (Dineutes) form the epineustone / supraneuston and the hyponeuston/ infraneuston includes the ‘larvae of mosquitoes’.

The animals such as fishes, amphibians, water snakes, terrapins, insects like ‘water scorpion’ (Ranatra), ‘back swimmer’ (Notonecta), ‘dividing beetles’ (Dytiscus), capable of swimming constitute the nekton.

The animals that are attached to / creeping on the aquatic plants such as the ‘water snails’, ‘nymphs of insects’, ‘bryozoans’, ‘turbellarians’, etc., constitute the ‘periphyton’.

The animals that rest on or move on the bottom of the lake constitute the ‘benthos’ e.g. red annelids, chironomid larvae, cray fishes, some isopods, amphipods, clams, etc.

Biota of the limnetic zone :
Limnetic zone is the largest zone of a lake. It is the region of rapid variations of the level of the water, temperature, oxygen availability, etc., from time to time. The limnetic zone has autotrophs (photosynthetic plants) in abundance. The chief autotrophs of this region are the phytoplankton such as the euglenoids, diatoms, cyanobacteria, dinoflagellates and green algae. The consumers of the limnetic zone are the zooplanktonic organisms such as the copepods. Fishes, frogs, water snakes, etc., form the limnetic nekton.

Biota of the profundal zone :
It includes the organisms such as decomposers (bacteria), chironomid larvae, Chaoborus (Phantom larva), red annelids, clams, etc., that are capable of living in low oxygen levels. The decomposers of this zone decompose the dead plants and animals and release nutrients which are used by the biotic communities of both littoral and limnetic zones.

The lake ecosystem performs all the functions of any ecosystem and of the biosphere as a whole, i.e., conversion of inorganic substances into organic material, with the help of the radiant solar energy by the autotrophs; consumption of the autotrophs by the heterotrophs; decomposition and mineralization of the dead matter to release them back for reuse by the autotrophs (recycling of minerals).

Question 5.
Give an account of the various types of ecosystems on the Earth.
Answer:
An ‘ecosystem’ is a functional unit of nature, where living organisms interact among themselves and also with the surrounding physical environment. Ecosystem varies greatly in size from a small pond to a large forest or a sea. Many ecologists regard the entire biosphere as a ‘global ecosystem’, as a composite of all local ecosystems on Earth. Since this system is too big and complex to be studied at one time, it is convenient to divide it into two basic categories, namely natural and artificial. The natural ecosystems include aquatic ecosystems of water and terrestrial ecosystems of the land. Both types of natural and artificial ecosystems have several subdivisions.

The Natural Ecosystems :
These are naturally occurring ecosystems and there is no role of humans in the formation of such types of ecosystems. These are categorized mainly into two types – aquatic and terrestrial ecosystems.

Aquatic Ecosystems :
Based on the salinity of water, three types of aquatic ecosystems are identified marine, freshwater, and estuarine.

i. The Marine Ecosystem :
It is the largest of all the aquatic ecosystems. It is the most stable ecosystems.

ii. Estuarine Ecosystem :
Estuary is the zone where river joins the sea. Sea water ascends up into the river twice a day (effect of high tides and low tides). The salinity of water in an estuary also depends on the seasons. During the rainy season out flow of river water makes the estuary less saline and the opposite occurs during the summer. Estuarine organisms are capable of withstanding the ‘fluctuations’ in salinity.

iii. The Freshwater Ecosystem :
The freshwater ecosystem is the smallest aquatic ecosystem. It includes rivers, lakes, ponds, etc. It is divided into two groups – the lentic and lotic. The still water bodies like ponds, lakes, reservoirs, etc., fall under the category of lentic ecosystems, whereas, streams, rivers and flowing water bodies are called lotic ecosystems. The communities of the above two types are called lentic and lotic communities respectively. The study of freshwater ecosystem is called as limnology.

The Terrestrial Ecosystems :
The ecosystems of land are known as terrestrial ecosystems. Some examples of terrestrial ecosystems are the forest, grassland and desert.

i. The forest Ecosystems :
The two important types of forests seen in India are i) tropical rain forest and ii) tropical deciduous forests.

ii. The Grassland Ecosystems :
These are present the Himalayan region in India. They occupy large areas of sandy and saline soils in Western Rajasthan.

iii. Desert Ecosystems :
The areas having less than 25 cm rainfall per year are called desert. They have characteristics flora and fauna. The deserts can be divided into two types – hot type and cold type deserts. Thar Desert in Rajasthan is the example for hot type of desert. Cold type desert is seen in Ladakh.

Artificial Ecosystems :
These are man-made ecosystems such as agricultural or agro-ecosystems. They include cropland ecosystems, aquaculture ponds and aquaria.

Question 6.
Describe different types of food chains that exist in an ecosystem. [March 2019, May 2017 – A.P.; May/June, Mar. 2014] (T.Q.)
Answer:
Energy flows into biological systems (ecosystems) from the Sun. The biological systems of environment include several food levels called trophic levels. A trophic level is composed of those organisms which have the same source of energy and having the same number of steps away from the sun. Thus a plant’s trophic level is one, while that of a herbivore – two, and that of the first level carnivore – three. The second and third levels of the carnivores occupy fourth and fifth trophic levels respectively.

A given organism may occupy more than one trophic level simultaneously. One must remember that the trophic level represents a functional level. A given species may occupy more than one trophic level in the same ecosystem at the same time; for example, a sparrow is a primary consumer when it eats seeds, fruits, and a secondary consumer when it eats insects and worms.
TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment 15

The food energy passes from one trophic level to another trophic level mostly from the lower to higher trophic leves. When the ‘path of food energy is ‘linear’, the components resemble the ‘links’ of a chain, and it is called ‘food chain’. Generally a food chain ends with decomposers. The three major types of food chains in an ecosystem are Grazing Food Ghain, Parasitic Food Chain and Detritus Food Chain.

I. Grazing Food Chain (GFC) :
It is also known as predatory food chain. It begins with the green plants (producers) and the second, third and fourth trophic levels are occupied by the herbivores, primary carnivores and secondary carnivores respectively. In some food chains there is yet another trophic level – the climax carnivores. The number of trophic levels in food chains varies from 3 to 5 generally. Some examples for grazing food chain (GFC) are given below.
TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment 5 TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment 6

II. Parasitic food chain :
Some authors included the ‘parasitic Food Chains’ as a part of the GFC. As in the case of GFCs, it also begins with the producers, the plants (directly or indirectly). However, the food energy passes from large organisms to small organisms in the parasitic chains. For instance, a tree which occupies the 1st trophic level provides shelter and food for many birds. These birds host many ecto-parasites and endo-parasites. Thus, unlike in the predator food chain, the path of the flow of energy includes fewer, large sized organisms in the lower trophic levels, and numerous, small sized organisms in the successive higher trophic levels.

III. Detritus Food Chain :
The detritus food chain (DFC) begins with dead organic matter (such as leaf litter, bodies of dead organisms). It is made up of Decomposers which are heterotrophic organisms, mainly the ‘fungi’ and ‘bacteria’. They meet their energy and nutrient requirements by degrading dead organic matter or detritus. These are also known as saprotrophs (sapro : to decompose)

Decomposers secrete digestive enzyme that breakdown dead and waste materials (such as faeces) into simple absorbable substances. Some examples of detritus food chains are :

  1. Detritus (formed from leaf litter) – Earthworms – Frogs – Snakes.
    Dead animals – Flies and maggots – Frogs – Snakes.

In an aquatic ecosystem, GFC is the major ‘conduit’ for the energy flow. As against this, in a terrestrial ecosystem, a much larger fraction of energy flows through the detritus food chain than through the GFC. Detritus food chain may be connected with the grazing food chain at some levels. Some of the organisms of DFC may form the prey of the GFC animals. For example, in the detritus food chain given above, the earthworms of the DFC may become the food of the birds of the GFC. It is to be understood that food chains are not ‘isolated’ always.

TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 7.
Write an essay on productivity of an ecosystem.
Answer:
The rate of production of biomass is called productivity. It can be divided into primary and secondary productivities.

I. Primary productivity is defined as the amount of biomass or organic matter produced per unit area over a period of time by plants, during photosynthesis. It can be divided into gross primary productivity (GPP) and net primary productivity (NPP).

a) Gross primary productivity of an ecosystem is the rate of production of organic matter during photosynthesis. A considerable amount of GPP is utilized by plants for their catabolic process (respiration).

b) Net primary productivity Gross primary productivity minus respiratory loss (R), is the net primary productivity (NPP). On average about 20 – 25 percent of GPP is used for the catabolic (respiratory) activity.

GPP – R = NPP
The net primary productivity is the biomass available for the consumption of the heterotrophs (herbivores and decomposers).

II. Secondary productivity is defined as the rate of formation of new organic matter by consumers.

Question 8.
Give an account of flow of energy in an ecosystem. [March 2015 – A.P.] (T.Q.)
Answer:
Energy Flow :
Except for the deep sea hydro-thermal ecosystem, sun is the only source of energy for all ecosystems on Earth. Of the incident solar radiation less than 50 per cent of it is photosynthetically active radiation (PAR). We know that plants and photosynthetic bacteria (autotrophs), fix Sun’s radiant energy to synthesise food from simple inorganic materials. Plants capture only 2-10 percent of the PAR and this small amount of energy sustains the entire living world. So, it is very important to know how the solar energy captured by plants flows through different organisms of an ecosystem.

All heterotrophs are dependent on the producers for their food, either directly or indirectly. The law of conservation of energy is the first law of thermodynamics. It states that energy may transform from one form into another form, but it is neither created nor destroyed. The energy that reaches earth is balanced by the energy that leaves the surface of the earth as invisible heart radiation.

The energy transfers in an ecosystem are essential for sustaining life. Without energy transfers there could be no life and ecosystems. Living beings are the natural proliferations that depend on the continuous inflow of concentrated energy.

Further, ecosystems are not exempted from the Second Law of thermodynamics. It states that no process involving energy transformation will spontaneously occur unless there is degradation of energy. As per the second law of thermodynamics – the energy dispersed is in the form of unavailable heat energy, and constitutes the entropy (energy lost or not available for work in a system). The organisms need a constant supply of energy to synthesize the molecules they require. The transfer of energy through a food chain is known as energy flow.

A constant input of mostly solar energy is the basic requirement for any ecosystem to function. The important point to note is that the amount of energy available decreases at successive trophic levels. When an organism dies, it is converted to detritus or dead biomass that serves as a source of energy for the decomposers. Organisms at each trophic level depend on those at the lower trophic level, for their energy demands.

Each trophic level has a certain mass of living material at a particular time, and it is called the standing crop. The standing crop is measured as the mass of living organisms (biomass) or the number of organisms per unit area. The biomass of a species is expressed in terms of fresh or dry weight (dry weight is more accurate because water contains no usable energy).
TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment 16

The 10 percent Law :
The 10 percent law for the transfer of energy from one trophic level to the next was introduced by Lindeman (the Founder of the modern Ecosystem Ecology).

According to this law, during the transfer of energy from one trophic level to the next, only about 10 percent of the energy is stored / converted as body mass / biomass. The remaining is lost during the transfer or broken down in catabolic activities (Respiration).

Lindeman’s rule of trophic efficiency /Gross ecological efficiency is one of the earliest and most widely used measures of ecological efficiency. For example, if the NPP (Net primary production) in a plant is 100 kJ, the organic substance converted into body mass of the herbivores which feeds on it is 10 kJ only. Similarly the body mass of the carnivore -1 is 1 kJ only.

Question 9.
List out the major air pollutants and describe their effects on human beings. [March 2018, 17 – A.P.] (T.Q.)
Answer:
The major air pollutants :
1. Carbon monoxide (CO) :
It is produced mainly due to incomplete combustion of fossil fuels. Automobiles are a major cause of CO pollution in larger cities and towns. Automobile exhausts, fumes from factories, emissions from power plants, forest fires and even burning of fire-wood contribute to CO pollution. Haemoglobin has greater affinity for CO and so CO competitively interferes with oxygen transport. Co causes symptoms such as headache and blurred vision at lower concentrations. In higher concentrations, it leads to coma and death.

2. Carbon Dioxide (CO2) :
Carbon dioxide is the main pollutant that is leading to global warming. Plants utilize CO2 for photosynthesis and all living organisms emit carbon dioxide in the process of respiration. With rapid urbanization, automobiles, aeroplanes, power plants, and other human activities that involve the burning of fossil fuels such as gasoline, carbon dioxide is turning out to be an important pollutant of concern.

3. Sulphur Dioxide (SO2) :
TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment 17
It is mainly produced by burning of fossil fuels. Melting of sulphur ores is another important source for SO2 pollution. Metal smelting and other industrial processes also contribute to SO2 pollution. Sulphur dioxide and nitrogen oxides are the major causes of acid rains, which cause acidification of soils, lakes and streams, and also accelerated corrosion of buildings and monuments. High concentrations of sulphur dioxide (SO2) can result in breathing problems in asthmatic children and adults. Other effects associated with long – term exposure to sulphur dioxide, include respiratory illness, alterations in the lungs’ defenses and aggravation of existing cardiovascular problems.

To control SO2 pollution, the emissions are filtered through scrubbers. Scrubbers are devices that are used to clean the impurities in exhaust gases. Gaseous pollutants such as SO2 are removed by scrubbers.

4. Nitrogen Oxides :
Nitrogen oxides are considered to to be major primary pollutants. The source is mainly automobile exhaust. The air polluted by nitrogen oxides is not only harmful to humans and animals, but also dangerous for the life of plants. Nitrogen oxide pollution also results in acid rains and formation of photochemical smog. The effect of nitrogen oxides on plants include the occurrence of necrotic spots on the surface of leaves. Photosynthesis is affected in crop plants and they yield is reduced. Nitrogen oxides combine with volatile organic compounds by the action of sunlight to form secondary pollutants called Peroxyacetyl nitrate (PAN) which are found especially in photochemical smog. They are powerful irritants to eyes and respiratory tract.

5. Particulate matter/Aerosols :
Tiny particles of solid matter suspended in a gas or liquid constitute the ‘particulate matter’. ‘Aerosols’ refer to particles and / or liquid droplets and the gas together (a system of colloidal particles dispersed in a gas). Combustion of ‘fossil fuels’ (petrol, diesel, etc.,), fly ash produced in thermal plants, forest fires, cement factories, asbestos mining and manufacturing units, spinning and ginning mills etc., are the main sources of particulate matter pollution. According to the Central Pollution Control Board (CPCB) particles of 2.5 micrometers or less in diameter are highly harmful to man and other air breathing organisms.

Question 10.
What are the causes of water pollution and suggest measures for control of water pollution?
Answer:
TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment 18
Inferior quality of water, caused by pollution of natural waters is a major problem world is facing today. It is posing all the rivers in India are grossly polluted either by sewage or discharge of industrial effluents.

The major water pollutants :
1. Domestic Sewage :
Sewage is the major source of water pollution in large cities and towns. It mainly consists of human and animal excreta and other waste materials. It is usually released into freshwater bodies or sea directly. As per the regulations the sewage has to be passed through treatment plants before it is released into the water courses. Only 0.1 percent of impurities from domestic sewage are making these water sources unfit for human consumption. In the treatment of sewage, solids are easy to remove. Removal of dissolved salts such as nitrates, phosphates and other nutrients and toxic metal ions and organic compounds is much more difficult. Domestic sewage primarily contains biodegradable organic matter, which will be readily decomposed by the action of bacteria and other microorganisms.
TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment 19

Effect of sewage discharge on some important characteristics of a river
Biological Oxygen Demand (BOD) :
BOD is measure of the content of biologically degradable substances in sewage. The organic degradable substances are broken-down by microorganisms using oxygen. The demand of oxygen is measured in terms of the oxygen consumed by microorganisms over a period of 5 days (BOD 5) or seven days (BOD 7). BOD forms an index for measuring pollution load in the sewage. Microorganisms involved in biodegradation of organic matter in water bodies consume a lot of oxygen, and as a result there is a sharp decline in dissolved oxygen causing death offish and other aquatic animals.

Algal blooms :
Presence of large amounts of nutrients in waters also causes excessive growth of planktonic algae and the phenomenon is commonly called ‘algal blooms’. Algal blooms impart distinct colour to the water bodies and deteriorate the quality of water. It also causes mortality of fish. Some algae which are involved in algal blooms are toxic to human beings and animals.

Excessive growth of aquatic plants such as the common water hyacinth (Eichhornia crassipes), the world’s most problematic aquatic weed which is also called ‘Terror of Bengal’ causes blocks in our water ways. They grow faster than our ability to remove them. They grow abundantly in eutrophic water bodies (water bodies rich in nutrients) and lead to imbalance in the ecosystem dynamics of the water body.

Sewage arising from homes and hospitals may contain undesirable pathogenic microorganisms. If it is released untreated into water resources, there is a likelihood of outbreak of serious diseases, such as dysentery, typhoid, jaundice, cholera etc.

2. Industrial Effluents :
Untreated industrial effluents released into water bodies pollute most of the rivers, fresh water streams, etc. Effluents contain a wide variety of both inorganic and organic pollutants such as oils, greases, plastics, metallic wastes, suspended solids and toxins. Most of them are non-degradable. Arsenic, Cadmium, Copper, Chromium, Mercury, Zinc, and Nickel are the common heavy metals discharged from industries.

Effects :
Organic substances present in the water deplete the dissolved oxygen content in water by increasing the BOD (Biological oxygen demand) and COD (Chemical oxygen demand). Most of the inorganic substances render the water unfit for drinking. Outbreaks of dysentery, typhoid, jaundice, cholera etc., are caused by sewage pollution.
TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment 10

b) Biomagnification :
Increase in the concentration of the pollutant or toxicant at successive trophic levels in an aquatic food chain is called Biological Magnification or Bio – magnification. This happens in the instances where a toxic substance accumulated by an organism is not metabolized or excreted and thus passes on to the next higher trophic level. This phenomenon is well known regarding DDT and mercury pollution.

As shown in the above example, the concentration of DDT is increased at successive trophic levels starting at a very low concentration of 0.003 ppb (ppb – parts per billion) in water, which ultimately reached an alarmingly high concentration of 25 ppm (ppm – parts per million) in fish-eating birds, through bio-magnification. High concentrations of DDT disturb calcium metabolism in birds, which causes thinning of egg shell and their premature breaking, eventually causing decline in bird populations.

Eutrophication :
Natural ageing of a lake by nutrient enrichment of its water is known as eutrophication. In a young lake, the water is cold and clear, supporting little life. Gradually nutrients such as nitrates and phosphates are carried into the lake via streams, in course of time. This encourages the growth of aquatic algae and other plants. Consequently the animal life proliferates, and organic matter gets deposited on the bottom of the lake. Over centuries, as silt and organic debris piles up, the lake grows shallower and warmer. As a result, the aquatic organisms thriving in the cold environment are gradually replaced by warm-water organisms. Marsh plants appear by taking root in the shallow regions of the lake. Eventually, the lake gives way to large masses of floating plants (bog) and finally converted into land.

Depending upon the climatic conditions, size of the lake and other factors, the natural ageing of a lake may span thousands of years. However, pollutants from human ativity (anthropogenic) radically accelerate the aging process. This phenomenon is called ‘Cultural or Accelerated eutrophication’.

During the past century, lakes in many parts of the earth have been severey eutrophied by sewage, agricultural and industrial wastes. The prime contaminants are nitrates and phosphates, which are the ’chief plant nutrients’. The dissolved oxygen which is vital to other aquatic clife is depleted. At the same time, other pollutants flowing into the lake may poison the whole population of fish, whose decomposing remains further deplete the dissolved oxygen content in the water.

Thermal pollution :
Water is used as a coolant in Thermal power plants and other industries. Hot water flowing out of industries also constitute an important category of pollutants. Thermal waste water eliminates sensitive organisms (Stenothermal organisms such as fish – especially the juveniles) downstream and may enhance the growth of plants and fish in extremely cold areas but, only after causing damage to the indigenous flora and fauna.

Ecological Sanitation – ’Ecosan Toilets’ :
Generally it is assumed that removal of wastes requires water, which means creation of sewage. If water is not used to dispose off human waste like excreta, and if one didn’t have to flush the tiolet after its use, a large amount of water can be saved. This is already a reality. Ecological sanitation is a substainable system for handling human excreta, using ‘dry composting toilets’. This is a practical, hygienic, efficient and cost-effective solution to human waste disposal. The key point to note here is that, with this composting method, human excreta can be recycled into a resource (as natural fertiliser), which reduces the need for chemical fertilizers. ‘EcoSan’ toilets are in use in many parts of Kerala and Sri Lanka.

TS Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 11.
Write an essay on soil pollution and measures to control soil pollution.
Answer:
Solid Wastes :
Any thing (substance/ material / articles / goods) that is thrown out as waste in solid form is referred to as solid waste. Municipal solid wastes are wastes from homes, offices, institutions, shops, hotels, restaurants etc., in towns and cities.

The municipal solid wastes generally consist of paper, food wastes, plastics, glass, metals, rubber, leather, textile, etc. The wastes are burnt to reduce the volume of the wastes. But generally wastes are not completely burnt and left as open dumps which often serve as the breeding grounds for rats and flies. As the substitute for open-burning dumps, sanitary landfills are adopted. In a sanitary landfill, wastes are dumped in a depression or trench after compaction, and covered landfill, wastes are dumped in a depression or trench after compaction, and covered with dirt everyday. These is a danger of seepage of chemicals and pollutants from these landfills, which may contaminate the underground water resources.

The best solution is to develop awareness in the society on these environmental issues. All wastes that we generate can be categorized into three types (a) biodegradable, (b) recyclable and (c) non-biodegradable. It is important that all garbage generated should be sorted out category wise. The reusable or recyclable material has to be separated out and utilised. (Rag-pickers in the streets are doing a great job of separation of materials for recycling.) The biodegradable materials can be put into deep pits in the ground and be left for natural breakdown. The remaining non-biodegradable waste left over is to be disposed off properly.

The prime goal should be to reduce our garbage generation. But we are increasing the use of non-biodegradable products. We are packaging products of our daily use such as milk and water also in polythene bags. In cities and towns, many purchased things are packed in polystyrene and plastic packets. Thus we are contributing heavily to environmental pollution. State Governments across the country are trying to educate people on the reduction in use of plastic and use of eco-friendly packaging. We can do our bit by using carry-bags made of cloth or other natural fibres when we go for shopping and by refusing polythene bags.

i) Hospital wastes :
Hospitals generate hazardous wastes that contain disinfectants, harmful chemicals and also pathogenic micro-organisms. Such wastes also require careful treatment and disposal. The use of incinerators (to burn wastes) is essential for disposal of hospital waste.

ii) Electronic wastes (e-wastes) :
Irreparable computers and other electronic goods constitute the modern day pollutants called electronic wastes (e-wastes), e – wastes are buried in landfills or incinerated. Over half of the e-wastes generated in the developed world are exported to developing countries, mainly to China, India and Pakistan, where metals like copper, iron, silicon, nickel and gold are recovered during recycling process.

Unlike developed countries, which have specifically built facilities for recycling of e-wastes, recycling in developing countries often involves manual participation thus exposing workers to toxic substances present in e – wastes. Eventually recycling is the only solution for the treatment of e – wastes provided it is carried out in an environmental friendly manner.

iii) Agro – chemicals and their effects :
In the wake of the Green Revoltuion, use of inorganic fertilisers and pesticides has increased many times, for enhancing crop production. Pesticides, herbicides, fungicides, etc., are being increasingly used. They are also toxic to non-target organisms such as earthworms, nitrogen fixing bacteria, etc., that are important components of soil eco-system. Moreover due to bio-magnification, the harmful chemicals pose a great threat to human health. Indiscriminate use of fertilizers will lead to increased drain of nutrients into the nearby aquatic ecosystems causing eutrophication and the consequent effects.

iv) Radioactive wastes :
Initially, nuclear energy was hailed as a non-polluting way for generating electricity. Later on, it was realised that the use of nuclear energy has two very serious inherent problems. The first is accidental leakages, as occurred in the Three Mile Island (USA) and Chernobyl (Russia) and the second is the safe disposal of radioactive wastes.

Radiation, that is released from nuclear waste is extremely dangerous to biological organisms, because it induces mutations. Exposure to high doses of nuclear radiation is lethal as it can lead to cancers (e.g. leukemia). Therefore, nuclear waste is an extremely potent pollutant and has to be dealt with utmost caution. Storage of nuclear wastes should be done in suitably shielded containers and buried deep in the soil or oceans (about 500 meters). Even when done so, geological upheavals can bring them up, some day and cause radiation.

TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach)

Telangana TSBIE TS Inter 1st Year Zoology Study Material 7th Lesson Type Study of Periplaneta Americana (Cockroach) Textbook Questions and Answers.

TS Inter 1st Year Zoology Study Material 7th Lesson Type Study of Periplaneta Americana (Cockroach)

Very Short Answer Type Questions

Question 1.
Why do you call cockroach a pest? (U).
Answer:
Cockroach is commonly found in kitchen and contaminates our food with its excreta. It can transmit a number of bacterial diseases by contaminating food. Hence it is called a pest.

Question 2.
Name the terga of thoracic segments of cockroach. (K)
Answer:
The terga of thoracic segments are one large pronotum covering prothorax. The terga on the mesothorax and metathorax are called mesonotum and metanotum.

Question 3.
What are the structures with which cockroach walks on smooth surfaces and on rough surfaces respectively? (U)
Answer:
The claws and the arolium help in locomotion on rough surfaces whereas plantulae are useful on smooth surfaces.

Question 4.
Name the chitinous tubes that support the wings of cockroach. (K)
Answer:
The wings of cockroach contain a network of hollow veins or nervures.

Question 5.
What is tegmen? What is its function? (KJ
Answer:
The fore wings are thick, opaque and leathery. They donot help in flight, but cover and protect the hind wings when they are not in use. They are called tegmina (singular: tegmen).

TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach)

Question 6.
Why is the head in cockroach called hypognathous? (U)
Answer:
The head of cockroach is called hypognathous because it lies hanging almost at right angles to the body with the posterior wider part upwards and the mouth parts directed downwards.

Question 7.
How is a tripod formed with reference to locomotion in cockroach? (U)
Answer:
Each tripod is formed by fore leg and hind leg of one side and the middle leg of the other side in cockroach.

Question 8.
Name the muscles that help in elevating and depressing the wings of a cockroach. (K)
Answer:
Wings are elevated by the contraction of dorsoventral muscles. Contraction of the dorso – longitudinal muscles depresses the wings.

Question 9.
Name the different blood sinuses in cockroach. (K)
Answer:
The blood sinuses are a) Pericardial haemocoel or the dorsal sinus b) Perivisceral haemocoel or the middle sinus, c) Sternal haemocoel or ventral sinus.

Question 10.
How are the fat bodies similar to the liver of the vertebrates? (A)
Answer:
The haemocoel of cockroach contains many large sized fat bodies called corpora adiposa. These are similar to the liver of the vertebrates in certain function like storing of food, secrete lipids, store uric acid and contain symbiotic bacteria.

Question 11.
What are the three regions of the alimentary canal in cockroach? (K)
Answer:
The three regions are foregut or stomodaeum, midgut or mesenteron and hindgut or proctodaeum.

Question 12.
How many denticulate plates are present in the gizzard of cockroach? (K)
Answer:
The chitinous inner lining of the gizzard of cockroach has six powerful teeth, which form an efficient grinding apparatus.

TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach)

Question 13.
Which part of the gut secretes the peritrophic membrane in cockroach? (U)
Answer:
The ‘bolus’ of food in the mesenteron is enveloped by a chitinous and porous membrane called peritrophic membrane, which is secreted by the funnel like stomodeal valve of the gizzard.

Question 14.
Which parts of cockroach help in locating the food? (U)
Answer:
Cockroach locates the food by the olfactory sensillae of antenna, labial palps and maxillary palps.

Question 15.
In which part of the gut of cockroach, water is reabsorbed? (K)
Answer:
The undigested food is passed into the ileum, colon and then reaches the rectum, where water is reabsorbed by rectal papillae.

Question 16.
Write the names of mouthparts in cockroach that help in biting and tasting the food. (K).
Answer:
Mandibles help in biting and chewing of food. Labrum helps in tasting the food.

Question 17.
What are alary muscles? [K)
Answer:
There is a series of paired triangular muscles, called “Alary muscles.” Every segment has one pair of these muscles. These are attached to the pericardial septum by their broad bases and to the terga by their pointed ends.

Question 18.
What is haemocoel? (K)
Answer:
The body cavity of an arthropod or a mollusk filled with haemolymph : derived from the blastocoels of the embryo, also called the “Primary body cavity.”

TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach)

Question 19.
The three sinuses in a cockroach are not equal in size. Why? (U)
Answer:
The middle sinus is very large as it contains most of the viscera. The dorsal and ventral sinuses are small as they have only heart and nerve cord, respectively.

Question 20.
Why is the blood of Periplaneta called haemolymph? (U)
Answer:
The blood of periplaneta is colourless and is called haemolymph. It consists of a fluid called plasma and free blood corpuscles or haemocytes which are phagocytic.

Question 21.
What is the function of haemocytes found in the blood of Periplaneta? (K)
Answer:
Haemocytes are phagocytic. They are large in size and can ingest foreign particles such as bacteria. Hence defensive in function.

Question 22.
Why does not the blood of Periplaneta help in respiration ? (A)
Answer:
There is no respiratory pigment in the blood and so it plays no major role in respiration.

Question 23.
Write important functions of blood in Periplaneta. (K)
Answer:
Blood functions

  1. It absorbs digested food from alimentary canal and distributes it to the rest of the body.
  2. It brings nitrogenous wastes from all parts of the body to the excretory organs for their elimination.
  3. Phagocytes of blood are defensive in function.
  4. It transports secretions of the ductless glands to the target organs.

Question 24.
The blood of Periplaneta is not red. Which pigment, do you think, is absent in it? (U)
Answer:
The blood of periplaneta is not red. There is no respiratory pigment called haemoglobin in the blood and so it plays no major role in respiration.

Question 25.
How many spiracles are present in cockroach? Mention their locations. (K)
Answer:
There are ten pairs of openings called stigmata or spiracles. The first two pairs of spiracles are present in the thoracic segments (2nd and 3rd). Remaining eight pairs are present in the first eight abdominal segments.

TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach)

Question 26.
What are trichomes? Write their functions. (K)
Answer:
All spiracles bear small hair like structures called ‘trichomes’ to filter the dust particles.

Question 27.
Why is the respiratory system of cockroach called polypneustic and holopneustic system? (U)
Answer:
The spiracles of cockroach are polypneustic (as they are more than 3 pairs) and holopneustic (as all of them are functional).

Question 28.
Name the chitinous ring that encircles the spiracle of cockroach. (K)
Answer:
All spiracles are valvular and each of them is surrounded by a chitinous ring called peritreme.

Question 29.
What is intima? (K)
Answer:
Trachea in cockroach is made up of three layers, outer basement membrane, a middle one cell thick epithelium and an inner layer of cuticle called intima. The intima is produced into spiral thickenings called taenidia.

Question 30.
Name the protein that lines the tracheole of the cockroach. (K)
Answer:
Tracheoles are formed of a protein called trachein.

TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach)

Question 31.
During inspiration which spiracles are kept open and which are kept closed? (K)
Answer:
During inspiration the thoracic spiracles are kept open and the abdominal spiracles are kept closed.

Question 32.
Which factors regulate the opening of the spiracles? (U)
Answer:
Opening and closing or spiracles is influenced by C02 tension in haemolymph and oxygen tension in the tracheae.

Question 33.
Inspiration in cockroach is a passive process and expiration is an active process. Justify. (U)
Answer:
During inspiration air is drawn in due to the relaxation of the muscles, the process is a “passive process”. Expiration involves the contraction of muscles, hence is described as ‘active process.’

Question 34.
The nitrogenous wastes in Periplaneta are removed from the body through alimentary canal. Why? (U)
Answer:
The removal of nitrogenous waste material through the alimentary canal helps in complete reabsorption of water from the wastes and formation of dry uric acid. It is an adaptation for conservation of water.

Question 35.
How does the cuticle of a cockroach help in excretion? (A)
Answer:
Some nitrogenous waste materials are deposited on the cuticle and eliminated during moulting.

Question 36.
How do fat bodies help in excretion? (A)
Answer:
Fat bodies absorb and store uric acid throughout the life. This is called storage excretion as they remain stored in the cells of the corpora adiposa.

Question 37.
What is ‘storage excretion’? (U)
Answer:
Fat bodies absorb and store uric acid throughout the life. This is called storage excretion as they remain stored in the cells of the corpora adiposa.

TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach)

Question 38.
In which part of the alimentary canal of Periplaneta more water is reabsorbed? (K)
Answer:
The part of the alimentary canal of Periplaneta more water is reabsorbed in rectum containing rectal papillae.

Question 39.
List out the organs associated with excretion in Periplaneta. (K)
Answer:
The organs associated with excretion are
1) Malpighian tubules of anterior end of hind gut 2) Fat bodies 3) Uricose glands 4) Cuticle.

Question 40.
Which part of malpighian tubules extract water, salts and nitrogenous wastes from the haemolymph? (K)
Answer:
The distal portion of malpighian tubules containing glandular cells absorb salts, water and nitrogenous wastes from the haemolymph.

Question 41.
Which structure of cockroach acts as sensory and endocrine centre? (K)
Answer:
Brain or cerebral ganglia is mainly a sensory and an endocrine centre.

Question 42.
Distinguish between scolopidia and sensillae. (U)
Answer:

  1. Sensillae are the units of cuticular receptors and chemoreceptors.
  2. Scolopidia are the subcuticular units of mechanoreceptors of chordo-tonal organs.

Question 43.
How is the ommatidium of cockroach different from that of a diurnal insect? (A]
Answer:
In cockroach a nocturnal insect ommatidia form superposition image in which overlapping of images occur and it is a blurred image.

In diurnal insects the image is called apposition image as it is formed by the juxtaposition of small parts of the visual field. This type of vision is called mosaic vision.

TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach)

Question 44.
How many segmental ganglia are present on the ventral nerve cord of cockroach? (K)
Answer:
On the ventral nerve cord segmental ganglia show 3 + 6 arrangement. 3 thoracic ganglia and 6 abdominal ganglia in first 7 segments except 5th segment.

Question 45.
Which of the abdominal ganglia is the largest and why? (U)
Answer:
The last or the 6th abdominal ganglion is the largest of all the abdominal ganglia. It is formed by the fusion of the ganglia of the 7th, 8th, 9th and 10th abdominal segments.

Question 46.
Name the structural and functional unit of compound eye of cockroach. How many such units are present in a single compound eye ? (K)
Answer:
Structural and functional unit of compound eye is ommatidium. In a single compound eye of cockroach 2000 ommatidia are present.

Question 47.
Why is the brain called the principal sensory centre in cockroach? (U)
Answer:
In brain protocerebrum receives sensory impulses from compound eyes through optic nerves, deutero cerebrum receives sensory impulses from antennae through antennel nerves and tritocerebrum receives sensory impulses from the labrum. Hence brain in principally sensory in nature.

Question 48.
Which parts of an ommatidium constitute dioptric region? (K)
Answer:
The region containing the cornea and crystalline cone constitute the dioptrical or focussing region of the ommatidium. ,

Question 49.
Distinguish between apposition image and superposition image. (U)
Answer:
In cockroach a nocturnal insect ommatidia form superposition image in which overlapping of images occur and it is a blurred image. In diurnal inserts the image is called apposition image as it is formed by the juxtaposition of small parts of the visual field. This type of vision is called mosaic vision.

Question 50.
List out the characters that help in understanding the difference between male and female cockroaches. (K)
Answer:
The female is different from the male in respect of short and broad abdomen, presence of brood pouches and absence of anal styles.

Question 51.
What is the function of mushroom gland in cockroach? (K)
Answer:
A characteristic mushroom shaped gland is present in the 6th and 7th abdominal segments which functions as an accessory reproductive gland.

Question 52.
Compare the utriculi majores and utriculi breviores of the mushroom gland functionally. (U)
Answer:
Secretion of utriculi majores forms the inner layer of the spermatophore while that of utriculi breviores nourishes the sperms.

TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach)

Question 53.
How many ovarioles are present in a single ovary of Periplaneta and what are the two parts of a single ovariole? (U)
Answer:
Each ovary consists of eight tubules called ovarioles. Each ovariole consists of a tapering anterior filament called germarium and a posterior wider vitellarium.

Question 54.
What are phallomeres? (K)
Answer:
Surrounding the male genital opening there are chitinous and asymmetrical structures called phallic organs or gonapophyses or phallomeres which help in copulation. These are the male external genitalia.

Question 55.
What are gonapophyses? (K)
Answer:
Three pairs of plate like chitinous structures called gonapophyses are present around the female genital aperture. These gonapophyses guide the ova into ootheca as ovipositors. These are female external genitalia.

Question 56.
How is colleterial gland helpful in reproduction of Periplaneta? (A)
Answer:
Secretion of the two collateral glands forms a hard egg case called ootheca around the eggs.

TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach)

Question 57.
What is paurometabolous development ? (U)
Answer:
Cockroach is paurometabolous, which means the development is gradual through nymphal stages.

Short Answer Type Questions

Question 1.
Draw a neat labelled diagram of the mouth parts of cockroach. (S)
Answer:
TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach) 1

Question 2.
Describe the physiology of digestion in cockroach. (K)
Answer:
Digestion :
After swallowing, the food passes through the pharynx and oesophagus, and reaches the crop. In the crop, food is mixed with digestive juices that are regurgitated into it through the grooves of the gizzard. Hence, most of the food is digested in the crop. The partly digested food is filtered by the bristles of the gizzard and later it passes through the stomodeal valve into the ventriculus.

The enzyme amylase of the salivary juice converts starches into disaccharides. Invertase or sucrase digests sucrose into glucose and fructose. Maltase converts maltose into glucose. The enzyme lipase digests lipids into fatty acids and glycerol. Proteases digest proteins into amino acids. Cellulose of the food is digested by the enzyme cellulase secreted by the microorganisms present in the hindgut of cockroach. Cellulose in converted into glucose.

In the ventriculus, the digested food is absorbed. The undigested food is passed into the ileum, colon, and then reaches the rectum, where water is reabsorbed by rectal papillae. Then the remaining material is finally defaecated as dry pellets, through the anus.

Question 3.
Draw a neat labelled diagram of the salivaary apparatus of cock roach. (S)
Answer:
TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach) 2

Question 4.
What is haemocoel? Describe it with reference to Periplaneta. (U)
Answer:
The blood filled body cavity of an organism is called Haemocoel.
Haemocoel in Periplaneta :
The haemocoel of cockroach is divided in to three sinuses by two muscular, horizontal membranes, called dorsal diaphragm or ‘pericardial septum’ and ventral diaphragm. Both the diaphragms have pores. There is a series of paired triangular muscles, called alary muscles. Every segment has one pair of these muscles situated on the lateral sides of the body.

These are attached to the pericardial septum by their broad bases and to the terga by their pointed ends or apices. The three sinuses of the haemocoel are known as pericardial haemocoel or the ‘dorsal sinus’, the perivisceral haemocoel or the ‘middle sinus’ and sternal haemocoel or ‘venral sinus’ or ‘perineural sinus’. The middle sinus is very large as it contains most of the viscera. The dorsal and ventral sinuses are small as they have only heart and nerve cord, respectively.

TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach)

Question 5.
Describe the structure and function of the heart in Periplaneta. (K)
Answer:
Heart :
The heart lies in the pericardial haemocoel or dorsal sinus. It is a long, muscular, contractile tube found along the mid dorsal line, beneath the terga of the thorax and abdomen. It consists of 13 chambers. Every chamber opens into the other present in front of it. Three of the thirteen chambers are situated in the thorax and ten in the abdomen. Its posterior end is closed while the anterior end is continued forward as the anterior aorta. At the posterior side of each chamber, except the last, there is a pair of small apertures called ‘ostia’ one on each side. Ostia have valves which allow the blood to pass only into the heart from the dorsal sinus.

Question 6.
Describe the process of blood circulation in Periplaneta. (K)
Answer:
Circulation of blood :
The blood flows forward in the heart by the contractions of its chambers. At the anterior end of the heart, the blood flows into the aorta and from there it enters the sinus of the head. From the head sinus, the blood flows into the perivisceral and sternal sinuses. On contraction of the alary muscles the pericardial septum is pulled down. This increases the volume of the pericardial sinus.

Hence blood flows from the perivisceral sinus into the pericardial sinus through the appertures of the pericardial septum. On relaxation of the alary muscles, the pericardial septum moves upwards to its original position. This forces the blood, to enter the chambers of the heart through the ostia from the pericardial sinus.
TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach) 3

Question 7.
How do contraction and relaxation of alary muscles help in circulation? (A)
Answer:
On contraction of the alary muscles the pericardial septum is pulled down. This increases the volume of the pericardial sinus. Hence blood flows from the perivisceral sinus into the pericardial sinus through the appertures of the pericardial septum. On relaxation of the alary muscles, the pericardial septum moves upwards to its original position. This forces the blood, to enter the chambers of the heart through the ostia from the pericardial sinus.

Question 8.
Describe the structure of trachea of cockroach. (K)
Answer:
Structure of Trachea :
The wall of the tracheae is made of three layers. They are an outer basement membrane, a middle one cell thick epithelium and an inner layer of cuticle called intima. It has protein / chitin layer and epicuticle towards lumen. The intima is produced into spiral thickenings called taenidia. In taenidia, protein / chitin layer is differentiated as exocuticle. The taenidia keep the tracheae always open.

TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach)

Question 9.
Explain the structure of malpighian tubules. (U)
Answer:
Malpighian tubule :
The malpighian tubules are long, unbranched yellowish tubules, attached at the extreme anterior end of the hindgut, lying freely in the haemolymph, but do not open into it, being blind at the free ends. They are 100 – 150 in number arranged in 6 – 8 bundles, each bundle having 15-25 tubules. Marcello Malpighi, described these tubules and called them vasa varicosa Meckel called them Malpighian tubules. Each tubule is lined by a single layer of glandular epithelium with a brush border on the inner surface. The ‘distal portion’ of the tubule is secretory and the ‘proximal part’ is absorptive in nature.

Question 10.
What are different excretory organs in Periplaneta? Describe the process of excretion in detail. (K)
Answer:
The structures associated with excretory function are
a) Malpighian tubules b) Fat bodies c) Uricose glands d) Cuticle.

Malpighian tubules :
The glandular cells of the malpighian tubules absorb water, salts, CO2 and nitrogenous wastes from the haemohymph and secrete them into the lumen of the tubules. The cells of the proximal part of the tubules reabsorb water and certain inorganic salts. By the contraction of the tubules urine is pushed into the ileum. More water is reabsorbed from it, when it moves in to the rectum and almost solid uric acid is excreted along with faecal matter.

The removal of nitrogenous waste material through the alimentary canal helps in complete reabsorption of water from the wastes and formation of dry uric acid. It is an adaptation for conservation of water as it is very important in terrestrial organisms.

Fat bodies :
Fat body is a lobed white structure. Urate cells present in these bodies are associated with excretion in a way. These cells absorb and store uric acid throughout the life. This is called storage excretion as they remain stored in the cells of the corpora adiposa.

Uricose glands :
Uric acid is stored in uricose gland or utriculi majores of the mushroom gland in male cockroach. It is discharged during copulation.

Cuticle :
Some nitrogenous waste materials are deposited on the cuticle and eliminated during moulting.
TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach) 4

Question 11.
How does Periplaneta conserve water? Explain it with the help of excretion in it. (A)
Answer:
The removal of nitrogenous waste material through the alimentary canal helps in complete reabsorption of water from the wastes and formation of dry uric acid. It is an adaptation for conservation of water as it is very important in terristrial organisms.

Fat bodies :
Fat body is a lobed white structure. Urate cells present in these bodies are associated with excretion in a way. These cells absorb and store uric acid throughout the life. This is called storage excretion as they remain stored in the cells of the corpora adiposa.

TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach)

Question 12.
Describe the structure of an ommatidium and label its parts. (K)
Answer:
Structure of an Ommatidium :
Each typical ommatidium is an elongated sub unit of the compound eye, consisting of the following parts.
TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach) 5

1. Cornea :
It is the outermost part of an ommatidium and corresponds to a ‘hexagonal facet’ of the compound eye. It is a biconvex, transparent part of the cuticle and allows light to pass through it. Cornea is secreted by specialized cells of epidermis. Cornea is the ‘refractive* 1 2 3 region of ommatidium.

2. Corneagen cells or lenticular cells :
These are two transparent specialized epidermal cells that secrete cornea. These cells later form the ‘primary pigment sheath’ or iris pigment sheath.

3. Vitrellae or cone cells (Semper cells) :
These are four transparent more or less conical cells that lie below the corneagen cells. They surround the transparent crystalline cone. Crystalline cone is secreted by the cone cells.

4. Crystalline cone :
It is the transparent conical structure that is secreted by the vitrellae and is surrounded by them. Light absorbing dark primary pigment cells surround the vitrellae. The region containing the cornea and crystalline cone constitute the dioptrical or focussing region of the ommatidium. Crystalline cone focuses the light on to the next part of the ommatidium.

5. Retinulae :
These are the innermost and elongated cell of an ommatidium. They are seven in number. They rest on the basement membrane. Each cell bears microvilli towards the inner surface. Microvilli of each retinular cell collectively form a rhabdomere that contains photoreceptor pigments. These rhabdomeres fuse along the axis of the ommatidium to form the rhabdome in the centre. Retinulae are the nerve cells from which sensory nerve fibres leave as the optic nerve to the protocerebrum. They are the photoreceptor cells of the ommatidium. Rhabdome and retinulae form the retinal or receptor region. Receptor region is surrounded by seven secondary pigment cells, which absorb light and serve to isolate each ommatidium from the rest (retinal pigment sheath).

Question 13.
Draw a neat and labelled diagram of ommatidium. (S)
Answer:
TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach) 5

Question 14.
How can you identify the male and female cockroaches? Explain it describing the chief features of the external and internal genetalia. (U)
Answer:
The sexual dimorphism is evident both externally and internally. The female is different from the male in respect of short and broad abdomen, presence of brood pouches and absence of anal styles.

Male :
Internally male reproductive system contains a pair of testes, vas deferens, ductus ejaculatorius, mushroom shaped gland, conglobate gland.

Surrounding the male genital opening there are chitinous and asymmetrical structures called phallic organs or phallomeres which help in copulation. These are the male external genitalia.

Female :
Internally female reproductive system consists of a pair of ovaries, a pair of oviducts, vagina, spermathecae, spermathecal papilla and colleteral glands.

Three pairs of plate like chitinous structures called gonopophyses are present around the female genital aperture. These gonopophyses guide the ova into ootheca as ovipositors. These are the female external genitalia.

Question 15.
Describe the male reproductive system of cockroach. (K)
Answer:
Male Reproductive System :
The male reproductive system .consists of a pair of testes. These are elongated and lobed structures lying one on each lateral side in the fourth to sixth abdominal segments. They are embedded in the fat bodies. From the posterior end of each testis, there starts a thin duct, the vas deferens. The two vasa deferntia run backwards and inwards to open into a wide median duct, the ductus ejaculatorius in the seventh segment.

A characteristic mushroom shaped gland is present in the 6th and 7th abdominal segments which functions as an accessory reproductive gland. The gland consists of two types of tubules, i) long slendertubules, the utriculi majores or ‘peripheral tubules’ ii) Short tubules, the utriculi breviores. Secretion of utriculi majores forms the inner layer of the spermatophore while that of utriculi breviores nourishes the sperms. These tubules open into the anterior part of the ejaculatory duct.
TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach) 6

The seminal vesicles are present on the ventral surface cf the ejaculatory duct. These sacs store the sperms in the form of bundles called spermatophores. The ejaculatory duct is a muscular tube that extends posteriorly and opens at the gonopore or the ‘male genital pore’. The duct of phallic or conglobate gland also opens near the gonopore. Its function is still not known. Surrounding the male genital opening there are chitinous and asymmetrical structures called phallic organs or gonapophyses or phallomeres which help in copulation. These are the male external genitalia.

TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach)

Question 16.
Describe the female reproductive system of cockroach. (K)
Answer:
Female Reproductive System :
The female reproductive system of Periplaneta consists of a pair of ovaries, a pair of oviducts, vagina, spermathecae, spermathecal papilla, and colleterial glands.

Ovaries :
A pair of large ovaries lies laterally in 2 to 6 abdominal segments. They are light yellow in colour surrounded by fat bodies. Each ovary consists of eight tubules called ovarian tubules or ovarioles. Each ovariole consists of a tapering anterior filament called germarium and a posterior wider vitellarium. The germarium contains various stages of developing ova, and the vitellarium contains mature ova with yolk. The tapering ends of the ovarioles of each ovary unite to form a single thread which attaches to the dorsal body wall. The ovarioles, at their posterior end unite to form a short wide oviduct.

The oviducts unite to form a very short median vagina. The vertical opening of the vagina is called female genital pore. It opens into a large genital pouch on the eighth sternum. A spermatheca or ‘receptaculium seminis’. consisting of a left-sac like and a right filamentous caecum, is present in the 6th segment which opens by a median aperture on a small spermathecal papilla in the dorsal wall of the genital pouch on the ninth sternum. In a fertile female, the spermatheca contains spermatophores, obtained during copulation.
TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach) 7

A pair of branched colleterial glands is present behind the ovaries. These glands open into the genital pouch separately, just above the spermathecal aperture. Secretion of the two collaterial glands forms a hard egg case called ootheca around the eggs.

Genital pouch is formed by 7th, 8th, and 9th abdominal sterna. The sternum of the seventh segment is boat shaped and forms the floor and side walls of the genital pouch. The sterna of the eighth and ninth segments, which are tucked into the seventh segment, constitute the anterior wall and the roof of the genital pouch, respectively. The genital pouch has two chambers the anterior ‘gynatrium’ or genital chamber and posterior ‘vestibulum’ or oothecal chamber.

Three pairs of plate like chitinous structures called gonapophyses are present around the female genital aperture. These gonapophyses guide the ova into ootheca as ovipositors. These are the female external genitalia.

Essay Answer Type Questions

Question 1.
Descirbe the structure of the head of cockroach, with the help of a neat labelled diagram. (K & S)
Answer:
Heat :
The head of cockroach is small and triangular. It is called hypognathous because it lies hanging almost at right angles to the body with the posterior wider part upwards, and the mouthparts directed downwards.
TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach) 8

The head of cockroach is formed by the fusion of six embryonic segments. It is movably attached to the thorax by a short neck or cervicum. it is covered by a number of sclerites which fuse to form a capsule. The top of the head between the eyes is called vertex. The vertex has two sclerites called ‘epicranial plates’ connected by an ‘epicranial suture’. Below the vertex, the sclerites covering the’ head in front are a large frons, a narrow rectangular clypeus and a movable labrum. Covering the sides of the head, below the compound eyes are.the ‘cheek sclerities’ or ‘genae. At the back of the head capsule there is a large opening called occipital foramen.

It is bordered by a sclerite called occiput. The occipital foramen forms a passage for the oesophagus, aorta; nerve cord and tracheae. At the base of each antenna, a small whitish speck called fenestra or ‘ocellar spot’ or ‘simple eye’ is present. Appendages are absent in the first and third segments of the head. The second segment bears a pair of long, slenife^ and segmented antennae, one on each side of the head. The antennae are tactile and olfactory in function. The fourth segment bears a pair of mandibles. The fifth segment has a pair of ‘first maxillae’. The sixth segment bears a pair of ‘second maxillae’, which fuse to form the labium (also called ‘lower lip’)

TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach)

Question 2.
Describe the abdomen of cockroach. (K)
Answer:
Abdomen :
The abdomen consists of ten segments. Each segment is covered by the dorsal tergum, the ventral sternum and the two lateral pleura or pleurites. There are ten terga but only nine sterna as the tenth sternum is absent. The eighth tergum in the male and both eighth and ninth terga in the female are not visible as they are overlapped by the seventh tergum. The tenth tergum extends beyond the posterior end of the body and has a deep notch/groove in the middle of its free end. In the male nine sterna are visible whereas in the female, only seven sterna are visible. The seventh, eighth and ninth sterna together form a brood pouch. The brood pouch has two parts the anterior genital chamber or gynatrium and posterior oothecal chamber.

The posterior end of the abdomen has a pair of anal cerci, a pair of anal styles and gonapophyses in the males. Anal cerci are jointed and arise from the lateral sides of the tenth tergum and are found in both the sexes. The anal styles are without joints and arise from the ninth sternum (seen only in the males). The gonapophyses are small chitinous processes arising from the ninth sternum in the males and eighth and ninth sterna in the females. They are the external genital organs. The anus is at the posterior end of the abdomen. The genital aperture in male is present just below the anus on one of the gonapophyses and in female it is located on the eighth sternum.

TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach)

Question 3.
Describe the digestive system of cockroach with the help of a neat labelled diagram. (K & S)
Answer:
The digestive system of cockroach consists of an alimentary canal and the associated glands. The preoral cavity, surrounded by the mouth parts, is present in front of the mouth. The hypopharynx divides it into two chambers called cibarium (anterior) and salivarium (posterior)

Alimentary canal :
The alimentary canal of cockroach is a long tube and is coiled at some places. It extends between the mouth and the anus. It is divided into three regions, namely, foregut or stomodaeum, midgut or mesenteron and hindgut or proctodaeum. The foregut and hindgut are internally lined by ectoderm. The mesenteron is lined by the endodermal cells.

Foregut of stomodaeum :
The foregut includes pharynx, oesophagus, crop, and gizzard. It is internally lined by a chitinous cuticle. Mouth opens into the pharynx, which in turn leads into a narrow tubular oesophagus. The oesophagus opens behind into a thin walled distensible sac called crop. The crop serves as a reservoir for storing food. Its outer surface is covered by a network of tracheae.
TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach) 9

Behind the crop there is a thick walled muscular proventriculus, or gizzard. The chitinous inner lining of the gizzard has six powerful teeth, which form an efficient grinding apparatus. Behind each tooth is a hairy pad, which bears backwardly directed bristles. Among these plates, food is thoroughly ground into fine particles. These food particles are filtered by the bristles. The gizzard thus acts both as a grinding mill and also as a sieve. There is a membranous projection of the gizzard into the mesenteron in the form of a funnel called stomodeal valve. This valve prevents the entry (regurgitation) of food from the mesenteron back into the gizzard.

Midgut (mesenteron or ventriculus) :
The midgut is a short and narrow tube behind the gizzard. It is also called mesenteron or ventriculus. Between the ventriculus and the gizzard, arising from ventriculus, there are six to eight finger like diverticula called hepatic caecae. They are helpful in digestion and absorption of the digested t food materials. Ventriculus is functionally divided into an anterior secretory part and a posterior absorptive part.

The secretory part of the ventriculus has many gland cells and it secretes several enzymes. The ‘bolus’ of food in the mesenteron is enveloped by a chitinous and porous membrane called peritrophic membrane, which is secreted by the funnel like stomodeal valve of the gizzard.

Digested food is absorbed into the blood through the peritrophic membrane in the posterior absorptive region of the ventriculus. The peritrophic membrane protects the wall of the ventriculus from hard food particles in the food. The opening of the ventriculus into the hindgut is controlled by a sphincter muscle. It prevents entry of undigested food and uric acid from the hindgut into the midgut.

Hindgut or proctodaeum :
The hindgut is a long coiled tube, consisting of three regions namely ileum, colon and rectum, it is internally lined by chitinous cuticle. The ileum that lies behind the mesenteron is a short tube. Six bundles of fine yellow, blind tubules called malpighiari tubules open into the ileum near the junction of mesenteron and ileum. Malpighian tubules are excretory in function.

Ileum collects uric acid from the malpighian tubules and undigested food from the mesenteron. Ileum opens behind into a long coiled tube called colon. Colon leads into a short and wide rectum, which opens out through the anus. Rectum bears on its inner side six longitudinal chitinous folds called rectal papillae. They are concerned with the reabsorption of water from the undigested food.

Digestive glands :
The digestive glands associated with the alimentary canal of cockroach are salivary glands, hepatic caecae and glandular cells of the mesenteron. Salivary glands : There is a pair of salivary glands attached to the ventrolateral sides of the crop, one on each side. Each salivary gland has two lobes. Each lobe of salivary gland has many lobules called acini. Each acinus is a group of secretory cells called zymogen cells with a small ductule. The ductules of both the lobes of a salivary gland unite to form a common salivary duct on each side.
TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach) 10

The two common salivary ducts are joined to form the median salivary duct. Between the two lobes of a salivary gland of each side is a sac called salivary receptacle that stores saliva. It leads into a receptacular duct, or ‘reservoir duct’. The receptacular ducts of both the sides are united to form a common receptacular duct, or ‘common reservoir duct’. The median salivary duct opens into the common receptacular duct. Later these two form an efferent salivary duct. The efferent salivary duct opens at the base of the hypopharynx. Acinar cells secrete saliva, which contains starch digesting enzymes such as amylase.

Question 4.
Describe the blood circulatory system of Periplaneta in detail and draw a neat labelled diagram of it. (K & S)
Answer:
Circulatory system of Periplaneta :
The circulatory system helps in the transportation of digested food, hormones etc., from one part to another in the body. Periplaneta has an open type of circulatory system as the blood, or haemohymph, flows freely within the body cavity or haemocoel. Blood vessels are poorly developed and open into spaces. Visceral organs located in the haemocoel are bathed in the blood. The three main parts associated with the blood circulatory system of Periplaneta are the haemocoel, heart, and blood.
TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach) 11

Haemocoel :
The haemocoel of cockroach is divided into three sinuses by two muscular, horizontal membranes, called dorsal diaphragm or ‘pericardial septum1 and ventral diaphragm. Both the diaphragms have pores. There is a series of paired triangular muscles, called alary muscles. Every segment has one pair of these muscles situated on the lateral sides of the body. These are attached to the pericardial septum by their broad bases and to the terga by their pointed ends or apices.

The three sinuses of the haemocoel are known as pericardial haemocoel or the ‘dorsal sinus’, the perivisceral haemocoel or the ‘middle sinus’ and sternal haemocoel or ‘vental sinus’ or ‘perineural sinus’. The middle sinus is very large as it contains most of the viscera. The dorsal and ventral sinuses are small as they have only heart and nerve cord, respectively.

Heart :
The heart lies in the pericardial haemocoel or dorsal sinus. It is along, muscular, contractile tube found along the mid dorsal line, beneath the terga of the thorax and abdomen. It consists of 13 chambers. Every chamber opens into the other present in front of it. Three of the thirteen chambers are situated in the thorax and ten in the abdomen. Its posterior end is closed while the anterior end is continued forward as the anterior aorta. At the posterior side of each chamber, except the labt, there a pair of small apertures called ostia’ one on each side. Ostia have valves which allow the blood to pass only into the heart from the dorsal sinus.

Blood :
The blood of Periplaneta is colourless and is called haemolymph. it consists of a fluid called plasma, and free blood corpuscles or haemocytes, which are ‘phagocytic1. The phagocytes are large in size and can’ingest1 foreign particles such as bacteria. There is no respiratory pigment in the blood and so it plays no major role in respiration. The important functions of the blood are :

  1. It absorbs digested food from the alimentary canal and distributes it to the rest of the body.
  2. It brings nitrogenous wastes from all parts of the body to the excretory organs for their elimination.
  3. It carries defensive phagocytes to the places of infection where they engulf the germs and disintegrating tissue parts.
  4. It transports secretions of the ductless glands to the target organs.

Circulation of blood :
The blood flows forward in the heart by the contractions of its chambers. At the anterior end of the heart, the blood flows into the aorta and from there it enters the sinus of the head. From the head sinus, the blood flows into the perivisceral and sternal sinuses. On contraction of the alary muscles, the pericardial septum is pulled down. This increases the volume of the pericardial sinus. Hence blood flows from the perivisceral sinus into the pericardial sinus through the appertures of the pericardial septum. On relaxation of the alary muscles, the pericardial septum moves upwards to its original position. This forces the blood, to enter the chambers of the heart through the ostia from the pericardial sinus.
TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach) 3

Question 5.
The blood circulatory system of Periplaneta is of open type. Illustrate the statement describing the course of circulation in it. (U)
Answer:
The circulatory system helps in the transportation of digested food, hormones etc. from one part to another in the body. Periplaneta has an open type of circulatory system as blood or haemolymph flows freely with in the body cavity or haemocoel.

Circulation of blood :
The blood flows forward in the heart by the contraction of its chambers. At the anterior end of the heart, the blood flows into the aorta and from there it enters the sinus of the head. From the head sinus, the blood flows into the perivisceral and sternal sinuses. On contraction of the alary muscles the pericardial septum is pulled down. This increases the volume of the pericardial sinus. Hence blood flows from the perivisceral sinus into the pericardial sinus through the appertures of the pericardial septum. On relaxation of the alary muscles, the pericardial septum moves upwards to its original position. This forces the blood, to enter the chambers of the heart through the ostia from the pericardial sinus.
TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach) 3

Question 6.
Describe the respiratory system of cockroach with the help of neat and labelled diagrams. (K & S)
Answer:
Respiratory System of Periplaneta :
Due to the absence of respiratory pigment, the blood of cockroach is colourless and it cannot carry oxygen to different tissues. Therefore a tracheal system is developed to carry the air directly to the tissues. The respiratory system of cockroach consists of stigmata, tracheae and tracheoles.
TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach) 12

Stigmata or spiracles :
The tracheal system communicates with the exterior by ten pairs of openings called stigmata or spiracles. The first two pairs of spiracles are present in the thoracic segments, one pair in mesothorax and one pair in the metathorax. The remaining eight pairs are present in the first eight abdominal segments. Spiracles are located in the pleura of their respective segments. The respiratory system in insects is classified on the basis of number and nature of spiracles.

The spiracles of cockroach are polypneustic (as they are more than 3 pairs) and holopneustic (as all of them are functional). All spiracles are valvular and each of them is surrounded by a chitinous ring called peritreme. All spiracles bear small hair like structures called trichomes to filter the dust particles. Each spiracle opens into a small chamber called atrium.

Tracheae :
From the atrium of each thoracic spiracle several horizontal tracheae run inside. They join with each other in the thorax to form many tracheal trunks like dorsal cephalic, ventral cephalic trunks and their branches. These branches enter all organs of the head. The thoracic region also contains lateral longitudinal trunks. The abdominal spiracles lead into atria. From the atrium of each abdominal spiracle three tracheal tubes arise. All these tracheal tubes of one side open into three separate longitudinal tracheal trunks.

They are lateral, dorsal and ventral longitudinal trunks. Lateral longitudinal trunks are the longest tracheal trunks. The three pairs of longitudinal tracheal trunks of both the sides are interconnected by many commissural tracheae. From all the tracheal trunks several branches are given out, which enter different organs. All tracheal branches entering into an organ end in a special cell called tracheole cell.

The wall of the tracheae is made of three layers. They are an outer basement membrane, a middle one cell thick epithelium and an inner layer of cuticle called intima. The intima is produced into spiral thickenings called taenidia. The taenidia keep the tracheae always open and prevent it from collapsing.

Tracheoles :
The terminal cell of trachea is called tracheoblast or tracheole cell. It has several intracellular tubular extensions called tracheoles. Tracheoles are devoid of intima and taenidia. They are formed of a protein called trachein. Tracheolar fluid is present inside the tracheoles. The level of the tracheolar fluid varies with the metabolic activity of the insect. It is more when the insect is inactive and completely reabsorbed into the tissues, when the insect is more active. Tracheoles penetrate the cell and are intimately associated with mitochondria (to supply oxygen to them).

Mechanism of respriation :
Respiration includes two events, viz., inspiration and expiration. The muscles helpful are dorsoventral muscles and ventral longitudinal musdes. Dorsoventral muscles are the principal muscles of respriation.

TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach) 13

Inspiration :
Taking in of air is inspiration. lifts effected by the relaxation of the dorsoventral muscles and ventral longitudinal muscles. Due to the relaxation of the dorsoventral muscles, tergal plates are elevated and the volume of the body cavity increases. Due to the relaxation of the ventral longitudinal muscles, the telescoped segments come to the normal position. So the volume of the body cavity increases in the longitudinal axis. As air is drawn in due to the relaxation of the muscles, the process is a ‘passive’ process. During inspiration the thoracic spiracles are kept open and the abdominal spiracles are kept closed.

Expiration :
Sending out air from the body is called expiration. On contraction the dorsoventral muscles depress the tergal plates. Body cavity decreases in size and pressure increases. Due to the contraction of the ventral longitudinal muscles, the segments are telescoped and the volume of the body cavity decreases in the longitudinal axis increasing the pressure further. As this process involves the contraction of muscles, expiration is described as active process. During expiration thoracic spiracles are closed and abdominal spiracles are kept open.

Exchange of gases :
As air enters the tracheoles, oxygen from the air is taken into the cells and CO2 is released into haemolymph. The CO2 from the haemolymph mostly goes out through the inter-segmental membranes of the body wall. Cockroach and some other insects such as grasshoppers and beetles exhibit the phenomenon of discontinuous ventilation. In this mode of respiration continuous exchange of gases is interrupted by extended periods during which spiracles remain closed. The expulsion of CO2 from the body occurs in bursts, when the spiracles are open.

The exchange of gases depends on the metabolic rate and temperature. When air enters the tracheoles, oxygen diffuses faster into the tissues due to its high partial pressure. At the same time the carbon dioxide of tissues, instead of passing into the tracheal system, goes into the haemolymph. Carbon dioxide is carried more quickly into the haemolymph due to its greater solubility in it. This CO2 accumulates near the spiracles and diffuses into the artial chambers near the spiracles and goes o.ut in bursts through the abdominal spiracles. Opening and closing of spiracles is influenced by CO2 tension in haemolymph and oxygen tension in the trachea.

TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach)

Question 7.
Describe the nervous system of Periplaneta and draw a labelled diagram of it. (K & S)
Answer:
The nervous sytem of cockroach consists of central nervous system, peripheral nervous system and autonomous nervous system.

Central Nervous System :
It consists of a nerve ring, and a ganglionated double ventral nerve cord.

Nerve ring :
The nerve ring, which is present around the oesophagus, is formed by the following.

Brain :
Brain lies above the oesophagus. The brain is mainly a sensory and an endocrine centre. Three lobes of the brain are protocerebrum,deutocerebrum and tritocerebrum. the protocerebrum receives sensory impulses from the compound eyes through optic nerves; deutocerebrum receives sensory impulses from antennae through antennal nerves; and tritocerebrum receives sensory impulses from the labrum. Hence brain is principally ‘sensory’ in nature.
TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach) 14

Sub-oesophageal ganglion :
It lies below the oesophagus. It is the motor center that controls the movements of mouthparts, legs and wings, it is formed by the fusion of paired ganglia of mandibular, maxillary and labial segments of the head. Circum-oesophageal connectives : A pair of circum-oesophageal connectives is present around the oesophagus, connecting the tritocerebrum with the sub – oesophageal ganglion/ sub oesophageal ganglia.

Ventral nerve cord :
The two ventral nerve cords are solid and ganglionated. They arise from the sub-oesophageal ganglion and extend upto the 7th abdominal segment. The two nerve cords remain separate except at the ganglia. Three thoracic ganglia are present, one in each thoracic segment. In addition, there are six abdominal ganglia. The first to the fourth abdominal segments have one abdominal ganglion each. The 5th abdominal segment has no ganglion . The serially 5th abdominal ganglion is present in the 6th segment. The serially 6th abdominal ganglion is present in the 7th segment. The last or the 6th abdominal ganglion is the largest of all the abdominal ganglia. It is formed by the fusion of the ganglia of the 7th, 8th, 9th, and 10th abdominal segments.

Peripheral Nervous System :
It consists of nerves arising from the central nervous system. It receives a pair of optic nerves, from the compound eyes, a pair of antennal nerves, from the antennae and a pair of labral nerves from the labrum. Motor neurons of the frontal nerve to the frontal ganglion join the sensory neurons of the labral nerve to form the larbro frontal nerve arising from the tritocerebrum. Sub – oesophageal ganglion gives off motor nerves to the mandibles, maxillae, labium, wings and legs. It is the principal ‘motor centre’ in the body. Thoracic ganglia supply nerves to the parts of their respective segments. Metathoracic ganglia send nerves to the first abdominal segment also.

Nerves from the first four abdominal ganglia supply to the organs of the segments 2 – 6 serially (the 1st to the 4th ganglia innervate segments 2nd to 5th respectively). The 5th ganglion present in the 6th segment innervates the organs of the 6th segment. All organs present in 7th to 10th segments receive nerves from the last abdominal ganglion (present in the 7th segment.) The organs include the reproductive organs, copulatory appendages besides anal cerci.

Autonomous Nervous System :
This system is also called stomatogastric nervous system or ‘visceral nervous system’. It controls the visceral organs, particularly the muscles of the alimentary canal, and the heart. Autonomous nervous system includes four ganglia, a frontal ganglion on the dorsal wall of the pharynx, in front of the brain, hypocerebral ganglion or occipital ganglion above the oesophagus, behind the brain, a visceral ganglion or ingluvial ganglion on the wall of the crop and a proventricular ganglion on the gizzard.

These ganglia contain the ‘somata’ of the post ganglionic motor neurons. Pregaglionic motor neurons of tritocerebrum go to the frontal ganglion as labrofrontal and frontal nerve. Frontal ganglion is connected to the hypocerebral ganglion by a ‘recurrent nerve’. Hypocerebral ganglion is connected to the visceral ganglion and in turn the visceral ganglion is connected to proventricular ganglion.

TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach)

Question 8.
Describe the reproductive system of Periplaneta and draw neat and labelled diagrams of it. (K & S)
Answer:
Reproductive System of Periplaneta :
Periplaneta is dioecious, or unisexual, and both the sexes have well developed reproductive organs. The sexual dimorphism is evident both externally and internally. The female is different from the male in respect of short and broad abdomen, presence of brood pouches and absence of anal styles.

Male Reproductive System :
The male reproductive system consists of a pair of testes. These are elongated and lobed structures lying one on each lateral side in the fourth to sixth abdominal segments. They are embedded in the fat bodies. From the posterior end of each testis, there starts a thin duct, the vas deferens. The two vasa deferentia run backwards and inwards to open into a wide median duct, the ductus ejaculatorius ip the seventh segment. A characteristic mushroom shaped gland is present in the 6th and 7th abdominal segments which functions as an accessory reproductive gland. The gland consists of two types of tubules, i) long slender tubules. The utriculi majores or ‘peripheral tubules’ ii) Short tubules, the utriculi breviores. Secretion of utriculi majores forms the inner layer of the spermatophore while that of utriculi breviores nourishes the sperms. These tubules open into the anterior part of the ejaculatory duct.
TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach) 6

The seminal vesicles are present on the ventral surface of the ejaculatory duct. These sacs store the sperms in the form of bundles called spermatophores. The ejaculatory duct is a muscular tube that extends posteriorly and opens at the gonopore or the ‘male genital ‘pore’. The duct of phallic or conglobate gland also opens near the gonopore. Its function is still not known. Surrounding the male genital opening there are chitinous and asymmetrical structures called phallic organs or gonapophyses or phallomeres which help in copulation. These are the male external genitalia.

Female Reproductive System :
The female reproductive system of Periplaneta consists of a pair of ovaries, a pair of oviducts, vagina, spermathecae, spermathecal papilla, and colleterial glands.

Ovaries :
A pair of large ovaries lies laterally in 2 to 6 abdominal segments. They are light yellow in colour surrounded by fat bodies. Each ovary consists of eight tubules called ovarian tubules or ovarioles. Each ovariole consists of a tapering anterior filament called germarium, and a posterior wider vitellarium. The germarium contains various stages of developing ova, and the vitellarium contains mature ova with yolk. The tapering ends of the ovarioles of each ovary unite to form a single thread which attaches to the dorsal body wall. The ovarioles, at their posterior end unite to form a short wide oviduct. The oviducts unite to form a very short median vagina.

The vertical opening of the vagina is called female genital pore. It opens into a large genital pouch on the eighth sternum. A spermatheca or receptaculum seminis, consisting of a left-sac like and a right filamentous caecum, is present in the 6th segment which opens by a median aperture on a small spermathecal papilla in the dorsal wall of the genital pouch on the ninth sternum. In a fertile female, the spermatheca contains spermatophores, obtained during copulation.

A pair of branched colleterial glands is present behind the ovaries. These glands open into the genital pouch separately, just above the spermathecal aperture. Secretion of the two collaterial glands forms a hard egg case called ootheca around the eggs.

Genital pouch is formed by 7th, 8th, and 9th abdominal sterna. The sternum of the seventh segment is boat shaped and forms the floor and side walls of the genital pouch. The sterna of the eighth and ninth segments, which are tucked into the seventh segment, constitute the anterior wall and the roof of the genital pouch, respectively. The genital pouch has two chambers the anterior ‘gynatrium’ or genital chamber and posterior ‘vestibuium’ or oothecal chamber.

Three pairs of plate like chitinous structures called gonapophyses are present around the female genital aperture. These gonapophyses guide the ova into ootheca as ovipositors. These are the female external genitalia.
TS Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach) 15

TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare

Telangana TSBIE TS Inter 1st Year Zoology Study Material 6th Lesson Biology in Human Welfare Textbook Questions and Answers.

TS Inter 1st Year Zoology Study Material 6th Lesson Biology in Human Welfare

Very Short Answer Type Questions

Question 1.
Define parasitism and justify this term.
Answer:
An intimate association between two organisms of different species in which, one is benefited and the other one is often adversely affected is called parasitism. The organism that obtains nourishment is called parasite and the organism from which the nourishment is obtained is called host.

Question 2.
Distinguish between facultative parasite and obligatory parasite.
Answer:
a) Facultative Parasite :
They lead parasitic life on host if available or may lead free living life in its absence.
Example : (Mycobacterium tuberculosus), Ascaris lumbricoides is facultative anaerobe.

b) Obligatory parasite :
They lead total parasitic life on host and in its absence, they die.
Example : (Blood fluke) Taenia solium is an obligatory anaerobe.

Question 3.
Distinguish between definitive host and intermediate host.
Answer:
Definitive host :
It is the host that harbours the adult stage or sexually mature stage of a parasite or the host in which the parasite undergoes sexual reproduction.

Intermediate host :
It is the host that harbours the developing larval or immature or asexual stages of a parasite or the host in which the parasite undergoes asexual reproduction.

Question 4.
Distinguish between vector and a reservoir host.
Answer:
a) Reservoir host :
It is the host that lodges the infective stages of a parasite in its body when the main host is not available. In the reservoir host, the parasite neither undergoes development nor causes any disease.

b) Vector :
It is an organism which transfers the infective stages of a parasite from one main host to another.

Question 5.
Distinguish between mechanical vector and biological vector.
Answer:
a) Mechanical vector :
It is the vector, which merely transfers the infective stages of a parasite but no part of the parasitic development takes place in it.

b) Biological vector :
It is the vector in which the parasite undergoes a part of the development before it gets transferred to another host.

TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare

Question 6.
What is a hyper-parasite? Mention the name of one hyper-parasite.
Answer:
Hyperparasite :
It is a parasite which lives in /on the body of another parasite.
eg : Nosema notabilis (cnidosporan parasite) lives in Sphaerospora polymorpha (also a cnidosporan parasite) which lives in the urinary bladder of toad fish.

Question 7.
What do you mean by parasitic castration? Give one example. [March 2020, 17]
Answer:
Parasitic castration :
Some parasites cause the degeneration of gonads of the host, making it sterile. This effect is called parasitic castration, eg : Sacculina (a crustacean) causes degeneration of ovaries in the crab carcinus.

Question 8.
What are the parasitic adaptations observed in Ascaris lumbricoides?
Answer:

  1. Developed protective cuticle to withstand the action of digestive enzymes of the host.
  2. Reproductive potential is high due to risky lifecycle. Ascaris female lays 2 lakhs of eggs per day.
  3. It is a facultative anaerobe.

Question 9.
What are the endo-parasitic adaptations observed in Fasciola hepatica?
Answer:
Life cycle of Fasciola hepatica is very complex involving many developmental stages and two intermediate hosts to increase the chances of reaching a new definitive host.

Question 10.
Define neoplasia. Give one example.
Answer:
Neoplasia :
Some cause an abnormal growth of the host cells in a tissue to form . new structures. This effect is called Neoplasia which leads to cancers, eg : Some viruses.

Question 11.
Define the most accurate definition of the term ‘health’ and write any two factors that affect the health.
Answer:
Health is a state of complete physical, mental and social well-being and not merely ‘absence of any disease’ or ‘absence of physical fitness’. Two factors that affect the health are 1) contaminated water 2) pollution of air.

Question 12.
Distinguish between infectious and non-infectious diseases. Give two examples each.
Answer:
a) Infectious diseases :
The diseases which are easily transmitted from one person to another are called infectious diseases. Eg. Amoebic dysentery, Malaria,

b) Non – infectious diseases :
The diseases which are not transmitted from one person to another and are not caused by pathogens are called non-infectious diseases.
eg : Kidney problems, genetic disorders, heart problems.

TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare

Question 13.
When can you diagnose a healthy person as unhealthy?
Answer:
When the functioning of one or more organs or systems of the body is adversely affected characterized by various signs and symptoms, we say that we are not healthy or we are unhealthy.

Question 14.
Write any two diagnostic features of trophozoite of Entamoeba histolytica.
Answer:
a) Cartwheel shaped nucleus.
b) Single blunt finger like pseudopodium called lobopodium.
c) Presence of RBC in food vacuole.

Question 15.
‘Entamoeba histolytica is an obligatory anaerobe.’ Justify.
Answer:
The absence of mitochondria indicates the obligate anaerobic nature of Entamoeba histolytica.

Question 16.
Distinguish between precystic stage and cystic stage of E. histolytica.
Answer:
Precystic stage :
It is the non-feeding and non-pathogenic stage of Entamoeba histolytica. It is a small, spherical or oval, non-motile form.

Cystic stage :
It is round in shape and is surrounded by a thin, delicate, and highly resistant cyst wall. This is infective stage.

Question 17.
What is the reserve food in the precystic and early cyst stages of Entamoeba histolytica?
Answer:
The reserve food in the precystic and early cyst stages of Entamoeba histolytica are glycogen granules and chromatoid bars (made up of ribonucleo protein).

TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare

Question 18.
What is a metacystic form with reference to Entamoeba histolytica?
Answer:
When the tetranucleate cysts of entamoeba histolytica enter a new human host, they pass into small intestine where the cyst wall gets ruptured by the action of the enzyme trypsin, releasing the tetranucleate amoebae. Such tetranucleate excystic amoebae are called metacysts.

Question 19.
A person is suffering from bowel irregularity, abdominal pain, blood and mucus in stool /etc. Based on these symptoms, name the disease and its causative organism.
Answer:
The disease is Amoebic dysentery or Amoebiasis. The causative organism is Entamoeba histolytica.

Question 20.
On the advice of a doctor, a patient has gone to a clinical laboratory for the examination of a sample of faeces. The lab technician, on observing the stool of the patient diagnosed that the patient was suffering from amoebiasis. Write any two characteristic features based on which the technician came to that conclusion.
Answer:

  1. Stools with blood and mucous.
  2. Stools containing the tetranucleate cysts.

Question 21.
Define ‘asymptomatic cyst passers’ with reference to Entamoeba histolytica.
Answer:
Some people do not exhibit any symptoms of amoebic dysentery or intestinal amoebiasis or tropical amoebiasis. Such people are called carriers or ‘asymptomatic cyst passers’ as their stool contains the tetranucleate cysts.

Question 22.
Distinguish between primary amoebiasis and secondary amoebiasis.
Answer:

  1. Stools with blood and mucous with acute abdominal pair is generally referred to as primary amoebiasis.
  2. When trophozoites rupture the wall of capillaries, enter the blood stream, and primarily reach the liver where they may cause abscesses – some call it secondary amoebiasis.

Question 23.
What are the stages of Plasmodium vivax that infect the hepatocytes of man?
Answer:
The stages of plasmodium vivax that infect the hepatocytes of man are

  1. Sporozoites primarily
  2. Cryptozoites Secondarily (They are also referred to as I generation merozoites)

Question 24.
What are the stages of Plasmodium vivax that infect the RBC of the intermediate host?
Answer:
The stages of Plasmodium vivax that infect the RBc of the intermediate host are a) Cryptozoites b) Micro-metacryptozoites.

Question 25.
Define prepatent period. What is its duration in the life cycle of Plasmodium vivax?
Answer:
The interval between the first entry of plasmodium into the blood in the form of sporozoites and the second entry of plasmodium into the blood in the form of cryptozoites is called “Prepatent period”. In Plasmodium vivax it is 8 days.

Question 26.
Define incubation period. What is its duration in the life cycle of Plasmodium vivax? [March 2018 – A.P.]
Answer:
The period between the entry of plasmodium into the blood in the form of sporozoites and the first appearance of symptoms of malaria in man is called incubation period. It is approximately 10 to 14 days in Plasmodium vivax.

Question 27.
What are Schuffner’s dots? What is their significance?
Answer:
In the signet ring stage’of malarial parasite small red coloured dots appear in the cytoplasm of the RBC known as Schuffner’s dots. These are believed to be the antigens released by the parasite.

TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare

Question 28.
What are haemozoin granules ? What is their significance?
Answer:
The malaria parasite digests the globin part of the ingested haemoglobin and converts the soluble haem into an insoluble crystalline haemozoin granules. It is called the malaria pigment which is a disposable product.

Question 29.
Distinguish between schizogony and sporogony.
Answer:

  1. In man the plasmodium reproduces by asexual reproduction called Schizogony. It occurs in liver cells as well as in RBC.
  2. The formation of sporozoites in the oocysts seen in the stomach wall of female Anopheles mosquito is called SPOROGONY.

Question 30.
What is exflagellation and what are the resultant products called?
Answer:
The formation of male gametes or microgametes by lashing movements like flagella from micro gametocyte is called exflagellation.

Question 31.
Why is the syngamy found in Plasmodium called anisogamy?
Answer:
In plasmodium the male and female gametes are dissimilar in size, the process of union or syngamy is called Anisogamy.

Question 32.
What is ookinete? Based on the ‘sets of chromosomes’ how do you describe it?
Answer:
The zygote remains inactive for some time and then transforms into a long, slender, motile, vermiform ookinete or vermicule within 18 – 24 hours. The ookinete is diploid.

Question 33.
What is tertian fever, with reference to the types of malaria you have learnt about? Give the name of the causative species of the pathogen concerned.
Answer:
Four species of Plasmodium cause four types of malaria in man. They are
a) Plasmodium vivax – benign tertian malaria.
b) Plasmodium falciparum – malignant tertian malaria or cerebral malaria.
c) Plasmodium ovale – mild tertian malaria.
d) Plasmodium malariae-quartan malaria.
Recurrent fever of Malaria is called tertian fever.

TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare

Question 34.
What is the significance of hypnozoites, with reference to malaria fever?
Answer:
Some of the stages of macro-metacryptozoites may survive for a long period in liver as dormant stages called HYPNOZOITES. Reactivation of these hypnozoites leads to the initiation of fresh erythrocytic cycles resulting in the new attacks of malaria. This is referred to as relapse of malaria.

Question 35.
A person is suffering from chills and shivering and high temperature. These symptoms arp cyclically followed by profuse sweating and return to normal body temperature. Based on these symptoms, name the disease and its causative organism.
Answer:
a) Malaria is the disease; b) Causative organism Plasmodium.

Question 36.
Describe the methods of biological control of mosquitoes.
Answer:
Biological control of mosquitoes :
Introduction of larvivorous fishes like Gambusia, insectivorous plants like utricularia into the places where mosquitoes breed.

Question 37.
The eggs of Ascaris are called ‘mammillated eggs’. Justify.
Answer:
In Ascaris each egg is surrounded by a protein coat with rippled surface. Hence the eggs of Ascaris are described as “mammillated eggs”.

Question 38.
Write the route of extra intestinal migration followed by the juveniles of Ascaris lumbricoides.
Answer:
Extra intestinal migration exhibited by 2nd stage larva of Ascaris.
TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare 1

Question 39.
Write any two differences between male and female worms of Wuchereria bancrofti.
Answer:
Wuchereria bancrofti male and female differences.
a) Male worm :
Its posterior end is curved with a cloacal aperture. A pair of unequal, chitinous pineal spicules or copulatory spicules is present in the cloacal region.

b) Female worm :
Its posterior end is straight. Anus is present near the posterior end. The female genital pore or vulva is present mid ventrally at about one third the length from the mouth. It is ovi viviparous.

Question 40.
What is meant by nocturnal periodicity with reference to the life history of a nematode parasite you have studied?
Answer:
The microfilaria larvae of Wuchereria bancrofti in man move in the peripheral blood circulation during the night time between 10 pm and 4 am. This tendency is referred to as NOCTURNAL PERIODICITY.

TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare

Question 41.
Distinguish between lymphadenitis and lymphangitis.
Answer:

  1. Inflammation in the lymph vessels is called lymphangitis.
  2. Inflammation in the lymph glands is called lymphadenitis. Both are caused by filarial worm injection.

Question 42.
‘Elephantiasis is the terminal condition of filariasis.’ Justify.
Answer:
Fibroblasts accumulate in these tissues and form the fibrous tissue. In severe cases, the sweat glands of the skin in the affected regions disintegrate and the skin becomes rough. This terminal condition is referred to as Elephantiasis.

Question 43.
Mention the pathogens that cause ringworm.
Answer:
Ringworm is one of the most common infectious diseases in man. It is caused by many fungi belonging to the genera Microsporum, Trichophyton and Epidermophyton.

Question 44.
Explain any three preventive measures to control microbial infections.
Answer:

  1. The immunization programmes by the use of vaccines have enabled us to completely eradicate a deadly disease like smallpox.
  2. Biotechnology is at the verge of making available newer and safer vaccines.
  3. Discovery of antibiotics and various other drugs has also enabled us to treat infectious diseases effectively.

Question 45.
“Maintenance of personal and public hygiene is necessary for prevention and control of many infectious diseases.” Justify the statement giving suitable examples.
Answer:
The following hygienic habits help prevent spread of this disease.
a) Using boiled and filtered water
b) Washing hands, fruits and vegetables properly
c) Using septic tank toilets
These habits prevent diseases like Amoebiasis and Ascariasis.

TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare

Question 46.
Diseases like amoebic dysentery, ascariasis, typhoid etc., are more common in overcrowded human settlements. Why?
Answer:
In overcrowded human settlements, poor sanitation facilities are seen. No septic toilets for defecation. Drinking water is only from available water sources like tanks, lakes and ponds where cleaning of cattle, washing of clothes is also done. Due to these conditions the contaminated water, food and air spread the diseases like typhoid, amoebic dysentery and ascariasis.

Question 47.
In which way does tobacco affect the respiration? Name the alkaloid found in tobacco : [May 2017 – A.P.]
Answer:
Tobacco is smoked or chewed as gutkha or used in the form of snuff. Smoking increases the carbon monoxide level and reduces the oxygen level in the blood. Alkaloid found in tobacco is NICOTINE.

Question 48.
Define drug abuse.
Answer:
When drugs are taken for a purpose other than the medicinal use or in excess amounts that impair one’s physical or psychological functions, it constitutes “drug abuse”.

Question 49.
From which substances ‘Smack’ and ‘Coke’ are obtained?
Answer:
Smack and coke are obtained from :
Smack is chemically diacetylmorphine obtained by the acetylation of morphine extracted from dried latex of unripe seed capsule of poppy plant. Coke or crack is obtained from coca plant Erythroxylum coca, commonly called cocaine.

Question 50.
‘Many secondary metabolites of plants have medicinal properties. It is their misuse that creates problems.’ Justify the statement with an example.
Answer:
Benzodiazepines (tranquilizers), Lysergic acid diethyl amides (LSD) and other similar drugs, normally used as medicines to treat patients with mental illnesses like depression are often abused.

Question 51.
Write the scientific names of any two plants with hallucinogenic properties.
Answer:
TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare 2

Question 52.
Why are cannabinoids and anabolic steroids banned in sports and games?
Answer:
Cannabinoids and anabolic steroids (AAS) increase protein synthesis within cells which result in the build up of cellular tissues, especially in muscles. Cannabinoids show their effects on cardio-vascular system of the body. Hence they are banned in sports and games (DOPING TEST). They give tremendous energy temporarily.

TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare

Question 53.
Mention the names of any four drugs which are used as medicines to treat patients with mental illness like depression, insomnia, etc.’ that are often abused.
Answer:

  1. Lysergic acid diethyl amides (LSD)
  2. Barbiturates (sleeping pills)
  3. Amphetamines
  4. Benzodiazepines (tranquilizers)

Short Answer Type Questions

Question 1.
What is the need for parasites to develop special adaptations? Mention some special adaptations developed by the parasites. [May 2017 – A.P.]
Answer:
Parasites have evolved special adaptations to meet the requirements and lead successful life in the hosts.

  1. In order to live in the host, some parasites have developed structures like hooks, suckers, rostellum, etc., for anchoring, e.g. Taenia solium.
  2. Some intestinal parasites have developed protective cuticle to withstand the action of the digestive enzymes of the host. e.g. Ascaris lumbricoides.
  3. Some intestinal parasites produce anti enzymes to neutralize the effect of host’s digestive enzymes, e.g. Taenia solium.
  4. Some parasites live as obligatory anaerobes as the availability of oxygen is very rare for them. e.g. Entamoeba histolytica, Taenia solium, etc.
  5. Some intestinal parasites live as facultative anaerobes i.e., if oxygen is not available, they live anaerobically and if oxygen is available, they respire aerobically, e.g. Ascaris lumbricoides.
  6. The morphological and anatomical features are greatly simplified while emphasizing their reproductive potential. For example, an Ascaris lays nearly two lakh eggs per day. In Taenia solium the body is divided into 700 to 900 proglottids of which each proglottid acts as a unit of reproductive system and releases approximately 35,000 eggs.
  7. The life cycles of endoparasites are more complex because of their extreme specialization. For example, life cycle of certain parasites like Fasciola hepatica (sheep liver fluke) is very complex involving many developmental stages and two intermediate hosts to increase the chances of reaching a new definitive host.
  8. Certain parasites like Entamoeba develop cysts to tide over the unfavourable conditions like desiccation while reaching the new host.
  9. Some parasites elude production of vaccines against them (smart parasites!) as they keep changing their surface antigens form time to time.
    e.g. Plasmodium, HIV, etc.

Question 2.
Describe the effects of parasites on the host.
Answer:
Effects of parasites on hosts :
In general, the parasites cause weakening of the body of their hosts by causing the deprivation of nutrients, fluids and metabolites as they compete with their hosts for the same. They may also cause pathological effects in their hosts suchas

1) Parasitic castration :
Some parasites cause the degeneration of gonads of the host, making it sterile. This effect is called parasitic castration.
e.g. Sacculina (root headed barnacle, a crustacean) causes the degeneration of ovaries in the crab Carcinus maenas.

2) Neoplasia :
Some cause an abnormal growth of the host cells in a tissue to form new structures. This effect is called neoplasia which leads to cancers, e.g. Some viruses

3) Gigantism :
Some parasites cause an abnormal increase in the size of the host. This effect is called gigantism, e.g. The larval stages of Fasciola hepatica cause gigantism in snail (an intermediate host).

4) Hyperplasia :
Some parasites cause an increase in the number of cells. This effect is called hyperplasia, e.g. Fasciola hepatica in the bile ducts of sheep.

5) Hypertrophy :
Some parasites cause an abnormal increase in the volume / size of the infected host cells. This effect is called hypertrophy, e.g. RBC of man infected by Plasmodium.

6) Most of the parasites cause various types of diseases like

  1. African sleeping sickness by Trypanosoma gambiense
  2. Delhi boils / Tashkent ulcers / Oriental sores by Leishmania tropica
  3. Kala azar/ Dum dum fever/Visceral leishmaniasis by Leishmania donovani
  4. Malaria by Plasmodium sps
  5. Elephantiasis by Wuchereria bancrofti.

Question 3.
Distinguish between hypertrophy and hyperplasia with an example [March 2020]
Answer:
Hyperplasia :
Some parasites cause an increase in the number of cells. This effect is called hyperplasia, e.g. Fasaola hepatica in the bile ducts of sheep Hypertrophy: Some parasites cause an abnormal increase in the volume / size of the infected host cells. This effect is called hypertrophy, e.g. RBC of man infected by Plasmodium

TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare

Question 4.
Write any four types of parasitic diseases. Mention the primary and secondary hosts of these parasites.
Answer:

  1. Malaria caused by Plasmodium. Primary host – female Anopheles mosquito. Secondary host-man.
  2. Elephantiasis caused by wuchereria bancrofti. Primary host – man. Secondary host – female culex mosquito.
  3. Amoebic dysentery or Anoebiasis caused by Entamoeba histolytica. Single host -man.
  4. Ascariasis caused by Ascaris lumbricoides. Single host – man.

Question 5.
Describe the structure of a trophozoite of Entamoeba histolytica.
Answer:
TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare 3
Trophozoite stage :
It is the most active, motile, feeding and pathogenic stage that lives in the mucosa and sub-mucosa membranes of the large intestine. It moves with the help of a single blunt finger like pseudopodium called lobopodium which is produced anteriorly. The body of the trophozoite is surrounded by plas- malemma. Its cytoplasm is differentiated into an outer clear, viscous, non-granular ectoplasm and the inner fluid like, granular endoplasm. Ribosomes, food vacuoles and a vesicular, cartwheel shaped nucleus are present in the endoplasm. However contractile vacuoles, endoplasmic reticulum, Golgi apparatus and mitochondria are absent.

The absence of mitochondria indicates the ‘obligate anaerobic nature’ of Entamoeba histolytica. It produces the proteolytic enzyme called histolysin due to which the species name histolytica’ was assigned to it. Due to the effect of this enzyme, the mucosa and sub-mucosa of the gut wall are dissolved releasing some amount of blood, tissue debris which are ingested by the trophozoites. Hence, the food vacuoles are with erythrocytes, fragments of epithelial cells and bacteria. The mode of nutrition is holozoic. Presence of ‘RBC in food vacuoles’ and cartwheel shaped nucleus are the characteristic features of the trophozoites of Entamoeba histolytica.

Question 6.
Explain the life cycle of Entamoeba histolytica.
Answer:
Life cycle :
The trophozoites undergo binary fissions in the wall of the large intestine and produce a number of daughter entamoebae. They feed upon the bacteria and the host’s tissue elements, grow in size and again multiply. After repeated binary fissions, when the trophozoites increase in number, some of the young ones enter the lumen of the large intestine and transform into precystic stages. Here, the precystic stages transform into cystic stages which in turn develop into tetranucleate cysts.

The entire process is completed only in a few hours. These tetranucleate cysts come out along with the faecal matter and can remain alive for about 10 days. These cysts reach new host through contaminated food and water. They pass into the small intestine of a new human host where the cyst wall gets ruptured by the action of the enzyme trypsin, releasing the tetranucleate amoebae. Such tetranucleate excystic amoebae are called metacysts.

The four nuclei of the metacyst undergo mitotic divisions and produce eight nuclei. Each nucleus gets a bit of the cytoplasm and thus eight daughter entamoebae or ‘metacystic trophozoites’ are produced. These young ones develop into feeding stages called trophozoites. They invade the mucous membrane of the large intestine and grow into mature trophozoites.

Question 7.
Write a short note on the pathogenicity of Entamoeba histolytica.
Answer:
The trophozoites ‘dissolve’ the mucosal lining by histolysin, go deep into sub¬mucosa and cause ulcers. These ulcers contain cellular debris, lymphocytes, blood corpuscles and bacteria. It leads to the formation of abscesses in the wall of large intestine. Ultimately it results in stool with blood and mucous. This condition is called amoebic dysentery or intestinal amoebiasisor tropical amoebiasis. Some people do not exhibit any symptoms. Such people are called ‘carriers or asymptomatic cyst passers’ as their stool contains the tetranucleate cysts. They help in spreading the parasites to other persons.

Extra-intestinal amoebiasis:
Sometimes, the trophozoites may rupture the wall of capillaries, enter the blood stream and primarily reach the liver where they may cause ‘abscesses’ (some call it ‘secondary amoebiasis’). From there, they may go to lungs, heart, brain, kidneys, gonads, etc., and cause abscesses in those parts leading to severe pathological conditions.

TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare

Question 8.
Describe the structure of sporozoite of Plasmodium vivax.
Answer:
TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare 4
Structure of sporozoite :
It is sickle shaped with a swollen middle part and pointed at both ends of its body. It measures about 15 microns in length and one micron in width. The body is covered by an elastic pellicle with microtubules which help in the wriggling movements of the sporozoite. The cytoplasm contains cell organelles such as Golgi complex, endoplasmic reticulum, mitochondria and a nucleus. Cytoplasm also shows many convoluted tubules of unknown function throughout the length of the body. It contains a cup like depression called apical cup at the anterior end into which a pair of secretory organelles opens. They secrete a cytolytic enzyme, which helps in the penetration of sporozoite into the liver cells.

Question 9.
What do you know about the exo-erythrocytic cycle of plasmodium vivax?
Answer:
This is a part of the life cycle of Plasmodium taking place in the liver of man. This starts after the liberation of cryptozoites from ruptured liver cells. Exo-erythrocytic cycle: If the cryptozoites enter the fresh liver cells, they undergo the changes similar to that of the pre-erythrocytic cycle and produce the second generation merozoites called metacryptozoites. These are of two types – the smaller micro-metacryptozoites and larger macro-metacryptozoites. This entire process is completed approximately in two days. The macro-metacryptozoites attack fresh liver cells and continue another exo-erythrocytic cycle, whereas the micro- metacryptozoites always enter blood stream and attack fresh RBC to continue erythrocytic cycle.

Question 10.
Describe the cycle of Golgi in the life history of Plasmodium vivax.
Answer:
Erythrocytic cycle :
Itwas first described by Camillo Golgi. Hence it is also called Golgi cycle. This cycle is initiated either by the cryptozoites of pre-erythrocytic cycle or the micro-metacryptozoites of exo-erythrocytic cycle. In the fresh RBC, these stages assume spherical shape and transform into trophozoites. It develops a small vacuole which gradually enlarges in size, pushing the cytoplasm and nucleus to the periphery. Now the Plasmodium looks like a finger ring. Hence this stage is called signet ring stage. Soon it loses the vacuole, develops pseudopodia and becomes amoeboid stage. With the help of pseudopodia, it actively feeds on the contents of the RBC and increases in size. As a result, the RBC grows almost double the size, this process is called hypertrophy.

The malaria parasite digests the globin part of the ingested haemoglobin and converts the soluble haem into an insoluble crystalline haemozoin. It is called the ‘malaria pigment’ which is a disposable product. During this stage, small red coloured dots appear in the cytoplasm of the RBC known as Schuffner s dots. These are believed to be the antigens released by the parasite. Now the Plasmodium loses the pseudopodia, further increases in size, occupies the entire RBC and becomes a schizont. It undergoes schizogony similar to that of the pre-erythrocytic cycle and produces 12 to 24 erythrocytic merozoites. They are arranged in the form of the petals of a rose in the RBC. Hence, this stage is called the rosette stage. Finally the erythrocyte bursts and releases the merozoites along with haemozoin into the blood. This cycle is completed approximately in 48 hours.
TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare 5

Question 11.
Describe the process of sporogony in the life cycle of Plasmodium vivax. What is the significance of sporozoites?
Answer:
TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare 6
Sporogony :
The formation of sporozoites in the oocysts is called sporogony. According to Bano, the nucleus of the oocyst first undergoes reduction division followed by repeated mitotic divisions resulting in the formation of about 1,000 daughter nuclei. Each bit of nucleus is surrounded by a little bit of the cytoplasm and transforms into a sickle shaped sporozoite. Oocyst with such sporozoites is called sporocyst. When this sporocyst ruptures, the sporozoites are liberated into the haemocoel of the mosquito. From there, they travel into the salivary glands and are ready for infection. The life cycle of Plasmodium in mosquito is completed in about 10 to 24 days.

TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare

Question 12.
Explain the pathogenicity of Wuchereria bancrofti in man.
Answer:
Pathogenicity of Wuchereria bancrofti in man :
Light infection causes filarial fever which is characterised by headache, mental depression and increase in the body temperature. In general, the infection of filarial worm causes inflammatory effect in lymph vessels and lymph glands. Inflammation in the lymph vessels is called lymphangitis and that of lymph glands is called lymphadenitis.

In the case of heavy infection, the accumulation of dead worms blocks the lymph vessels and lymph glands resulting in immense swelling. This condition is called lymphoedema which is noticed in the extremities of limbs, scrotum of males and mammary glands in females. Fibroblasts accumulate in these tissues and form the fibrous tissue. In severe cases, the sweat glands of the skin in the affected regions disintegrate and the skin becomes rough. This terminal condition is referred to as elephantiasis.

Question 13.
Write short notes on typhoid fever and its prophylaxis.
Answer:
Typhoid fever :
It is caused by Salmonella typhi which is a Gram negative bacterium. It mainly lives in the small intestine of man and then migrates to other organs through blood. It can be confirmed by Widal test.

Mode of infection :
Contamination through food and water.

Symptoms :
Sustained fever with high temperature upto 104 F, weakness, stomach pain, constipation, headache and loss of appetite. Intestinal perforation and death may also occur in severe cases.

Prophylaxis :
The immunization programmes by the use of vaccines have enabled us to completely eradicate a deadly disease like smallpox. Biotechnology is at the verge of making available newer and safer vaccines. Discovery of antibiotics and various other drugs has also enabled us to treat infectious diseases effectively.

Question 14.
Write short notes on pneumonia and its prophylaxis.
Answer:
Pneumonia :
It is caused by Gram positive bacteria such as streptococcus pneumoniae and Haemophilus influenzae. They infect the alveoli of lungs in human beings.

Mode of infection :
Contamination by inhaling the droplets / aerosols released by an infected person or even by sharing the utensils with an infected person.

Symptoms :
The alveoli get filled with fluid leading to severe problems in respiration. In severe cases, the lips and finger nails may turn gray to bluish in colour.

Prophylaxis :
The immunization programmes by the use of vaccines have enabled us to completely eradicate a deadly disease like smallpox. Biotechnology is at the verge of making available newer and safer vaccines. Discovery of antibiotics and various other drugs has also enabled us to treat infectious diseases effectively.

TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare

Question 15.
Write short notes on common cold and its prophylaxis.
Answer:
Common cold :
It is caused by Rhino virus group of viruses. They infect nose and respiratory passage but not lungs.

Mode of infection :
Contamination by direct inhalation of the droplets resulting from cough or sneezes of an infected person or indirectly through contaminated objects such as pens, books, cups, door-knobs, computer keyboard or mouse etc. Generally all the medicines that are used against cold cause drowsiness.

Symptoms :
Nasal congestion, discharge from nose, sore throat, hoarseness, cough, headache, tiredness, etc., which usually last for 3 – 7 days.

Prophylaxis :
The immunization programmes by the use of vaccines have enabled us to completely eradicate a deadly disease like smallpox. Biotechnology is at the verge of making available newer and safer vaccines. Discovery of antibiotics and various other drugs has also enabled us to treat infectious diseases effectively.

Question 16.
Write short notes on ‘ringworm’ and its prophylaxis.
Answer:
Ringworm :
It is one of the most common infectious diseases in man. It is caused by many fungi belonging to the genera Microsporum, Trichophyton, and Epidermophyton. Heat and moisture help these fungi grow in the skin folds such as those in the groin or between the toes.

Mode of infection :
Contamination by using towels, clothes or combs of the infected persons or even from soil.

Symptoms :
Appearance of dry, scaly, usually round lesions accompanied by intense itching on various parts of the body such as skin, nails and scalp.

Prophylaxis :
Using of antifungal drugs like Griesoven orally, and antifungal creams like Zole, Itchguard. Ring guard for external tise’cures the disease. Hygiene habits prevent the disease.

Question 17.
What are the adverse effects of tobacco?
Answer:
Tobacco has been used by human beings for more than 400 years. It contains a large number of chemical substances including nicotine, an alkaloid. While buying cigarettes one cannot miss the statutory warning present on the packet Smoking is injurious to health.

Mode of abuse :
It is smoked or chewed as gutkha or used in the form of snuff.

Effect :
Smoking increases the carbon monixide (CO) level and reduces the oxygen level in the blood. Nicotine stimulates the adrenal gland to release adrenaline and nor-adrenaline into blood. These hormones raise the blood pressure and increase the heart rate. Smoking is associated with bronchitis, emphysema, coronary heart disease, gastric ulcer and increases the incidence of cancers of throat, lungs, urinary bladder etc. Smoking also paves the way to hard drugs. Yet, smoking is very prevalent in society, both among young and old. Tobacco chewing is associated with increased risk of cancer of the oral cavity.

TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare

Question 18.
Write short notes on opioids.
Answer:
Opioids :
These are the drugs obtained from opium poppy plant Papaver somniferum (vernacular name : Nallamandu mokka).

They bind to specific opioid receptors present in our central nervous system and gastrointestinal tract. Some of them are morphine, heroin, etc.

i) Morphine :
It is extracted from the dried latex of the unripe seed capsule (pod) of poppy plant. It occurs as colourless crystals or a white crystalline powder.

Mode of abuse :
Generally it is taken orally or by injection.

Effect :
It is a very effective sedative and painkiller. It is very useful in patients who have undergone surgery.

ii) Heroin :
It is a white, bitter, odourless and crystalline compound, obtained by the acetylation of morphine. Chemically it is diacetylmorphine. It is commonly called ‘smack’.

Mode of abuse :
Generally it is taken by snorting and injection.

Effect :
Heroin is a depressant and slows down the body functions.

Question 19.
Write short notes on Cannabinoids.
Answer:
Cannabinoids :
These are a group of chemicals obtained from Indian hemp plant Cannabis sativa (vernacular name: Ganjai mokka). They interact with cannabinoid receptors present in the brain. The flower tops, leaves and the resin of this plant are used in various combinations to produce marijuana, hashish, charas and ganja. These days, cannabinoids are being abused by even some sports-persons (doping).

Mode of abuse :
These are generally taken by inhalation and oral ingestion.

Effect :
Show their effects on cardiovascular system of the body.

Question 20.
Write short notes on Cocaine.
Answer:
Coca alkaloid or Cocaine :
It is a white, crystalline alkaloid that is obtained from the leaves of Coca plant Erythroxylum coca, native to South America. It is commonly called coke or crack.

Mode of abuse :
It is usually snorted.

Effect :
It has a potent stimulating action on the central nervous system as it interferes with the transport of the neuro transmitter dopamine. Hence it produces a sense of euphoria and increased energy. Its excessive dosage causes hallucinations. Other well-known plants with hallucinogenic properties are Atropa belladonna and Datura. Certain drugs like ‘Barbiturates (sleeping pills), Amphetamines (cause sleeplessness), Benzodiazepines (tranquilizers), Lysergic acid diethyl amides (LSD) and other similar drugs, normally used as medicines to treat patients with mental illnesses like depression, insomnia, etc.,1 are often abused.

Question 21.
Why is adolescence considered vulnerable phase?
Answer:
Adolescence :
It is the time period between the beginning of puberty and the beginning of adulthood. In other words, it is the bridge linking childhood and adulthood. The age between 12 -18 years is considered adolescence period. It is both ‘a period and a process’ during which a child becomes mature. It is accompanied by several biological and behavioural changes. Thus, adolescence is a very vulnerable phase of mental and psychological development of an individual.

TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare

Question 22.
Why do some adolescents start taking drugs? How can this be avoided?
Answer:
Curiosity, desire for adventure and excitement, experimentation, are the common causes for the motivation of youngsters towards the use of tobacco, drugs. The first use of drugs or alcohol may be out of curiosity or experimentation, but later the person starts using them to escape facing problems. Recently ‘stress from the pressure to excel in academics or examination’s has played a significant role in alluring the youngsters to try certain drugs. Television, movies, newspapers and internet also help promoting this wrong perception. Other factors that are associated with tobacco, drug and alcohol abuse among adolescents are unstable or unsupportive family structures and peer pressure.

A lot of help is available in the form of highly qualified psychologists, psychiatrists and de-addiction and rehabilitation programmers. Educating and counselling the adolescents to face problems, stress and failures as a part of life.

Question 23.
Distinguish between addiction and dependence. [March 2017 -A.P.]
Answer:
Addiction :
It is a psychological attachment to certain effects such as euphoria. The most important thing one fails to realise is, the inherent addictive nature of tobacco, drugs and alcohol. With the repeated use of TDA, the tolerance level of the receptors present in our body increases. Consequently the receptors respond only to higher doses leading to greater intake and addiction.

However it should be clearly borne in mind that use of TDA even once, can be a fore-runner to addiction. Thus, the addictive potential of tobacco, drugs’and alcohol pull the users into a vicious circle leading to their regular use (abuse) from which they may not be able to get out. In the absence of any guidance or counseling, people get addicted and become dependent on them.

Dependence :
It is the tendency of the body to manifest a characteristic and unpleasant condition (withdrawal syndrome) if the regular dose of drugs or alcohol is abruptly discontinued. The withdrawal syndrome is characterised by anxiety, shakiness (tremors), nausea and sweating which may be relieved when the regular use is resumed again. Dependence leads the patients to ignore all social norms.

Question 24.
“Prevention is better than cure.” Justify with regard to TDA abuse. [March 2018 – A.P.]
Answer:
Prevention and Control: The age-old adage of Prevention is better than cure holds true here also. Some of the measures useful for prevention and control of TDA abuse among the adolescents are :

i) Avoid undue parental pressure :
Every child has his / her own choice, capacity and personality. The parents should not force their children to perform beyond their capacity by comparing them with others in studies, games, etc.

ii) Responsibility of parents and teachers :
They should look for the danger signs and counsel such students who are likely to get into the ‘trap’.

iii) Seeking help from peers :
If peers find some one abusing drugs or alcohol, immediately it should be brought to the notice of their parents or teachers so that they can guide them appropriately.

iv) Education and counselling :
Educating and counselling the children to face problems, stress and failures as a part of life.

v) Seeking professional and medical help :
A lot of help is available in the form of highly qualified psychologists, psychiatrists and de-addiction and rehabilitation programmers.

Essay Answer Type Questions

Question 1.
Explain the structure and life cycle of Entamoeba histolytica with the help of neat and labelled diagrams.
Answer:
Structure: Entamoeba histolytica passes through three distinct stages in its life cycle, namely:

  1. Trophozoite stage
  2. Precystic stage and
  3. Cystic stage

TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare 3
Trophozoite stage :
It is the most active, motile, feeding and pathogenic stage that lives in the mucosa and sub-mucosa membranes of the large intestine. It moves with the help of a single blunt finger like pseudopodium called lobopodium which is produced anteriorly. The body of the trophozoite is surrounded by plasmalemma. Its cytoplasm is differentiated into an outer clear, viscous, non-granular ectoplasm and the inner fluid like, granular endoplasm. Ribosomes, food vacuoles and a vesicular, cartwheel shaped nucleus are present in the endoplasm. However contractile vacuoles, endoplasmic reticulum, Golgi apparatus and mitochondria are absent.

The absence of mitochondria indicates the ‘obligate anaerobic nature’ of Entamoeba histolytica. It produces the proteolytic enzyme called histolysin due to which the species name histolytica’ was assigned to it. Due to the effect of this enzyme, the mucosa and sub-mucosa of the gut wall are dissolved releasing some amount of blood, tissue debris which are ingested by the trophozoites. Hence, the food vacuoles are with erythrocytes, fragments of epithelial cells and bacteria. The mode of nutrition is holozoic. Presence of ‘RBC in food vacuoles’ and cartwheel shaped nucleus are the characteristic features of the trophozoites of Entamoeba histolytica.

Precysticstage :
TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare 7
It is the non-feeding and non-pathogenic stage of Entamoeba histolytica that is found in the lumen of the large intestine. It is a small, spherical or oval, non- motile form. The cytoplasm of the precystic stage stores glycogen granules and chromatoid bars (made of ribonucleo protein) which act as reserve food.

Cystic stage :
TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare 8
It is round in shape and is surrounded by a thin, delicate and highly resistant cyst wall. It is found in the lumen of the large intestine. The process of development of cyst wall is called encystation which is a means to tide over the unfavourable conditions that the parasite is going to encounter while passing to a new host. Soon after the encystation, the nucleus undergoes two successive mitotic divisions to form four daughter nuclei. This type of cystic stage is called tetra nucleate cyst or mature cyst which is ‘the stage infective to man’.

Life cycle :
The trophozoites undergo binary fissions in the wall of the large intestine and produce a number of daughter entamoebae. They feed upon the bacteria and the host’s tissue elements, grow in size and again multiply. After repeated binary fissions, when the trophozoites increase in number, some of the young ones enter the lumen of the large intestine and transform into precystic stages. Here, the precystic stages transform into cystic stages which in turn develop into the tetranucleate cysts.

The entire process is compelted only in a few hours. These tetranucleate cysts come out along with the faecal matter and can remain alive for about 10 days. These cysts reach new host through contaminated food and water. They pass into the small intestine of a new human host where the cyst wall gets ruptured by the action of the enzyme trypsin, releasing the tetranucleate amoebae. Such tetranucleate excystic amoebae are called metacysts.

The four nuclei of the metacyst undergo mitotic divisions and produce eight nuclei. Each nucleus gets a bit of the cytoplasm and thus eight daughter entamoebae or ‘metacystic trophozoites’ are produced. These young ones develop into feeding stages called trophozoites. They invade the mucous membrane of the large intestine and grow into mature trophozoites.

Pathogenicity :
The trophozoites ‘dissolve’ the mucosal lining by histolysin, go deep into sub¬mucosa and cause ulcers. These ulcers contain cellular debris, lymphocytes, blood corpuscles and bacteria. It leads to the formatjon of abscesses in the wall of large intestine . Ultimately it results in stool with blood and mucous. This condition is called amoebic dysentery or intestinal amoebiasis or tropical amoebiasis. Some people do not exhibit any symptoms. Such people are called ‘carriers or asymptomatic cyst passers’ as their stool contains the tetranucleate cysts. They help in spreading the parasites to other persons.

Extra-intestinal amoebiasis :
Sometimes, the trophozoites may rupture the wall of capillaries, enter the blood stream, and primarily reach the liver where they may cause ‘abscesses’ (some call it ‘secondary amoebiasis’). From there, they may go to lungs, heart, brain, kidneys, gonads, etc., cause abscesses in those parts leading to severe pathological conditions.

TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare

Question 2.
Describe the life cycle of Plasmodium vivax in man. [March 2017 – A.P.]
Answer:
Life cycle of Plasmodium in man (The human phase): In man, the Plasmodium reproduces by asexual reproduction called schizogony. It occurs in liver cells (hepatocytes) as well as in RBC. In liver cells, it is called hepatic schizogony and in RBC it is called erythrocytic schizogony.

Hepatic schizogony :
This was discovered by Shortt and Garnham. Whenever, a mosquito infected by Plasmodium bites a man, nearly 2000 sporozoites are released into the blood of man through its saliva. Within half an hour, they reach the hepatocytes where they undergo pre-erythrocytic and exo-erythrocytic cycles.

Pre-erythrocytic cycle :
Whenever the sporozoites reach the liver cells, they transform into trophozoites. They feed on the contents of the hepatic cells, assume spherical shape and attain the maximum size. This stage is called schizont stage. Its nucleus divides several times mitotically, followed by the cytoplasmic divisions resulting in approximately 12,000 daughter individuals called cryptozoites or the regeneration merozoites. They enter the sinusoids of the liver by rupturing the cell membrane of the schizont and the liver cells. This entire process is completed approximately in 8 days. Now these first generation merozoites have two options i.e., they can enter either fresh liver cells and continue exo-erythrocytic cycle or they can enter RBC and continue erythrocytic cycle.

Exo-erythrocytic cycle :
If the cryptozoites enter the fresh liver cells, they undergo the changes similar to that of the pre-erythrocytic cycle and produce the second generation merozoites called metacryptozoites. These are of two types -the smaller micro-metacryptozoites and larger macro-metacryptozoites. This entire process is completed approximately in two days. The macro-metacryptozoites attack fresh liver cells and continue another exo – erythrocytic cycle, whereas the micro- metacryptozoites always enter blood stream and attack fresh RBC to continue erythrocytic cycle.

Prepatent period :
The interval between ‘the first entry of Plasmodium into the blood in the form of sporozoites and the second entry of Plasmodium into the blood in the form of cryptozoites is called prepatent period. It lasts approximately 8 days. During this period, the host does not show any clinical symptoms of the disease. It is only a means of multiplication.

Erythrocytic cycle :
It was first described by Camillo Golgi. Hence it is also called Golgi cycle. This cycle is initiated either by the cryptozoites of pre-erythrocytic cycle or the micro-metacryptozoites of exo – erythrocytic cycle. In the fresh RBC, these stages assume spherical shape and transform into trophozoites. It develops a small vacuole which gradually enlarges in size, pushing the cytoplasm and nucleus to the periphery. Now the Plasmodium looks like a finger ring. Hence this stage is-” called signet ring stage. Soon it loses the vacuole, develops pseudopodia and becomes amoeboid stage.

With the help of pseudopodia, it actively feeds on the contents of the RBC and increases in size. As a result, the RBC grows almost double the size. This process is called hypertrophy. The malaria parasite digests the globin part of the ingested haemoglobin and converts the soluble haem into an insoluble crystalline haemozoin. It is called the ‘malaria pigment’ which is a disposable product. During this stage, small red coloured dots appear in the cytoplasm of the RBC known as Schuffner s dots.

These are believed to be the antigens released by the parasite. Now the Plasmodium loses the pseudopodia, further increases in size, occupies the entire RBC and becomes a schizont. It undergoes schizogony similar to that of the pre-erythrocytic cycle and produces 12 to 24 erythrocytic merozoites. They are arranged in the form of the petals of a rose in the RBC. Hence, this stage is called the rosette stage. Finally the erythrocyte bursts and releases the merozoites along with haemozoin into the blood. This cycle is completed approximately in 48 hours.

Incubation Period :
The period between ‘the entry of Plasmodium into the blood in the form of sporozoite and the first appearance of symptoms of malaria in man’ is called incubation period. It is approximately 10 to 14 days.

Formation of gametocytes :
After repeated cycles of erythrocytic schizogony, when the number of fresh RBC decreases, some merozoites enter the RBC and transform into gametocytes instead of continuing the erythrocytic cycle. This process generally takes place when the RBCs are present in spleen and bone marrow.

The gametocytes are of two types namely, smaller microgametocytes or male gametocytes and larger macrogametocytes or female gametocytes. The gametocytes cannot undergo further development in man as the temperature and pH of the blood of man are not suitable for further development. These gametocytes reach the blood circulation and wait to reach the next host. They degenerate and die if they are not transferred to mosquito within a week.
TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare 9

Question 3.
Describe the life cycle of Plasmodium vivax in mosquito.
Answer:
Life cycle of Plasmodium in mosquito (The mosquito phase)- Ross cycle: When a female Anopheles mosquito bites and sucks the blood of a malaria patient, the gametocytes along with the other stages of the erythrocytic cycle reach the crop of mosquito. Here all the stages are digested except the gametocytes. Further part of the life cycle consists of:

  1. Gametogony
  2. Fertilization
  3. Formation of Ookinete & Oocysts
  4. Sporogony

i) Gametogony :
The formation of male and female gametes from the gametocytes is called gametogony. It occurs in the lumen of the crop of mosquito.

Formation of male gametes :
During this process, the nucleus of microgametocyte divides into eight daughter nuclei called pronuclei which reach the periphery. The cytoplasm is pushed out in the form of eight flagella like processes. Into each flagellum like process, one pronucleus enters and forms a micro gamete or male gamete. These male gametes show lashing movements like flagella and get separated from the cytoplasm of microgametocyte. This process is called exflagellation.

Formation of female gamete :
The female gametocyte undergoes a few changes and transforms into a female gamete. This process is called maturation. The nucleus of the female gamete moves towards the periphery and the cytoplasm at that point forms a projection. This projected region is called the fertilization cone.

ii) Fertilization :
The fusion of male and female gametes is called fertilization. It also occurs in the lumen of the crop of the mosquito. When an actively moving male gamete comes into contact with the fertilization cone of the female gamete, it enters it. The pronuclei and cytoplasm of these two gametes fuse with each other, resulting in the formation of a synkaryon. Since the two gametes are dissimilar in size, this process is known as anisogamy. The female gamete that bears the synkaryon is called the zygote which is round and non – motile.
TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare 10

iii) Formation of ookinete and oocysts :
The zygote remains inactive for some time and then transforms into a long, slender, motile, vermiform ookinete or vermicide within 18 to 24 hours. It pierces the wall of the crop and settles beneath the basement membrane. It becomes round and secretes a cyst around its body. This encysted ookinete is now called oocyst. About 50 to 500 oocysts are formed on the wall of the crop and appear in the form of small nodules. (Sir Ronald Ross identified these oocysts for the first time).

iv) Sporogony :
TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare 6
The formation of sporozoites in the oocysts is called sporogony. According to Bano, the nucleus of the oocyst first undergoes reduction division followed by repeated mitotic divisions resulting in the formation of about 1,000 daughter nuclei. Each bit of nucleus is surrounded by a little bit of the cytoplasm and transforms into a sickle shaped sporozoite. Oocyst with such sporozoites is called sporocyst. When this sporocyst ruptures, the sporozoites are liberated into the haemocoel of the mosquito. From there, they travel into the salivary glands and are ready for infection. The life cycle of Plasmodium in mosquito is completed in about 10 to 24 days.

TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare

Question 4.
Describe the structure and life cycle of Ascaris lumbricoides with the help of a neat and labelled diagram.
Answer:
TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare 11
Structure :
Sexes are separate and the sexual dimorphism is distinct. In both males and females, the body is elongated and cylindrical. Mouth is present at the extreme anterior end and is surrounded by three chitinous lips. Close to the mouth mid ventrally, there is a small aperture called excretory pore.

Male :
It has a curved posterior end which is considered the tail. The posterior end possesses a cloacal aperture and a pair of equal sized chitinous ‘pineal spicules or ‘pineal setae which serve to transfer the sperms during copulation.

Female :
It has a straight posterior end, the tail. The female genital pore or vulva is present mid ventrally at about one third the length from mouth. The anus is present a little in front of the tail end.

Life history :
Copulation takes place in the small intestine of man. After copulation, the female releases approximately two lakh eggs per day. Each egg is surrounded by ‘a protein coat with rippled surface. Hence the eggs of Ascaris are described as mammillated eggs. The protein coat is followed by a chitinous shell and a lipid layer internally. These eggs come out along with faecal matter. In the moist soil, development takes place inside the egg so that the 1st stage rhabditiform larva is produced. It undergoes the 1st moulting and becomes the 2nd stage rhabditiform larva which is considered ‘the stage infective to man’. They reach the alimentary canal of man through contaminated food and water.
TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare 12

In the small intestine, the shell gets dissolved so that the 2nd stage larva is released. Now it undergoes extra intestinal migration. First it reaches the liver through the hepatic portal vein. From there it reaches the heart through the post caval vein. It goes to the lungs through the pulmonary arteries. In the alveoli of lungs it undergoes the 2nd moulting to produce the 3rd stage larva. It undergoes the 3rd moulting so that the 4th stage larva is produced in the alveoli only. It leaves the alveoli and reaches the small intestine again, through bronchi, trachea, larynx, glottis, pharynx, oesophagus and stomach. In the small intestine, it undergoes the 4th and final moulting to become a young one which attains sexual maturity within 8 to 10 weeks.

Pathogenicity :
The disease caused by Ascaris lumbricoides is called ascariasis. The disease is asymptomatic if the number of worms is less. A heavy infection causes nutritional deficiency and severe abdominal pain. It also causes stunted growth in children.
TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare 13

Question 5.
Describe the life cycle of Wuchereria bancrofti.
Answer:
Life Cycle :
Wuchereria bancrofti completes its life cycle in two hosts namely man and female culex mosquito.

In man :
Both male and female worms are found coiled together in the lymphatic vessels of man. After copulation the female releases the sheathed microfilaria larvae into the lymph of man. Each sheathed microfilaria larva measures 0.2 to 0.3 mm in length. It is surrounded by a loose cuticular sheath which is supposed to be the modified shell. They migrate to the blood circulation and reside in the deeper blood vessels during the day time. They move to the peripheral blood circulation during the night time between 10.00 pm and 4.00 am. This tendency is referred to as nocturnal periodicity. When a female Culex mosquito sucks the blood of an infected person, they enter the gut of mosquito. They die if they are not transferred to mosquito with in 70 days.

In mosquito :
In the mid gut of mosquito, the sheath of the larva is dissolved within 2 to 6 hours of the infection. The ex-sheathed microfilaria larva penetrates the gut wall and reaches the haernocoel of mosquito. From there, it reaches the thoracic muscles and transforms into a ‘sausage shaped larva within two days. It is called the first stage larva or first stage microfilaria. This undergoes two moultings within 10 to 20 days and transforms into a long, infective 3rd stage microfilaria. It reaches the labium of the mosquito.
TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare 14

In man after the infection :
When an infected mosquito bites a man, the 3rd stage microfilaria larvae enter the blood circulation of man and finally reach the lymphatic vessels. Here they undergo the 3rdand the 4th moultings to produce young filarial worms. They attain sexual maturity within 5 to 18 months.

TS Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare

Question 6.
Write an essay on adolescence and TDA abuse.
Answer:
Adolescence and TDA abuse:
Adolescence :
It is the time period between the beginning of puberty and the beginning of adulthood. In other words, it is the bridge linking childhood and adulthood. The age between 12 -18 years is considered adolescence period. It is both ‘a period and a process’ during which a child becomes mature. It is accompanied by several biological and behavioural changes. Thus, adolescence is a very vulnerable phase of mental and psychological development of an individual.

TDA abuse :
Curiosity, desire for adventure and excitement, experimentation, are the common causes for the motivation of youngsters towards the use of tobacco, drugs and alcohol. The first use of drugs or alcohol may be out of curiosity or experimentation, but later the person starts using them to escape facing problems. Recently ‘stress from the pressure to excel in academics or examinations’ has played a significant role in alluring the youngsters to try certain drugs. Television, movies, newspapers and internet also help promoting this wrong perception. Other factors that are associated with tobacco, drug and alcohol abuse among adolescents are unstable or unsupportive family structures and peer pressure.

Addiction and Dependence :
The TDA abuse leads to addiction and dependence.

Addiction :
It is a psychological attachment to certain effects such as euphoria. The most important thing one fails to realise is, the inherent addictive nature of tobacco, drugs and alcohol. With the repeated use of TDA, the tolerance level of the receptors present in our body increases. Consequently the receptors respond only to higher doses leading to greater intake and addiction. However it should be clearly borne in mind that use of TDA even once, can be a fore-runner to addiction. Thus, the addictive potential of tobacco, drugs and alcohol pull the users into a vicious circle leading to their regular use (abuse) from which they may not be able to get out. In the absence of any guidance or counselling, people get addicted and become dependent on them.

Dependence :
It is the tendency of the body to manifest a characteristic and unpleasant condition (withdrawal syndrome) if the regular dose of drugs or alcohol is abruptly discontinued. The withdrawal syndrome is characterised by anxiety, shakiness (tremors), nausea and sweating which may be relieved when the regular use is resumed again. Dependence leads the patients to ignore all social norms.

Adverse effects of drugs and alcohol abuse :
The immediate adverse effects of drugs and alcohol abuse are manifested in the form of reluctant behaviour, vandalism and violence. Excessive doses of drugs may lead to coma and death due to respiratory or heart failure or cerebral haemorrhage. A combination of drugs or their intake along with alcohol generally results in overdosing and even death. The most common warning signs of drug and alcohol abuse among the youth include drop in academic performance, lack of interest in personal hygiene, depression, fatigue, aggressive behaviour, loss of interest in hobbies, change in sleeping and eating habits, fluctuations in weight, appetite, etc.

Those who take drugs intravenously are much more likely to acquire serious infections such as HIV, HBV (Hepatitis – B virus) etc., as the viruses are transferred from one person to another by the sharing of infected needles and syringes. The chronic use of drugs and alcohol damages nervous system and liver. The use of drugs and alcohol during pregnancy is also known to affect the foetus adversely.

Some sports-persorTs take drugs such as anabolic steroids to enhance their performance. The side-effects of the use of these drugs in females include masculinisation, increased aggressiveness, mood swings, depression, abnormal menstrual cycles, excessive hair growth on the face and body and the enlargement of clitoris. In males it includes acne (pimples), increased aggressiveness, mood swings, depression, reduction in the size of testicles, decreased sperm production, kidney and liver dysfunction, enlargement of breasts, premature baldness and the enlargement of the prostate gland.

Prevention and Control :
The age-old adage of Prevention is better than cure holds true here also. Some of the measures useful for prevention and control of TDA abuse among the adolescents are :
i) Avoid undue parental pressure :
Every child has his / her own choice, capacity and personality. The parents should not force their children to perform beyond their capacity by comparing them with others in studies, games, etc.

ii) Responsibility of parents and teachers :
They should look for the danger signs and counsel such students who are likely to get into the ‘trap’.

iii) Seeking help from peers :
If peers find someone abusing drugs or alcohol, immediately it should be brought to the notice of their parents or teachers so that they can guide them appropriately.

iv) Education and counselling :
Educating and counselling the children to face problems, stress and failures as a part of life.

v) Seeking professional and medical help:
A lot of help is available in the form of highly qualified psychologists, psychiatrists and de-addiction and rehabilitation programmers.

TS Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa

Telangana TSBIE TS Inter 1st Year Zoology Study Material 5th Lesson Locomotion and Reproduction in Protozoa Textbook Questions and Answers.

TS Inter 1st Year Zoology Study Material 5th Lesson Locomotion and Reproduction in Protozoa

Very Short Answer Type Questions

Question 1.
Draw a labelled diagram of T.S. of flagellum. [March 2018-A.P.]
Answer:
TS Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa 1

Question 2.
List any two differences between a flagellum and a cilium. [Mar. 2020, ’19, 17 – A.P ; May/June, Mar. 2014]
Answer:

FlagellumCilium
1) The long whip-like locomotor organelles are called flagella. Found in Mastigophoran Protozoans.1) These are small hair-like structures found in ciliate protists, genital ducts, respiratory tract.
2) Flagellum helps only in locomotion.2) Cilia serve as organelles or locomotion, food collection and also act as Sensory structures.
3) Flagella perform undular movement.3) Cilia perform pendular movement.

TS Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa

Question 3.
What are dynein arms? What is their significance? [May 2017 – A.P.]
Answer:
In cross section of flagellum / cilium Dynein arms are seen. The ‘A’ tubule of flagellum/ cilium of each peripheral doublet bears paired arms along its length, called dynein arms

Bending movement of a flagellum is brought about by the sliding of microtubules past each other due to the functioning of ‘dynein arms’.

Question 4.
What is a Kinety? [March 2014]
Answer:
A longitudinal row of kinetosomes together with Kinetodesmata constitute a unit called ‘Kinety’.

Question 5.
Distinguish between synchronous and metachronous movements.
Answer:
1) Synchronous movement :
Cilia in a transverse row beat simultaneously in one direction. It is called synchronous movement.

2) Metachronous movement :
The sequential movement of Cilia, in a longitudinal row, one after the other in one direction is called metachronous movement.

Question 6.
Why do we refer to the offspring, formed by asexual method of reproduction, a clone? [March 2020, 19]
Answer:
The term clone is used to describe morphologically and genetically similar daughter individuals formed from single parent by asexual reproduction. They are not only identical to one another but also exact copies of their parent.

Question 7.
Distinguish between proter and opisthe. [Mar. 2017, 13 – A.P ; Mar. 2015 A.P & T.S]
Answer:
During Transverse binary fission of paramecium, two daughter paramecia are formed. The upper or anterior one is proter which receive upper contractile vacuole, cytopharynx and cytostome of its parent. The lower or posterior daughter is opisthe which receives the posterior contractile vacuole only.

Question 8.
How is sexual reproduction advantageous in evolution?
Answer:
Because of the fusion of male and female gametes, sexual reproduction results in offspring that are not identical to the parents or amongst themselves. This leads gradually to origin of new species in course of evolution.

TS Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa

Question 9.
Distinguish between lobopodium and filopodium. Give an example to each of them. [May ’17 ; March 2015 – A.P.]
Answer:
a) Blunt finger like pseudopodia are called lobopodium and are seen in Amoeba and Entamoeba.
b) Fibre like pseudopodia are called filopodiurn and are seen in Euglypha.

Question 10.
Define conjugation with reference to ciliates. Give two examples. [March 2018 – A.P.; March 2015 – T.S ; May/June 2014]
A. Conjugation is a temporary union between two senile ciliates that belong to two different mating types for the exchange of nuclear material and its reorganization. This is observed in Paramecium and Vorticella.

Short Answer Type Questions

Question 1.
Name the system that controls the fastest swimming movement of protozoans and write its components.
Answer:
The system that controls the fastest swimming movement of protozoans is infraciliary system.

It is located just below the pellicle in the ectoplasm of a ciliate. It includes kinetosomes, kinetodesmal fibrils and kinetodesmata. The kinetosomes are present at the bases of cilia in transverse and longitudinal rows. The kinetodesmal fibrils are connected to the kinetosomes and run along the right side of each row of kinetosomes as a ‘cord of fibres’ called kinetodesmata. A longitudinal row of kinetosomes together with kinetodesmata constitute a unit called ‘kinety’.

All the kineties together form an infraciliary system, which is connected to a ‘motorium’, located near the cytopharynx. The infraciliary system and motorium form the ‘neuromotor system’ that controls and coordinates the movement of cilia.

Question 2.
Write the mechanism of bending of flagellum and explain effective and recovery strokes.
Answer:
Bending movement of flagella: Dynein arms show a complex cycle of movements using energy provided by ATP (dynein arms are the sites of ATPase activity in the cilia and flagella). The dynein arms of each doublet attach to an adjacent doublet and pull the neighbouring doublet. So the doublets slide past each other in opposite directions. The arms release and reattach a little farther on the adjacent doublet and again ‘puli’. As the doublets of a flagellum or cilium are physically held in place by the radial spokes, the doublets cannot slide past much. Instead they curve and cause bending of flagellum or cilium. Such bending movements of flagella and cilia play an important role in the flagellar and ciliary locomotion.
TS Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa 2

i) Effective stroke :
Flagellum becomes rigid and starts bending to one side beating against the water. This beating against water is at right angles to the body axis and the organism moves forwards.

ii) Recovery stroke :
Flagellum becomes comparatively soft so as to offer least resistance to water and moves backwards to its original position. It is called ‘recovery stroke’.

TS Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa

Question 3.
What are lateral appendages? Based on their presence and absence, write the various types of flagella giving at least one example for each type. [Mar. 17 – A.P ; Mar. 15 – A.P & T.S]
Answer:
Lateral appendages :
Some flagella bear one or two or many rows of short, lateral hair like fibrils called lateral appendages. They are of two types namely ‘mastigonemes’ and ‘flimmers’.

Types of Flagella :
Based on the presence or absence and / or the number of rows of lateral appendages, five types of flagella are recognised.

a. Stichonematic :
This flagellum bears one row of lateral appendages on the axoneme. Eg : Euglena and Astasia.

b. Pantonematic :
This flagellum has two or more rows of lateral appendages on the axoneme. Eg: Peranema and Monas.

c. Acronematic :
This type of flagellum does not bear lateral appendages and the terminal part of the axoneme is naked without the outer sheath at its tip.
Eg: Chlamydomonas and Polytoma.
TS Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa 3

d. Pantacronematic :
This type of flagellum is provided with two or more rows of lateral appendages and the axoneme ends in a terminal naked filament.
Eg. Urceolus.

e. Anematic or simple :
In this type of flagellum, lateral appendages and terminal filament are absent. Hence, it is called anematic (a – no; nematic – threads).
Eg: Chilomonas and Cryptomonas.

Question 4.
Describe the process of transverse binary fission in paramecium. [May 2017 – A.P.; May/June 2014]
Answer:
TS Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa 4
During favourable conditions, Paramecium stops feeding after attaining its maximum growth. At first the micronucleus divides by mitosis and the macronucleus divides into two daughter nuclei by amitosis. The oral groove disappears. After karyokinesis, a transverse constriction appears in the middle of the body, which deepens and divides the parent cell into two daughter individuals, the anterior proter and the posterior opisthe. The proter receives the anterior contractile vacuole, cytopharynx and cytosome from its parent individual.

It develops posterior contractile vacuole and a new oral groove. The opisthe receives the posterior contractile vacuole of its parent. It develops a new anterior contractile vacuole, cytopharynx, cytostome and a new oral groove. Binary fission is completed in almost two hours, in favourable conditions and paramecium can produce four generations of daughter individuals by binary fission in a day.

The transverse binary fission is also called homothetogenic fission, because the plane of fission is at right angles to the longitudinal axis of the body. As it occurs at right angles to the kineties, it is also called perkinetai fission.

Question 5.
Describe the process of longitudinal binary fission in Euglena. [March 2018 – A.P.; March 2014]
Answer:
During the process of binary fission, the nucleus, basal granules, chromatophores, cytoplasm undergo division. The nucleus divides by mitosis into two daughter nuclei. Then the kinetosomes and the chromatophores also divide. At first, a longitudinal groove develops in the middle of the anterior end. This groove extends gradually towards the posterior end until the two daughter individuals are separated.

One daughter Euglena retains the parental flagella. The other daughter individual develops new flagella from the newly formed basal granules. The stigma, paraflagellar body and contractile vacuole of the parent disappear. They develop afresh in both the daughter euglenae. The longitudinal binary fission is known as symmetrogenic division, because the two daughter euglenae resemble each other like ‘mirror images’.
TS Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa 5

Question 6.
Write a short note on multiple fission.
Answer:
Multiple Fission :
It is the division of the parent body into many smaller individuals (Multi – many; Fission – splitting). Normally multiple fission occurs during unfavourable conditions. During the multiple fission, the nucleus first undergoes repeated mitotic divisions without cytokinesis. This causes the formation of many daughter nuclei. Then the cytoplasm also divides into as many-number of bits as there are nuclei. Each cytoplasmic bit encircles one daughter nucleus. This results in the formation of many smaller individuals from a single parent organism.

There are different types of multiple fissions in protozoans such as schizogony, male gametogony, sporogony in Plasmodium, sporulation in Amoeba, etc.

TS Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa

Question 7.
Give an account of pseudopodia.
Answer:
Pseudopodia :
These are found in rhizopods. The pseudopodia are temporary extensions of cytoplasm that develop in the direction of the movement. These temporary structures are useful to move on the substratum as our legs do, hence the name ‘pseudopodia’. There are four kinds of pseudopodia namely lobopodia (blunt finger-like) as in Amoeba and Entamoeba, filopodia (fibre-like) as in Euglypha, reticulopodia (net-like) as in Elphidium and axopodia or heliopodia (sun ray-like) as in Actionophrys.
TS Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa 6

The Pseudopodium is formed by the conversion of gel (viscous outer endoplasmic part) to sol (fluid-like inner endoplasmic part) and vice-versa. There are different theories on the formation of pseudopodium. The sol-gel transformation theory is the most accepted theory. Pseudopodial movement is performed by Amoeba, Polystomella, Actionophrys.

Question 8.
Give an account of the ultra structure of an axoneme.
Answer:
The structural components of a typical flagellum include axoneme, microtubules, dynein arms, inner sheath, outer sheath, radial spokes, lateral appendages (such as mastigonemes or flimmers) and a basal granule (kinetosome).
i) Axoneme / axial filament :
It is the central, longitudinal, microtubular structure of cilium and flagellum. It is surrounded by a membrane which is continuous with plasma membrane. All the components of the axoneme are embedded within the matrix.

ii) Microtubules :
An axoneme is made up of 2 central ‘singlets’ and 9 peripheral ‘doublets (9+2 array)’. These are formed by the protein, tubulin. Each peripheral doublet consists of an outer ‘A’ (alpha) and the inner ‘B’ (beta) tubules, so, the peripheral tubules are actually 9 microtubular doublets. (The microtubule ‘A’ is smaller but complete whereas ‘B’ is larger and incomplete). The peripheral doublets are interconnected by linkers called nexins.

iii) Dynein arms :
The ‘A’ tubule of each peripheral doublet bears paired arms along its length, called dynein arms (Dyne – pulling like a dynamo). The dynein arms of ‘A’ tubule face the tubule ‘B’ of the adjacent doublet. They are oriented in the same direction (clockwise) in all microtubules, when the axoneme is viewed from the base to the top. The dynein arms are considered ‘protein motor molecules’. They are made of the protein, dynein.

iv) Inner and outer sheaths :
The two central singlets are enclosed by a fibrous inner sheath and the peripheral doublets are enclosed by an outer sheath (an extension of plasma membrane). The central singlets do not reach below the level of the pellicle or plasmalemma.
TS Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa 1

v) Radial spokes :
They are elastic fibres that connect the inner sheath, with the ‘A’ tubule of each doublet. They resemble the spokes that connect the rim of a bicycle wheel with the centre, hence the name ‘radial spokes / radial bridges’. The nine radial spokes limit the extent of sliding past of the doublets, during bending movements.

Question 9.
Draw a neat labelled diagram of Euglena.
Answer:
TS Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa 7

TS Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa

Question 10.
Draw a neat diagram of Paramecium and label its important structures/components.
Answer:
TS Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa 8

TS Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II: Phylum Chordata

Telangana TSBIE TS Inter 1st Year Zoology Study Material 4th Lesson Animal Diversity-II: Phylum Chordata Textbook Questions and Answers.

TS Inter 1st Year Zoology Study Material 4th Lesson Animal Diversity-II: Phylum Chordata

Very Short Answer Type Questions

Question 1.
List out the characters shared by chordates and echinoderms.
Answer:
Chordates share deuterostomeate condition, radial and indeterminate type of cleavage and enterocoelom with the echinoderms.

Question 2.
Write four salient features of cyclostomes.
Answer:
Salient features of cyclostomes

  1. These are jawless aquatic forms.
  2. Body is scaleless, long, slender and eel-like.
  3. Endoskeleton is cartilaginous
  4. Mouth is circular and suctorial. Tongue bears horny teeth.

Question 3.
What is the importance of endostyle in lancelets and ascidians?
Answer:
In ascidians and lancelets, ventral side of the pharynx possesses an endostyle which is believed to be the fore runner of the thyroid gland of a vertebrate. It helps in nutrition.

Question 4.
Name the type of caudal fin and scales that are present in a shark and Catla, respectively.
Answer:

  1. Shark → Heterocercal caudal fin and placoid scales are seen.
  2. Catla → Homocercal caudal fin, ctenoid or cycloid scales are seen.

Question 5.
What is the importance of air bladder in fishes?
Answer:
An air bladder is present with or without connection to the gut. It is either helpful in gas exchange or in maintaining buoyancy.

Question 6.
How do you justify the statement-“heart in fishes is a branchial heart”?
Answer:
Fish heart is known as ‘branchial heart’ as it supplies blood only to the gills.

TS Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II: Phylum Chordata

Question 7.
What are claspers? Which group of fishes possesses them?
Answer:
In the group chondrichthyes males possess claspers on pelvic fins to facilitate internal fertilization, (example : shark).

Question 8.
How does the heart of an amphibian differ from that of a reptile? [March 2015 – T.S.]
Answer:

  1. Heart in amphibians is 3 chambered with sinus venosus and conus arteriosus.
  2. Heart in reptiles is completely four chambered except in the crocodiles. Sinus venosus is present but conus arteriosus is absent.

Question 9.
Name the structures that appeared for the first time in amphibians, in the course of evolution.
Answer:
Tympanum, Lacrimal and Harderian glands (associated with eye) appeared for the first time in the amphibians in the course of evolution.

Question 10.
How do you distinguish a male frog from a female frog? [March 2018 – A.P.]
Answer:
Male frog can be distinguished by the presence of sound amplifying vocal sacs and also a copulatory pad on the first digit of each forelimb. They are absent in female frogs.

Question 11.
What is ‘force pump’ in frog? Why is it named so?
Answer:
In frog during the pulmonary respiration, the bucco – pharyngeal cavity acts like a “force pump”. Due to the elevation of bucco – pharyngeal cavity the air forces the glottis to open and enter the lungs, where exchange of gases takes place.

Question 12.
What are corporabigemina? Mention their chief function.
Answer:
In amphibians Mid brain is represented by a pair of optic lobes called corpora bigemina. The optic lobes are associated with the sense of sight.

TS Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II: Phylum Chordata

Question 13.
Distinguish between mesorchium and mesovarium.
Answer:

  1. In male frog testes are attached to kidney and dorsal body wall by a double fold of peritoneum called mesorchium.
  2. In female frog ovaries are attached to the kidneys and dorsal body wall by a double fold of peritoneum called mesovarium.

Question 14.
Distinguish between milt and spawn. [Mar. 2017 – A.P ; May/June 2014,]
Answer:

  1. Mass of eggs released by a female frog into water is called Spawn.
  2. Mass of sperms released by a male frog into water is called Milt.

Question 15.
What are the “Golden ages” of the first jawed vertebrates and the first amniotes?
Answer:

  1. Golden age of first jawed vertebrates is “DEVONIAN PERIOD” (Fishes)
  2. Golden age of first amniotes is MESOZOIC ERA (Reptiles)

Question 16.
Name two poisonous and non-poisonous snakes found in South India. [March 2014]
Answer:

  1. Two poisonous snakes found in South India are a) Naja naja (cobra), b) Bungarus (Krait)
  2. Two non-poisonous snakes found in South India are a) Ptyas (rat snake) b) Tropidonotus (pond snake)

Question 17.
In which features does the skin of a reptile differ from that of a frog?
Answer:

  1. Skin of a frog is thin, scaleless and moist.
  2. Skin of a reptile is rough and dry. The exoskeleton occurs in the form of horny epidermal scales, shields and claws.

Question 18.
Describe a cat and a lizard on the basis of their chief nitrogenous wastes excreted.
Answer:

  1. Reptiles excrete uric acid as nitrogenous waste (uricotelic).
  2. Cat excrete urea as nitrogenous waste (ureotelic).

Question 19.
Name the four extra embryonic membranes. [March 2020]
Answer:
The four extra embryonic membranes are a) Amnion b) Allantois c) Chorion and d) Yolk sac.

TS Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II: Phylum Chordata

Question 20.
What are Jacobson’s organs? What is their function?
Answer:
Jacobson’s organs, the specialized olfactory structures are highly developed in lizards and snakes.

Question 21.
What are pneumatic bones? How do they help birds?
Answer:
In birds bones are hollow or air filled and referred to as pneumatic bones. They help in reducing the weight, thus help in effective flying (aerial adaptation).

Question 22.
What is ‘wish bone’? What are the skeletal components that form it?
Answer:
In birds both clavicles are fused with the interclavicle to form a ‘V’- shaped bone, called furcula or ‘wish bone’ or merry thought bone.

Question 23.
What is continuous oxygenation of the blood? How is it made possible in birds?
Answer:
In birds the compact spongy lungs are associated with air sacs. Air sacs facilitate continuous oxygenation (oxygen supplied to lungs uninterruptedly and heavily) of blood and pneumacity of bones.

Question 24.
Distinguish between the crop and the gizzard in birds.
Answer:
a) In birds oesophagus is often dilated into crop for the storage of food.
b) Stomach is usually divided into glandular proventriculus and muscular gizzard (grinding mill).

Question 25.
Distinguish between altricial and precocial hatchlings.
Answer:

  1. Altricial hatchlings are seen in flying birds which are incapable of moving around on its own soon after hatching.
  2. PrecOcial hatchlings are seen in flightless birds which are capable of moving around on its own soon.

TS Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II: Phylum Chordata

Question 26.
In which group of animals do we find three ear ossicles on each side and what are their names from the innermost to the outermost?
Answer:
In mammals, we find three ear ossicles on each side. From innermost to outer¬most they are stapes, incus and malleus.

Question 27.
How does a mature RBC of a mammal differ from that of other vertebrates?
Answer:
Mature RBC in mammals is unique because it is circular, biconcave and enucleate.

In other vertebrates RBC are oval, biconvex and nucleated.

Question 28.
Name the characteristic type of vertebrae found in reptiles, birds and mammals.
Answer:
Characteristic vertebrae

  1. In Reptiles → Procoelous.
  2. In Birds → Heterocoelous
  3. In Mammals → Amphiplatyan.

Question 29.
Name the three meninges. In which group of animals do you find all of them?
Answer:
The three meninges which are brain layers are
a) Dura Mater b) Arachnoid membrane c) Piamater.
The group mammals possess all the three layers.

TS Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II: Phylum Chordata

Question 30.
Name the vertebrate groups in which ’renal portal system’ is absent.
Answer:
Renal portal system is absent in the group Mammals.

Short Answer Type Questions

Question 1.
Give three major differences between chordstes and non-chordates and draw the sketch of a chordate’s body showing those features.
Answer:
Major differences between chordates and non-chordates:

ChordatesNon-chordates
1. Notochord is presentNotochord is absent
2. Pharynx is perforated by gill slitsGill slits are absent
3. Heart is ventralHeart is dorsal (if present).
4. A post-anal tail is present.Post-anal tail is absent

TS Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II Phylum Chordata 1

Question 2.
Name the four ‘hallmarks’ of chordates and explain the principal function of each of them. [March 2020, 18]
Answer:
The four principal chordate characters and their functions are
a) Notochord :
Flexible rod like structure placed along the mid dorsal line between gut and nerve cord. It is a supporting structure.

b) Dorsal tubular nerve cord :
A single hollow tubular and fluid filled nerve cord is situated above notochord and below dorsal body wall. It is sensory in function. In higher chordates anteriorly it becomes brain and the rest becomes the spinal cord.

c) Pharyngeal gill slits :
These are a series of lateral perforations in the wall of pharynx which are ecto-endodermal in origin. They are helpful in the exchange of respiratory gases.

d) Post-anal tail :
Chordates have a tail extending posterior to the anus. It is lost in many species during embryonic development. It provides the propelling force in the locomotion of aquatic forms and act as a balancing organ in many terrestrial animals.

Question 3.
Describe the features of a tunicate that reveals its chordate identity.
Answer:
SUBPHYLUM – UROCHORDATA OR TUNICATA : All urochordates are marine and occur from the surface water to greater depths. They are either sessile (ascidians) or pelagic (Salpa, Doliolum) and solitary (Ascidia) or colonial (Pyrosoma).
TS Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II Phylum Chordata 2

Body is un-segmented and covered by a ‘test’ or ‘tunic’ composed of cellulose1 which is uncommon in animals. Coelom is absent. However, an ectoderm – lined atrial cavity surrounds the pharynx, into which gill slits, anus and genital ducts open. Ventral side of the pharynx possesses an endostyle, (see Glossary) which is believed to be the ‘forerunner’ of the thyroid gland of a vertebrate. Atrium leads to the exterior by a dorsal or posterior atrial aperture. Digestive tract is ‘complete’.

Two to numerous gills slits are found in the pharyngeal wall. Circulatory system is of ‘open type’. There is a simple, tubular, ventral heart which alternately reverses the direction of the flow of blood. Nervous system is represented in the adult by a single dorsal ‘ganglion’. They are bisexual or hermaphrodites. Development generally includes a free-swimming tadpole larva with a tail, a dorsal hollow nerve cord, and a notochord confined to the tail, hence the name urochordata. e.g.: Ascidia, Salpa, Doliolum, Pyrosoma and Oikopleura.

Question 4.
Compare and contrast sea squirts and lancelets.
Answer:

Sea squirtsLancelets
1) Marine and occur from the surface water to greater depths. Sessile or pelagic and solitary or colonial.1) Marine forms lead a burrowing mode of life in shallow sea waters.
2) Body is unsegmented and covered by a test or tunic made up of cellulose.2) Body is fish like, translucent with median fins but without paired fins.
3) Larva stage exhibits a tail, dorsal hollow nerve cord and a notochord confined to the tail, hence the name urochordata.3) Often described as typical chordate because they possess principal chordate. Chordates such as notochord, tubular nerve cord and pharyngeal gill slits throughout their life.
4) Circulatory system is of open type. Simple, tubular, ventral heart which alternately reverses the direction of the flow of blood.
Example: Ascidia (sea squirt)
4) Circulatory system is of closed type and heart, blood corpuscles and respiratory pigment are absent.
Example: Branchiostoma (lancelet)

TS Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II: Phylum Chordata

Question 5.
List out eight characteristics that help distinguish a fish from the other vertebrates. [March 2019]
Answer:
TS Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II Phylum Chordata 3
Characteristics of group PISCES.

  1. Fishes are completely aquatic poikilo- thermic (cold blooded ) animals.
  2. Body of a fish is usually streamlined and differentiated into head, trunk and tail.
  3. Exoskeleton consists of mesodermal scales or bony plates. A few are scaleless.
  4. Endoskeleton may be cartilaginous or bony. Skull is monocondylic. Vertebrae are amphicoelous (centrum is concave at both anterior and posterior faces).
  5. Locomotion is assisted by unpaired (median and caudal) fins along with paired (pectoral and pelvic) fins.
  6. Mouth is ventral or terminal. Teeth are usually acrodont, homodont and polyphyodont.
  7. Exchange of respiratory gases is performed by the gills.
  8. Heart is ‘two chambered’ and is described as ‘branchial heart’ as it supplies blood only to the gills.
  9. Lateral line sensory system is well developed.

Question 6.
Compare and contrast cartilaginous and bony fishes. [May 2017 – A.P.; May/June. Mar. 2014 ; March 2013]
Answer:

Cartilaginous fishesBony fishes
1) Caudal fin is heterocercal1) Caudal fin is diphycercal or homocercal.
2) Scales if present placoid scales.2) Scales are ganoid, cycloid or ctenoid scales.
3) Endoskeleton is entirely cartilage.3) Endoskeleton is bony.
4) Mouth and nostrils are ventral.4) Mouth is usually terminal.
5) Digestive tract opens into cloaca, if present.5) Digestive tract opens out by anus.
6) Air bladder is absent.6) Air bladder is often present.
7) Fertilization is internal. Mostly viviparous.
eg: Scoliodon.
7) Fertilization is external. Mostly oviparous.
Eg: Catla catla

TS Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II: Phylum Chordata

Question 7.
Describe the structure of the heart of frog.
Answer:
TS Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II Phylum Chordata 4
The heart is a muscular organ situated in the upper part of the body cavity. If has two separate atria and a single undivided ventricle. It is covered bye double layered membrane called pericardium. A triangular chamber called sinus venosus joins the right atrium on the dorsal side. It receives blood through three vena cavae (caval veins). The ventricle opens into the conus arteriosus on the ventral side. The conus arteriosus bifurcates into two branches and each of it divides into three aortic arches namely carotid, systemic and pulmocutaneous. Blood from the heart is distributed to all parts of the body by the branches of aortic arches. Three major veins collect blood from the different parts of the body and carry it to the sinus venosus.

Question 8.
Write eight salient features of the class Amphibia.
Answer:

  1. Body is divided into distinct’head’ and’trunk’. Tail may or may not be present.
  2. Skin is soft, scale-less (except the members of Apoda), moist and glandular.
  3. The body bears two pairs of equal or unequal pentadactyle limbs (caecilians are limbless).
  4. Skull is dicondylic as in mammals. Vertebrae are mostly procoelous (centrum is concave at its anterior face only) in the anurans, amphicoelous in the caecilians and usually opisthocoelous (centrum is concave at its posterior face) in the urodeles. Sternum appeared for the first time in the amphibians.
  5. Mouth is large; teeth are acrodont, homodont and polyphyodont.
  6. Respiratory gaseous exchange is mostly cutaneous; pulmonary and bucco pharyngeal respirations also occur. Branchial respiration is performed by larvae and some adult urodeles.
  7. Heart is three-chambered with sinus venosus and conus arteriosus. Three pairs of aortic arches and well-developed portal systems are present; erythrocytes are nucleate.
  8. Kidneys are mesonephric; ureotelic.
  9. Meninges are the inner piamater and outer duramater; cranial nerves are 10 pairs.
  10. Middle ear consists of a single ear ossicle, the columella auris which is the modified ‘hyomandibula’ of the fishes. Tympanum, lacrimal and harderian glands appeared for the first time in the amphibians.
  11. Sexes are separate and fertilization is mostly external. Development is mostly indirect, e.g.: Bufo (toad), Rana (frog), Hyla (tree frog), Salamandra (salamander), Ichthyophis (limbless amphibian), Rhacophorus (flying frog).

Question 9.
Describe the male reproductive system of frog with the help of a labelled diagram.
Answer:
Male Reproductive System of frog:
TS Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II Phylum Chordata 5
The male reproductive system consists of a pair of yellowish and ovoid testes, which are attached to the kidneys and dorsal body wall by a double fold of peritoneum called mesorchium. Each testis is composed of innumerable seminiferous tubules which are connected to form 10 to 12 narrow tubules, the vasa efferentia. They enter the kidneys and open into the Bidder’s canal which is connected to the ureter through transverse canals of the kidney. The urino-genital ducts of both the sides open into the cloaca.

Question 10.
Write short notes on organs of special senses in frog.
Answer:
Organs of special senses : Frog has sense organs such as the organs of touch, taste, smell, sight and hearing. The well-organised structures among them are eyes, internal ears and the rest are ‘cellular aggregations’ around nerve endings. The receptors of touch occur in the skin. Organs of taste are called taste buds that lie on small papillae of tongue. The organs of smell are a pair of nasal chambers.

The organs of sight are a pair of eyes located in the orbits of the skull. Eyes are protected by eyelids. The upper eyelid is immovable. The lower eyelid is folded into a transparent nictitating membrane, which can be drawn across the surface of the eye. The retina of the eye contains both rods and cones. Cones provide ‘colour vision’ and rods are helpful in ‘dimlight vision’.

Ear is useful for hearing and balance. It consists of a middle ear closed externally by a large tympanic membrane (ear drum) and a columella that transmits vibrations to the inner ear. The inner ear consists of a utriculus with three semicircular canals and a small sacculus.

TS Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II: Phylum Chordata

Question 11.
List out the salient features of exo and endoskeleton in reptiles.
Answer:
Skin is rough and dry. The exoskeleton occurs in the form of horny epidermal scales, shields, and claws (which appeared for the first time in reptiles).

Dentition is acrodont, homodont and polyphyodont (thecodont in corcodiles as seen in the mammals). Chelonians are ‘edentate’.

Skull is monocondylic and many have temporal fossae. Each half of the lower jaw is formed by six bones. Vertebrae are mostly procoelous. The first two cervical vertebrae are specialized into atlas and axis to facilitate independent movement of the head from the rest of the body; sacral vertebrae are two in number.

Question 12.
List out the extant orders of the Class Reptilia. Give two examples for each order.
Answer:
The extant reptiles are grouped into four orders.
1. Chelonia –
Chelone (marine green turtle), Testudo (terrestrial form), Trionyx (fresh water form)

2. Rhynchocephalia –
Sphenodon (a ’living fossil’, endemic to New Zealand)

3. Crocodilia –
Crocodylus palustris (Indian crocodile or maggur), Alligator (alligator), Gavialis gangeticus (Indian gavial or gharial)

4. Squamata
a) Lizards – Hemidactylus (wall lizard), Chameleon, Draco (flying lizard)
b) Snakes
i) Poisonous snakes – Naja naja (cobra), Ophiophagushannah (King cobra), Bungarus(Krait), Daboia/Vipera russelli (chain viper).
ii) Non-poisonous snakes- Ptyas (rat snake), Tropidonotus (grass snake or pond snake)

Question 13.
What are the modifications that are observed in birds that help them in flight? [Mar. 2017 – A.P : Mar. 2015 – A.P & T.S]
Answer:
TS Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II Phylum Chordata 6

  1. Body is streamlined
  2. The fore limbs are modified into ‘wings for flight.
  3. Skin is dry and devoid of glands, except the oil or preen gland or uropygial gland at the base of the tail to protect tail feathers.
  4. Exoskeleton consists of epidermal feathers (a unique feature), with interlocking system.
  5. Long bones are hollow with air cavities (pneumatic). Skull is monocondylic. Vertebrae are heterocoelous. The last thoracic, lumbar, sacral and anterior few caudal vertebrae are fused to form a synsacrum. it is fused with pelvic girdle to provide support to hind limbs. A few posterior most caudal vertebrae are fused to form the pygostyle that provides support to the tail feathers. Sternum has a keel/ carina for the attachment of flight muscles (except in the ratite birds). Both the clavicles are fused with the interclavicle to form a V – shaped bone, called furcula or ‘Wish bone1 or ‘Merry thought bone’.
  6. All modern flying birds are provided with powerful breast muscles (flight muscles), chiefly the pectoralis major and pectoralis minor.

Question 14.
What are the features peculiar to ratite birds? Give two examples of ratite birds.
Answer:
Ratitae / Palaeognathae :
They are modern flightless running birds. They are ‘discontinuous’ in their distribution like the lung fishes and marsupials. They are characterized by the presence of reduced wings, a raft like sternum without keel and males with penis. They do not possess syrinx, clavicles and usually preen gland, e.g. Struthio camelus (African ostrich), Kiwi (National bird of N6w Zealand), Rhea (American ostrich), Dromaeus (Emu), Casuarius.

TS Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II: Phylum Chordata

Question 15.
Mention the most important features of nervous system and sense organs in mammals.
Answer:
Mammals have a relatively large brain when compared to that of other animals in relation to body size. The four optic lobes constitute corpora quadrigemina. The two halves of cerebrum are connected by corpus callosum. The CNS is enveloped by three meninges. The middle menix called, arachnoid membrane, is present in mammals only.

Cranial nerves are twelve pairs. Eyes have movable eye lids with ‘eye lashes’. External ear has a large fleshy and cartilaginous flap called pinna. Middle ear possesses three ear ossicles. They are malleus, incus and stapes. Cochlea of the internal ear is spirally coiled and bears the ‘organ of Corti’ which is the receptor of sound.

Question 16.
Write short notes on the following features of the eutherians.
a) Dentition
b) Endoskeleton.
Answer:
a) Dentition :
Teeth are thecodont, heterodont and diphyodont. Four pairs of salivary glands are present in association with the buccal cavity (3 pairs in man).

b) Endoskeleton :
Skull is dicondylic. Each halfofthe lowerjaw consists of a ‘single’ bone, the dentary. Most mammals have seven cervical vertebrae, six in Choloepus (two toed sloth) and Trichechus (manatee) and nine in Bradypus (three-toed sloth).Sacral vertebrae are two to five. Vertebrae are of the amphiplatyan type (centrum is flat at both faces). Ribs are double – headed. Buccal cavity is separated from the upper nasal cavity by a secondary palate.

Question 17.
Give an example for each of the following, a) A viviparous fish, b) A fish possessing electric organs, c) A fish possessing poison sting, d) An organ which regulates buoyancy in the body of fish, e) An oviparous animal with milk producing glands.
Answer:
a) Viviparous fish – Scoliodon (dog fish /shark)
b) Fish possessing electric organs – Torpedo (electric ray)
c) Fish possessing poison sting – Trygon (Sting ray)
d) Organ which regulates buoyancy in the body of fish – Air bladder.
e) Oviparous animal with milk producing glands – Ornithorhynchus (Duck-billed platypus)

Question 18.
Mention two similarities between a) Aves and mammals b) A frog and a crocodile c) A lizard and a snake.
Answer:
A) Aves and mammals :

  1. Ribs are double headed in both gorups.
  2. Heart is 4 chambered.
  3. Kidneys are metanephric.
  4. Cranial nerves are twelve pairs.

B) A frog and a crocodile :

  1. Both are amphibians and can lead life on land and in water.
  2. Cloaca is present.
  3. Middle ear consists of a single ear ossicle, the columella auris.

C) A lizard and a snake :

  1. Jacobson’s organs, the specialized olfactory structures are highly developed in lizards and snakes.
  2. In lizards and snakes also some are oviparous and some are viviparous.

TS Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II: Phylum Chordata

Question 19.
Name the following animals, a) A limbless amphibian b) The largest of all living animals c) An animal possessing dry and cornified skin d) ‘National Animal’ of India.
Answer:
a) A limbless amphibian : Ichthyophis
b) The largest of all living animals : Balaenoptera (Blue whale)
c) An animal possessing dry and cornified skin : Reptiles (eg : Sphenodon)
d) National animal of India : Panthera tigris (tiger)
Ornithorhynchus (Duck-billed platypus)

Question 20.
Write the generic names of the following.
a) An oviparous mammal
b) Flying fox
c) Blue whale
d) Kangaroo
Answer:
a) An oviparous mammal – Ornithorhynchus (Duck-billed platypus)
b) Flying fox – Pteropus
c) Blue whale – Balaenoptera.
d) Kangaroo – Macropus.

TS Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I: Invertebrate Phyla

Telangana TSBIE TS Inter 1st Year Zoology Study Material 3rd Lesson Animal Diversity-I: Invertebrate Phyla Textbook Questions and Answers.

TS Inter 1st Year Zoology Study Material 3rd Lesson Animal Diversity-I: Invertebrate Phyla

Very Short Answer Type Questions

Question 1.
What physical feature pertaining to the organism and its medium do you notice in a sponge body from which sponges can be/were identified as animals and not plants? What do you call the region in the sponge body in which you noticed that feature?
Answer:
Sponge body possesses two types of pores. Small pores are called ostia which are numerous and large one or two pores are called oscula. The osculum propels out water coming from body cavity called spongocoel. The presence of osculum confirm the sponge is an animal. It is situated at anterior free end.

Question 2.
What are the different structures that makeup the internal skeleton of a sponge? What are the chemicals involved in the formation of these structures?
Answer:
The body of sponge is supported by skeleton made up of calcareous or siliceous spicules or spongin fibres or both. Calcareous spicules are made up of CaCO3 Siliceous spicules are made up of glass (silicon dioxide). Spongin is fibre material.

Question 3.
What are the functions of canal system of sponges? [March 2018 – A.P.; March 2013]
Answer:
Sponges have a water transport system or canal system that constantly conducts water. It helps in gathering food (filter feeders), exchange of gases (respiration) and removal of wastes (excretion).

Question 4.
What are the two chief morphological ‘body forms’ of cnidarians? What are their chief functions? [March 2020]
Answer:
The two chief morphological body forms of cnidarians are
1. Polyp 2. Medusa.

Polyp is sessile cylindrical form and produce medusae asexually by budding. Medusa is umbrella-shaped and free swimming form and sexually produce polypoid forms.

Question 5.
What is metagenesis? Animals belonging to which phylum exhibit metagenesis?
Answer:
The cnidarians which exist in both forms namely polyp and medusa exhibit alternation of generations called metagenesis. Polyps by asexually method called budding produce medusae. Medusae by sexual method called syngamy give rise to polypoid forms.

TS Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I: Invertebrate Phyla

Question 6.
What is the cnidarian group with quantitatively / relatively large mesoglea? What is the significance of such a well developed mesoglea pertaining to the aquatic life of that group?
Answer:
In cnidarians, the animals of class scyphozoa contain relatively large mesoglea containing amoebocytes. The mesoglea is important in buoyancy.

Question 7.
What is the chief difference between the hydrozoans and the rest of the cnidarians regarding the germinal layer(s) in which its ‘defencive structures or cells of defence’ occur?
Answer:
The chief difference between the hydrozoans and the rest of the cnidarians regarding the germinal layer in which its defencive structures or cells of defence called cnidocytes occur is only in ectoderm. In others, cnidocytes are present both in ectoderm and endoderm.

Question 8.
What are the excretory cells of flatworms called? What is the other important function of these specialized cells?
Answer:
The excretory cells of flatworms are called Flame cells (protonephridia). They help in osmoregulation and excretion.

Question 9.
Distinguish between amphids and phasmids [March 2019]
Answer:
a) Amphids are cuticular depressions around oral region performing chemoreceptor function.
b) Phasmids are posterior glandulo – sensory structures. Both are found in Nematodes.

Question 10.
What is the essential difference between a ‘flat worm’ and a ’round worm’ with reference to the perivisceral area of their bodies?
Answer:
Flat worms are the first bilaterally symmetrical triploblastic and acodomate animals. Round worms are bilaterally symmetrical, triploblastic and pseudocoelomate animals.

Question 11.
How do you account for the origin of the perivisceral space in the body of a nematode and an annelid?
Answer:

  1. In nematodes during embryonic development, mesoderm occupies only a part of the blastocoel adjoining the ectoderm. The unoccupied portion of the blastocoel persists as pseudocoelom.
  2. In annelids the perivisceral space is a true coelom formed by splitting of mesodermal blocks between ectoderm and endoderm which is a schizocoelom.

TS Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I: Invertebrate Phyla

Question 12.
What is metamerism? What is the essential difference between the mode of formation of individual morphological body units of a tapeworm and those of an earthworm?
Answer:
Metamerism is division of body into segments divided by septa and are referred to as metameres. This is unique of annelids. Segments are formed at posterior tip.

Pseudometamerism is found in tape worm. There is no actual division of segments intervally. But gives false appearance because segments called proglottids are attached to each other. They are formed at anterior end.

Question 13.
How do you distinguish a ‘hirudinean’ from the rest of the annelids, based on the morphological features pertaining to metamerism? How does the coelom of a leech differ from the coelom of an earthworm with reference to its contents?
Answer:
In Hirudineans the body is with definite number of segments. The segments are externally subdivided into annuli. Internal segmentation is absent. Coelom of leech is filled with a characteristic tissue called botryoidal tissue. This is absent in earthworm.

Question 14.
What do you call the locomotor structures of Nereis? Why is Nereis called a polychaete?
Answer:
In polychaetes parapodia are the locomotory structures. Which hear many setae. Hence the name polychaeta. Nereis is a polychaete because of parapodia with many setae are seen.

Question 15.
What is botryoidal tissue?
Answer:
Botryoidal tissue is a characteristic tissue filled in the coelom of leeches (hirudineans). This tissue is believed to be excretory in funtion.

Question 16.
What is the difference between the epidermis of a nematode and that of an annelid? How does a nematode differ from an annelid with reference to the musculature of the body wall?
Answer:

  1. In nematodes the unsegmented body is covered by a transparent tough and protective collagenous cuticle. In some epidermis is synctial. Only longitudinal muscles are present (circular muscles are absent).
  2. In annelids the segmented body possesses dermo-muscular body wall containing cuticle, epidermis, dermis, longitudinal and circular muscles.

Question 17.
What do you call the first and second pairs of cephalic appendages of a scorpion?
Answer:
The first and second pairs of cephalic appendages of a scorpion are-called chelicerae and pedipalpi respectively.

TS Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I: Invertebrate Phyla

Question 18.
What is the uniqueness about the first two pairs of cephalic appendages of a crustacean compared to those of the other extant arthropods?
Answer:
In Crustacea Head and Thorax are fused to form cephalothorax. The first two pairs of cephalic appendages are antennules and antennae which is unique feature.

Question 19.
What is the sub-phylum to which ‘ticks’ and ‘mites’ belong? How do you distinguish them from the insects with reference to their walking legs?
Answer:
Ticks and mites belong to sub-phylum chelicerata. They are differed from insects in having four pairs of walking legs. Insects possess three pairs of walking legs.

Question 20.
What are the respiratory structures of Limulus and Palamnaeus respectively?
Answer:

  1. Respiratory structures of Limulus (King crab) → Book – gills
  2. Respiratory structures of palaemnaeus (scorpion) → Book-lungs.

Question 21.
What are antennae? What is the arthropod group without antennae?
Answer:
Antennae are cephalic appendages of arthropods which are sensory in function (sense of touch). They are absent in the sub-phylum chelicerata.)

Question 22.
What do you call the perivisceral cavity of an arthropod? Where from is it derived during development?
Answer:
The perivisceral cavity of arthropods is called Haemocoel (blood cavity) derived from mostly the embryonic blastocoel.

Question 23.
Which arthropod, you have studied, is called a ‘living fossil’? Name its respiratory organs. [March 2015 – A.P.]
Answer:
Limulus (King crab) belonging to class Xiphosura of sub-phylum chelicerata is called living fossil. Its respiratory organs are book gills.

Question 24.
How do you identify a Chiton from its external appearance? How many pairs of gills help in the respiration of Chiton?
Answer:
Chiton is identified by a dorsal shell consisting of 8 transverse plates. 6 – 88 pairs of gills help in respiration of Chiton.

TS Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I: Invertebrate Phyla

Question 25.
What is the function of radula? Give the name of the group of molluscs which do not possess a radula. [May/June, March 2014]
Answer:
The buccal cavity of molluscs contain a file like rasping organ called radula for feeding. It is absent in the class bivalve or pelecypoda.

Question 26.
What is the other name for the gill of a mollusc? What is the function of osphradium?
Answer:
Ctenedia is the other name of gill of a mollusc. Osphradium in molluscs tests the purity of water. It is present in the bivalves and gastropods.

Question 27.
What is Aristotle’s lantern? Give one example of an animal possessing it. [May 2017 ; Mar. ’17 – A.P ; Mar. 2015 – T.S]
Answer:
In the mouth of a sea urchin a complex five jawed masticatory apparatus called Aristotle’s lantern is present. Example : Echinus (sea urchin).

Question 28.
What is the essential difference between the juveniles and adults of echinoderms, symmetry wise?
Answer:

  1. Adult of echinoderms exhibit pentamerous radial symmetry.
  2. Juveniles of echinoderms exhibit bilateral symmetry.

Question 29.
What are blood glands in Pheretima?
Blood glands in pheretima are present in the 4th, 5th, and 6th segments. They produce blood cells and haemoglobin which is dissolved in plasma. Blood cells are phagocytic in nature.

TS Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I: Invertebrate Phyla

Question 30.
What are spermathecae on the body of pheretima?
Answer:
4 pairs of spermathecae are located in the segments 6th to 9th (one pair in each segment). They receive and store sperms during copulation.

Short Answer Type Questions

Question 1.
Write short notes on the salient features of the anthozoans.
Answer:
TS Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I Invertebrate Phyla 1
Class – Anthozoa or Actinozoa :

  1. They are commonly referred to as sea anemones.
  2. They are sedentary and only polypoid in form.
  3. Coelenteron is divided into several compartments by vertical septa called mesenteries.
  4. Mesoglea contains connective tissue.
  5. Cnidocytes occur both in the ectoderm and endoderm.
  6. Germ cells are derived from the endoderm.
    Examples : Adamsia (sea anemone), Corallium rubrum (precious red stone coral), Gorgonia (sea fan), Pennatula (sea pen).

Question 2.
What is the class to which the flukes belong? Write short notes on the chief characters of that group.
Answer:
Flukes belong to class Trematoda of phylum platyhelminthes.
TS Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I Invertebrate Phyla 2

Class – Trematoda:

  1. They are commonly called flukes.
  2. They are parasitic on other animals.
  3. Body is covered by a thick cuticle (tegument); bears two suckers, an oral and a ventral (acetabulum).
  4. Mouth is anterior and the intestine is bifurcated.
  5. They are bisexual (monoecious).
  6. Life history is complex with many hosts and different types of stages (miracidium, sporocyst, redia, cercaria etc.).

Examples : Fasciola (liver fluke), Schistosoma or Bilharzia (blood fluke).

Question 3.
What are the salient features exhibited by polychaetes?
Answer:
TS Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I Invertebrate Phyla 3
Class – Polychaeta salient features :

  1. Polychaeta (poly : many; chaetae : setae) includes marine annelids commonly called ‘bristle worms’.
  2. Some are free-moving and others burrowing or tubicolous (tube dwelling).
  3. Head is distinct with sense organs such as eyes, tentacles and palps.
  4. Parapodia which bear many setae (hence the name polychaeta) help in locomotion and respiration.
  5. Clitellum is absent.
  6. They are dioecious (unisexual); gonoducts are absent; gametes are shed into the coelom and passed out through the nephridiopores.
  7. Fertilization is external; development includes a trochophore larva.
    Examples : Nereis (sandworm or ragworm or clam worm), Aphrodite (sea mouse), Arenicola (lugwarm).

Question 4.
How do the hirudineans differ from the polychaetes and oligochaetes?
Answer:
Class: Hirudinea.

  1. Hirudinea (hirudo : leech) includes leeches.
  2. All are ectoparasites; majority live in fresh water; some are marine and others live on moist land (terrestrial).
  3. They have dorso-ventrally flattened body with a ‘definite number1 of segments; the segments are externally subdivided into ‘annuli’; internal segmentation is absent.
  4. Suckers help in locomotion.
  5. Clitellum is conspicuous during the breeding season only.
  6. Coelom is filled with a characteristic tissue called botryoidal tissue.
  7. They are monoecious (hermaphroditic); males possesses a copulatory organ, the cirrus.
  8. Fertilization is internal and development is direct.
    Examples: Hirudinaria (freshwater leech), Pontobdella (marine leech), Haemadipsa (land leech).

Question 5.
What are the chief characters of the crustaceans? [May – 2017 – A.P.; March 2015 T.S.]
Answer:
TS Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I Invertebrate Phyla 4
Class – Crustacea – Chief characters :

  1. They are aquatic.
  2. Head and thorax fuse forming the cephalothorax (covered by chitinous carapace). In some the exoskeleton is hardened by calcium carbonate (crabs and lobsters).
  3. Cephalic region bears two pairs of antennae (antennules and antennae – unique feature), one pair of mandibles and two pairs of maxillae.
  4. Thoracic and abdominal appendages are ‘biramous’
  5. Respiratory organs are gills (branchiae).
  6. Excretory organs are green glands or antennary glands.
  7. Sense organs include antennae compound eyes, statocysts, etc.
  8. Development is indirect and includes different larval forms.
    Examples : Palaemon (freshwater prawn), Cancer (crab), Balanus (rock barnacle),
    Sacculina (root-headed barnacle), Astacus (cray fish), Daphnia (water flea).

TS Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I: Invertebrate Phyla

Question 6.
Mention the general characters of Arachnida. [May/June ’14]
Answer:
TS Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I Invertebrate Phyla 5
Class – Arachnida general characters :

  1. They are terrestrial.
  2. Prosoma bears a pair of chelicerae, a pair of pedipalpi and four pairs of walking legs.
  3. Mesosomal appendages are modified into book lungs.
  4. Four pairs of posterior abdominal appendages are modified into spinnerets in spiders.
  5. Respiratory organs are book – lungs (scorpions and some spiders), tracheae (some spiders) or both (some spiders).
  6. Respiratory pigments is ‘copper1 containing haemocyanin.
  7. Excretory organs are malpighian tubules and coxal glands.
  8. Development is direct; scorpions are viviparous.
    Example : Palamnaeus (scorpion), Aranea (spider), Sarcoptes (itch mite).

Question 7.
Compare briefly a centipede and a millipede.
Answer:

Centipede (chilopod)Millipede (Diplopod)
1) Commonly called hundred leggers.1) Commonly called thousand leggers.
2) Terrestrial, air breathing carnivorous animals.2) Terrestrial air breathing animals feeding on decaying plant material.
3) Body consists of head and trunk3) Body consists of head and trunk.
4) Each segment of trunk bears a pair of clawed appendages.4) Head bears paired antennae, mandibles and maxillae, (Maxillae modified into gnathochilarium)
5) First pair of trunk appendages bear poison claws.5) Each trunk segment bear two pairs of legs.
6) Respiratory structures are tracheae6) Respiratory structures are tracheae.
7) Excretory organs are Malpighian tubules.
Ex : Scolopendra, Scutigera
7) Excretory organs are Malpighian tubules
Ex: Spirostreptus, Julus.

TS Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I: Invertebrate Phyla

Question 8.
Cephalopods show several unique or advanced features when compared to the other molluscs. Discuss briefly.
Answer:
Cephalopoda or Siphonopoda is an advanced group of Mollusca. This class includes cuttle fishes, squids, octopuses, nautili etc. Head is distinct with conspicuous eyes similar to those of vertebrates, a pair of horny beak like jaws and a radula occur in the buccal cavity. Shell may be external and multi – chambered (Nautilus) or internal (Sepia, Loligo) or absent (Octopus). The shell of Sepia is commonly called ‘cuttle bone’ and that of Loligo is commonly called ‘pen’. Foot is modified into eight (octopus) to ten arms (Sepia, Loligo) provided with suckers present around the mouth and a part of the foot is modified into a ‘siphon’ (useful in swift darting movements).

Some possess ink gland and eject a cloud of ink to escape from the predators (defensive adaptation). Ctenidia, atria and nephridia are two in dibranchiates (Sepia) and four in tetrabranchiates (Nautilus). Circulatory system is a closed type (unique feature of cephalopoda); heart’has two to four atria and a ventricle. Nervous system is well developed with a well developed brain enclosed in a cartilaginous cranium (braincase). They are dioecious (sexes are separate); development is direct.
examples : Sepia (cuttle fish), Architeuthis (giant squid – the largest living invertebrate), Nautilus, Octopus (devil fish).

Question 9.
Which class of Mollusca represents the primitive molluscs? What are their chief features?
Answer:
Class Aplacophora represents the primitive molluscs. They are primitive ‘worm like’ marine molluscs without mantle, shell, foot and nephridia. Head is poorly developed; a radula is present. Cuticle contains calcareous spicules. In some there is a mid-ventral groove which is homologous to the foot of the other molluscs. Examples: Neomenia, Chaetoderma.

Question 10.
What are the salient features of the echinoids? [March 2019, ’17 – A.P.]
Answer:

  1. The class Echinoidea includes sea urchins, heart urchins, sand dollars, sea biscuits, etc.
  2. The body is ovoid or discoid and covered with morable spines.
  3. Arms are absent; tube feet bear suckers.
  4. Calcarious ossicles of the body unite to form a rigid test or corona or case.
  5. Madreporite and anus are aboral in position.
  6. Ambulacral grooves are closed.
  7. Pedicellariae are three jawed.
  8. In the mouth of sea urchin a complex five jawed masticatory apparatus called Aristotle’s lantern is present (absent in the heart urchins).
  9. Development includes echinopluteus larava.
    Examples: Echinus (sea urchin), Echinocardium (heart urchin) Echinodiscus (sand dollar).

Question 11.
Mention the salient features of Holothuroidea. [March 2015 – A.P.]
Answer:
Holothuroidea-salient features.

  1. This class includes sea cucumbers.
  2. Body is elongated in the oro-aboral axis.
  3. Skin is leathery (coriaceous) and dermis contains loose spicules.
  4. Arms, spines and pedicellariae are absent. .
  5. Mouth is surrounded by retractile tentacles (modified tube feet useful for feeding).
  6. Ambulacral grooves are ‘closed1; tube feet bear suckers.
  7. Madreporite is internal (occurs in coelom).
  8. Respiratory organs are a pair of cloacal ‘respiratory trees’.
  9. Development is indirect and includes auricularia larva.
    Examples : Holothuria, Synapta, Thyone.

Question 12.
What is the function of Nephridia?
Answer:
The excretory organs occur as segmentally arranged called tubules, called nephridia. Earthworms are mostly ureotelic animals. The chief nitrogenous excretory waste is urea enteronephric nephridia have a role in asmoregulation/conservation/ homeostasis of water. Exonephric nephridia collect nitrogenous wastes and send them out through nephridiopores. They are ciliated coelomic ducts.

Question 13.
How many types of nephridia occur in pheretima and how do you distinguish them?
Answer:
There are 3 types of nephridia. 1) Septal nephridia present on both sides of the inter segmental septa of segments 15th/16th to the last. They open into the intestine. 2) Integumentary nephridia attached to inner body wall from 3rd segment to last. Open out by nephridiopores. 3) Pharyngeal nephridia present as three paired tufts in the segments 4th, 5th and 6th. They open into buccal cavity and pharynx. Septal nephridia with nephrostomes but without nephridiopore. Pharyngeal and integumentary without nephrostomes. Pharyngeal nephridia without Nephrostome and nephridiopore.

TS Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I: Invertebrate Phyla

Question 14.
Give an account of the hearts on the circulatory system of pheretima.
Answer:
In pheretima there are 4 pairs of hearts which connect the dorsal and ventral blood vessels in 7, 9, 12,13 segments. They are muscular and valvular. They allow blood from dorsal blood vessel to ventral blood vessel only. The 12th, 13th segment lateral hearts are also called as lateral oesophageal hearts as they receive blood from supra oesophageal blood vessel and give to ventral blood vessel.

Essay Answer Type Questions

Question 1.
Draw a neat labelled diagram of the reproductive organs of pheretima.
Answer:
TS Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I Invertebrate Phyla 6

Question 2.
Describe the digestive system and process of digestion in pheretima.
Answer:
Digestive system :
The alimentary canal is a straight tube and runs from the first to the last segment of the body (Fig). The mouth opens into the buccal cavity (1-3 segments) which leads into the muscular pharynx (4th segment). A small narrow tube. Oesophagus (5-7 segments), continues into a muscular gizzard (8th segment). It helps in grinding the small particles of food the decaying leaves (grinding mill). The stomach extends fpm the segments 9 glands, present in the stomach, neutralise the humic acid present in the humus of the soil. The intestine starts from the 15th segment and continues till the last segment.

A pair of short and conical intestinal caecae project from the intestine in the 26th segment. An internal median fold of the dorsal wall of the intestine called typhlosole, helping in increasing the area of absorption, is poorly developed in Pheretima (between the 26th and the rectum, which occupies the last 23 to 28 segments). The alimentary canal opens to the exterior by a small rounded aperture called anus. The ingested soil rich in organic mater passes through the digestive tract where digestive enzymes breakdown complex food into smaller absorbable units. These simpler molecules are absorbed through intestinal membranes and are utilised for various metabolic activities.
TS Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I Invertebrate Phyla 7

TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Telangana TSBIE TS Inter 1st Year Zoology Study Material 2nd Lesson Structural Organisation in Animals Textbook Questions and Answers.

TS Inter 1st Year Zoology Study Material 2nd Lesson Structural Organisation in Animals

Very Short Answer Type Questions

Question 1.
The body of sponges does not possess tissue level of the organisation, though it is made up of thousands of cells. Comment on it.
Answer:
Sponges belonging to group Parazoa are example of cellular level of organisation. The cells are arranged as loose cells aggregates and do not form tissues. There is division of labour among the cells.

Question 2.
What is ’tissue’ level of organisation among animals? Which metazoans do exhibit this organisation?
Answer:
This is the lowest level of organisation among the eumetazoans, exhibited by diploblastic animals like the cnidarians. In these animals, the cells which perform the same function are arranged into tissues. There is a co-ordination between functioning of cells because of nerve cells and sensory cells.

Question 3.
Animals exhibiting which level of the organisation lead relatively more efficient way of life when compared to those of the other levels of organisation? Why?
Answer:
Animals exhibiting organ system level of organisation lead relatively more efficient way of life when compared to those of the other levels of organisation because highly specialized sensory and nerve cells bring about a higher level of co-ordination and integration among the various organ systems.

Question 4.
What is monaxial heteropolar symmetry? Name the group of animals in which it is the principal symmetry.
Answer:
When any plane passing through the central axis (oro-aboral axis) of the body divides an organism into two identical parts, it is called monaxial heteropolar symmetry. In cnidarians it is the principal symmetry.

Question 5.
Radial symmetry is an advantage to the sessile or slow moving organisms. Justify this statement.
Answer:
Animals showing radial symmetry live in water and they can respond equally to stimuli that arrive from all directions. Thus, radial symmetry is an advantage to sessile or slow moving animals.

TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 6.
What is cephalization? How is it useful to its possessors? [May 2017 – A.P.; March 2015 – T.S.; March 2013]
Answer:
Cephalization is concentration of nerve and sensory cells at the anterior end. As a result of cephalization, bilaterally symmetrical animals can sense the new environment into which they enter and respond more efficiently and quickly.

Question 7.
Mention the animals that exhibited a ‘tube-within-a-tube’ organisation for the first time. Name their body cavity.
Answer:
A tube-with-in-a-tube organisation for the first time is seen in the group Nematoda. Their body cavity is named as Pseudocoelam.

Question 8.
Why is the true coelom considered a secondary body cavity? [March 2015 – T.S.]
A. During the embryonic development of the eucoelomates, the blastocoel is replaced by true coelom derived from the mesoderm. So, the true coelom is also called ‘Secondary body cavity’.

Question 9.
What are retroperitoneal organs? [March 2018 – A.P.]
Answer:
Certain organs such as the kidneys of the vertebrates are covered by the parietal peritoneum only on their ventral side. Such a peritoneum is called retroperitoneum and the organs lined by it are called ‘retroperitoneal organs’.

Question 10.
If the mesentoblast cell is removed in the early embryonic development of protostomes, what would be the fate of such animals?
Answer:
The 4d blastomere or mesentoblast cell Of protostomes divides to form mesodermal blocks between the ectoderm and endoderm. The split that appears in each mesodermal block leads to the formation of schizocoel (true body cavity). If mesentoblast cell is removed, the true coelom will not be formed.

Question 11.
What is enterocoelom? Name the enterocoelomate phyla in the animal kingdom. [March 2014]
Answer:
Animals in which the body cavity is formed from the mesodermal pouches of archenteron are called enterocoelomates. Echinoderms, hemichordates and chordates are the enterocoelomates.

Question 12.
Stratified epithelial cells have limited role in secretion. Justify their role in our skin.
Answer:
Stratified epithelium is made up of more than one layer of cells. Its main function is to provide protection against chemical and mechanical stress. It covers the dry surface of the skin. Hence it has limited role in secretion.

TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 13.
Distinguish between exocrine and endocrine glands with examples. [March 2014]
Answer:

  1. Exocrine glands are provided with ducts : secrete mucus, saliva, earwax, oil, milk, digestive enzymes and other cell products.
  2. Endocrine glands are ductless and their products are hormones which are not sent out via ducts but are carried to the target organs by blood, eg : Thyroid glands secreting hormone Thyroxine.

Question 14.
Distinguish between holocrine and apocrine glands.
Answer:

  1. Apocrine glands (eg : mammary glands) in which the apical part of the cell is pinched off along with the secretory product.
  2. Holocrine glands (e.g : sebaceous glands) in which the entire cell disintegrates to discharge the contents.

Question 15.
Mention any two substances secreted by mast cells and their functions. [May/June, March 2014]
Answer:

  1. Heparin – an anticoagulant (prevents blood clotting)
  2. Histamine – vasodilators (cause inflammation in response to injury and infection.)

Question 16.
Distinguish between a tendon and a ligament. [May 2017-A.P.; March 2017, 15, March 2019]
Answer:

  1. Tendons attach the skeletal muscles to bones.
  2. Ligaments attach bones to other bones.

Question 17.
Distinguish between brown fat and white fat.
Answer:
1) White adipose tissue (WAT) :
It is mostly in adults. Adipocyte has a single large lipid droplet (monolocular). White fat is metabolically not active.

2) Brown adipose tissue (BAT) :
It is found in foetuses and infants. Adipocyte of BAT has several small lipid droplets and numerous mitochondria. Brown fat is metobolically active and generates heat to maintain body temperature required by infants.

TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 18.
What is the strongest cartilage? In which regions of the human body, do you find it? [March 2020 ; Mar 15 – T.S.]
Answer:
Fibrous cartilage is the strongest of all types of cartilage. It occurs in the intervertebral discs and pubic symphysis of the pelvis.

Question 19.
Distinguish between osteoblasts, and osteoclasts. [March 2017 – A.P.]
Answer:
Osteoblasts (Immature bone cells) secrete the organic components (collagen fibres) of matrix and also play a major role in ‘mineralization of bone’.

Osteoclasts are phagocytic cells involved in resorption of bone.

Question 20.
Define osteon. [March 2015 – A.P.]
Answr:
In a dense bone a Haversian canal and the surrounding lamillae (rows of osteocytes) and lacunae are collectively called a Haversian system or Osteon.

Question 21.
What are Volkmann’s canals? What is their role?
Answer:
The Haversian canals communicate with one another, with the periosteum and also with the marrow cavity by transverse or oblique canals called Volkmann’s canals.

Question 22.
What is a Sesamoid bone? Give an example. [May/June 2014]
Answer:
Sesamoid bones are formed by ossification in tendons. Eg : Patella (Knee cap)

Question 23.
What is lymph? How does it differ from plasma?
Answer:

  1. Lymph is a colourless fluid. It lacks RBC, platelets and large plasma proteins, but has more number of Leucocytes.
  2. Plasma is the fluid matrix of blood. It consists of 92% of water and 8% of solutes.

Question 24.
What is the haematocrit value? [Mar. 2019, ’17, May ’17 – A.P ; May/June 2014]
Answer:
The percentage of total volume occupied by RBCs is called haematocrit value.

Question 25.
What are intercalated discs? What is their significance?
Answer:
The dark lines across cardiac muscle are called intercalated discs (IDs). These discs are highly characteristic of the cardiac muscle. They act as boosters of muscle contraction.

TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 26.
“Cardiac muscle is highly resistant to fatigue.” Justify. [March 2020]
Answer:
The cardiac muscle js highly resistant to fatigue because it has numerous sarcosomes, many molecules of myoglobin (oxygen storing pigment) and copious supply of blood which facilitate “Continuous aerobic respiration”.

Question 27.
Distinguish between ‘nucleus’ and ‘ganglion’ with respect to the nervous system.
Answer:
A group of cell bodies in the central nervous system is called a ‘nucleus’ and in the peripheral nervous system, it is called a ‘ganglion’.

Question 28.
Distinguish between tracts and nerves with respect to the nervous system.
Answer:
Groups of axons (nerve fibres) in the central nervous system (CNS) are called tracts and in the peripheral nervous system (PNS) they are called nerves.

Question 29.
Name the glial cells that form myelin sheath around the axons of central nervous system and peripheral nervous system respectively.
Answer:
Myelin sheath around the axons of central nervous system is formed from ‘oligodendrocytes.‘ In peripheral nervous system myelin sheath around axon is formed from ‘Satellite cells’ a kind of glial cells.

Question 30.
Distinguish between white matter and grey matter of ‘CNS’
Answer:
Myelinated nerve fibres occur in the white matter of CNS. Non-myelinated nerve fibres occur in the grey matter of the CNS and autonomus nervous system.

TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 31.
What are microglia and what is their origin and add a note on their function. [March 2018 – A.P.]
Answer:
Microglial cells which are phagocytic cells, of mesodermal origin. They become activated into ‘phagocytes’ when there is infection or injury in the nervous system.

Question 32.
What are pseudounipolar neurons? Where do you find them?
Answer:
Unipolar neurons are also called pseudounipolar neurons. They are found jn the ‘dorsal root ganglion’ of spinal nerve.

Short Answer Type Questions

Question 1.
Describe the four different levels of organization in metazoans.
Answer:
The levels of organisation in metazoans are as follows.
TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals 1

Cellular level of organisation :
It is the lowest level of organisation among the metazoans and is exhibited by the sponges (parazoans). Different types of cells are functionally isolated due to the absence sensory and nerve cells. The cells are arranged as ‘loose cell aggregates’ and do not form tissues. There is division of labour among the cells.

Tissue level of organisation :
This is the lowest level of organization among the eumetazoans, exhibited by diploblastic animals like the cnidarians. In these animals, the cells which perform the same function are arranged into tissues. The cells of a tissue together perform their common function as a highly coordinated unit and this coordination is due to the presence of nerve cells and sensory cells.

Organ level of organisation :
An aggregation of different kinds of tissues which is specialized for a particular function is called an organ. Organ level of organisation appeared for the first time in the members of the Phylum Platyhelminthes.

Organ – system level of organisation :
It is the highest level of organisation among the animals and is exhibited by the triploblastic animals such as the flat worms, nematodes, annelids, arthropods, molluscs, echinoderms and chordates. In the triploblastic animals, the evolution of ‘mesoderm’ resulted in structural complexity. In these animals, the tissues are assembled to form organs and complex organ – systems. Highly specialized sensory and nerve cells bring about a higher level of coordination and integration among the various organ systems to lead an efficient way of life.

Question 2.
In which group of bilaterians do you find solid bauplan? Why it is called so?
Answer:
Solid bauplan is seen in acoelomate bilaterians eg : Platyhelminthes.
TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals 3

The bilaterian animals in which the body cavity is absent are called acoelomates. In these animals, the mesenchyme derived from the third germinal layer, called mesoderm, occupies the entire blastocoel, between the ectoderm and the endoderm, so that the adults have neither the primary cavity (blastocoelom) nor the secondary cavity (coelom). As there is no body cavity, the acoelomates exhibit solid body plan. Problems faced by the acoelomates due to absence of perivisceral cavity are – their internal organs cannot move freely, as they are embedded in the mesenchyme, diffusion of material from the gut to the body wall is made slow and less efficient.

TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 3.
Mention the advantages of coelom over pseudocoelom.
Answer:
Advantages of coelom over pseudocoelom :

  1. Visceral organs of eucoelomates are muscular (because of their association with mesoderm) and so they can contract and relax freely independent of the muscular movements of the body wall in the coelomic space, e.g. peristaltic movements of alimentary canal.
  2. Gametes are released into the coelom in some invertebrates (which do not have gonoducts) and in the female vertebrates.
  3. Coelomic fluid receives excretory products and stores them temporarily before their elimination.
  4. In the eucoelomates, the mesoderm comes into contact with the endoderm of the alimentary canal, and it causes ‘regional specialization of the gut, such as the development of gizzard, stomach etc… This is referred to as ‘primary induction’. In the case of the pseudocoelomates, due to the absence of such a contact between the gut and the mesoderm, the wall of the gut does not show complex and highly specialized organs.

Question 4.
Describe the formation of schizocoelom and enterocoelom.
Answer:
Formation of Schizocoelom: Animals in which the body cavity is formed by ‘splitting of mesoderm’ are called schizocoelomates. Annelids, arthropods and molluscs are schizocoelomates in the animal kingdom. All the schizocoelomates are protostomians and they show ‘holoblastic’, ‘spiral’ and ‘determinate’ cleavage. The 4d blastomere or mesentoblast cell of the early embryo divides to form mesodermal blocks between the ectoderm and the endoderm and replaces the blastocoel. The split that appears in each mesodermal block leads to the formation of Schizocoelom. Eg : Annelida, Arthropoda, Mollusca.
TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals 4

Formation of enterocoelom :
Animals in which the body cavity is formed from the mesodermal pouches of archenteron are called enterocoelomates. Echinoderms, hemichordates and chordates are the enterocoelomates. In these animals, mesodermal pouches that evaginate from the wall of the archenteron into the blastocoel are fused with one another to form the enterocoelom. All the enterocoelomates are deuterostomes and they show radial and indeterminate cleavage.
TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals 5

Question 5.
Describe briefly about the three types of intercellular junctions of epithelial tissues.
Answer:
There are three types of intercellular junctions of epithelial tissues.

A. Tight junctions :
These junctions between epithelial cells prevent ‘leakages’ of body fluids. For example, they prevent leakage of water into the surrounding cells in our sweat glands (making our skin water-tight).

B. Desmosomes :
Muscle cells are provided with ‘desmosomes (anchoring junctions) which act as ‘rivets’ binding the cells together into strong sheets. Intermediate filaments made of the protein ‘keratin’, anchor desmosomes in the cytoplasm.
TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals 6

C. Gap junctions :
They provide continuous ‘cytoplasmic channels’ between adjacent cells (comparable to the ‘plasmodesmata’ between adjacent plant cells). Various types of ions, sugar molecules, amino acids etc., can pass from a cell to an adjacent cell through ‘gap junctions’. They occur in many types of tissues including the ‘cardiac muscles’, where they allow rapid conduction of impulses or depolarisation.

Question 6.
Give an account of glandular epithelium. [March 2015 – A.P.]
Answer:
Some of the columnar or cuboidal cells that get specialised for the production of certain secretions, form glandular epithelium. The glands are of two types – unicellular glands consisting of isolated glandular cells such as goblet cells of the gut, and multicellular glands consisting of clusters of cells such as salivary glands.
TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals 7

On the basis of the mode of pouring of their secretions, glands are divided into two types namely exocrine and endocrine glands. Exocrine glands are provided with ducts; secrete mucus, saliva, earwax (cerumen), oil, milk, digestive enzymes and other cell products. In contrast, endocrine glands are ductless and their products are ‘hormones’, which are not sent out via ducts, but are carried to the target organs by blood. Based on the mode of secretion, exocrine glands are further divided into i. Merocrine glands (e.g. pancreas) which release the secretory granules without the loss of other cellular material ii. Apocrine glands (e.g. mammary glands) in which the apical part of the cell is pinched off along with the secretory product and iii. Holocrine glands (e.g. sebaceous glands), in which the entire cell disintegrates to discharge the contents.

TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 7.
Give a brief account of the cells of areolar tissue.
Answer:
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Cells of the areolar tissue are fibroblasts, mast cells, macrophages, adipocytes and plasma cells.

  1. Fibroblasts are the most common cells which secrete fibres. The inactive cells are called fibrocytes.
  2. Mast cells secrete heparian (an anticoagulant), histamine, bradykinin (vasodilators), and serotonin (vasoconstrictor). Vasodilators cause inflammation in response to injury and infection.
  3. Macrophages are amoeboid cells, phagocytic in function and act as internal scavengers. They are derived from the monocytes of blood. ‘Tissue fixed macrophages’ are called histiocytes and others are ‘wandering macrophages’.
  4. Plasma cells are derived from the B-lymphocytes and produce antibodies.
  5. Adipocytes are specialized cells for the storage of fats.

Question 8.
Describe the three types of cartilage. [March 2020, ’17]
Answer:
The three types of cartilage are 1. Hyaline cartilage 2. Elastic cartilage 3. Fibrous cartilage.

1. Hyaline cartilage :
It is bluish-white, translucent and glass – like cartilage. Matrix is homogeneous and shows delicate collagen fibres. It is the weakest and the most common type of all the cartilages. Perichondrium is present except in articular cartilages. It forms the embryonic endoskeleton of bony vertebrates, endoskeleton of cyclostomes and cartilaginous fishes. It forms the articular cartilages (free surfaces of long bones that form joints), costal cartilages (sternal parts of ribs), and the epiphyseal plates. It also forms the nasal septal cartilage, cartilaginous rings of trachea, bronchi and cartilages of larynx.
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2) Elastic cartilage :
It is yellowish due to elastic fibres. Matrix has abundance of yellow elastic fibres in addition to collagen fibres. It provides strength and elasticity. Perichondrium is present. It is found in the pinnae of the external ears, Eustachian tubes and epiglottis.

3) Fibrous cartilage :
Matrix has bundles of collagen fibres. Perichondrium is absent. It is the strongest of all types of cartilages. It occurs in the intervertebral discs and pubic symphysis of the pelvis.

Question 9.
Explain Haversian system. [May 2017 – A.P; March 2014]
Answer:
In a bone between the outer and inner circumferential lamellae, there are many Haversian systems. The spaces between the Haversian systems are filled with interstitial lamellae. Haversian system consists of a Haversian canal that runs parallel to the marrow cavity. It contains an artery, a vein and a lymphatic vessel. Haversian canal is surrounded by concentric lamellae, small fluid filled spaces called ‘lacunae’ provided with minute canaliculi lie in between the lamellae. Canaliculi connect the lacunae with one another and with Haversian canal. Each lacuna encloses one osteocyte (inactive form of osteoblast).

The cytoplasmic processes of osteocytes extend through canaliculi. A Haversian canal and the surrounding lamellae and lacunae are collectively called a Haversian system or osteon. The Haversian canals communicate with one another, with the periosteum and also with the marrow cavity by transverse or oblique canals called Volkmanns canals. Nutrients and gases diffuse from the vascular supply of Haversian canals.
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Question 10.
Write short notes on Lymph.
Answer:
Lymph :
Lymph is a colourless fluid. It lacks RBC, platelets and large plasma proteins, but has more number of leucocytes. It is chiefly composed of plasma and lymphocytes. When compared to the tissue fluid, it contains very small amounts of nutrients and oxygen but has abundant C02 and other metabolites. The most important site of formation of lymph is interstitial space. As blood passes through the blood capillaries, some portion of blood that includes water, solutes and proteins of low molecular weight passes through the walls of capillaries, into the interstitial spaces due to hydrostatic pressure at the arteriolar ends.

This fluid forms the interstitial fluid (tissue fluid). Most of the interstitial fluid is returned directly to the capillaries due to osmotic pressure at the venular ends. Little amount of this tissue fluid passes through a system of lymphatic capillaries (lymph capillaries of the intestinal villi are called ’lacteals’), vessels, ducts and finally reach the blood through the subclavian veins. The extracecellular ’tissue fluid’ that passes into the lymph capillaries and lymph vessels is called ‘lymph’. Lymphatic system represents an ‘accessory route’ by which interstitial fluid flows from tissue spaces into blood.

Question 11.
Describe the structure of a skeletal muscle. [March 2018 – A.P.; March 2015 – T.S.]
Answer:
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Skeletal (striped and voluntary) muscle: It is usually attached to skeletal structures by ‘tendons’. In a typical muscle such as the ‘biceps’ muscle, skeletal muscle fibre is surrounded by a thin connective tissue sheath, the endomysium. A bundle of muscle fibres is called a fascicle. It is surrounded by a connective tissue sheath called perimysium. A group of fascicles form a ‘muscle’ which is surrounded by an epimysium (outermost connective tissue sheath). These connective tissue layers may extend beyond the muscle to form a chord-like tendon or sheet-like aponeurosis.

A skeletal muscle fibre is a long, cylindrical and unbranched cell. It is a multinucleated cell with many oval nuclei characteristically in the “peripheral” cytoplasm (a syncytium formed by fusion of cells). Sarcoplasm has many myofibrils which show alternate dark and light bands. So it is called striped or striated muscle. Skeletal muscle usually works under the conscious control of an organism (a voluntary muscle). Skeletal muscle contracts quickly and undergoes fatigue quickly. They are innervated by the ‘somatic nervous system’. Satellite cells are quiescent (quiet and inactive), mononucleate and myogenic cells and help in regeneration, which is ‘limited’.

TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 12.
Describe the structure of a cardiac muscle. [March 2013]
Answer:
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Cardiac (striped and involuntary) muscle :
The cardiac muscle is striated like the skeletal muscle (shows sarcomeres). Cardiac muscle is found in the ‘myocardium’ of the heart of vertebrates. The cardiac muscle cells or the ‘myocardial cells’ are short, cylindrical, mononucleate or binucleate cells whose ends branch and form junctions with other cardiac muscle cells. Each myocardial cell is joined to adjacent myocardial cells by ‘electrical synapses’ or ‘gap junctions’. They permit ‘electrical impulses’ to be conducted along the long axis of the cardiac muscle fibre. The dark lines across cardiac muscle are called intercalated discs (IDs). These discs are highly characteristic of the cardiac muscle.

The cardiac muscle is highly resistant to fatigue, because it has numerous sarcosomes, many molecules of myoglobin (oxygen storing pigment) and copious supply of blood which facilitate ‘continuous aerobic respiration’.

Question 13.
Give an account of the supporting cells of Nervous tissue.
Answer:
Neuroglia (supporting cells):
These are the supporting and non-conducting cells that provide a microenvironment suitable for neuronal activity. Unlike neurons, they continue to divide throughout the life. Neuroglial cells of the CNS include oligodendrocytes; astrocytes (star shaped cells) that form interconnected network and bind neurons and capillaries (helping in providing blood-brain barrier); ependymal cells, which are ciliated cells that line the cavities of brain and spinal cord to bring movements in the cerebrospinal fluid; microglial cells, which are phagocytic cells, of mesodermal origin. Neuroglial cells of the peripheral nervous system include the satellite cells and Schwann cells. Satellite cells surround the cell bodies in ganglia, and Schwann cells form neurilemma around axons.

Question 14.
Describe the structure of a multipolar neuron.
Answer:
A neuron usually consists of a “cell body” with one to many dendrites and a single axon.
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Cell body :
It is also called perikaryon, cyton or soma. It contains abundant granular cytoplasm and a large spherical nucleus. The cytoplasm has Nissl bodies (they represent RER, the sites of protein synthesis), neurofibrils and lipofuscin granules (the products of cellular wear and tear, accumulating in lysosomes with age). A group of cell bodies in the central nervous system is called a ‘nucleus’, and in the peripheral nervous system, it is called a ‘ganglion’.

Dendrites :
Several short, branched processes which arise from the cyton are called dendrites. They also contain Nissl bodies and neurofibrils. They conduct nerve impulses towards the cell body
(afferent processes).

Axon :
An axon is a single, long, cylindrical process that originates from a region of the cyton called axon hillock. Plasmalemma of an axon is called axolemma, and the cytoplasm is called axoplasm, which contains neurofibrils. However, Nissl bodies are absent. An axon may give rise to collateral branches. Distally it branches into many fine filaments called telodendria, (axon terminals), which end in bulb like structures called synaptic knobs or terminal boutons. Synaptic knobs possess ‘synaptic vesicles’ containing chemicals called neurotransmitters. Axon transmits nerve impulse away from the cyton (efferent process) to an interneuronal or neuromuscular junction called synapse.

TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 15.
Write short notes on (A) Platelets (B) Synapse.
Answer:
TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals 14
A) Blood platelets (Thrombocytes) :
These are colourless non-nucleated, round or oval biconvex discs. Number of platelets per cubic mm of blood is about 2,50,000 – 4,50,000. They are formed from giant megakaryocytes produced in the red bone marrow by fragmentation. The average life-span of blood platelets is about 5 to 9 days. They secrete thromboplastin and play an important role in blood clotting. They adhere to the damaged endothelial lining of capillaries and seal minor vascular openings.

B) Synapse :
An axon distally branches into many fine filaments called telodendria (Axon terminals) which end in bulb like synaptic knobs or terminal boutons. Synaptic knobs possess synaptic vesicles containing chemicals called neurotransmitters. Axon transmits nerve impulse away from the cyton into an interneuronal or neuromuscular junction called synapse. Synapse is the smallest gap between telodendria of a neuron and dendrites of next neuron.

Essay Answer Type Questions

Question 1.
What is a coelom? Explain the different types of coelom with suitable examples and neat labelled diagrams.
Answer:
Coelom :
The term ‘coelom’ was coined by Haeckel. The body cavity, which is lined by mesoderm, is called coelom. More elaborately, coelom is a fluid-filled space between the body wall and visceral organs and lined by mesodermal epithelium, the peritoneum. Animals possessing coelom are called coelomates/ eucoelomates. Evolution of efficient organ systems was not possible until the evolution of coelom for supporting the organs and distributing material.

Acoelomate bilaterians:
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The bilaterian animals in which the body cavity is absent are called acoelomates, e.g. Platyhelminthes (lowest bilaterians). In these animals, the mesenchyme derived from the third germinal layer, called mesoderm, occupies the entire blastocoel, between the ectoderm and the endoderm, so that the adults have neither the primary cavity (blastocoelom). nor the secondary cavity (coelom) As there is no body cavity, the acoelomates exhibit solid body plan. Problems faced by the acoelomates due to absence of perivisceral cavity are – their internal organs cannot move freely, as they are embedded in the mesenchyme, diffusion of material from the gut to the body wall is made slow and less efficient.

Pseudocoelomate bilaterians :
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In some animals, the body cavity is not lined by mesodermal epithelia. Such animals are called Pseudocoelomates. They include the members of phylum Aschelminthes (Nematoda, Rotifera and some minor phyla). During the embryonic development mesoderm (mesenchyme) occupies only a part of the blastocoel adjoining the ectoderm. The unoccupied portion of the blastocoel persists as pseudocoelom, which is filled with pseudocoelomic fluid. Pseudocoelomates are the first animals to exhibit a ‘tube-within-a-tube’ organisation.

As the gut wall is made of only endodermal epithelium, diffusion of digested food from the lumen of the gut into the surrounding pseudocoelomic fluid becomes easier and the absence of circulatory system is thus compensated. Though it is called pseudocoelom (false coelom), it performs almost all the functions of a regular coelom. Pseudocoelomic fluid of pseudocoelomates serves as a hydrostatic skeleton and a ‘shock absorber’. It allows the free movements of visceral organs, helps in the circulation of nutrients, and storage of nitrogenous wastes.

Eucoelomate bilaterians:
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Coelom or ‘true coelom’ is a fluid-filled cavity, that lies between the body wall and the visceral organs and is lined by mesodermal epithelium, the peritoneum. The portion of the peritoneum that underlines the body wall is the parietal peritoneum or somatic peritoneum. The portion of the peritonieum that covers the visceral organs is the splanchnic peritoneum or visceral peritoneum. In coelomates, the visceral organs are suspended in the coelom by the peritoneum. A double layered peritoneum that connects some visceral organs to the body wall is called mesentery.

In some eucoelomates such as the annelids, the dorsal and ventral mesenteries divide the coelom into paired compartments. Certain organs such as the kidneys of the vertebrates are covered by the parietal peritoneum only on their ventral side. Such a peritoneum is called the ‘retroperitoneum’ and the organs lined by it are called ‘retroperitoneal organs’.

During the embryonic development of the eucoelomates, the blastocoel is replaced by true coelom derived from the mesoderm. So, the true coelom is also called ‘secondary body cavity’. In the eucoelomates, mesodermal epithelium (peritoneum) lines both the body wall and the walls of the visceral organs. So, visceral organs become ‘muscular’ and exhibit free movements ‘independent’ of the movement of the body wall in the coelomic fluid. As the wall of the gut becomes thick and muscular, digested food and other nutrients cannot diffuse from the lumen of the gut into the coelom. Circulatory system (blood vascular system) is developed in the eucoelomates to overcome this problem. Based on the mode of formation of coelom, the eucoelomates are classified into two types :

I. Schizocoelomates:
Animals in which the body cavity is formed by spilitting of mesoderm’ are called schizocoelomates. Annelids, arthropods and molluscs are schizocoelomates in the animal kingdom. All the schizocoelomates are protostomians and they show ‘holoblastic’, ‘spiral’ and ‘determinate’ cleavage. The 4d blastomere or mesentoblast cell of the early embryo divides to form mesodermal blocks between the ectoderm and the endoderm and replaces the blastocoel. The split that appears in each mesodermal block leads to the formation of’schizocoelom’ (split coelom).
TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals 18

In annelids, the functional body cavity (perivisceral cavity) is schizocoelom. It is in the form of a series of paired coelomic cavities, but in arthropods and molluscs, the functional body cavity that lies around visceral organs is filled with blood (haemolymph) and is called haemocoel. It is formed by the fusion of the embryonic blastocoel with some coelomic spaces and the tissues are directly bathed in the blood (haemolymph).

II. Enterocoelomates:
Animals in which the body cavity is formed from the mesodermal pouches of archenteron are called enterocoelomates. Echinoderms, hemichordates and chordates are the enterocoelomates. In these animals, mesodermal pouches that evaginate from the wall of the archenteron into the blastocoel are fused with one another to form the enterocoelom. All the enterocoelomates are deuterostomes and they show radial and indeterminate cleavage.
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TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 2.
What is symmetry? Describe the different types of symmetry in the animal kingdom with suitable examples.
Answer:
Importance of Symmetry :
The concept of symmetry is fundamental in understanding the organisation of an animal. Symmetry in animals is balanced distribution of paired body parts. The body plan of a vast majority of metazoans exhibits some kind of symmetry. However, most of the sponges and snails show asymmetry (lack of symmetry). The symmetry of an animal and its mode of life are correlated.

Asymmetry :
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The animals, which cannot be cut into two equal parts (antimeres) in any plane passing through the centre of the body are called asymmetrical, e.g. most sponges and adult gastropods. In the asymmetrical animals, the body lacks a definite form. Asymmetry cannot be said to be an adaptation or advantage to an organism. Most of the asymmetrical organisms do not develop complex sensory and locomotor functions.

Symmetry :
The regular arrangement of body parts in a geometrical design relative to the axis of the body is called symmetry. The animals, which can be cut into two equal parts, or antimeters in one or more planes passing through the ‘principal axis’ of the body are called symmetrical animals. In a symmetrical animal, paired body parts are arranged on either side of the plane passing through the principal axis, such that they are equidistant from the plane. The unpaired body parts are located mostly on the plane, passing through the principal axis. Basically, the symmetry in animals is of two kinds.
i) Radial symmetry and
ii) Bilateral symmetry

i) Radial Symmetry or Monaxial heteropolar (axis is single; poles are different) Symmetry :
TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals 21 TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals 22
When any plane passing through the central axis (oro-aboral axis/ principal axis) of the body divides an organism into two identical parts, it is called radial symmetry. The animals with radial symmetry are either sessile or planktonic or sluggish forms. It is the principal symmetry of the diploblastic animals such as the cnidarians and ctenophores (considered as biradial animals by some authors).

Animals showing radial symmetry live in water and they can respond equally to stimuli that arrive from all directions. Thus, radial symmetry is an advantage to sessile or slow moving animals. However, triploblastic animals such as echinoderms are ‘secondarily radially symmetrical’ (as it is five angled, it is also called pentamerous radial symmetry). Radially symmetrical animals have many planes of symmetry, whereas pentamerous radially symmetrical animals have five planes of symmetry.

ii) Bilateral symmetry :
When only one plane (median sagittal plane) that passes through the central axis (anterior – posterior axis) divides an organism into two identical parts, it is called bilateral symmetry. It is the ‘principal type of symmetry’ in the triploblastic animals. Among the triploblastic animals, some gastropods become secondarily asymmetrical, though they have primarily bilaterally symmetrical larvae.
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Question 3.
Classify and describe the epithelial tissues on the basis of structural modification of cells with examples.
Answer:
There are two types of epithelial tissues namely ‘simple epithelia’ and ‘compound epithelia’ based on the number of layers or strata. Various glands in the body involved in secretions are made up of epithelial tissue (glandular epithelium).

A) Simple epithelium :
Simple epithelium is composed of a single layer of cells and forms the lining of body cavities, ducts and vessels. It helps in diffusion, absorption, filtration and secretion of substances. On the basis of the shape of the cells, it is further divided into three types :

i) Simple squamous epithelium (Pavement epithelium) :
TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals 24
It is composed of a single layer of flat and tile-like cells, each with a centrally located ‘ovoid nucleus’. It is found in endothelium of blood vessels, mesothelium of body cavities (pleura, peritoneum, and pericardium), wall of Bowman’s capsule of nephron, lining of alveoli of lungs, etc.

ii) Simple cuboidal epithelium :
TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals 25
It is composed of a single layer of cube-like cells with centrally located spherical nuclei. It is found in germinal epithelium, proximal and distal convoluted tubules of nephron. Cuboidal epithelium of proximal convoluted tubule of nephron has ‘microvilli’.

iii) Simple columnar epithelium :
It is composed of a single layer of tall and slender cells with oval nuclei located near the base. It has mucus – secreting’ goblet cells’ in some places. It is of two types.

a) Ciliated columnar epithelium :
TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals 26
Columnar epithelial cells have cilia on their free surface. It is mainly present in the inner surface of hollow organs like fallopian tubes, ventricles of brain, central canal of spinal cord, bronchioles etc.

b) Non-ciliated columnar epithelium :
TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals 27
Columnar cells are without cilia. It is found in the lining of Stomach and intestine. Microvilli are present in the columnar epithelium of intestine to increase the surface area of absorption.
TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals 28

B) Compound epithelium (stratified epithelium):
It is made up of more than one layer of cells. Its main function is to provide protection against chemical and mechanical stress. It covers the dry surface of the skin as stratified, keratinized, squamous epithelium. It covers the moist surface of buccal cavity, pharynx, oesophagus and vagina as stratified non – keratinized squamous epithelium. It forms the inner lining of the larger ducts of salivary glands, sweat glands and pancreatic ducts as stratified cuboidal epithelium. It forms the wall of the urinary bladder as transitional epithelium.

c) Glandular epithelium :
Some of the columnar or cuboidal cells that get specialised for the production of certain secretions, form glandular epithelium. The glands are of two types – unicellular glands consisting of-isolated glandular cells such as goblet cells of the gut, and multicellular glands, consisting of clusters of cells such as salivary glands. On the basis of the mode of pouring of their secretions, glands are divided into two types namely exocrine and endocrine glands. Exocrine glands are provided with ducts; secrete mucus, saliva, earwax (cerumen), oil, milk, digestive enzymes and other cell products. In contrast, endocrine
TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals 29
glands are ductless and their products are ‘hormones’, which are not sent out via ducts, but are carried to the target organs by blood. Based on the mode of secretion, exocrine glands are further divided into i. Merocrine glands (e.g. pancreas) which release the secretory granules without the loss of other cellular material ii. Apocrine glands (e.g. mammary glands) in which the apical part of the cell is pinched off along with the secretory product and iii. Holocrine glands (e.g. sebaceous glands), in which the entire cell disintegrates to discharge the contents.
TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals 30

TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 4.
Describe the various types of connective tissue proper with suitable examples.
Answer:
Connective tissue proper:
It is of two types.

A) Loose connective tissue :
Cells and fibres are loosely arranged in a semi fluid ground substance. There are three types of loose connective tissues – areolar tissue, adipose tissue and reticular tissue.

i. Areolar tissue :
It is one of the most widely distributed connective tissues in the body. It forms the packing tissue in almost all the organs. Areolar tissue forms subcutaneous layer of the skin. It has cells and fibres. Cells of the areolartissue are fibroblasts, mast cells, macrophages, adipocytes and plasma cells.

  1. Fibroblasts are the most common cells which secrete fibres. The inactive cells are called fibrocytes.
  2. Mast cells secrete heparin (in anticoagulant), histamine, bradykinin (vasodilators), and serotonin (vasoconstrictor). Vasodilators cause inflammation in response to injury and infection.
  3. Macrophages are amoeboid cells, phagocytic in function and act as internal scavengers. They are derived from the monocytes of blood. ‘Tissue fixed macrophages’ are called histiocytes and others are ‘wandering macrophages’.
  4. Plasma cells are derived from the B- lymphocytes and produce antibodies.
  5. Adipocytes are specialized cells for the storage of fats.

TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals 31

Fibres of areolar tissue are of three types. They are collagen, reticular and elastic fibres. Collagen and reticular fibres are composed of the protein collagen, elastic fibres are made of the protein elastin. Collagen (white) fibres occur in bundles, and they are strong and stretch resistant. Reticular fibres are thin and form a network and they provide strength and support to certain tissues such as bone marrow. Elastic (yellow) fibres are branched and form a network. They are also found in elastic cartilages, elastic ligaments, etc.

ii. Adipose tissue :
It is specialized for fat storage. It consists of a large number of adipocytes and few fibres. Adipose tissue which is found beneath the skin provides thermal insulation. It forms blubber of aquatic mammals such as whales and sea cows and the hump of camel. It acts as shock absorber in palms and soles. Adipose tissue is of two types; white adipose tissue, brown adipose tissue. Excess nutrients which are not used immediately are converted into fats and stored in this tissue.

White adipose tissue (WAT) :
It is the predominant type in the adults, and the adipocyte has a single large lipid droplet (monolocular). White fat is metabolically not active.

Brown adipose tissue (BAT) :
It is found in foetuses and infants. Adipocyte of BAT has several small ‘lipid droplets’ (multilocular) and numerous mitochondria. Brown fat is metabolically active and generates ‘heat’ to maintain body temperature required by infants.

iii. Reticular tissue :
It has specialized fibroblasts called reticular cells. They secrete ‘reticular fibres’ that form an inter connecting network. It forms the ‘supporting frame work’ of lymphoid organs such as bone marrow, spleen and lymph nodes and forms the reticular lamina of the ‘basement membrane’.

B) Dense connective tissue :
This tissue consists of more fibres, but fewer cells. It has very little ground substance. Based on the arrangement of fibres, dense connective tissue is of three types.

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i. Dense regular connective tissue :
In this tissue, collagen fibres are arranged parallel to one another in bundles. Tendons which attach the skeletal muscles to bones and ligaments which attach bones to other bones are examples of this type of connective tissue,

ii. Dense irregular connective tissue :
TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals 33
In this type of connective tissue, bundles of collagen fibres are irregularly arranged. Periosteum, endosteum, pericardium, heart valves, joint capsule and deeper region of dermis of skin contain/ are made up of this type of connective tissue.

iii. Elastic connective tissue :
It is mainly made of yellow elastic fibres, capable of considerable extension and recoil. This tissue can recoil to its original shape, when the forces of stretch are released. It occurs in the wall of arteries, vocal cords, trachea, bronchi and ‘elastic ligaments’ present between vertebrate.

In addition to the above mentioned connective tissues, mucous connective tissue occurs as foetal or embryonic connective tissue. It is present in the umbilical cord as Wharton’s jelly.

Question 5.
What is a skeletal tissue? Describe the various types of skeletal tissue.
Answer:
Skeletal tissue (supporting tissue): It forms the endoskeletonof the vertebrates. It supports the body, protects various organs, provides surface for the attachment of muscles and helps in locomotion. It is of two types.

A) Cartilage (Gristle) :
Cartilage is a solid, but semi-rigid (flexible) connective tissue. It resists compression. Matrix is firm, but somewhat pliable. It has collagen fibres, elastic fibres (only in the elastic cartilage) and matrix-secreting cells called chondroblasts. These cells are enclosed in fluid filled spaces called lacunae, chondrocytes are the inactive cells of a cartilage. Cartilage is surrounded by a fibrous connective tissue sheath called perichondrium. Cartilage is ‘avascular’ and it is nourished by ‘diffusion of nutrients’ from the blood capillaries of the perichondrium. Growth, regeneration and repair of cartilage take place by the activity of perichondrial cells. Cartilage is of three types, which differ from each other chiefly in the composition of the matrix.

1. Hyaline cartilage :
It is bluish-white, translucent and glass-like cartilage. Matrix is homogeneous and shows delicate collagen fibres. It is the weakest and the most common type of all the cartilages. Perichondrium is present except in articular cartilages. It forms the embryonic endoskeleton of bony vertebrates, endoskeleton of cyclostomes and cartilaginous fishes. It forms the articular cartilages (free surfaces of long bones that form joints), costal cartilages (sternal parts of. lbs), and the epiphyseal plates. It also forms the nasal septal cartilage, cartilaginous rings of trachea, bronchi and cartilages of larynx.
TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals 34

2) Elastic cartilage :
It is yellowish due to elastic fibres. Matrix has abundance of yellow elastic fibres in addition to collagen fibres. It provides strength and elasticity. Perichondrium is present. It is found in the pinnae of the external ears, Eustachian tubes and epiglottis.

3) Fibrous cartilage :
Matrix has bundles of collagen fibres. Perichondrium is absent. It is the strongest of all types of cartilages. It occurs in the intervertebral discs and pubic symphysis of the pelvis.

B. Bone (osseous) tissue :
Bone is highly calcified (mineralized), solid, hard and rigid connective tissue. It is the major component of the endoskeleton of most adult vertebrates. It is the main tissue that provides structural frame work to the body. It supports the soft tissues, protects the delicate organs. Limb bones of animals serve the weight bearing functions. Bones also interact with muscles attached to them to bring about movements. Bones have a hard and non-pliable matrix, rich in calcium salts and collagen fibres which give the bone its strength. During ageing, the proportion of inorganic materials increases in a bone, making it more brittle. Bone forms homeostatic reservoir of calcium, magnesium, phosphorus, etc. Bone is highly vascular.

Bone has an outer fibrous connective tissue sheath called periosteum, the inner connective tissue sheath that lines the marrow cavity called endosteum, non-living extra cellular matrix, living cells and bone marrow. Bone cells include osteoblasts, osteocytes and osteoclasts. Osteoblasts (immature bone cells) secrete the organic components (collagen fibres) of matrix and also play an important role in ‘mineralization of bone’ and become osteocytes (mature bone cells). Osteocytes are enclosed in fluid filled lacunae. Osteoclasts are phagocytic cells involved in resorption of bone.

Types of bones based on the method of formation :

  1. Cartilage bones (replacing bones or endochondral bones) are formed by ossification within the cartilage e.g. bones of limbs, girdles and vertebrae.
  2. Investing bones (membrane bones or dermal bones) are formed by the ossification in the embryonic mesenchyme e.g. most of the bones of cranium.
  3. Sesamoid bones are formed by ossification in tendons e.g. patella (knee cap) and pisiform bone of the wrist of a mammal.
  4. Visceral bones are formed by ossification in the soft tissues, e.g. Oscordis (Inside the heart of ruminants), Os penis (inside the glans-penis of many mammals such as the rodents, bats and carnivores).

Types of bones based on the structure :
1) Spongy bone (Cancellous bone or trabecular bone) :
It occurs in the epiphyses and metaphyses of long bones. It looks spongy and contains columns of bone called ‘trabeculae’ with irregular interspaces filled with red bone marrow.

2) Compact bone :
The diaphysis of a long bone is made up of ‘compact bone’. It has dense continuous lamellar matrix between periosteum and endosteum.

Structure of a compact bone :
Diaphysis (shaft) is a part of a long bone that lies in between expanded ends (epiphyses). In a growing bone there is a region called metaphysis between the diaphysis and epiphysis. It consists of an epiphyseal plate (formed by hyaline cartilage). It helps in the elongation of the bone. In adults it is represented by a bony epiphyseal line. Diaphysis is covered by a dense connective fibrous tissue called periosteum.

Diaphysis of a long bone has a hollow cavity called marrow cavity which is lined or surrounded by the endosteum. In between periosteum and endosteum, the matrix of the bone is laid down in the form of ‘lamellae’. Outer circumferential lamellae are located immediately beneath the periosteum; inner circumferential lamellae are located around the endosteum. Between the outer and inner circumferential lamellae, there are many Haversian systems (osteons – units of bone).

The spaces between the Haversian systems are filled with interstitial lamellae. Haversian system consists of a Haversian canal that runs parallel to the marrow cavity. It contains an artery, a vein and a lymphatic vessel. Haversian canal is surrounded by concentric lamellae. Small fluid filled spaces called ‘lacunae’ provided with minute canaliculi lie in between the lamellae. Canaliculi connect the lacunae with one another and with Haversian canal. Each lacuna encloses one osteocyte (inactive form of osteoblast).

The cytoplasmic processes of osteocytes extend through canaliculi. A Haversian canal and the surrounding lamellae and lacunae are collectively called a Haversian system or osteon. The Haversian canals communicate with one another, with the periosteum and also with the marrow cavity by transverse or oblique canals called Volkmann’s canals. Nutrients and gases diffuse from the vascular supply of Haversian canals.
TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals 35

TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 6.
Give an account of the “formed elements” of Blood.
Answer:
Formed elements :
They include erythrocytes (red blood corpuscles), leucocytes (white blood corpuscles) and platelets. The process of formation of blood cells is called haemopoiesis or haematopoiesis. In the earliest stages of embryogenesis, blood cells are formed from the yolk sac mesoderm. Later on, the liver and the spleen serve as temporary haemopoietic tissues”. In the final stage of embryonic development and after birth, the red bone marrow is the primary site of haemopoiesis.

Red blood corpuscles (Erythrocytes) :
TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals 36
Erythrocytes of mammals are circular (elliptical in camels and Llamas), biconcave and enucleate. The biconcave shape provides a large surface area to – volume ratio, thus providing more area for the exchange of gases. These are 7.8 mm in diameter. The number of RBC per cubic millimeter of blood is about 5 million in a man, and 4.5 million in a woman. Decrease in the number of erythrocytes is called erythrocytopenia and it leads to anaemia. An abnormal rise in RBC count is called polycythemia.Shortage of oxygen stimulates the kidneys to secrete a hormone called erythropoietin into the blood. Erythropoietin stimulates the bone marrow to increase the production of RBC. Vitamin B12 and folic acid are required for maturation of RBC.

Mammalian RBC is surrounded by plasma membrane. Nucleus and other cell organelles are lost in the reticulocyte stage of its development. Cytoplasm of RBC contains a chromo protein, the ‘haemoglobin’. Each Haemoglobin molecule consists of 4 polypeptide (2α & 2β) chains and 4 haeme molecules. In the centre of each haeme group is one Fe2+, which can combine with one molecule of O2. Life span of RBC in humans is about 120 days. The worn out RBC are destroyed in the ‘spleen’ and ‘liver’.

White blood corpuscles (Leucocytes) :
These are.nucleate, colourless, complete cells. They are spherical or irregular in shape, and are capable of exhibiting amoeboid movement into the extravascular areas by diapedesis. They are larger than RBC in size, and less than RBC in number. The total leucocyte count is 6,000 – 10,000 per cubic millimeter of blood under normal conditions. The process of formation of WBC is called Leucopoiesis. Slight increase in the WBC count is called Leucocytosis (during infection and allergy).

An abnormal increase in the number of WBC is indicated in a type of cancer called Leukemia. Fall in WBC count is called Leucocytopenia. WBC are of two main types : Granulocytes and Agranulocytes. Granulocytes : They possess cytoplasmic granules that may take three different types of stains, neutral or acidic or basic. Nucleus of the granulocytes is divided into lobes and assumes different shapes, hence, these are also called Polymorph – nuclear leucocytes. Based on the staining properties these are of three types.

TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals 37
Eosinophils (acidophils) :
They constitute about 2.3% of the total leucocytes. Nucleus is distinctly bilobed. Cytoplasm has large granules which stain with acidic dyes such as ‘eosin’. They play a role in allergic reactions. Their number increases during ‘allergic reactions’ and ‘helminth infections’. They remove ‘antigen – antibody complexes’.

Neutrophils :
They constitute about 62% of the total leucocytes. Nucleus is many lobed (2-5). Specific cytoplasmic granules are small and abundant. They stain with ‘neutral dyes’. These are active phagocytic cells commonly described as ‘microscopic policemen’. Certain neutrophils of female mammals have sex chromatin body or Drumstick body (an extra ‘X’ chromosome) attached to the nucleus.

Agranulocytes :
TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals 38
Cytoplasmic granules are absent in agranulocytes. Nucleus of these cells is not divided into lobes. These are of two types; a) Lymphocytes: They constitute about 30% of the total leucocytes. They are small, spherical cells with large spherical nucleus and scanty peripheral cytoplasm. There are functionally two types of lymphocytes – ‘B’ lymphocytes, which produce ‘antibodies’ and ‘T’ lymphocytes which play the key role in the immunological reactions of the body.

Some lymphocytes live only a few days while others survive for many years.

Monocytes :
TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals 39
They constitute about 5.3% of the leucocytes. The nucleus is kidney shaped (reniform). These are the largest, motile phagocytes. They engulf bacteria and cellular debris. They differentiate into macrophages, when they enter the connective tissues.

Blood platelets (Thrombocytes) :
TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals 40
These are colourless non- nucleated, round or oval biconvex discs. Number of platelets per cubic mm of blood is about 2,50,000 – 4,50,000. They are formed from giant megakaryocytes produced in the red bone marrow by fragmentation. The average life-span of blood platelets is about 5 to 9 days. They secrete thromboplastin and play an important role in blood clotting. They adhere to the damaged endothelial lining of capillaries and seal minor vascular openings.

TS Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 7.
Compare and contrast the three types of muscular tissues.
Answer:
Muscular tissue is mesodermal in origin except iris and ciliary body muscles which are ectodermal in origin. Muscles are 3 types- ‘Skeletal, smooth and cardiac muscles.

Skeletal muscleSmooth muscleCardiac muscle
1. Otherwise called as volunatry or striated or striped muscle.Otherwise called as unstriped or involuntary or visceral muscle.Otherwise called as striped involuntary muscle.
2. Long cylindrical unbranched multinucleated cell.Arranged in layers /sheets. Spindle shaped uninucleate cell.Short, cylindrical mononucleate or binucleate cells whose ends branch and form junctions with other cardiac muscle cells.
3. Skeletal muscle usually works under the conscious control of an organism.Smooth muscles do not work under the conscious control and so they are called involuntary muscles.Cardiac muscles are also involuntary in function. However the rate of beat is regulated by an autonomic innervation and hormones like epinephrine/ adrenaline
4. Contracts quickly and undergoes fatigue quickly. They are innervated by somatic nervous system.Smooth muscles exhibit slow and prolonged contractions. They may remain contracted for long periods without fatigue. The contractions are under the control of autonomons nervous system.Contractions are very fast. Cardiac muscle is highly resistant to fatigue because of numerous sarcosomes and many molecules of myoglobin. Intercalated discs are characteristic.
5. They are always attached to bones to bring movements of the body.They help in internal movements to regulate daily metabolic activities like digestion, micturition etc.They help in pumping of blood by contraction (systole) and expansion (diastole) of heart and facilitate continuous “aerobic respiration”.

TS Inter 1st Year Zoology Study Material Chapter 1 Diversity of Living World

Telangana TSBIE TS Inter 1st Year Zoology Study Material 1st Lesson Diversity of Living World Textbook Questions and Answers.

TS Inter 1st Year Zoology Study Material 1st Lesson Diversity of Living World

Very Short Answer Type Questions

Question 1.
Define the term metabolism. Give any one example.
Answer:
The sum total of all the chemical reactions occurring in the bodies of organisms constitute metabolism and it is a defining feature of all living organisms without exception. Eg : Photosynthesis and respiration.

Question 2.
How do you differentiate between growth in a living organism and non living object?
Answer:
Growth in living beings is ‘growth from inside’ whereas growth in the non-living things is by accumulation of material on the surface.

Question 3.
What is biogenesis?
Answer:
‘Life comes only from life and not from non-living substances’ is known as biogenesis.

Question 4.
Define the term histology. What is it otherwise called? [March 2019]
Answer:
Histology is the study of microscopic structure of different tissues. This branch is also referred to as “Microanatomy”.

Question 5.
Distinguish between embryology and ethology.
Answer:

  1. Embryology deals with the study of events that lead to fertilization, cleavages, early growth and differentiation of zygote into an embryo.
  2. The study of the animal behaviour based on the systematic observation, recording, analysis of functions of animals, with special attention to ecological, physiological and evolutionary aspects is called Ethology.

TS Inter 1st Year Zoology Study Material Chapter 1 Diversity of Living World

Question 6.
‘In a given area, remains of an animal that lived in the remote past are excavated for study. Which branch of science is it called?
Answer:
Palaeontology, (specifically palaeozoology – study of fossils of animals)

Question 7.
“Zoos are tools for classification” Explain.
Answer:
Zoos are the places where wild animals, taken out of their natural habitat, are placed in protected environment under human care. This enables us to study the various aspects of animal living. Thus it enables us to systematise the organism and position it in the animal world.

Question 8.
Where and how do we preserve skeletons of animals, dry specimens etc?
Answer:
In Museums animal specimens may also be preserved as dry specimens. Museur often have collections of skeletons of animals too.

Question 9.
What is trinominal nomenclature ? Give an example. [March 2018 – A.P.; March 2015 – T.S.]
Answer:
Trinominal nomenclature is the extension of the binominal system of nomenclature. This system permits the designation of subspecies with a three-worded name called ‘trinomen’. Subspecies is a category below the level of species. Eg : Corvus splendens splendens.,

Question 10.
What is meant by tautonymy. Give two examples.[ May 2017 – A.P.; May/June, Mar. 2014]
Answer:
The practice of naming the animals, in which the generic name and species name are the same, is called tautonymy. So the name is called tautonym. Eg : Naja naja (the Indian cobra), Axis axis (spotted deer).

Question 11.
Differentiate between Protostomia and Deuterostomia.
Answer:
a) Protostomia :
The eumetazoans in which blastopore develops into mouth are referred to as the protostomians. (eg : Annelida)

b) Deuterostomia :
These are eucoelomates in which anus is formed from or near the blastopore, (eg: Echinodermata)

Question 12.
‘Echinoderms are enterocoelomates’. Comment.
Answer:
Enterocoel is a true coelom formed from the archenteron. In phylum Echinodermata, true coelom is formed from the primitive gut called archenteron. Hence Echinoderms are enterocoelomates.

TS Inter 1st Year Zoology Study Material Chapter 1 Diversity of Living World

Question 13.
What does ICZN stand for? [March 2015 – A.P.]
Answer:
ICZN stands for International Code of Zoological Nomenclature.

Question 14.
Give the names of any four protostomian phyla.
Answer:
1) Platyhelminthes 2) Nematoda 3) Annelida 4) Arthropoda.

Question 15.
Nematode is a protostomian but not a eucoelomate. Justify the statement.
Answer:
In Nematodes the blastopore becomes mouth and the group belongs to protostomia. However the body cavity is not a true coelom as it is not lined by mesodermal epithelial layers. Their body cavity is a pseudocoel.

Question 16.
What is ecological diversity? Mention the different types of ecological diversities.
Answer:
Diversity at the ecosystem level is called “Ecological diversity”. The different types of ecological diversities are Alpha, Beta and Gamma diversities.

Question 17.
Define species richness. [March 2017 – A.P.]
Answer:
Species richness in simple terms, it is the number of species per unit area. The more the number of species in an area the more is the species richness.

Question 18.
Mention any two products of medicinal importance obtained from Nature.
Answer:

  1. Quinine (drug of Malaria) obtained from the bark of cinchona officinalis.
  2. Vin blastin (anti cancer drug) from Vinca rosea.
  3. Digitalin from ‘fox glove’ plant (Digitalis purpurea).

Question 19.
Invasion of an Alien species leads to extinction of native species. Justify this with two examples.
Answer:

  1. Nile perch introduced into Lake Victoria, in East Africa lead to the extinction of 200 species of cichlid fish in the lake.
  2. Illegal introduction of exotic African cat fish, Clarias gariepinus, for aquaculture purpose is posing a threat to the indigenous cat fishes.

TS Inter 1st Year Zoology Study Material Chapter 1 Diversity of Living World

Question 20.
List out any four sacred groves in India.
Answer:

NameState
1) Khasi and Jaintia Hills ………….Meghalaya.
2) Aravalli Hills ………….Rajasthan and Gujarat.
3) Sarguja, Bastar ………….Chhattisgarh.
4) Western Ghat region ………….Karnataka and Maharastra.

Question 21.
Write the full form of IUCN. in which book threatened species are enlisted. [March 2020]
Answer:

  1. International Union for the Conservation of Nature and Natural Resources (IUCN)
  2. All the threatened species are listed in the Red Data Books published by the IUCN.

Short Answer Type Questions

Question 1.
Explain the Phylogenetic system of biological classification.
Answer:
Carolus Linnaeus introduced the system of hierarchical classification. Phylogenetic system of biological classification (cladistic classification).

It is an evolutionary classification based on how a common ancestry was shared. Cladistic classification summarizes the ‘genetic distance’ between all species in the ‘phylogenetic tree’. In cladistic classification characters such as analogous characters (Characters shared by a pair of organisms due to convergent evolution e.g. wings in sparrows and patagia (wing like structures in flying squirrels) and homologous characters (characters shared by a pair of organisms, inherited from a common ancestor e.g., wings of sparrows and finches) are followed/ taken into consideration. Ernst Haeckel introduced the method of representing phylogeny by ‘trees’ or branching diagrams.

Question 2.
Explain the hierarchy of classification.
Answer:
Linnaeus was the first taxonomist to establish a definite hierarchy of taxonomic categories called taxa (singular: taxon) like kingdom, class, order, genus and species. Haeckel introduced the taxon phylum. A species sometimes may have more subspecies, which shows some morphological variations (intra – specific variations).

Taxonomic Categories :
Nowadays the three Domain classification is followed. CARL WOESE and co – workers observed that many prokaryotes previously classified under ‘Prokaryota/ Monera1 are more closely related to the ‘eukaryotes’ and classified them under a separate Domain the ARCHAEA. This type of study is called ‘MOLECULAR SYSTEMATICS’.

Now there is a general agreement on the THREE DOMAIN CLASSIFICATION of the living organisms namely DOMAIN -1: BACTERIA, DOMAIN – II: ARCHAEA and ; DOMAIN – III: EUKARYA. (Note : DOMAIN is a taxon higher than ‘Kingdom’.)

TS Inter 1st Year Zoology Study Material Chapter 1 Diversity of Living World

Question 3.
What is meant by classification? Explain the need for classification.
Answer:
NEED FOR CLASSIFICATION :
It is impossible to study all living organisms. So, it is | necessary to devise some means to make this possible. This process is called ‘classification’. Classification is defined as the process by which anything is grouped into convenient categories based on some easily observable characters. The scientific term used for these categories is ‘TAXA’ (singular: taxon). Taxa can indicate categories at different levels e.g. Animalia (which includes multicellular animals), chordata, mammalia, etc., represent taxa at different levels.

Hence, based on characteristics, all the living organisms can be classified into different taxa. This process of classification is called taxonomy.

Question 4.
Define species. Explain the various aspects of ‘species’. [March 2020, ’13, ’14]
Answer:
Species :
Species is the ‘basic unit’ of classification. Species is a Latin word meaning ‘kind’ or ‘appearance’. John Ray in his book ‘Historia Generalis Plantarum’, used the term ‘species’ and described it on the basis of common descent (origin from common ancestors) as a group of morphologically similar organisms. Linnaeus considered species, in his book ‘Systema Naturae’, as the basic unit of classification.

Buffon’s biological concept of species explains that species is an interbreeding group of similar individuals sharing the common ‘gene pool’, and producing fertile offspring. Species is considered as a group of individuals which are :

  1. Reproductively isolated from the individuals of other species – a breeding unit.
  2. Sharing the same ecological niche – An ecological unit.
  3. Showing similarity in the karyotype – a genetic unit.
  4. Having similar structure and functional characteristics – an evolutionary unit.
  5. Species is dynamic.

Question 5.
What is genetic diversity and what are the different types of genetic diversity?
Answer:
Genetic diversity :
It is the diversity of genes with in a species. A single species may show high diversity at the genetic level over its distributional range. For e.g. Rauwolfia vomitoria, a medicinal plant growing in the Himalayan ranges shows great genetic variation, which might be in terms of potency and concentration of the active chemical (reserpine extracted from it is used in treating high blood pressure) that the plant produces. India has more than 50,000 different strains of rice, and 1,000 varieties of mangoes. Genetic diversity increases with environmental variability and is advantageous for its survival.

TS Inter 1st Year Zoology Study Material Chapter 1 Diversity of Living World

Question 6.
What are the reasons for greater biodiversity in the tropics?
Answer:
Reasons for greater biodiversity in the tropics:

Reason 1 :
Tropical latitudes have remained relatively undisturbed for millions of years and thus had a long ‘evolutionary time’. As long duration was available in this region for speciation, it led to the species diversification.

Reason 2 :
Tropical climates are relatively more constant and predictable than that of the temperate regions. Constant environment promotes niche specialization (how an organism responds, behaves with environment and other organisms of its biotic community), and this leads to greater species diversity.

Reason 3 :
Solar energy, resources like water etc., are available in abundance in this region. This contributed to higher productivity in terms of food production, leading to greater diversity.

Question 7.
What is the “evil quartet”? [March 2018 – A.P.; March 2015 – A.P. & T.S.]
Answer:
The following are the four major causes for accelerated rates of species extinction in the world. These causes are referred to as evil quartet,
a) Habitat loss and fragmentation :
These are most important reasons for the loss of biodiversity.

b) Over exploitation :
When need turns to greed, it leads to over exploitation.

c) Invasion of Allen species :
When Alien species are introduced into a habitat, they turn invasive and establish themselves at the cost of indigenous species.

d) Co – extinctions :
In an obligate association between a plant and an animal, if a plant becomes extinct, the animal also becomes extinct as seen in a parasitic and host association.

Question 8.
Explain in brief “Biodiversity Hot Spots”. [March 2019]
Answer:
Conservationists identified certain regions by name ‘Biodiversity hot spots’ for maximum protection as they are characterized by very high levels of species richness & high degree of endemism. By definition ‘Biodiversity hot spot’ is a ‘Biogeographic Region’ with a significant reservoir of biodiversity that is under threat of extinction from humans. They are Earth’s biologically ‘richest’ and ‘most threatened’ terrestrial Ecoregions.

Biodiversity hot spots :
The concept of biodiversity originated by Norman Myers. There are about 34 biodiversity hot spots in the world. As these regions are threatened by destruction, habitat loss is accelerated e.g. I) Western Ghats and Srilanka; II) Indo Burma; III) Himalayas in India. Ecologically unique and biodiversity rich regions are legally protected as in
1. Biosphere Reserves -14, 2. National Parks – 90, 3. Sanctuaries -448.

Question 9.
Explain ‘Rivet Popper” hypothesis. [May ’17; Mar. ’17 – A.P; May/June ’14]
Answer:
This hypothesis is mainly for a reason, what happens if we lose a few species? Will it affect man’s life?

Paul Ehrlich’s experiments ‘The RIVET POPPER’ hypothesis, taking an aeroplane as an ecosystem, explains how removal of one by one ‘rivets’ (species of an ecosystem) of various parts can slowly damage the plane (ecosystem) – shows how important a ‘species’ is in the overall functioning of an ecosystem. Removing a rivet from a seat or some other relatively minor important parts may not damage the plane, but removal of a rivet from a part supporting the wing can result in a crash. Likewise, removal of a critical species’ may affect the entire community and thus the entire ecosystem.

TS Inter 1st Year Zoology Study Material Chapter 1 Diversity of Living World

Question 10.
Write short notes on In-situ conservation.
Answer:
In-situ conservation is the process of protecting an animal species in its natural habitat. The following are the types.

Biosphere Reserves
An area which is set aside, minimally disturbed for the conservation of the resources of the biosphere is ‘Biosphere reserve’. Latest biosphere reserve (17th biosphere reserve in India) is Seshachalam hills.

National Parks
A National Park is a natural habitat strictly reserved for protection of natural life. National Parks, across the country, offer a fascinating diversity of terrain, flora, and fauna. Some important National Parks in India are – Jim Corbett National Park (the first National Park in India located in Uttarakhand), Kaziranga National Park (Assam), Kasu Brahmananda Reddy National Park, Mahavir Harina Vanasthali National Park (Telangana), Keoladeo Ghana National Park (Rajasthan), etc.

Sanctuaries:
Specific endangered faunal species are well protected in wildlife sanctuaries which permits eco-tourism (as long as animal life is undisturbed). Some important sanctuaries in India (AP) include – Koringa Sanctuary, Eturnagaram Sanctuary, Papikondalu Sanctuary.

Sacred Groves:

  1. A smaller group of trees than a forest is called grove.
  2. A grove of trees of special religious importance to a particular culture is called sacred grove.
  3. In these regions all the trees of wild life were venerated (respected) and given total protection.

The following is a list of Sacred Groves in INDIA.

NameState
Khasi and Jaintia HillsMeghalaya
Aravalli HillsRajasthan and Gujarat
Western Ghat regionKarnataka and Maharashtra
Sarguja, BastarChhattisgarh
ChandaMadhya Pradesh

In Meghalaya, Sacred Groves are the last refuge for a large number of rare and threatened species.

TS Inter 1st Year Botany Study Material Chapter 13 Ecological Adaptations, Succession and Ecological Services

Telangana TSBIE TS Inter 1st Year Botany Study Material 13th Lesson Ecological Adaptations, Succession and Ecological Services Textbook Questions and Answers.

TS Inter 1st Year Botany Study Material 13th Lesson Ecological Adaptations, Succession and Ecological Services

Very Short Answer Type Questions

Question 1.
Climax stage is achieved quickly in secondary succession as compared to primary succession. Why? [May ’14]
Answer:

  1. Secondary succession begins in areas where natural communities have destroyed.
  2. It is achieved quickly than primary succession, it occurs in abandoned farm lands burned or cut forests with some soil or sediment.

Question 2.
Among bryophytes, lichens and ferns which one is a pioneer species in a xeric succession?
Answer:

  1. Lichens are pioneer species that growth on rocks in rocks in xeric succession.
  2. They lead to soil formation through weathering of rocks by secreating acids.

Question 3.
Give any two examples of xerarch succession.
Answer:

  1. Xerach succession takes place is dry areas and series progress from xeric to mesic conditions.
  2. Formation of plant communities on dry rocks, on cooled and hardened lava from volcanoes.

Question 4.
Name the type of land plants that can tolerate the salinities of the sea. [Mar. ’14]
Answer:

  1. Halophytes
  2. Ex : Rhizophora

Question 5.
Define heliophytes and sciophytes. Name a plant from your locality that is either heliophyte or sciophyte.
Answer:

  1. Plants which grow in direct sunlight are called heliophytes Ex : Tridax, grass plants
  2. Plants which grow in shady places are called sciophytes Ex : Ferns, Mosses

TS Inter 1st Year Botany Study Material Chapter 13 Ecological Adaptations, Succession and Ecological Services

Question 6.
Define population and community. [Mar. – 2019, May ’17, Mar. ’15 – T.S. ; Mar. ’13]
Answer:
Population :
It is a group of similar individuals belonging to the same species found in an area.

Community :
It is an assemblage of all the populations (belonging to different species) occurring in an area.

Question 7.
Define communities. Who classified plant communities into hydrophytes, mesophytes and xerophytes?
Answer:

  1. Community : An assemblage of all populations (belonging to different species) occurring in an area is called community.
  2. Warming classified plant communities into hydrophytes, mesopytes and xerophytes.

Question 8.
Hydrophytes show reduced Xylem. Why? [Mar. ’20, ’18, ’17, ’15]
Answer:
In hydrophytes, the absorption of water takes place through all over the surface of the plant body. So xylem is reduced.

Short Answer Type Questions

Question 1.
What are hydrophytes ? Briefly discuss the different kinds of hydrophytes with examples. [Mar. ’15 – T.S.]
Answer:
Plants growing in water or in very wet conditions are called hydrophytes. Hydrophytes are 5 types based on the way of developing in water.

1. Free floating hydrophytes :
They float freely on water. They have no contact with soil. Ex : Pistia, Eichhornia.

2. Rooted hydrophytes with floating leaves :
Roots of these plants are fixed in mud. Leaves have long petioles and float on water surface. Ex : Nelumbo, Nymphaea.
TS Inter 1st Year Botany Study Material Chapter 13 Ecological Adaptations, Succession and Ecological Services 1
TS Inter 1st Year Botany Study Material Chapter 13 Ecological Adaptations, Succession and Ecological Services 2

3. Submerged suspended hydrophytes :
These are not rooted. But they are completely submerged and suspended in water.
Ex : Hydrilla, Ceratophyllum,

4. Submerged rooted hydrophytes :
They are rooted at the bottom of the pond and remain submerged in water.
Ex : Vallisneria, Potamogeton

5. Amphibious plants :
They live partly in air and partly in water.
Ex : Ranunculus, Limnophila

Some plants growing around the water bodies are occasionally touched by water currents. They also survive in dry periods.
Ex : Cyperus, Typha.

Question 2.
Enumerate the morphological adaptations of hydrophytes. [Mar. – 2018]
Answer:
Morphological adaptations in hydrophytes :

  1. Roots may be absent or poorty developed. In some plants submerged leaves compensate for roots.
  2. Root caps are usually absent. However in some amphibious plants which grow in mud, roots are well developed with distinct root caps. In some plants, root caps are replaced by root pockets.
  3. Roots, if present, are generally fibrous, adventitious, reduced in length, unbranched or poorly branched.
  4. Stem is long, slender and flexible.
  5. Leaves are thin, and either long and ribbon-shaped or long and linear or finely dissected. Floating leaves are large and flat with their upper surfaces coated with wax.

TS Inter 1st Year Botany Study Material Chapter 13 Ecological Adaptations, Succession and Ecological Services

Question 3.
List out the anatomical adaptations of hydrophytes. [Mar. ’14, ’13]
Answer:
Anatomical Adaptations:

  1. Cuticle is totally absent in the submerged parts of the plants. It may be present as very fine fiim on the surface of parts that are exposed to atmosphere.
  2. Thin walled epidermal cells help in absorption and assimilation by having chloroplasts.
  3. Stomata are totally absent in submerged forms or non-functional as in Potamogeton. The floating leaves of Nelumbo possess stomata only on the upper surface (epistomatous).
  4. Mechanical tissues like collenchyma and sclerenchyma are absent, because the plants are not exposed to stress and strain.
  5. Water is absorbed through all over the surface of plant body, so the xylem is poorly developed. But phloem is relatively better developed.
  6. Aerenchyma present in all parts of all hydrophytes, provides buoyancy.

Question 4.
Write a brief account on classification of Xerophytes. [Mar. ’20, ’17; May. ’17, ’14]
Answer:
Plants growing in water deficient or physiologically dry habitats are called xerophytes.

Xerophytes are 3 types based on their morphology, physiology and life cycle pattern.
TS Inter 1st Year Botany Study Material Chapter 13 Ecological Adaptations, Succession and Ecological Services 3

1. Ephemerals :
These are annuals which complete their life cycle within 6 – 8 weeks. They are also called drought evaders or drought escapers.
Eg : Tribulus.

2. Succulents :
These are drought avoiding plants. They absorb large amounts of water during rainy season and store in their parts in the form of mucilage. Those parts become fleshy and succulent. Stored water is used during dry periods.
Eg : (a) Root succulents – Ceiba parvifiora, Asparagus;
(b) Stem succulents – Opuntia;
(c) Leaf succulents – Bryophyllum, Aloe

3. Non succulents –
These are true xerophytes. These are perennial and withstand prolonged period of drought. Eg : Casuarina, Nerium.

Question 5.
Enumerate the morphological adaptations of xerophytes. [Mar. – 2019]
Answer:
Morphological adaptations in xerophytes :

  1. Roots are long with extensive branching spread over wide areas.
  2. Root hairs and root caps are very well developed.
  3. Stems are stunted, woody, hard and covered with thick bark.
  4. Stems are usually covered by hairs and or waxy coatings.
  5. Leaves are very much reduced, small, scale like and sometimes modified into spines to reduce the rate of transpiration.

TS Inter 1st Year Botany Study Material Chapter 13 Ecological Adaptations, Succession and Ecological Services

Question 6.
Give in detail the anatomical adaptations shown by xerophytes. [Mar. ’15 – A.P]
Answer:
Anatomical adaptations :

  1. Epidermis is covered by thick cuticle to reduce transpiration.
  2. Epidermal cells are thick walled and may have silica crystals.
  3. Multilayered (multiple) epidermis is present as leaves of Nerium.
  4. Leaves and stem of Calotropis consists of waxy coating.
  5. Stomata are generally confined to lower epidermis of leaves (hypostomatous) and present in pits (sunken stomata) lined with hairs. Eg : Nerium.
  6. Mesophyll in leaves is very well differentiated into palisade and spongy paren¬chyma.
  7. Mechanical tissues are very well developed.
  8. Vascular tissues are very well developed.

Question 7.
Define Plant succession. Differentiate primary and secondary successions.
Answer:
The gradual change in structure and composition of all communities constantly in response to the changing environment till it reaches equilibrium is called plant succession.

Primary successionSecondary succession
1. Succession that starts where no living organisms ever existed is called Primary succession.
Ex : Bare rocks
1. Succession that starts in the areas where somehow all living organisms that existed are lost is called secondary succession. Ex: Abandoned farmlands, burned forests
2. It occurs in biologically sterile area.2. It occurs in biologically fertile area.
3. It takes long time to reach the climax stage.3. It occurs faster than primary succession because of the presence of soil.

TS Inter 1st Year Botany Study Material Chapter 13 Ecological Adaptations, Succession and Ecological Services

Question 8.
Define ecosystem/ecological services. Explain in brief with regard to pollination.
Answer:

  1. Natural ecosystem performs fundamental life support services called ecosystem services or ecological services.
  2. Transfer of pollen grain from anther to stigma is called pollination.
  3. Most flowering plants require help from pollinators to produce fruits and seeds. So Pollinators play a significant role in the production of more food crops in the world.
  4. The most important pollinators for agricultural purpose is the honey bee.
  5. Predicting the effects of the loss of a particular pollinator is extremely difficult, but it is important to remember that no species exists in isolation.
  6. Each is part of an ecological web, and as we lose more and more pieces of that web, the remaining structure must eventually collapse.
  7. If pollination services are lost due to the loss of some species, then those remaining are unable to compensate the loss.
  8. Decline in pollinator activity disturbs the entire ecological system.

Question 9.
Write about the measures to be taken to sustain ecological functions.
Answer:

  1. Choose products produced with methods that conserve resources, minimize waste and reduce or eliminate environmental damage.
  2. Prefer products made with methods that reduce or eliminate the use of pesticides and artificial fertilizers.
  3. Reduce consumption and waste production.
  4. Support usage of renewable energy alternatives.
  5. Use public transit, cycle or walk to conserve natural resources and to reduce pollution and enjoy the health benefits.
  6. Participate in developing community garden and tree plantation programmes.
  7. Avoid the usage of pesticides and follow methods of natural pest control.
  8. Use native plants in the garden and provide habitat for wild life.

Question 10.
What measures do you suggest to protect the pollinators?
Answer:

  1. Creating own pollinator – frendly garden using a wide variety of native flowering plants. Encourage the planting of native flowers in open spaces and outside public buildings.
  2. Reducing the level of pesticides used in and around your home.
  3. Encouraging local clubs or school groups to build artificial habitats such as butterfly gardens, bee boards and bee boxes.
  4. Supporting agriculture enterprises with pollinator-friendly practices such as forms that avoid or minimize pesticide use.
  5. Encouraging government agencies to take into account the full economic benefits of wild pollinators when formulating policies for agriculture and other land uses.
  6. Stress the need to develop techniques for cultivating native pollinator species for crop pollination.

Long Answer Type Questions

Question 1.
Give an account of ecosystem services with reference to carbon fixation and oxygen release.
Answer:
Ecosystem Service – Carbon Fixation :
Trees are essential to all living organisms. Exchange of CO2 and O2 is done by photosynthesis. Forests provide a vast bank for CO2 and a huge amount of CO2 is deposited in its timber. This cuts down the CO2 concentration in atmosphere and plays an essential role in maintaining a dynamic balance between CO2 and O2 in atmosphere. So CO2 fixation has an obvious indirect economic value that can be estimated by taking account alternative methods of fixing CO2.

According to photosynthesis equation, to produce 180 g glucose and 193 g O2, plant will absorb 264 g CO2 and 108 gm water and consume 677.2 k.cal of solar energy. Then 180 g glucose can be transformed to 162 g polysaccharide inside the plant. Therefore, whenever plant produces 162 gm of dry organic water, 264 g CO2 will be fixed, i.e., production of every 1 gm dry organic matter can fix 1.63 g CO2.

The economic value of CO2 fixation can be estimated by the total fixed CO2 amount multiplied by a standard opportunity cost for per unit CO2 fixation.

Natural ecosystems may have helped to stabilize climate and prevent overheating of the Earth by removing more of the greenhouse gas, CO2 from the atmosphere. Ecosystem service-Oxygen release : Plants take CO2 and give CL during photosynthesis. The amount of O2 produced by a tree depends upon its age, health and also on the tree’s surroundings. According to Research findings, “a mature leafy tree produces as much as oxygen in a season as 10 people inhale in a year”.

Another quotes “A single mature tree can absorb CO2 at a rate of 48 lbs / year and release enough oxygen back into the atmosphere to support 2 human beings.”

Other quote “One acre of trees annually consumes the amount of carbon dioxide equivalent to that produced by driving an average car for 26000 miles. That same acre of trees also produces O2 for 18 people to breathe for a year.”

Submerged water plants release O2 and enrich dissolved oxygen in water.

Plants and Planktons are described as “the lungs of the world”.

Micro-organisms also produce O2 directly or indirectly. Cyanobacteria produce O2 directly; some bacteria indirectly.

For example, the degradation of organic compounds by bacteria can make the compounds capable of being used as a food source by another organism. This subsequent utilization can both consume and produce oxygen at various stages of the digestive process.

Intext Question Answers

Question 1.
Categorise the following plants into hydrophytes, halophytes, mesophytes and xerophytes and give reasons.
a) Salvinia b) Opuntia C) Rhizophora d) Mangifera
Answer:
a) Salvinia is a hydrophyte, it grows on the surface of water.
b) Opuntia is a xerophyte, grows in dry areas.
c) Rhizophora is a halophyte, it tolerates the salinities of the sea.
d) Mangifera is a mesophyte, it grows in habitats where water availability is normal.

Question 2.
In a pond, we see plants which are free-floating ; rooted-submerged ; rooted emergent ; rooted with floating leaves. Write the type of plants against each of them.
Answer:

Plant nameType
a) HydrillaSubmerged suspended hydrophyte
b) TyphaAmphibious plant
c) NymphaeaRooted with floating leaves
d) LemnaFree floating hydrophytes
e) VallisnariaSubmerged rooted hydrophyte

TS Inter 1st Year Botany Study Material Chapter 13 Ecological Adaptations, Succession and Ecological Services

Question 3.
Undertake the following a part of learning process :
a) Identify and assess ecological services found in your area.
b) Think of measures or means to sustain such ecological services.
c) Observe the type of plants or crops grown in your area.
d) Enumerate ecological services of your area.
e) Find out the ecological goods of natural forests commonly used in your area.
f) Observe the biotic agents of pollination for ornamental flowering plants and or agricultural crops in your locality.
Answer:
a) Ecological services :

  1. Purification of air and water
  2. Decomposition of water
  3. Detoxification of water

b) Measures to sustain Ecological services :

  1. Reduce consumption and waste production
  2. Avoid the usage of pesticides

c) Crops grown in our area :

  1. Paddy
  2. Maize
  3. Vegetables
  4. Blackgram

d) Ecological services:

  1. Purification of air and water
  2. Decomposition of wastes

e) Ecological goods:

  1. Clean air
  2. Fresh air
  3. Fibre
  4. Timber
  5. Medicines

f) Biotic agents of pollination :

  1. Insects
  2. Birds
  3. Animals like bats, snails, etc.

TS Inter 1st Year Botany Study Material Chapter 12 Histology and Anatomy of Flowering Plants

Telangana TSBIE TS Inter 1st Year Botany Study Material 12th Lesson Histology and Anatomy of Flowering Plants Textbook Questions and Answers.

TS Inter 1st Year Botany Study Material 12th Lesson Histology and Anatomy of Flowering Plants

Very Short Answer Type Questions

Question 1.
The transverse section of a plant material shows the following anatomical features –
a) the vascular bundles are conjoint, scattered and surrounded by a sclerenchymatous bundle sheaths.
b) phloem parenchyma is absent. What will you identify it as?
Answer:
Monocotyledonous stem with closed vascular bundles.

Question 2.
Why are xylem and phloem called complex tissues?
Answer:

  1. Complex tissues is made of more than one type of cells that work together as a unit.
  2. Xylem and phloem are made of more than one type of cells i.e., parenchyma, fibers etc.

Question 3.
How is the study of plant anatomy useful to us?
Answer:

  1. Anatomy is useful to known the internal structure of the plant. It is useful in classification of plants based on natural relations.
  2. It is useful to understand the plant functions, habitat of the plant and evolution of plants.

Question 4.
Protoxylem is the first formed xylem. If the protoxylem lies radially next to phloem, what kind of arrangement of xylem would you call it? Where do you find it?
Answer:

  1. Radial arrangement
  2. They are found in roots.

Question 5.
What is the function of phloem parenchyma?
Answer:
Phloem parenchyma stores food materials and other substances like resins, latex and mucilage.

Question 6.
a) What is present on the surface of the leaves which helps the plant to prevent loss of water but is absent in roots?
b) What is the epidermal cell modification in plants which prevents water loss?
Answer:
a) Cuticle
b) Bulliform cells is Isobilateral (monocotyledonous) leaf.

TS Inter 1st Year Botany Study Material Chapter 12 Histology and Anatomy of Flowering Plants

Question 7.
Which part of the plant would show the following?
a) Radial vascular bundle
b) Polyarch xylem
c) Well developed pith
d) Exarch xylem
Answe:
a) Radid vascular bundle – Root
b) Polyarch xylem – Monocot root
c) Well developed pith – Monocot root
d) Exarch xylem – Root

Question 8.
What are the cells that make the leaves cur! in plants during water stress? Give an example.
Answer:

  1. Large, colourless Bulliform ceils
  2. Ex : Monocot (Grass) leaves

Question 9.
What constitutes the vascular cambial ring?
Answer:

  1. Intrafascicular cambium and interfasicular cambium.
  2. Cambial ring is formed in dicot stem during secondary growth.

Question 10.
Give one basic functional difference between phellogen and phelloderm.
Answer:

  1. Phellogen (cork cambium) is a meristematic tissue, formed from primary cortex.
  2. Phelloderm (secondary cortex) is a permanent tissue formed by inner cells that cuts off from phellogen.

TS Inter 1st Year Botany Study Material Chapter 12 Histology and Anatomy of Flowering Plants

Question 11.
If one debarks a tree, what parts of the plant are removed?
Answer:

  1. Periderm and secondary phloem are removed.
  2. All those tissues exterior to the vascular cambium.

Short Answer Type Questions

Question 1.
State the location and function of different types of Meristems. [Mar. ’20, ’17, ’15, ’13]
Answer:
Based on location. Meristems are three types.
TS Inter 1st Year Botany Study Material Chapter 12 Histology and Anatomy of Flowering Plants 1

1. Apical meristems :
They are present at the growing tips of roots, stem,, branches etc., They help in the linear growth of the plant body. These are primary meristems because they appear early in life and contribute to the formation of primary plant body.

2. Intercalary meristem :
They are found in between permanent tissues. This meristem is separated from the apical meristems during the course of plant growth. They help in linear growth of the stem and leaves. Growth of flowers and fruits after their initiation at the apex also occurs due to this meristems. They are active only for a short period. These are also primary meristems.
Eg : Meristems seen at the base of internodes and leaf bases of monocotyledons (particularly grasses).

3. Lateral meristems :
They are found at the lateral sides of the plant body. The cells divide periclinally and increase the thickness of the organs like stem and root. These are secondary meristems.
Eg : Vascular cambium that help in secondary growth by producing secondary xylem and secondary phloem; phellogen (cork cambium) that helps in the formation of periderm.

Question 2.
Cut a transverse section of young stem of a plant from your garden and observe it under the microscope. How would you ascertain whether it is a monocot stem or a dicot stem ? Give reasons.
Answer:

Dicot StemMonocot Stem
1. In Epidermis, multicellular hairs trichomes are present.1. Trichomes are absent.
2. Hypodermis is collenchymatous.2. Hypodermis is sclerenchymatous.
3. Endodermis, pericycle, medulla medullary rays are present.3. Endodermis, pericycle, medulla, medullary rays are absent.
4. Vascular bundles are few in number and arranged as a circular ring (eustele).4. Vascular bundles are numerous and arranged in a scattered manner (atactostele).
5. Vascular bundle is top shaped or wedge shaped.5. Vascular bundle is oval in shape.
6. Vascular bundle is not enclosed by a bundle sheath.6. Vascular bundle is enclosed by fibrous sheath, (fibrovascular bundle)
7. Open vascular bundle.7. Closed vascular bundle.
8. Xylem vessels are more in number.8. Xylem vessels are few in number.
9. Protoxylem lacunae are absent.9. Protoxylem lacunae are present.
10. Vessels are in serial order.10. Vessels are in Y shape.
11. Phloem parenchyma is present.11. Phloem parenchyma is absent.

TS Inter 1st Year Botany Study Material Chapter 12 Histology and Anatomy of Flowering Plants

Question 3.
What is periderm? How does periderm formation take place in the dicot stems? [Mar. – 2018]
Answer:
Phellogen, Phellem and Phelloderm are collectively known as periderm.

  1. Due to the formation of more secondary vascular tissues a pressure is exerted on the epidermis causing its rupture. So a secondary protective layer (periderm) is formed.
  2. Parenchyma cells in middle or inner cortex dedifferentiate into a ring of secondary meristem. This is called cork cambium or phellogen. It cuts off new cells on both sides.
  3. Tissue produced on outside is called cork tissue or phellem. Tissue produced inside is called secondary cortex or phelloderm.
  4. The phellogen, phellem, and phelloderm together constitute periderm.
    TS Inter 1st Year Botany Study Material Chapter 12 Histology and Anatomy of Flowering Plants 2
  5. To facilitate gaseous exchange in the cork tissue certain bulged lens shaped structures are formed. They are called lenticels.

Question 4.
A transverse section of the trunk of a tree shows concentric rings which are known as annual rings. How are these rings formed? What is the significance of these rings?
Answer:

  1. In temperate and cold regions, the activity of cambium is influenced by seasonal variations.
    TS Inter 1st Year Botany Study Material Chapter 12 Histology and Anatomy of Flowering Plants 3
  2. In favourable conditions growth will be active. So plants require large amounts of water and minerals. During unfavourable conditions plants are less active.
  3. In spring, wood formed shows more number of xylem vessels having wide lumens. This is called spring wood or early wood.
  4. During autumn, wood formed shows less number of xylem vessels with narrow lumens. This is called autumn wood or late wood.
  5. Spring wood and autumn wood appear alternately in the form of circles in the T.S. of a tree trunk. These are called Growth rings or annual rings.
  6. By counting the number of annual rings the age of tree can be estimated approximately.

Question 5.
What is the difference between lenticels and stomata? [Mar. – 2019, ’15, May ’17]
Answer:
Lenticels :
Lens shaped openings in the cork of woody trees are called lenticels. They show closely arranged parenchymatous cells. The lenticels permit the exchange of gases between the outer atmosphere and the internal tissues of the woody organs. There is no opening and closing mechanism.

Stomata :
Stomata are present in the upper epidermis and lower epidermis of leaves. They help in exchange of gases. In dicot leaves, on either side of stomata kidney shaped guard cells are present. In monocot leaves, dumb bell shaped guard cells are present. Guard cell contains chloroplast. They help in opening and closing of stomata. Stomata helps in the gaseous exchange and also promote transpiration.

Question 6.
Write the precise function of
a) Sieve tube
b) Interfasicular cambium
c) Collenchyma
d) Sclerenchyma
Answer:
a) Sieve tube :
The functions of sieve tube are controlled by the nucleus of companion cells. The companion cells help in maintaining the pressure gradient in the seive tubes. It transports food materials from leaves to other parts.

b) Interfasicular cambium :
The cells of medullary cells adjoining the intrafascicular cambium becomes meristematic and forms interfasicular cambium.
Thus a continuous ring of vascular cambium is formed.

c) Collenchyma :
The collenchyma cells which contain chloroplast are green in colour. Photosynthetic in function. Intercellular spaces are absent as the corners are thickened with pectin. So they provide tensile mechanical strength. It helps in movement of young stem, petiole of leaf, pedicel of flower.

d) Sclerenchyma :
They are dead cells. Cell walls are thickened with legnin. Intercellular spaces are absent. So they, give mechanical strength to organs.

Question 7.
The stomatal pore is guarded by two kidney shaped guard cells. Name the epidermal cells surrounding the guard cells. How does a guard cell differ from an epidermal cell? Use a diagram to illustrate your answer.
Answer:

  • The epidermal cells surrounding the guard cells are called subsidary cells or accessory cells.
  • The stoma is bounded by two kidney shaped guard cells in dicots and dumbbell-shaped guard cells in monocots.
  • Unlike that of other epidermal cells, guard cells posses chloroplast.
  • The wall of the guard cells towards the stomatal pore is thick, while the outer wall is thin. The stoma, guard cell and subsidiary cells together constitutes stomatal complex.

TS Inter 1st Year Botany Study Material Chapter 12 Histology and Anatomy of Flowering Plants 4

Question 8.
Point out the differences in the anatomy of leaf of peepal [Ficus religiosa] and maize [Zea mays]. Draw the diagrams and label the differences.
Answer:

Dicot leaf Eg : PeepalMonocot Leaf Eg : Maize
1. Stomata are more on the lower epidermis.1. Stomata are in equal numbers on both sides.
2. Bulliform cells are absent.2. Bulliform cells are present on upper epidermis.
3. Mesophyll is differentiated into palisade and spongy tissues.3. Mesophyll is undifferentiated.
4. Bundle sheath extensions are generally parenchymatous.4. Bundle sheath extensions are sclerenchymatous.

TS Inter 1st Year Botany Study Material Chapter 12 Histology and Anatomy of Flowering Plants 5
TS Inter 1st Year Botany Study Material Chapter 12 Histology and Anatomy of Flowering Plants 6

Question 9.
Cork cambium forms tissues that form the cork. Do you agree with this statement? Explain.
Answer:
Yes, Cork cambium or phellogen is a secondary meristematic tissue. It has the capability to divide. It divides and forms new cells on both sides. The tissue produced outside is called cork tissue or phellem. The tissue produced towards in innerside in secondary cortex or phelloderm.

TS Inter 1st Year Botany Study Material Chapter 12 Histology and Anatomy of Flowering Plants

Question 10.
Name the three basic tissue systems in the flowering plants. Give the tissue names under each system.
Answer:
The three basic tissue systems in the flowering plants are

  1. Epidermal tissue sytem
  2. The ground or fundamental tissue system
  3. The vascular or conducting tissue system.

1. Epidermal tissue system consists of parenchymatous tissue. They are epidermis, stomata and out growths.
2. The ground or fundamental tissue system consists of simple tissues such as parenchyma, collenchyma and sclerenchyma.
3. The vascular or conducting tissue system consists of complex tissues, the phloem and the xylem.

Long Answer Type Questions

Question 1.
Explain the process of secondary growth in the stems of woody angiosperms with the help of schematic diagrams. What is its significance?
Answer:
Growth of stem or root in thickness due to the formation of secondary tissues due to the activity of primary and secondary meristems is called secondary growth.

Changes during secondary growth of a dicot stem are divided into two groups. They are
I. Intrastelar secondary growth.
II. Extrastelar secondary growth.

I. Intrastelar secondary growth :
The changes that occur inside the stele are called Intrastelar secondary growth. They are

A) Formation of vascular cambial ring :

1. Indicot stem, vascular bundles are in a circular ring. They are open type with fascicular cambiam.
2. Parenchyma in medullary rays dedifferentiate into secondary meristem connecting fascicular cambiam. These are called interfascicular cambiam.
3. Fascicular and interfascicular cambia fuse to form vascular cambial ring.

B) Activity of vascular cambium :
4. Vascular cambium has 2 types of initials.
a) Fusiform initials :
They give rise to secondary xylem towards centre and secondary phloem to outside.

b) Ray initials :
They produce phloem rays towards outside and xylem rays towards inside.
5. More secondary xylem is formed than secondary phloem.
6. Secondary xylem is called wood and secondary phloem bast.
TS Inter 1st Year Botany Study Material Chapter 12 Histology and Anatomy of Flowering Plants 7

7. As stem increases in thickness primary phloem and primary xylem are crushed and removed.
8. Secondary xylem has vessels, fibres and xylem perenchyma. Vessels are pitted.
9. Secondary phloem has sieve tubes, companion cells, fibres and phloem parenchyma.
10. Xylem ray and phloem ray are also called vascular rays. They are helpful in lateral conduction and storage.

II. Extrastelar secondary growth :
The changes which occur outside the stele are called Extrastelar secondary growth.

  1. Due to the formation of more secondary vascular tissues a pressure is exerted on the epidermis causing its rupture. So a secondary protective layer (periderm) is formed.
  2. Parenchyma cells in middle or inner cortex dedifferentiate into a ring of secondary meristem. This is called cork cambium or phellogen. It cuts off new cells on both sides.
    TS Inter 1st Year Botany Study Material Chapter 12 Histology and Anatomy of Flowering Plants 2
  3. Tissue produced on outside is called cork tissue or phellem. Tissue produced inside is called secondary cortex or phelloderm.
  4. The pheilogen, phellem, and phelloderm together constitute periderm.
  5. To facilitate gaseous exchange in the cork tissue certain bulged lens shaped structures are formed. They are called lenticels.

Question 2.
Draw illustrations to bring out the anatomical differences between
a) Monocot root and Dioofc mot
b) Monocot stem and Dirot stem
Answer:
TS Inter 1st Year Botany Study Material Chapter 12 Histology and Anatomy of Flowering Plants 8
TS Inter 1st Year Botany Study Material Chapter 12 Histology and Anatomy of Flowering Plants 9

Question 3.
What are simple tissues? Describe various types of simple tissues.
Answer:
The tissues which are made of only one type of cells are called simple tissues.

The various types of simple tissues are parenchyma, collenchyma and sclerenchyma.

Parenchyma :

  1. The Parenchyma is a living tissue. It occupies a major part of the plant body. So it is known as fundamental tissue or ground tissue.
  2. The cells are isodiamtetric. They may be spherical, oval round, polygonal or elongated in shape.
  3. Cell walls are thin, made up of cellulose.
  4. Intercellular spaces may be present or absent.
  5. Parenchyma performs functions like photosynthesis, storage, and secretion.

Collenchyma :

  1. Collenchyma is a living mechanical tissue.
  2. It is present below the epidermis in dicot plants.
  3. It is present as a continuous hypodermal ring (Eg : Helianthus annus) or as a discontinuous ring (cucurbita)
  4. Corners are thickened due to cellulose, hemicellulose and pectin.
  5. Intercellular spaces are absent.
  6. It provides mechanical support to the growing parts of the plant parts such as young stem, petiole, pedicle etc.
  7. In the cytoplasm, if chloroplast is present, photosynthetic in function.

Scelerenchyma :

  1. Sclerenchyma is a dead mechanical tissue.
  2. Cells are long and narrow.
  3. Cell walls are thickened with legnin with pits.
  4. They are dead cells. Protoplast is absent.
  5. Basing upon the structure, origin and development sclerenchyma are two types – fibres, sclereids.
  6. Fibres are thick walled, elongated and pointed cells.
  7. Sclereids (stone cells) are spherical, oval or cylindrical shape.
  8. Cell walls are highly thickened, lumen is very narrow.
  9. Sclereides are found in fleshy fruits like guava, pear and sapota. Seed coat of legumes, leaves of tea, fruit wall of nuts etc.
  10. Their main function is to give mechanical support.

TS Inter 1st Year Botany Study Material Chapter 12 Histology and Anatomy of Flowering Plants

Question 4.
What are complex tissues? Describe various types of complex tissues.
Answer:
Tissues which are made up of more than one type of cells and work together as a unit are called complex tissues.

Xylem and phloem are complex tissues.

Xylem :
The main function of xylem is conducting water and minerals from roots to the stem and leaves.

Xylem also provides mechanical strength to plant parts. Xylem consists of (1) Tracheids (2) Vessels (3) Xylem fibres (4) Xylem Parenchyma.

Tracheids :
Tracheids are elongated or tube like cells with thick and lignified walls and tapering ends. These are dead cells without protoplasm. Its main function is water transport.

Vessels :
Presence of vessels is an important character found in angiosperms. Vessels are absent in Gymnosperms. Vessels are dead cells without protoplasm. The cells are elongated or tube like cell thickened with lignin. Its main function is water transport.

Xylem fibres :
These are highly thickened with legnin with narrow central lumen. They may be septate or aseptate.

Xylem Parenchyma :
These are living cells. Cell walls are thickened with cellulose. They store food materials like starch, fats, and tannins. The ray parenchyma cell helps in radial conduction of water.

Primary xylem is of two types – protoxylem and metaxylem.

The first formed primary xylem elementsare called protoxylem. The laterformed primary xylem are called metaxylem.

In stems, protoxylem is towards centre and metaxylem is towards periphery. It is called endarch.

In roots, protoxylem is towards periphery and metaxylem is towards centre. It is called exarch.

Phloem :
The main function of phloem transports food materials usually from leaves to the other parts of the plant body.

Phloem contains (1) Sieve tube elements (2) Companion cells (3) Phloem parenchyma (4) Phloem fibres.

Sieve tube elements :
These are long, tube like structures arranged longitudinally and are associated with companion cells. Their end walls are perforated in sievelike manner to form sieve plates.

A mature sieve tube element possesses a peripheral cytoplasm and a large vacuole but lacks a nucleus. The function of sieve tubes are controlled by the nucleus of companion cells.

Companion cells :
These are specialized parenchymatous cells which are closely associated with sieve tube elements. Both are connected by pit fields present between their common longitudinal walls.

Phloem parenchyma :
These are parenchyma cells in phloem with tapering cylindrical cells which have dense cytoplasm and nucleus. They store food materials and other substances like resin, latex etc.

Phloem fibres (bast fibres) :
These are sclerenchymatous cells. The cell wall is thick. At maturity they lose their protoplasm and become dead.

Question 5.
Describe the internal structure of dorsiventral leaf with the help of labelled diagram.
Answer:
Transverse section of a dicot leaf or dorsiventral leaf shows three parts – epidermis, mesophyll and vascular bundles.

I. Epidermis :
1) It is the outermost layer of leaf with one cell thickness. Cells are barrel shaped. They are arranged compactly without intercellular spaces.
2) Epidermis present on upper (adaxial) side is called upper epidermis. Epidermis on lower (abaxial) side is called lower epidermis.
TS Inter 1st Year Botany Study Material Chapter 12 Histology and Anatomy of Flowering Plants 10
3) The outer surface of epidermis is covered by a waxy layer called Cuticle.
4) Epidermis shows multicellular hairs.
5) Stomata are present. They are more on lower surface than upper surface.
6) Epidermis gives protection to the inner tissues. Cuticle regulates transpiration. Stomata help in exchange of gases.

II. Mesophyll :

  1. Ground tissue present in between the two epidermal layers is called Mesophyll. It is Chlorenchymatous.
  2. In dorsiventral leaf, mesophyll is differentiated into two parts – palisade parenchyma and spongy parenchyma.
  3. Palisade parenchyma is found beneath the upper epidermis. Elongated and columnar cells are arranged in 1 – 3 rows. Intercellular spaces are narrow. Cells have large number of chloroplasts nearer to the cell wall. (So upper side of leaf is dark green in colour). Palisade tissue is mainly concerned with assimilation of carbohydrates.
  4. Spongy parenchyma is found towards the lower epidermis. It has 3-5 rows of irregularly shaped and loosely arranged cells. Intercellular spaces are large. Air cavities are found below the stomata. Cells have less number of chloroplasts. (So lower side of leaf is pale green in colour) Spongy Parenchyma facilitates gaseous exchange. They also help in the synthesis of food materials.

III. Vascular Bundle :

  1. Vascular bundles are extended in the mesophyl! in the form of veins.
  2. Vascular bundles are bigger at the base of the leaf blade and gradually becomes smaller towards margins and apex.
  3. Vascular bundles are conjoint, collateral and closed. Xylem is present towards upper side and phloem towards lower side.
  4. Vascular bundles help in conduction of water, mineral salts and food materials.
  5. They also provide mechanical strength to the leaf.
  6. Each vascular bundle is enclosed by a layer of special mesophyll cells arranged compactly. This layer is called Bundle sheath or Border parenchyma.
  7. Ceils of bundle sheath divide and extend towards both epidermal layers. These are called bundle sheath extensions. They help in the conduction of food materials from mesophyll cells to vascular bundles.

TS Inter 1st Year Botany Study Material Chapter 12 Histology and Anatomy of Flowering Plants

Question 6.
Describe the internal structure of an isobilateral leaf with the help of labelled diagram.
Answer:
Transverse section of monocot or isobilateal leaf shows three parts – epidermis, mesophyll and vascular bundles.
TS Inter 1st Year Botany Study Material Chapter 12 Histology and Anatomy of Flowering Plants 11

I. Epidermis :

  1. This is the outermost layer on both sides of the leaf. Cells are one cell in thickness. They are barrel shaped and closely packed without intercellular spaces.
  2. It is covered by a waxy layer called cuticle.
  3. Epidermis on adaxial (upper) surface is called upper epidermis. Epidermis on abaxial (lower) surface is called lower epidermis.
  4. Hairs are absent. Stomata are present on both sides in equal numbers.
  5. In grasses specialised cells are present in upper epidermis. They are called bulliform cells or motor cells. They are thin walled and filled with water. They help in rolling and unrolling of the leaf.
  6. Epidermis gives protection to inner tissues. Cuticle regulates transpiration. Stomata help in exchange of gases.

II. Mesophyll :

  1. Ground tissue present between two epidermal layers is called mesophyll. It is chlorenchymatous.
  2. Mesophyll is undifferentiated.
  3. Cells have chloroplasts and perform assimilation of carbohydrates.
  4. Sometimes patches of sclerenchyma are found beneath the epidermis. They provide mechanical strength.

III. Vascular Bundies :

  1. Vascular bundles are present in the mesophyll in the form of veins.
  2. Vascular bundles are conjoint, collateral and endarch.
  3. Xylem is present on the upper side and phloem is present on the lower side.
  4. Veins help in conduction of water, mineral salts and food materials. They also provide mechanical strength.
  5. Each vascular bundle is enclosed by a layer of special mesophyll cells called bundle sheath or border parenchyma.
  6. Cells present on eitherside of vascular bundles towards upper and lower epidermis are called bundle sheath extensions. In many monocots, they are sclerenchymatous and provide mechanical support.

Question 7.
Distinguish between the following :
a) Exarch and endarch condition of protoxylem
b) Stele and vascular bundle,
c) Protoxylem and metaxylem
d) Interfasicuiar cambium and intrafasicular cambium
e) Open and closed vascular bundles
f) Stem hair and root hair
g) Heart wood and sap wood,
h) Spring wood and autumn wood.
Answer:
a) In roots, the protoxylem lies towards periphery and metaxylem lies towards the centre. Such type of protoxylem is called Exarch, In stems, the protoxylem lies towards the centre (pitch) and the metaxylem lies towards the periphery of the organ. Such type of protoxylem is called Endarch.

b) Stele :
Stele is the central conducting cylinder. Generally it may have pericycle, vascular bundle, medulla and conjunctive tissue or medullary rays.

Vascular bundles :
Xylem and phloem are present in vascular bundles. Xylem conducts water and phloem conducts food materials.

c) Protoxylem :
The first formed primary xylem elements are called protoxylem. Metaxylem : Later formed primary xylem elements are called metaxylem.

d) Intrafasicular cambium :
Cambium present between primary xylem and primary phloem is called Intrafasicular cambium. It is present inside vascular bundle (Intra = Inside; fasicular = vascular bundle)

Interfasicuiar cambium :
The cells in medullary rays become meristematic and forms interfasicuiar cambium (Inter = in between; fasicular = vascular bundle)

e) Open vascular bundle :
The vascular bundle which have cambium betwen xylem and phloem are called open vascular bundle.

Closed vascular bundle :
In these vascular bundles cambium is absent between xylem and phloem.

f) Stem hair and root hair :
Stem hair :
Multicellular hairs present on the stem are stem hair or trichomes. Their main function is to prevent the entry of pathogens.

Root hair :
Unicellular hairs present on the root are root hair. Their main function is absorption of water.

g) Heart wood and sap wood :
Heart wood :
The dark brown coloured central part of secondary xylem comprising of dead elements with highly lignified walls is called heart wood. It is infiltrated with various organic compounds like tannins, resins, oils, gums, aromatic substances and essential oils. The heart wood does not conduct water but it gives mechanical support to the stem.

Sap wood :
The peripheral region of the secondary xylem is lighter in colour and is known as the sap wood. It conducts water and minerals from root to leaf.

h) Spring wood and Autumn wood :

Spring woodAutumn wood
1. It is produced during spring (favourable) season.1 It is produced during autumn (unfavourable) season.
2. Xylem vessels have wide lumens.2. Xylem vessels have narrow lumens.
3. More number of xylem vessels are produced.3. Less number of xylem vessels are produced.

TS Inter 1st Year Botany Study Material Chapter 12 Histology and Anatomy of Flowering Plants

Question 8.
What is stomata! apparatus? Describe the structure of stomata with a labelled diagram.
Answer:
Structure of stomata :

  1. Tiny pores in the epidermis of young aerial parts of the plant are called stomata. Stomata are more abundant in the leaf epidermis.
  2. The stoma is bounded by two kidney shaped guard cells in dicots and dumbbellshaped guard cells in monocots. (Eg : grasses).
  3. Unlike that of other epidermal ceils, guard cells posses chloroplasts. The wall of the guard cell towards the stomata! pore is thick, while the outer wall is thin.
  4. Epidermal cells surrounding guard cells are called subsidiary or accessory cells. They differ from other epidermal cells in their shape and position.
  5. The stoma is followed by an air cavity called substomatal cavity in the mesophyll.
  6. The stoma, guard cells and subsidiary cells together constitutes stomatal apparatus.

TS Inter 1st Year Botany Study Material Chapter 12 Histology and Anatomy of Flowering Plants 12

Question 9.
Describe the T.S of a dicot stem. [Mar. ’17 – A.P. ; Mar. ’15 – T.S ; Mar. ’13]
Answer:
TS Inter 1st Year Botany Study Material Chapter 12 Histology and Anatomy of Flowering Plants 13
Internal structure of primary dicot stem (Helianthus)
A) Ground plan B) Sector enlarged

Internal structure of young dicot stem – Ex : Helianthus annuus

Transverse section of a dicot stem shows three distinct zones – epidermis, cortex and stele.

I. Epidermis :

  1. The outermost layer in young dicot stem is called epidermis. It is one cell in thickness.
  2. Cells are tubular or rectangular. They are arranged compactly without intercellular spaces.
  3. Outer surface of epidermis is covered by a waxy subtance called cutin. This layer is called cuticle.
  4. Minute pores found in the epidermis are called stomata.
  5. Multicellular hairs developing on the epidermis are called trichomes.
  6. Epidermis gives protection to the inner tissues.
  7. Stomata facilitates exchange of gases and promotes transpiration.
  8. Cuticle and trichomes check transpiration. They also protect the stem from high temperature.
  9. Trichomes also help in preventing the entry of pathogenic micro-organisms.

II. Cortex :
It is extrastelar ground tissue. It shows three subzones – hypodermis, general cortex and endodermis.
A) Hypodermis :

  1. The layer present below the epidermis is called hypodermis.
  2. It consists of 3 – 6 layers of Collenchyma.
  3. Cells are arranged compactly without intercellular spaces. They show excessively thickened corners.
  4. Hypodermis helps in providing tensile strength to the stem.
  5. Hypodermis also helps in production of food materials by having chloroplasts.

B) General Cortex :

  1. It is found below the hypodermis.
  2. It consists of 5-10 rows of parenchyma.
  3. Cells are isodiametric or oval or spherical.
  4. Resin or latex ducts may be present in it.
  5. Outer layers of cells have chloroplasts and perform assimilation of food materials.
  6. Inner layers are concerned with storage of food.

C) Endodermis :

  1. It is the innermost layer of cortex.
  2. It is in one layer with barrel shaped, compactly arranged cells.
  3. Radial and transverse walls show casparian bands.
  4. Endodermal cells store starch grains. So it is known as starch sheath.

III. Stele :
Central conducting cylinder is called stele. It occupies major portion of stem. It shows four parts.
A) Pericycle :

  1. It is the outermost layer of stele.
  2. It lies between endodermis and vascular bundles.
  3. It has alternate patches of sclerenchyma and parenchyma.

B) Vascular Bundles :

  1. Each vascular bundle is wedge or top shaped.
  2. Limited number of vascular bundles are arranged in the shape of a circular ring. Such arrangement is called Eustele.
  3. In each vascular bundle the phloem is present outside and xylem towards inside on the same radius. So vascular bundle is conjoint and collateral.
  4. Meristematic tissue is present in between xylem and phloem. It is called fascicular cambium. Vascular bundle with cambium is called open type.
  5. Xylem is endarch (protoxylem towards centre).
  6. Xylem has vessels and xylem parenchyma. Tracheids and fibres are also present. Xylem conducts water and salts.
  7. Phloem has sieve tubes, companion cells, phloem parenchyma and fibres. It conducts food materials.

C) Medulla :
Central part of stele is called medulla. It is filled with parenchyma. It is well developed and extensive. It stores food materials.

D) Medullary rays :
Medulla extends to the periphery in between the vascular bundles forming medullary rays. Parenchymatous cells are living, thin walled and elongate radially.

Medullary rays connect stele and cortex. They hlep in lateral conduction.

TS Inter 1st Year Botany Study Material Chapter 12 Histology and Anatomy of Flowering Plants

Question 10.
Describe the T.S of monocot stem. [Mar. ’15 – A.P.]
Answer:
Internal structure of Monocot Stem :
The anatomy of Monocot stem shows four distinct parts – Epidermis, hypodermis, ground tissue, vascular bundles. A distinct cortex is absent. Endodermis, pericycle, medulla, medullary rays are absent.

I. Epidermis :

  1. The outermost layer is called epidermis. It is made up of living, rectangular or tabular cells. They are arranged compactly without intercellular spaces.
  2. A waxy layer is deposited on the outer surface of epidermis. This is called cuticle.
  3. Trichomes are absent. Numerous stomata are found in the epidermis.
  4. Epidermis gives protection to inner tissues. Stomata help in exchange of gases. Cuticle prevents evaporation of water.

TS Inter 1st Year Botany Study Material Chapter 12 Histology and Anatomy of Flowering Plants 14

II. Hypodermis :
It is present beneath the epidermis. It is made up of 1 – 4 rows of thick walled sclerenchymatous fibres. Intercellular spaces are absent. It gives mechanical strength to the stem.

III. Ground Tissue :

  1. Tissue next to the hypodermis filling the remaining part of the stem (except vascular bundles) is called ground tissue.
  2. It is parenchymatous.
  3. Cells are thin walled with or without chloroplasts. They are loosely packed with intercellular spaces.
  4. It is mainly concerned with synthesis and storage of food materials.

IV. Vascular Bundles :

  1. Numerous bundles are irregularly scattered in the ground tissue. Such an arrangement is called Atactostele.
  2. Inner bundles are bigger. Peripheral bundles are small in size. They are oval in shape.
  3. Each vascular bundle is enclosed by a sclerenchymatous sheath. So it is called fibro vascular bundle.
  4. Each bundle has phloem towards outside and xylem towards inside of the bundle on the same radius. So it is described as conjoint and collateral.
  5. Vascular bundles are concerned with conduction of water, salts and food materials.
  6. Cambium is absent between xylem and phloem. So the vascular bundle is closed type.
  7. Xylem is endarch (protoxylem towards centre). Xylem consists of tracheids, vessels, fibres and xylem parenchyma.
  8. Xylem is arranged in the shape of Y. Out of four xylem vessels, two are metaxylem and two are protoxylem vessels.
  9. One or two protoxylem vessels are crushed forming lysigenous cavity. It is called protoxylem lacuna. It stores water.
  10. Phloem has sieve tubes and companion cells. Phloem parenchyma is absent.

TS Inter 1st Year Botany Study Material Chapter 12 Histology and Anatomy of Flowering Plants 15

Question 11.
Describe the internal structure of a dicot root. [Mar. – 2018, May ’14]
Answer:
Internal structure of primary dicot root has three zones- epidermis, cortex, stele. Cortex is bigger than stele.

I. Epidermis :

  1. It is the outermost layer of thin walled rectangular living cells arranged compactly without intercellular spaces.
  2. Cuticle and stomata are absent.
  3. Some epidermal cells (trichoblasts) produce tubular extensions called root hairs. Cells giving rise to root hairs are smaller than other epidermal cells. Epidermis of root is called rhizodermis or piliferous layer or epiblema due to the presence of root hairs.
  4. Root hairs absorb capillary water. The epidermis gives protection to the inner tissues.

TS Inter 1st Year Botany Study Material Chapter 12 Histology and Anatomy of Flowering Plants 16

II. Cortex :
Ground tissue system extending from epidermis to stele is called cortex. It is differentiated into three parts.
A) Exodermis :

  1. The outermost layer of cortex with 2 or 3 rows of suberised thick walled cells is called exodermis.
  2. It acts as a protective layer when epidermis is removed.
  3. It prevents the exit of water from cortex.

B) General cortex :

  1. It is present beneath the exodermis.
  2. It has several layers of loosely arranged thin walled parenchyma. Intercellular spaces are present.
  3. Cells store food materials.
  4. General cortex helps in the lateral conduction of water from epidermis to xylem vessels.

C) Endodermis :

  1. It is the innermost layer of cortex.
  2. It is made up of single layer of barrel shaped cells.-
  3. Radial and transverse walls of endodermal cells show thickenings due to deposition of lignin and suberin. These are called casparian thickenings. It is characteristic feature of endodermis.
  4. Endodermal cells present opposite to protoxylem are thin walled without casparian strips. These cells are called passage cells.
  5. Passage cells help in the entry of water and salts from cortex into stele.

III. Stele :
The central conducting cylinder is called stele. It shows three parts.
A) Pericycle :
It is the outer layer of the stele. It is uniseriate with thin walled rectangular parenchymatous cells. Pericycle gives rise to lateral roots. It also helps in secondary growth.

B) Vascular bundles :

  1. Xylem and phloem are arranged alternately on separate radii. So vascular bundles are separate or radial. Xylem and phloem conduct water and food materials respectively.
  2. Protoxylem is towards pericycle and metaxylem towards centre. So xylem is exarch.
  3. Xylem is variable from monarch to octarch (xylem groups 1-8) usually tetrarch (4 xylem groups alternating with 4 phloem bundles). Monarch – Trapa, Tetrarch – Gossypium. Octarch – Castanea.
  4. Cambium is absent.
  5. Parenchyma tissue extending between xylem and phloem strands is called conjunctive tissue. It helps in the storage of food materials.

C) Pith or Medulla :
The central portion of stele is called medulla or pith. It may be completely absent in dicot root. When it is present, it is parenchymatous. It helps in the storage of food and water.

TS Inter 1st Year Botany Study Material Chapter 12 Histology and Anatomy of Flowering Plants

Question 12
Describe the internal structure of a monocot root. [Mar. ’20, May. ’17]
Answer:
Internal structure of monocot root is differentiated into three zones – epidermis, cortex and stele.

I. Epidermis:

  1. The outermost layer is called epidermis. Cells are living, rectangular, thin walled. They are compactly arranged without intercellular spaces.
  2. Cuticle and stomata are absent.
  3. Some cells (trichoblasts) show tubular extensions – root hairs. Epidermis of root having root hairs is called epibiema or piliferous layer or rhizodermis.
  4. Root hairs help in the absorption of capillary water form the soil. Epidermis gives protection to the inner tissues.

II. Cortex :
This is the ground tissue system extending from epidermis to stele. It is differentiated into three parts.
A) Exodermis :

  1. It is the outermost layer of the cortex with 2 or 3 rows of cells. Cell walls are thick and suberised.
  2. It acts as a protective layer when epidermis is removed.
  3. It prevents the exit of water from cortex.

B) General cortex :

  1. It is present beneath the exodermis.
  2. It has several layers of loosely arranged thin walled parenchyma. Intercellular spaces are present,
  3. Cells store food materials.
  4. General cortex helps in the lateral conduction of water from epidermis to xylem vessels.

C) Endodermis :

  1. It is the innermost layer of the cortex.
  2. It is made up of a single layer of barrel shaped cells.
  3. Radial and transverse walls of endodermal cells show thickenings due to the deposition of lignin and suherin. These are called Casparian thickenings.
  4. Endodermal cells present opposite to protoxylem are thin walled without casparian strips. These cells are called passage cells.
  5. Passage cells help in the entry of water and salts from cortex into stele.

III. Stele :
The central conducting cylinder is called stele. It is very prominent and bigger in size. It shows three parts,
A) Pericycle :

  1. It is the outermost layer of the stele. It is uniseriate with thin walled parenchymatous cells.
  2. Pericycle gives rise to lateral roots.
  3. In old roots it becomes sclerenchymatous and gives mechanical strength.

B) Vascular bundles :

  1. Xylem and phloem are arranged alternately on separate radii. So vascular bundles are radial or separate.
  2. Protoxylem is towards pericycle and metaxylem towards centre. So xylem is exarch.
  3. Xylem is polyarch (numerous xylem groups).
  4. Cambium is absent.
  5. Xylem is concerned with conduction of water and salts. Phloem conducts organic solutes.
  6. Parenchyma tissue extending between xylem and phloem strands is called conjunctive tissue. Cells are rarely thick walled. It is helpful in storage of food and provides mechanical strength.

C) Medulla or pith – The central part of stele is called medulla or pith. It is conspicuous. It is parenchymatous. It helps in storage of food. In some plants cell walls are lignified providing mechanical strength.

Intext Question Answers

Question 1.
Name the various kinds of cell layers which constitute the bark.
Answer:
Periderm and secondary phloem.

Question 2.
Every 50 years, for 200 years, a nail was drilled into a tree to the same depth and at exactly 1m above the sail surface (assuing the ground level has not changed). What will be the pattern of the four nails on the tree ? Do you know the reason for your answer ? If yes give the reason.
Answer:

  1. The heads of all the four nails are at same level.
  2. Stem of plant undergoes later growth due to cambial activity. Hence, growth (circumference) of stem increases.
  3. All the four nails will be seen in the xylem portion of the stem.
  4. There will not be change in nail position with respect to vertical position from ground level. Because the vertical growth is reduced after some period and lateral growth is promoted in plant.

Question 3.
Why is wood made of xylme and not a phloem?
Answer:

  1. Cambial ring produces more xylmen than phloem during secondary growth.
  2. Xylem with the exception of parenchyma, consists of dead tissues i.e., tracheids, vessels and fibres.
  3. Phloem is living complex tissue, with the exception of fibres (bast).
  4. Hence wood is made of xylem.

TS Inter 1st Year Botany Study Material Chapter 12 Histology and Anatomy of Flowering Plants

Question 4.
A student estimated the age of a tree to he about 300 years. How did he anatomically estimate the age of this tree?
Answer:
The age of a plant can be estimated by counting the number of annual rings.

Question 5.
Assume that you have removed the duramen part of a tree. Will the tree survive or die?
Answer:
The plant survives because of the presence of sapwood which is meant for the conduction of water and minerals. Duramen is not useful for conduction.