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Telangana TSBIE TS Inter 2nd Year Botany Study Material 14th Lesson Microbes in Human Welfare Textbook Questions and Answers.

TS Inter 2nd Year Botany Study Material 14th Lesson Microbes in Human Welfare

Very Short Answer Type Questions

Question 1.
Why does ‘Swiss cheese’ have big holes? Name the bacteria responsible for it. [Mar. 2020, 18; May 14]
Answer:

1. Large holes in ‘Swiss cheese’ are due to the production of a large amount of CO₂ by a Bacterium.
2. Propionibacterium sharmanii is responsible for it.

Question 2.
What are fermentors? [May 17; Mar. 14]
Answer:

1. Very large vessels that are used to grow microbes for production of valuable products on an industrial scale.
2. Beverages like wine, beer, whisky, brandy and rum are produced through fermentation of malted careals and fruit juices by yeast.

Question 3.
Name a microbe used for statin production. How do statins lower blood cholesterol level?
Answer:

1. Statins are produced by the yeast, Monascus purpureus.
2. The act by competitively inhibiting the enzyme responsible for the synthesis of cholesterol.

Question 4.
Why do we prefer to call secondary waste water treatment as biological treatment?
Answer:

1. During secondary waste water treatment, the aeration allows vigorous growth of useful aerobic microbes into floes and reduces BOD (Biochemical Oxygen Demand) of the effluent.
2. The microbes consume the major part of the organic matter in the effluent and hence secondary waste water treatment is called Biological treatment.

Question 5.
What is Nucleopolyhedrovirus is being used for nowadays?
Answer:

1. Nucleopolyhedrovirus are Baculoviruses, used as biocontrol agents.
2. They attack insect and other arthropods. They are species-specific,’narrow spectrum insecticides and have no negative impact on plants, birds, mammals, fish and even non-target insects.

Question 6.
How has the discovery of antibiotics helped mankind in the field of medicine?
Answer:

1. Penicillin was extensively used to treat American soldiers wounded in World War – II.
2. Antibiotics have greatly improved our capacity to treat dreadly diseases like plague, diphtheria, whooping cough and leprosy.

Question 7.
Why is distillation required for producing certain alcoholic drinks?
Answer:

1. Strong alcoholic drinks such as whisky, brandy and rum with a higher concentration of ethanol are produced by the distillation of the fermented broth.
2. Because, yeasts poison themselves to death when the concentration of alcohol reaches about 13 percent.

Question 8.
Write the most important characteristic that Aspergillus niger, Clostridium butylicum and Lactobacillus share.
Answer:

1. These are organic acid producers.
2. Aspergillus niger (a fungus) produce citric acid, Clostridium butylicum (a bacterium) produce butyric, and acid Lactobacillus share (a bacterium) produce lactic acid.

Question 9.
Give any two microbes that are useful in biotechnology. [March 2014]
Answer:

1. Streptiococcus to produce stroptokinase, a clot buster for removing clots from the blood vessels of patients with myocardial infection.
2. Trichoderma Polysporum to produce cyclosporin – A, an immuno oppressive agent in organ transplant patients.

Question 10.
Name any two genetically modified crops.
Answer:

1. Bt-cotton
2. Bt-brinjal.

Question 11.
Why are blue green algae not popular as biofertilisers?
Answer:

1. Blue green algae (cyanobacteria) mostly live in aquatic environment and are commercialized during recent days.
2. Hence, BGA are used as biofertilizers in paddy fields with water for most of the crop period.

Question 12.
Which species of Penicillium produces Roquefort cheese?
Answer:

1. Penicillium roquefort is used for ripening the Roquefort cheese.
2. This gives a particular flavour to cheese.

Question 13.
Name any two industrially important enzymes. [Mar. 2019, 17]
Answer:

1. Pectinases and proteases used to clarify bottled juices.
2. Streptokinase, a clot buster for removing clots from the blood vessels.

Question 14.
Name an immunosuppressive agent.
Answer:

1. Cyclosporin A is used as immunosuppressive agent.
2. It is produced from a fungus, trichoderma polysporum

Question 15.
Give an example of a rod shaped virus.
Answer:
Tobacco mosaic virus is rod shaped virus.

Question 16.
What is the group of bacteria found in both the rumen of cattle and sludge of sewage treatment?
Answer:
Methanogens – Methanbacterium.

Question 17.
Why are cyanobacteria considered useful in paddy fields?
Answer:

1. Cyanobacteria (Blue Green Algae) are autotrophic microbes, they require aquatic environment.
2. In paddy fields water is available for most of the time, hence BGA are considered as useful Nitrogen fixers that add organic matter to the soil and increase soil fertility.

Question 18.
In which food would you find lactic acid bacteria? Name the bacterium.
Answer:

1. Lactic acid bacteria grow in milk and convert it into curd.
2. Bacteria name is Lactobacillus.

Question 19.
Name any two fungi which are used in the production of antibiotics.
Answer:

1. Penicillin notatum – Penicillin
2. Penicillin griseoflavus – Griepseofulvin.

Question 20.
Name the scientists who were credited for showing the role of penicillin as an antibiotic.
Answer:

1. Alexander Fleming discovered penicillin, the first antibiotic from pencillium notatum.
2. Ernest Chain and Howard Florey used penicillin to treat American soldiers wounded in World War-II.

Short Answer Type Questions

Question 1.
Why are the floes important in the biological treatment of waste water?
Answer:

1. The primary effluent is passed into large aeration tanks where it is constantly agitated mechanically and air is pumped into it.
2. This allows vigorous growth of useful aerobic microbes into floes.
3. Floes are masses of bacteria associated with fungal filaments to form mesh like structures.
4. While growing, these microbes consume the major part of the organic matter in the effluent.
5. This significantly reduces the BOD (Biochemical Oxygen Demand) of the effluent.
6. When BOD of sewage is reduced, the effluent is passed into a settling tank, where the floes are allowed to form the activated sludge.

Question 2.
How is Bacillus thuringiensis helpful in controlling insect pests?
Answer:

• Bacillus thuringiensis often written as Bt.
• Dried spores of Bt are used to control butterfly Caterpillars.
• Insect larva, after eating these are killed by the toxin, released in their gut.
• The bacterial disease will kill the caterpillars but leave other insects unharmed.
• Bacillus thuringiensis toxin genes have been introduced into plants to provide resistance to pests.

Question 3.
How do mycorrhizal fungi help the plants harbouring them?
Answer:
The symbiotic association between fungi members and roots of vascular plants is called mycorrhizae.

Many members of the genus Glomus forms mycorrhiza. The fungal symbiont in these associations facilitates absorption of phosphorus by the plant from the soil.

Plants having such associations shows other benefits also such as resistance to root-borne pathogens, tolerance to salinity and drought and over all increase in plant growth and development.

Question 4.
How was penicillin discovered?
Answer:
Alexander Fleming while working on Staphylococci bacteria, once observed a mould growing in one of his unwashed culture plates around which staphylococci could not grow.

He found out that it was due to a chemical produced by the mould and he named it as penicillin after the mould penicillium notatum. However, its full potential as an effective antibiotic was established only much later by Ernest Chain and Howard Florey. Fleming, Chain and Florey were awarded Nobel Prize in 1945 for this discovery.

Question 5.
How do bioactive molecules of fungal origin help in restoring good health of humans?
Answer:

1. Bioactive molecule, cyclosporin A, is used as in immunosuppressive agent in organ- transplant patients. It is produced by fungus Trichoderma polysporum.
2. Statin is produced by the yeast Monascus purpurens. It has been commercialised as blood-cholesterol lowering agents. They act by competitively inhibiting the enzyme responsible for the synthesis of cholesterol.

Question 6.
What is the chemical nature of biogas? Explain the process of biogas production.
Answer:
Biogas comprises methane (CH₄), carbondioxide (CO₂), traces of hydrogen sulphide (H₂S) and moisture. Biogas is generated by the decomposition of excreta or dung of cattle (commonly called as gobar), domestic waste material, industrial and agriculture sewage due to the activity of anaerobic bacteria present in them.

Biogas formation from activated sludge :

• A small part of activated sludge is pumped into the aeration tank to serve as inoculum.
• In aeration tank anaerobic bacteria called methnogens digest the bacteria and fungi of the sludge.
• During the digestion the bacteria produce a mixture of gases like C02, CH3 and H2S which forms biogas.

Biogas formation from dung :

1. The biogas plant consists of a concrete tank in which bio wastes are collected and a slurry of a dung is fed.
2. A floating cover is placed over the slurry, which keeps on raising as the gas is produced in the tank due to the microbial activity.
3. The biogas plant has an outlet which is connected to a pipe to supply biogas to near by houses.

Question 7.
Which bacterium has been used as a clot buster? What is its mode of action?
Answer:
Bacterium Streptococcus has been used on a clot buster.

Steptokinase, produced by the bacterium streptococcus and modified by genetic engineering is used as a clot buster for removing clots from the blood vessels of patients who have undergone mycocardial infection leading to heart attack.

Question 8.
What are biofertilisers? Give two examples and discuss their role as biofertilisers.
Answer:
The organisms that enrich the nutrient quality of the soil are called biofertilisers. The main sources of biofertilisers are bacteria, fungi and cyanobacteria.

1. Rizobium bacteria present as a symbiotic association in the nodular roots of leguminous plants. They fix atmospheric nitrogen into organic forms, which are used by the plant as a nutrient.
2. In paddy fields cyanobacteria serve as an important fertiliser. They fix atmospheric nitrogen. They also add organic matter to the soil and increase its fertility.
   Eg : Anabaena, Nostoc, Oscillatoria etc.

Question 9.
What role do enzymes play in detergents that we use for washing clothes? Give examples.
Answer:
Microbes are used for producing enzymes. Enzyme Lipases are used in detergents formulations and are helpful in removing oily stains from laundry Example is Lipases.

Long Answer Type Questions

Question 1.
a) What would happen if a large volume of untreated sewage is discharged into a river?
b) In what way is anaerobic sludge digestion important in sewage treatments?
Answer:
a) The untreated sewage discharged directly into rivers, leading to their pollution and an increase in water born diseases.

b) Treatment of sewage involves two steps. 1) Primary treatment 2) Secondary treatment

Primary treatment :
It is a physical process of removal of small and large particles through filtration and sedimentation.

• The sewage is allowed to go into the primary setting tank, where the suspended material settle down to form primary sludge.
• The effluent is taken for secondary treatment. The anaerobic sludge digestion is important in secondary sewage treatment.

Secondary sewage treatment:

• It is a biological process that employs the heterotrophic bacteria naturally present in the sewage.
  The effluent from the primary treatment is passed into large aeration tanks, where it is constantly agitated and air is pumped in it.
• This allows the rapid growth of aerobic bacteria into floes, which consume the organic matter of sewage and reduce the BOD.
• The effluent is passed into a settling tank, where the floes are allowed to sediment forming the activated sludge.
• A small part of the activated sludge is pumped back into aeration tank as inoculum.
• The remaining major part of the sludge is pumped into sludge digestors, where the anaerobic bacteria digest the organic matter and produce a mixture of gases such as methane, hydrogen sulphide and CO₂. These gases form biogas which can be used as a source of energy as it is. inflammable.

Question 2.
Which type of food would have lactic acid bacteria? Discuss their useful application.
Answer:

1. Curd is formed by adding lactobacillus bacteria as Lactic Acid Bacteria (LAB) in milk.
2. A small amount of curd is added to the fresh milk as starter which contains millions of LAB.
3. LAB at suitable temperature multiply and convert milk into curd which also increase nutritional quality by increasing vitamin B₁₂.
4. During growth, LAB produce acids that coagulate and partially digest the milk proteins.
5. LAB also check disease-causing microbes in the stomach.
6. The use of such friendly bacteria for therapeutic purposes and for the betterment of human health has led to the concept of Probiotics.
7. Lactic acid bacteria is also used in bakery products, beverages, meat products, confectionery, dairy products etc.

Question 3.
Write a brief essay on Microbes as biocontrol agents.
Answer:
Biocontrol can be defined as the use of biological methods of controllingthe plant diseases and pests.

1. The use of insecticides and pesticides although useful, but are toxic and extremely harmful to humans, animals and polluting our natural resources.
2. Agriculture relies in natural control of pests i.e., natural predation rather than introduced chemicals.
3. Farmers are in view that the eradication of the creatures that are often described as pests is undesirable because without them beneficial predatory and parasitic insects would not survive.
4. Some approaches for having biocontrol agents.
   i) Familiarity with various life forms inhabiting the field.
   ii) Understanding of their life cycles, patterns of feeding and habitat of predators and pests.

5) Examples of biocontrol agents are
i) The lady bird and dragonflies are useful to get rid of aphids and mosquitoes, respectively.
ii) The bacteria Bacillus thuringiensis (Bt) are used to control butterfly, catterpillars.

• Dried spores of Bt are mixed with water and sprayed onto vulnerble plants such as brassicas and fruit trees, where these are eaten by the insect larvae.
• In the gut of the larvae, the toxin is released and the larvae get killed.
• The bacterial disease will kill the catterpillars but leave other insects unharmed.
• B thuringiensis toxin genes are introduced into plants by genetic engineering. Such plants are resistant to attack by insect pests. For example Bt. Cotton.

iii) Baculoviruses are pathogens that attack insects and other arthropods.

• Majority of baculoviruses used as biological control agents belong to the genus Nucleopoly – hedrovirus.
• These are species, specific, narrow spectrum insecticides.
• They do not harm plants, mammals, birds, fish and other nontarget insects.
• Baculoviruses are beneficial in Integrated Pest Management (IPM) programme, in which beneficial insects are conserved.

Question 4.
What is organic farming? Discuss the role of plant microbes in organic farming with examples.
Answer:
Organic farming is a form of agriculture that works in hormony with nature rather than against it. It is done by using only natural and organic materials. Organic farming refers to biofertilizers and biopesticides.

The role of plant microbes in organic farming :
Microbes are biofertilizers enrich the nutrients (nitrogen, phosphorus, etc) quality of the soil.
i) The main source of biofertilizers are bacteria, fungi and cyanobacteria.

ii) Bacteria as biofertilizer:
a) The nodules on the roots of leguminous plants are formed by the symbiotic association of Rhizobium’bacteria.
b) These bacteria fix atmospheric nitrogen into organic forms, which is used by the plants as nutrient.
c) Other bacteria, such as Azospirillum and Azatobacter fix atmospheric nitrogen while free living in Jhe soil. They enrich the nitrogen content of the soil.

iii) Fungi as biofertilizers:
a) Fungi form symbiotic association with plants (Mycorrhiza).
b) The fungal hyphae absorb phosphorus from soil and passes into the plant.
c) Mycorrhiza shows the following benefits.

• Resistance to root borne pathogens.
• Tolerance to salinity and drought.
• Overall increase in plant growth and development.

iv) Cyanobacteria as biofertilizers :
a) These are autotrophic microbes, many of them fix atmospheric nitrogen.
b) Examples of cyanobacteria are Anabaena, Nostoc, Oscillatoria etc.
c) Blue-green algae also add organic matter to the soil and increase its fertility.

v) A number of biofertilizers are available commercially in the market. Farmers use these in fields and replenish soil nutrients and to reduce dependence on chemical fertilizers.

Intext Question Answers

Question 1.
Bacteria cannot be seen with the naked eye, but these can be seen with the help of a microscope. If you have to carry a sample from your home to your biology laboratory to demonstrate the presence of microbes under a microscope. Which sample would you carry and why?
Answer:
Curd can be used as a sample for the study of microbes. Curd contains numerous lactic acid bacteria (LAB) or lactobacillus. These bacteria produce acids that coagulate and digest milk proteins. A small drop of curd contains millions of bacteria which can be easily observed under a microscope.

Question 2.
Give examples to prove that microbes release gases during metabolism.
Answer:
The examples of bacteria that release gases during metabolism are :
a) Bacteria and fungi carry out the process of fermentation and during this process, they release carbon dioxide. Fermentation is the process of converting a complex organic substance into simpler substance with the action of bacteria or yeast. Fermentation of sugar produces alcohol with the release of carbon dioxide and very little energy.
b) The dough used for making idli and dosa gives a puffed appearence. This is because of the action of bacteria which releases carbon dioxide. This CO₂ released from the dough gets trapped in the dough thereby giving it a puffed appearance.

Question 3.
Name the states involved in Ganga action plan.
Answer:
The states involved are Bihar, Uttar Pradesh, Uttarakhand.

Question 4.
Name some traditional Indian foods made of wheat, rice and bengal gram (or their products). Which of these foods involve the use of microbes?
Answer:
Wheat product: Bread, Cake etc.
Rice product: Idli, dosa.
Bengal gram product: Dhokla, Khandvi.

Question 5.
In which way have microbes played a major role in controlling diseases caused by harmful bacteria?
Answer:
Several microorganisms are used for preparing medicines. Antibiotics arfe (medicines produced by certain microorganisms to kill other (disease – causing) microorgansims. These medicines are commonly obtained from bacteria and fungi. They either kill or stop the growth of disease causing microorganisms Streptomycin, tetracycline and penicillin are common antibiotics.

Penicillin notatum produces chemical penicillin which checks the growth of staphylococci bacteria in the body. Antibiotics are designed to destroy bacteria by weakening their cell walls. Asa result of this weakening, certain immune cells such as the white blood cells enters the bacterial cell and cause cell lysis. Cell lysis is the process of destroying cells such as blood cells and bacteria.

Question 6.
Do you think microbes can also be used as a source of energy. If yes, how?
Answer:
Yes, microbes can be used as a source of energy. Bacteria such as Methane bacterium is used for the generation of gobar gas or biogas. The generation of biogas is an anaerobic process in a biogas plant which consists of a concrete tank (10-15 feet deep) with sufficient outlets and inlets. The dung is mixed with water to form the slurry and thrown into the tank.

The digester of the tank is filled with numerous anaerobic methane producing bacteria, which produce biogas from the slurry. Biogas can be removed through the pipe which is then used as a source of energy while the spent slurry is removed from outlet and is used as a fertilizer.

Question 7.
Microbes can be used to decrease the use of chemical fertilizers and pesticides. Explain how this can be accomplished.
Answer:
Microbes play an important role in organic farming which is done without the use of chemical fertilizers and pesticides. Biofertilizers are living organisms, which can help to increase the fertility of soil. It involves the selection of beneficial microorganisms that help in improving plant growth through the supply of plant nutrients. Biofertilizers are introduced in seeds, roots or soil to mobilize the availability of nutrients. Thus they are extremely beneficial in enriching the soil with organic nutrients.

Many species of bacteria and cyanobacteria have the ability to fix free atmospheric nitrogen. Rhizobium is a symbiotic bacteria found in the root nodules of leguminous plants. Azospirillum and Azatobacter are free living nitrogen fixing bacteria wheareas Anabena, Nostoc, Oscillitoria are examples of Nitrogen- fixing cyanobacteria. Biofertilizers are cost effective and eco-friendly. Microbes can also act as bio pesticides to control insect pests in plants. An example of bio-pesticides is Bacillus thuringiensis, which produces a toxin that kills the insect pests.

Dried bacterial spores are mixed in water and sprayed in agricultural fields. When larvae of insects feed on crops, these bacterial spores enter the gut of larvae and release toxins, there by it, similarly Trichoderma are free living fungi. They live in the roots of higher plants and protect them from various pathogens. Baculoviruses is another biopesticide that is used as a biological control agent against insects aod other arthropods.

Question 8.
Three water samples namely river water, untreated sewage water and secondary effluent discharged from a sewage treatment plant were subjected to BOD test. The samples were labelled A,B and C; but the laboratory attendant did not note which was which. The BOD values of the three samples A,B and C were recorded as 20mg/L, 8mg/L and 400 gm/L respectively. Which sample of the water is most polluted? Can you assign the correct label to each assuming the river water is relatively clean?
Answer:
Biological Oxygen Demand (BOD) is the method of determining the amount of oxygen required by micro-organisms to decompose the waste present in the water supply. If the quantity of organic wastes in the water supply is high, then the number of decomposing bacteria present in the water will also be high. As a result, the BOD value will increase. Therefore it can be concluded that if the water supply is more polluted, then it will have a higher BOD value out of the above three samples, sample C is the most polluted since it has the maximum BOD value of 400mg/L.

After untreated sewage water, secondary effluent discharge from a sewage treated plant is most polluted. Thus sample A is secondary effluent discharge from a sewage treatment plant and has the BOD value of 20mg/L. While sample B is river water and has the BOD value of 8mg/L.

Question 9.
Name the microbes from which Cyclosporin A (an immunosuppressive drug) and statins (blood cholesterol lowering agents) are obtained.
Answer:

Question 10.
Find out the role of microbes in the following and discuss it with your teacher, (a) Single Cell Protein (SCP). (b) Soil.
Answer:
a) Single cell protein (SCP) is a cell protein obtained from certain microbes, which form an alternated source of proteins in animal feeds. The microbes involved in the preparation of single cell proteins are algae, yeast or bacteria. These microbes are grown on an industrial scale to obtain the desired protein. For example Spirulina can be grown on waste material obtained from molasses, sewage, and animal manures. It serves as a rich supplement of dietary nutrients such as proteins, carbohydrate, fats, minerals and vitamins. Similarly microorganisms such as Methylophilus and Methylotrophus have a large rate of biomass production. Their growth can produce a large amount of proteins.

b) Soil microbes play an important role in maintaining soil fertility. They help in the formation of nutrient rich humus by the process of decomposition. Many species of bacteria and cyanobacteria have the ability to fix atmospheric nitrogen into usuable form. Rhizobium is a symbiotic bacteria found in the root nodules of leguminous plants. Azospirillium and Azotobacter are free living nitrogen fixing bacteria whereas Anaebena, Nostoc and Oscillitoria are examples of nitrogen fixing cyanobacteria.

Question 11.
Arrange the following in the decreasing order (most important first) of their importance, for the welfare of human society. Give reasons for your answer. Biogas, Citric acid, Penicillin and Curd.
Answer:
The order of arrangement of products according to their decreasing importance is Penicillin – Biogas – Citric acid – Curd. Penicillin is the most important product for the welfare of human society. It is an antibiotic, which is used for controlling various bacterial diseases. The second most important product is biogas. It is an eco-friendly source of energy. The next important product is citric acid which is used as a food preservative. The least important product is curd a food item obtained by the action of lactobacillus bacteria on milk. Hence the products in the decreasing order of their importance are as follows Penicillin – Biogas – Citric acid – Curd.

Question 12.
What is sewage? In which way can sewage be harmful to us?
Answer:
Sewage is the municipal waste matter that is carried away in sewers and drains. It includes both liquid and solid wastes, rich in organic matter and microbes. May of these microbes are pathogenic and can cause several waste borne diseases. Sewage water is the major cause of polluting drinking water. Hence it is essential that sewage water is properly collected, treated and disposed.

Question 13.
What is the key difference between primary and secondary sewage treatment?
Answer:

Telangana TSBIE TS Inter 2nd Year Botany Study Material 13th Lesson Strategies for Enhancement in Food Production Textbook Questions and Answers.

TS Inter 2nd Year Botany Study Material 13th Lesson Strategies for Enhancement in Food Production

Very Short Answer Type Questions

Question 1.
What is meant by ‘hidden hunger’?
Answer:

1. About 3 billion people over the globe are suffering from deficiencies of micro nutrients, protein and vitamins. This situation is said to be ‘Hidden hunger’.
2. More than 840 million people in the world do not have adequate food to meet their daily food and nutritional requirements.

Question 2.
Name two semi-dwarf varieties of rice developed in India. [Mar. 2020]
Answer:

1. Jaya and Ratna, are two better yielding semi – dwarf varieties of rice developed in India.
2. They are derivatives of semi – dwarf varieties derived from crossing IR – 8 with TN – 1.

Question 3.
Give two examples of wheat varieties introduced in India, which are high yielding and disease resistant. [March 2018]
Answer:

1. Sonalika and Kalyan Sona,
2. They were introduced in 1963, all over the wheat growing belt in India.

Question 4.
Give two examples of fungi used in SCP production. [March 2019, Mar. ’17; May ’17]
Answer:

1. Candida utilis (Torula yeast)
2. Saccharomyces cerevisiae (Baker’s yeast)
3. Chaetomium cellulolyticum

Question 5.
Why are plants obtained by protoplast fusion called somatic hybrids?
Answer:

1. Isolated protoplast from two different varieties of plants each having a desirable character- can be fused to get hybrid protoplasts which can be further cultured to form a novel plant.
2. Hybrid derives from fusion of vegetative (somatic) or body cells unlike that of fusion of sex cells is called somatic hybrid.

Question 6.
What is protoplast fusion?
Answer:

1. Nacked plant cells produced by digesting cell walls using cellulose and pectinose are called protoplasts.
2. The fusion of isolated protoplast from two different varieties of plants is called protoplast fusion.

Question 7.
Why is it easier to culture meristems compared to permanent tissues’?
Answer:

1. Meristems are undifferentiated, embryonal living plant cells which have the capacity of cell divisions and hence easy to culture.
2. Permanent tissues consists of differentiated cells that have to undergo dedifferentiation to undergo cell division and hence difficult to culture.

Question 8.
Why are protein synthesized from Spirulina called single cell proteins?
Answer:

1. Spirulina is a unicellular alga and is rich source of protein.
2. Dried biomass of a single species of microbe used as a protein source in the diet is •known as Single Cell Protein.

Question 9.
A person who is allergic to pulses are advised to take a capsule of Spirulina dialy. Give reasons for the advice. [May 2014]
Answer:
Pulses contain proteins, Similarly Spirulina also contains proteins. So if a person is allergic to pulses he can take Spirulina for proteins.

Question 10.
Would it be wrong to refer to plants obtained through micro propagation as ‘Clones’? Explain.
Answer:

1. No, since the formation of plants through micro propogation does not involve fusion of sex cells from two parents, they can be called as Clones.
2. The plants produced by Micropropagation are genetically identical to the origin or source plant and hence they are called Somaclones.

Question 11.
How is somatic hybrid different from a hybrid?
Answer:

1. Two different plants of desired characters are crossed to form a hybrid. It is a product of sexual reproduction.
2. Isolated protoplasts from vegetative cells of two different plants, each having a desirable character – can be fused to get hybrid, protoplast, and is cultured to form somatic hybrid.

Question 12.
What is emasculation? Why and when is it done?
Answer:

1. Removal of anthers from bisexual flowers of female parent, when the flowers are still in bud condition is called emasculation.
2. It prevents self-pollination and ensures artificial cross pollination.

Question 13.
Discuss the two main limitations of plant hybridization programme.
Answer:

1. It is not necessary that the hybrids do combine the desirable characters. Usually only one in a few hundred to a thousand crosses show the desirable combination.
2. Selection of superior recombinants and subjecting them to self pollination to achieve homozygosity so that the characters will not segregate in the progeny.

Question 14.
Give two important contributions of Dr. M.S. Swaminathan.
Answer:

1. Swaminathan and his team developed short duration high-yielding varieties of Rice including scented Basmati.
2. He introduced Mexican varieties of wheat in India.

Question 15.
Which two species of sugarcane were crossed for better yield?
Answer:

1. Saccharum barberi of North India had poor sugar content and Saccharum officinarum of South India had higher sugar content were crossed to produce a new varety with desirable qualities.
2. New variety has high yield, thick stems, high sugar and grow well in North India.

Short Answer Type Questions

Question 1.
What is meant by germplasm collection? What are its benefits?
Answer:
The entire collection of plants / seeds, having all the diverse alleles for all genes in a given crop is called germplasm collection.

1. Cell and tissue cultures of many plant species can be preserved maintained in a viable state for several years and used when required.
2. Plant materials from endangered species can be conserved using this method.
3. It is an ideal method for long term conservation of cell cultures producing secondary metabolites such as antibiotics.
4. Seeds which loose their viability or storage can be maintained for a long period of time.
5. Disease free plant material can be frozen and propagated whenever required.
6. Conservation of Somaclonal variations in cultures.
7. Rare germplasms developed by using Somatic hybridisation other genetic manipulation techniques can be stored.
8. Pollen conservation for enhancing longevity.
9. Germplasm banks to facilitate the exchange of information of international level.

Question 2.
Name the improved characteristics of wheat that helped India to achieve green revolution.
Answer:

1. Green revolution is the dramatic increase in food production due to plant breeding „ techniques.
2. During the period 1960 to 2000, wheat production increased from 11 million tonnes to 75 million tonnes this is due to development semi-dwarf variety of wheat.
3. Semi-dwarf wheat was developed by Norman E.Borlaug at International Centre for Wheat and Maize improvement in Mexico.
4. In 1963, several varieties such as Sonalika and Kalyan Sona which were high yielding and disease resistant, were introduced all over the wheat growing belt of India.

Question 3.
Suggest some of the features of plants that prevent insect and pest infestation.
Answer:

1. Major cause for large scale destruction of crop plant and crop produce is insect and pest infestation.
2. Insect resistance in host crop plants may be due to morphological, biochemical or physiological characteristics.
3. Hairy leaves in several plants are associated with resistance to insect pests, e.g.: resistance to jassids in cotton and cereal leaf beetle in vyheat.
4. In wheat, solid stems lead to non-preference by the stem sawfly.
5. Smooth leaved and nectar-less cotton varieties do not attract bollworms.
6. High aspartic acid, low nitrogen and sugar content in maize leads to resistance to maize stem borers.

Question 4.
The culture medium (nutrient medium) can be referred to as a ‘highly enriched laboratory soil’. Justify the statement.
Answer:
Culture medium must provide a carbon source such as sucrose and also in organic salts, vitamins, amino acids and growth regulators like auxins, cytokinins etc.

The culture medium is rich in nutrients.

Question 5.
Plants raised through tissue cultures are clones of the ‘parent’ plant.’ Discuss the utility of these plants.
Answer:

1. Plants produced through tissue culture are genetically identical to the original or source plant and hence they are called somaclones.
2. Many economically important plants like tomato, banana, apple, teak, eucalyptus, bamboo etc. have been produced on a commercial scale by the use of this method.
3. Recovery of healthy plants from diseased plants can be done.
4. One can remove the meristem and growin in vitro to obtain virus free plants.
5. Scientists have succeeded in culturing meristems of banana, sugarcane, potato etc.

Question 6.
Discuss the importance of testing of new plant varieties in a geographically vast country like India.
Answer:

1. The newly selected lines are evaluated for their yield and other agronomic traits of quality, disease resistance etc.
2. This evaluation is done by growing these in research fields and recording their performance under ideal fertiliser application, irrigation and other crop management practices.
3. The evaluation in research fields is followed by testing the materials in farmer’s fields, for at least three growing seasons at several locations in the country, representing all the agroclimatic zones where the crop is usually grown.
4. The material is evaluated in comparison to the best available local crop cultivar-a check or reference cultivar.

Question 7.
Give few examples of biofortified crops. What benefits do they offer to the society?
Answer:
Biofortified crops are with higher levels of vitamins and minerals, or higher protein and healthier fats. It is the most practical means to improve public health of the society.

Examples of fortified crops :

1. Atlas 66 used as a donor for developing wheat varieties with improved protein content.
2. Maize hybrids have increased amount of aminoacid lysine and tryptophan.
3. Iron fortified rice have increased iron content.
4. India Agricultural Research Institute, New Delhi have released some fortified crop

varieties:
a) Carrot, spinach and pumpkin – Vitamin A
b) Bitter gourd, bathua, mustard, tomato – Vitamin C
c) Spinach and bathua – Iron and Calcium.
d) Broad bean, lablab, fresh bean and garden pea – Protein.

Question 8.
Mutations are beneficial for plant breeding. Taking an example, justify the statement.
Answer:

1. Mutations is defined as sudden heritable change in the character of an organism, due to change in the sequence of bases of the gene.
2. Few mutations may create numerous new varieties of economic value.
3. Due to mutation breeding the farmers need not depend on nature for variation.
4. It is possible to induce mutation artificially through the use of chemicals or radiations (like gamma radiations) and selecting and using the plants that have a desirable character as a source in breeding. This is called mutation breeding.
   Eg: In mung bean, resistance to yellow mosaic virus and powdery mildew were induced by mutations.

Question 9.
Discuss briefly the technology that made us self-sufficient in food production.
Answer:

• Technology called Tissue Culture made us self sufficient in food production.
• By applying this technology, it is possible to produce a large number of plants in a very short time and limited space. Hence this technique is called micro propagation.
• Plants thus produced are genetically identical to the original or source plant and hence they are called somaclones.
• Many economically important plants like tomato, banana, apple, teak, eucalyptus, bamboo etc., have been produced on a commercial scale by the use of this method.
• Plant breeding as a technology has helped increased yield to a large extent. The green revolution was dependent on plant breeding techniques for developing high yielding and disease resistant varieties like rice, maize.
• Mutation breeding and rDNA technique also play important role in increasing food production.
• By mutation breeding, disease resistant crops were developed which prevented from large destruction.

Long Answer Type Questions

Question 1.
You are a Botanist working in the area of plant breeding. Describe the various steps that you will undertake to release a new variety. [Mar. ’18; May ’17, ’14]
Answer:
The steps in breeding a new genetic variety of a crop are

1. Collection of variability.
2. Evaluation and Selection of parents.
3. Cross hybridisation among selected parents.
4. Selection and testing of superior recombinants.
5. Testing, release and commercialization of new cultivars.

i) Collection of variability :

1. Genetic variability is important for any breeding programme.
2. Pre-existing genetic variability available in wild varieties, species and relatives of crop species is collected and preserved.
3. Evaluation of their characteristics is a pre-requisite for the effective exploitation of natural genes available in the population.
4. The entire collection of plants / seeds having all diverse alleles for all genes in a given crop is called germplasm collection.

ii) Evaluation and selection of parents :
a) It is carried out by evaluating germplasm to identify plants with desirable combination of characters.
b) The selected plant is multiplied and hybridized.
c) By self-pollination, purelines are created wherever desired.

iii) Cross hybridization among selected parents :

1. Cross hybridization is carried out to combine desired genetic characters from two different plants (parents).
2. Cross hybridization is a time consuming and tedious process as it involves collection of pollen grains from the desired plants and other pollination techniques to incorporate desired traits.
3. It is also not certain the hybrids combine desired characters. The chances of desirable combination is usually only one in few hundred to a thousand crosses carried out.

iv) Selection and testing of superior combinants :

1. This step consists of selection of plants among the progeny of the hybrids with desired combination of characters.
2. It yields plants that are superior to both of the parents. This is known as hybrid vigour / heterosis.
3. They are self pollinated for several generations till they reach a state of uniformity or homozygosity so that characters will not segregate in the progeny.

v. Testing, release and commercialization of new cultivars :

1. Evaluation is done for newly selected lines for their yield and other agronomic traits of quality, disease resistance etc.
2. Selected plants are grown in research fields and their performance is recorded under ideal fertilizer application irrigation and other crop management practices.
3. Testing of hybrid line is done in farmer’s field after evaluation.
4. After testing, the crop is grown at different locations in the country with different agroclimatic zones for at least three growing seasons.
5. The tested material is evaluated in comparison to the best available local crop cultivar used as reference cultivar.
6. Release of tested material is done in bulk after selection and certification.

Question 2.
Describe the tissue culture technique and what are the advantages of tissue culture over conventional method of plant breeding in crop improvement programmes? [Mar. 2020, 2019, 17, 14]
Answer:
The technique of growing, culturing and maintaining plant cells, tissues and organs in vitro is called tissue culture. Tissue culture is done by following methods :
1) Preparation of nutrient culture medium :
The nutrient medium must provide a carbon source such as sucrose and also inorganic salts, vitamins, amino acids and growth regulators like auxins, cytokinins etc.

2) Sterilization of the culture medium :

1. The medium is rich in nutrients and therefore, attracts the growth of microorganisms. The medium should be sterilized.
2. Sterilization is carried out in a steam sterilizer called autoclave.
3. The culture medium is autoclaved for 15 min, at 121°C and 15 lbs pounds of pressure.

3) Preparation of explant:
Any part of the plant body which is used as inoculum is called explant.

4) Inoculation of explants :
The transfer of explants on to the sterilized nutrient culture medium is called inoculation and is carried out in an aseptic condition.

5) Incubation of growth :

1. The cultures are incubated for 3 to 4 weeks during which period the cells of the explant absorb the nutrients, grow and undergo repeated divisions to produce undifferentiated mass of cells known as Callus.
2. Sometimes sheets or roots may be produced directly. ,
3. The explants or callus cultured on different combinations of auxins and cytokinins will produce shoots or roots and this process is called organogenesis.
4. Alternately, embryo like structures develop from the callus and this phenomenon is known as somatic embryogenesis, the embryo-like structure which develop from the callus are called embryoids. Since these embryoids develop from somatic tissue they are also called Somatic embryos.

6) Acclimatization of plantlets and transfer to pots :
The plants generated through organogenesis (or) somatogenesis need to be acclimatized before they are transferred to pots.

Advantages of tissue culture :

1. More number of plants can be produced in a short time.
2. Disease free plants can be developed from diseased plants.
3. Seedless plants can be multiplied.
4. Female plants are selectively produced through tissue culture.
5. Somatic hybrids can be raised by tissue culture, where sexual hybridization is not possible.

Flow chart showing Plant Tissue culture Technique

Question 3.
Modern methods of breeding plants can alleviate the global food ‘shortage’. Comment on the statement and give suitable examples.
Answer:
The development of several high yielding varieties of wheat and rice in the mid 1960’s as a result of various plant breeding techniques led to dramatic increase in food production in our country. This phase is often referred as green revolution.

High yielding and disease resistant varieties were introduced in India e.g. Sonalika and Kalyan Sona.

• Semi dwarf varities of rice e.g.: IR-8 Taichung Nalive -1 were introduced in 1966.
• Better yielding semi-dwarf varieties i.e., Jaya and Ratna were developed in India.

Successful developed high yield varieties of maizy jowar and bajra were obtained by hybrid breeding.

Plant breeding for disease resistance had enhance food production breeding is carried out by convention breeding techniques or by mutation breeding, e.g. In mung bean, resistance to yellow mosaic virus and powdery mildew were induced by mutations.

Plant breeding for developing resistance to insect pests :
Major cause for large scale destruction of crop plant and crop produce is insect and pest infestation. Insect resistance in lost crop plants may be due to morphological, biochemical or physiological characteristics.

Examples:

1. Wheat – Hairy leaves – resistance to cereal leaf beetle
2. Maize – High aspartic acid and low nitrogen and sugar contents – resistance to stem borer
3. Wheat – Solid stem – resistance to sawfly
4. Cotton – Smooth leaves and nectar-less condition – resistance to bollworm
5. Cotton – Hairy leaves – resistance to. sawfly

Plant breeding for improved food quality :
Breeding crops with higher levels of vitamins and minerals or higher protein and healthier fats is called Biofortification. It is the practical means to improve public health.

Examples:

1. Lysine and tryptophan rich maize.
2. High protein rich wheat.
3. Iron fortified rice.
4. Vitamin enriched bitter gourd, tomato, mustard, bathua.
5. Iron and Calcium enriched spinach and bathua.
6. Vitamin A rich carrots, spinach and pumpkin.

Single Cell Protein :
Microbes are grown on an industrial scale and used as nutrient rich food. E.g.: Spirulina.

Tissue Culture:
Technique of regeneration of whole plant from any part of the plant by growing it on suitable culture medium under aseptic conditions in vitro.

Advantages:

1. 1) A number of plants can be grown in a short period of time, e.g.: Tomato, banana, apple, teak, eucalyptus etc.
2. Healthy disease free plant can be grown by meristem culture, e.g.: Banana, sugarcane, potato etc.
3. Somatic hybrid can be raised by tissue culture where sexual hybridisation is not possible.
   e.g.: Pomato

Question 4.
Discuss how the property of plant cell totipotency has been utilized for plant propagation and improvement.
Answer:
Totipotency of a cell can be defined as the capacity of a cell to generate into a whole plant.
Requirements:
i) Explant :
It is any part of a plant taken out for growing a new planfin special nutrient medium under sterile / aseptic conditions.

ii) Nutrient medium :
It must have a carbon source such as sugar, inorganic salts, vitamins, amino acids, growth regulators like auxins, cytokinins etc.

iii) Suitable conditions of light and temperature. This is a tissue culture method by which a number of genetically similar plants called somadones are grown.

Utilization for plant propagation and improvement:

1. By utilizing this property of totipotency thousands of plants can be grown in a short period. Hence this technique is called micropropagation.
2. Plants are genetically identical, so certain desirable characters can be continued through generations.
3. Many economically important plants like tomato, banana, apple, teak, eucalyptus, bamboo etc. have been produced on a commercial scale by the use of this method.
4. Meristem Culture : This method is useful in recovery of healthy plant from diseased plants. Although plant is infected with a virus, meristems are free of virus. One can remove the meristem and grow it in vitro to obtain virus free plants.
5. Scientists have succeeded in culturing meristems of banana, sugarcane, potato etc.
6. Plants are genetically identical, so certain desirable characters can be continued through generations.
7. Hybrids can be produced by hybridisation.
8. Somatic hybridization offer vast potential for manipulation of plants in vitro to produce new varieties, e.g.: Pomato

Question 5.
What are the three options to increase food production? Discuss each giving the salient features, merits and demerits.
Answer:
Mutation breeding, tissue culture and r DNA technique, are going to play a pivotal role in enhancing food production.

Salient features of mutations :

1. Mutation is a phenomenon which results in alteration of genes (= DNA sequence).
2. It results in changes in the genotype and the phenotype of an organism.
3. In addition to recombination, mutation is another phenomenon that leads to variation in DNA.
4. The process of breeding by artificially inducing mutations using chemicals or radiations.
5. Breeding for disease resistance is carried out by mutation breeding.

Merits:

1. Mutation generates a large amount of variability in a population.
2. Plant breeder can select the desirable types.
3. Improved varieties of crop plants with desirable characters can be obtained after careful selection & hybridization.

Demerits:
1) It results in changes in the genetic makeup which could be lethal and results in the death of an individual.

Tissue culture:

1. Tissue culture is a technique of regeneration of a whole plant from any part of a plant by growing it on culture medium under aseptic conditions.
2. The capacity of a cell explant to grow into whole plant is called totipotency.
3. Explant is the part of plant taken for tissue culture.
4. The method of growing or producing thousands of plants through tissue culture is called micropropagation.
5. The plants produced from tissue culture are genetically identical to the original plant from which they are grown, so they are called somaclones.

Merits:

1. More number of plants can be produced in a short time.
2. Disease free plants can be developed from diseased plant.
3. Seedless plants can be multiplied.

Demerits:

1. No new combinations of traits.
2. All the plants that are produced have the same genetic material. So they are equally vulnerable to environmental factors, infections and pests.

r DNA technology:

1. Genetic engineering is a laboratory technique of gene manipulation which brings about novel combination of genes.
2. Recombinant DNA technology enables us to isolate a gene of interest in an organism which can be inserted into a vector and make it express its native characteristics.

Merits:

1. r DNA technologies will play a key role in preventing genetic diseases.
2. It produces targeted medicines (like insulin for diabetic patients).
3. It will also impart agriculture to resists diseases.

Demerits:

1. Genetically modified plants spreading beyond control and driving out local species.
2. Expensive lab facilities.
3. Maintaining sterile environment.

Intext Question Answers

Question 1.
Describe in brief various steps involved in plant breeding.
Answer:
Plant breeding is the process in which two genetically dissimilar varieties are purposely crossed to produce a new hybrid variety. As a result, characteristics from both parents can be obtained in the hybrid plant variety. Thus it involves the production of a new variety with the desired characteristics such as resistance to diseases, climatic adaptability and better productivity. The various steps involved in plant breeding are as follows.

a) Collection of genetic variability :
Genetic variability from various wild relatives of the cultivated species are collected to maintain the genetic diversity of a species. The entire collection of the diverse alleles of a gene in a crop is called the germplasm collection.

b) Evaluation of germplasm and selection of parents :
The germplasm collected is then evaluated for the desirable genes. The selected plants with the desired genes are then used as parents in plant breeding parents in plant breeding experiments and are multiplied by the process of hybridization.

c) Cross-hybridization between selected parents :
The next step in plant breeding is to combine the desirable characters present in two different parents to produce hybrids. It is a tedious job as one has to ensure that the pollengrains collected from the male parent reach the stigma of the female parent.

d) Selection of superior hybrids :
The progenies of the hybrids having the desired characteristics are selected through scientific evaluation. The selected progenies are then self-pollinated for several generations to ensure homozygosity.

e) Testing, release and commercialization of new cultivers :
The selected progenies are evaluated for characters such as yield resistance to disease performance etc., by growing them in research fields for at least three growing seasons in different parts of the country. After thorough testing and evaluation, the selected varieties are given to the farmers for growing in fields for a large scale production.

Question 2.
What is meant by biofortification?
Answer:
Biofortification is a process of breeding crops with higher levels of vitamins, minerals, proteins and fat content. This method is employed to improve public health. Breeding of crops with improved nutritional quality is undertaken to improve the content of proteins, oil, vitamins, minerals and micro nutrients in crops. It is also undertaken to upgrade the quality of oil and proteins. An example of this is a wheat variety known as Atlas 66, which has high protein content in comparison to the existing wheat, in addition, there are several other improved varieties of crop plants such as rice, carrots, spinach etc. Which have more nutritious value and more nutrients than the existing varieties.

Question 3.
Which part of the plant is best suited for making virus-free plants and why?
Answer:
Apical and axillary meristems of plants is used for making virus-free plants in a diseased plant. Only this region is not infected by the virus as compared to the rest of the plant region. Hence the scientists remove axillary and apical meristems of the diseased plant and grow it in vitro to obtain a disease-free and healthy plant. Virus-free plants of banana, sugarcane and potato have been obtained using this method of scientists.

Question 4.
What is the major advantage of producing plants by micropropagation?
Answer:
Micropropagation is a method of producing new plants in a short duration using plant tissue culture. Some major advantages of micropropagation are as follows.
a) Micropropagation helps in the propagation of a large number of plants in a short span of time.
b) The plants produced are identical to the mother plant.
c) It leads to the production of healthier plantlets which exhibit better disease resisting powers.

Question 5.
Find out what are the various components of the medium used for propagation of an explant in vitro?
Answer:
The major components of medium used for propagation of explants in vitro are carbon sources such as sucrose, inorganic salts, vitamins, amino acids, water, agar-agar and certain growth hormones such as auxins and gibberellins.

Question 6.
Name any five hybrid varieties of crop plants which have been developed in India.
Answer:
The five hybrid varieties of crop plants which have been developed in India are

Question 7.
The term ‘desirable trait’ can mean different things for different plants. Justify the statement with suitable examples.
Answer:
In case of rice, high yield and resistant to unfavourable conditions (like strom etc) is necessary. Hence the desirable trait is rice is high yield and resistant to unfavourable conditions. So semi-dwarf varieties are developed. In sugarcane desirable qualities are high yield, thick stem & high sugar. In wheat which are attached by rust the desirable trait in disease resistance. Similarly in case of lady finger plant insect resistance is a desirable character. This desirable traits are different for different plants.

Question 8.
Is there any relationship between dedifferentiation and higher degree of success achieved in plant tissue culture experiments?
Answer:
No, dedifferentiation occurs when a cell line is transformed, that means it becomes cancerous.

Question 9.
“Give me a living cell of any plant and I will give you a thousand plants of the same type”. Is this only a slogan or is it scientifically possible? Write your comments and justify them.
Answer:
It is not a slogan. It is possible scientifically by tissue culture techniques. The capacity to generate a whole plant from any cell is called totipotency. Large number of plants can be produced in a very short time and in limited space. Hence the technique is called Micro-propagation.

Question 10.
What are the physical barriers of a cell in the protoplast fusion experiment? How are the barriers overcome?
Answer:
The cell wall acts as a physical barrier of a cell in the protoplast fusion experiment. The barriers can be overcome by using hydrolyzing enzymes like cellulase and pectinase.

Telangana TSBIE TS Inter 2nd Year Botany Study Material 12th Lesson Biotechnology and its Applications Textbook Questions and Answers.

TS Inter 2nd Year Botany Study Material 12th Lesson Biotechnology and its Applications

Very Short Answer Type Questions

Question 1.
Expand GMO. How is it different from a hybrid?
Answer:

1. GMO means Genetically Modified Organisms.
2. Hybrids are new crop varieties produced by crossing two genetically different parents whereas GMO are plants, bacteria, fungi or animals whose genes have been altered by manipulation.

Question 2.
Give different types of cry genes and pests which are controlled by the proteins encoded by these genes.
Answer:

1. Genes Cry I Ac and Cry II Ab control the cotton bollworms.
2. Genes Cry I Ab controls corn borer.

Question 3.
Can a disease be detected before its symptoms appear? Explain the principle involved. [Mar. 2019, 2017]
Answer:

1. Yes, very low concentration of Bacteria or virus can be detected by amplification of their nucleic acids through PCR.
2. ELISA is used to detect infection by a pathogen based on the principle of antigen-antibody interaction.

Question 4.
Many toxic proteins are produced in their inactive form by micro-organisms. Explain how the mechanism is useful for the organism producing the toxin.
Answer:

1. The Bt toxin proteins exist as inactive protoxins but once an insect ingests the inactive toxin, it is converted into an active form of toxin due to the alkaline pH of the gut which solubilises the crystals.
2. The activated toxin binds to the surface of midgut epithelial cells and creates pores that cause swelling and lysis and eventually cause the death of the insect.

Question 5.
Why has the Indian parliament cleared the second amendment of the country’s patents bill?
Answer:

1. Amendment of the Indians patents bill by the Indian parliament, prevents unauthorized exploitation of their bioresources and traditional knowledge.
2. It also considers patent terms, emergency. Provisions and research and development initiatives.

Question 6.
Give any two reasons why the patent on Basmati should not have gone to an American company. [May 2014]
Answer:

1. Basmatic rice is distinct for its Unique aroma and flavour and 27 documented varieties of Basmati are grown in India.
2. There is reference to Basmati in ancient texts, forklore and poetry, as it has been grown in India for centuries. The new variety of Basmati produced by American company had actually been derived from Indian farmers variety.

Question 7.
PCR is a useful tool for early diagnosis of an infectious disease. Elaborate.
Answer:

1. The very low concentration of a bacteria or virus can be detected by amplification of their nucleic acid through PCR.
2. PCR is now used to detect HIV in suspected AIDS patient. It is used to detect mutations in genes, in suspected cancer patients too. It is a powerful technique to identify many genetic disorders.

Question 8.
What is GEAC and what are its objectives? [Mar. ’18, ’14; May ’17]
Answer:

1. GEAC – Genetic Engineering Approval Committee, established by Govt, of India.
2. It make decisions regarding the validity of GM research and the safety of introducing GM organisms for public services.

Question 9.
Name the nematode that infects the roots of tobacco plants. Name the strategy adopted to prevent this infestation.
Answer:

1. A nematode, Meloidegyne incognitia infects the roots of tobacco plants and causes a great reduction in yield.
2. This infestation can be prevented based on the process of RNA interference (RNAi).

Question 10.
For which variety of Indian rice, has a patent been filed by a USA company.
Answer:

1. Basmati variety of rice with unique aroma and flavour.
2. Indian Basmati was crossed with semi dwarf varieties and claimed as an invention (a novelty) by an American company for patent in 1997.

Question 11.
Give one example for each of transgenic plants which are suitable for food processing and those with improved nutritional quality. [Mar. 2020]
Answer:

1. Transgenic tomato ‘Flavr Savr’ is bruise resistant, i.e., suitable for storage and transport due to delayed ripening and offers longer shelf life.
2. Transgenic golden rice obtained from “Taipei” is rich in vitamin A and prevents blindness.

Short Answer Type Questions

Question 1.
List out the beneficial aspects of transgenic plants. [March 2019, Mar. ’18; May ’17, ’14]
Answer:
Beneficial aspects of transgenic plants are
I. Transgenic crop plants having resistance to pathogens and pests :

1. Transgenic papaya is resistant to papaya ring spot virus.
2. Bt. cotton is resistant to insects.
3. Transgenic tomato plants are resistant to the bacterial pathogen pseudomonas.
4. Transgenic potato plants are resistant to the fungus phytophthora.

II. Transgenic plants suitable for food processing technology :
Transgenic tomato ‘Flavr Savr’ is bruise resistant i.e., suitable for storage and transport due to delayed ripening and offers longer shelf life.

III. Transgenic plants with improved nutritional value :
Transgenic golden rice obtained from ‘Taipei’ is rich in vitamin A and prevents blindness.

IV. Transgenic plants useful for hybrid seed production :
Male sterile plants of Brassica napus are produced. This will eliminate the problem of manual emasculation and reduce the cost of hybrid seed production.

V. Transgenic plants tolerant to abiotic stresses caused by chemicals, cold, drought, salt, heat etc.

1. Basmati variety of rice was made resistant against biotic and abiotic stresses.
2. Round up ready soyabean is herbicide tolerant.

Question 2.
What are some bio-safety issues concerned with genetically modified crops? [March 2017, 2014]
Answer:
Biosafety issues concerned with genetically modified crop are :

1. There is fear of transferring allergins or toxins to humans and animals as side effects.
2. There is a rick of changing the fundamental nature of vegetables.
3. They may pose a harmful effect on biodiversity and have an adverse impact on environment.
4. There is a risk of gene pollution due to transfer of the new genes into related wild species through natural out-crossing. This may result in the development of super weeds which may be fast growing than the crops and may be resistant to weedicides.
5. They may bring about changes in natural evolutionary pattern.

Question 3.
Give a brief account of A) Bt. cotton [Mar. 2020]
B) Pest resistant plants
Answer:
A) Bt Cotton :
Bt cotton is created by using some strains of a bacterium, Bacillus thuringiensis (Bt is short form).

1. This bacterium produces protein that kill certain insects such as lepidopterans (tobacco, bud worm, army worm), coleopterans (beetles) and dipterans (flies, mosquitoes).
2. B.thuringiensis forms protein crystals during a particular phase of growth. These crystals contain a toxic insecticidal protein.
3. Bt toxin protein exists as inactive protoxins but once an insect ingests the inactive toxin, it is converted into an active form of toxin due to alkaline pH of the gut which solubilises the crystals. .
4. The activated toxin binds to the surface of midgut epithelial cells and create pores that cause cell swelling and lysis leading to death of an insect.
5. Specific Bt toxin genes were isolated from Bacillus thuringiensis and incorporated into several crop plants.
6. Most Bt toxins are insect group specific. Hence, the toxin is coded by a gene named ‘cry’. For example, the proteins encoded by the genes Cry I Ac and Cry II Ab control the cotton bollworms and Cry I Ab controls corn borer.

B) Pest resistant plants are developed by using biotechnology processes :

1. A nematode Meloidegyne incognitia infects the roots of tobacco plants which reduces the production of tabaco.
2. RNA interference (RNA i) process is used to check by silencing of a specific mRNA due to a complementary RNA molecule. It occurs in all eukaryotic organisms as a method of cellular defense.
3. RNA binds and prevents translation of the mRNA (silencing).
4. Agrobacterium vectors are used to introduce – specific genes into the host plant. It produces both sense and anti – sense RNA in the host cells.
5. These two RNAs are complementary to each other and formed a double stranded RNA (ds RNA) that initiate RNAi and hence silenced the specific mRNA of the nematode.
6. The parasite cannot survive in transgenic host, so prevents the plants from pest. The transgenic plant, thus gets itself protected from the parasite.

Question 4.
Write notes on green revolution and gene revolution.
Answer:
Green revolution :
i) Around 1960s, several countries including India experienced substantial and dramatic increase in agricultural production which was termed as green revolution by William Gaud.
Norman Borlaug is regarded as “Father of Green Revolution”.

ii) Green revolution increased food production by the following ways.

1. Use of improved crop varieties
2. Use of agro-chemicals (fertilizers and pesticides)
3. Use of better management practices
4. By land reforms

Gene Revolution :
With the advent of biotechnology, especially the genetic engineering, has increased the food production and decreased the use of chemical fertilizers and pesticides which lead to other type of revolution called Gene revolution.

Gene revolution provided new plants that would to lead to a more environmentally sound agricutural production.
GR plants are useful in many ways. Their

1. production of high yeilding and disease resistant varieties.
2. made crops more tolerant to abiotic stresses (cold, drought; salt, heat).
3. reduced reliance on chemical pesticides (Pest-resistant crops) eg : Bt Cotton.
4. helped to reduce past harvest losses.
5. increased efficiency of mineral usage by plants. This prevents early exhaustion of fertility of soil.
6. enhanced nutritional value of food. Eg : Vitamin A enriched rice.

Long Answer Type Questions

Question 1.
Give an account of bio-technological applications in agriculture and other fields.
Answer:
Bio-technology application in agriculture : Involves following three options

1. Agro-chemical based agriculture
2. Organic agriculture
3. Genetically engineered crop based agriculture.

Genetically modified organisms (GMO) are plants, animals, bacteria and fungi whose genes have been altered by manipulation.

Genetic modification in organisms lead to following results :

1. Crops become more tolerant to abiotic stresses, such as cold, drought, salt, heat, etc.
2. Dependence on chemical pesticides reduced i.e., pest resistant crop.
3. Post harvest losses reduced.
4. Efficiency of mineral usage increased in plants (preventing loss of soil fertility).
5. Nutritional value of food enhanced eg: Vitamin A enriched rice.
6. Tailor-made plants are created to supply alternative resources to industries, in the form of starches, fuels and pharmaceuticals.

Some of the application of biotechnology in agriculture are the production of pest resistant plants, eg : Bt Cotton, Bt Corn, etc.

Bt. Cotton :
It is created by using some strains of a bacterium, Bacillus thuringiensis (Bt is short form).

1. This bacterium produces protein that kill certain insects such as lepidopterans (tobacco budworm, armyworm), coleopterans (beetles) and dipterans (flies, mosquitoes).
2. Bt toxin proteins exists as inactive protoxins but once insect ingest the inactive toxin it becomes active and leads to death of an insect.
3. Most Bt toxins are insect group specific, hence the toxin is coded by a gene named cry. For example, the proteins encoded by the genes Cry II Ac and Cry I Ab control the cotton bollworms and Cry I Ab controls corn borer.

Pest resistant Plants are developed by using biotechnology processes.

1. A nematode Meloidegyne incognitia infects the roots of tobacco plants which reduces the production of tobacco.
2. Agro bacterium vectors are used to introduce nematode-specific genes into the host plant. It produces both sense and anti – sense RNA in the host cells.
3. These two RNA are complementary to each other and forms a double stranded RNA (ds RNA) that initiate RNAi and hence silenced the specific mRNA of the nematode.
4. The parasite cannot survive in transgenic host, so prevents the plants from pest. The transgenic plant thus gets itself protected from the parasite.

Biotechnology and Environment:
Bio remediation :
In the process of using microbes and plants to break down or recycle environmental pollutants.

Utilization of sewage and agrowastes to produce biogas and vermicompost.

Biotechnological application in medicine :
Have made immense impact in the area of health care by enabling the mass production of safe and more effective therapeutic drugs.

Vitamins (A, B₁₂ etc.) and antibiotics (pencillin) are produced at low cost using microorganisms.

Genetically engineered insulin leads to sufficient availability of insulin for the management of adult onset diabetes.

Gene therapy is a collection of methods that allows correction of gene defects diagnosed in a child or embryo. Genes are inserted into a person’s cells or tissue to treat a disease. Molecular diagnosis helps to solve the problem of early diagnosis and treatment of diseases.

i) Using conventional methods of diagnosis (serum and urine analysis) early detection of diseases is not possible.
ii) To overcome this problem, some molecular diagnosis techniques provide early detection of diseases. These are
a) Recombinant DNA technology
b) Polymerase Chain Reaction
c) Enzyme Linked Immuno – Sorbent Assay (ELISA)

Intext Question Answers

Question 1.
Crystals of Bt toxin produced by some bacteria do not kill the liacteria themselves because –
a) bacteria are resistant to the toxin
b) toxin is immature;
c) toxin is inactive;
d) bacteria encloses toxin in a special sac.
Answer:
Toxin is inactive.

In bacteria the toxin is present in an inactive form, called pro toxin, which gets converted into active form when it enters the body of an insect.

Question 2.
What are transgenic bacteria? Illustrate using any one example.
Answer:
Transgenic bacteria contain foreign gene that is intentionally introduced into its genome. They are manipulated to express the desirable gene for the production of various commercially important products. An example of transgenic bacteria is E.Coli. In the plasmid of E.coli, the two DNA sequences corresponding to A and B chain of human insulin are inserted, so as to produce the respective human insulin chains. Hence after the insertion of insulin gene into the bacterium, it becomes transgenic and starts producing chains of human insulin. Later on these chains are extracted from E.coli and combined to form human insulin.

Question 3.
Compare and contrast the advantages and disadvantages of production of genetically modified crops.
Answer:
The production of genetically modified (GM) or transgenic plants have many advantages.

1. Most of the GM crops have been developed for pest resistance, which increases the crop productivity and therefore reduces the reliance on chemical pesticides.
2. Many varieties of GM food crops have been developed, which have enhanced nutritional quality.
   For example : Golden rice is a transgenic variety of rice which is rich in vitamin A.
3. These plants prevent the loss of fertility of soil by increasing the efficiency of mineral usage.
4. They are highly tolerant to unfavourable abiotic conditions.
5. The use of GM crops decreases the post harvesting loss of crops. However there are certain controversies regarding the use of genetically modified crops around the world. The use of these crops can affect the native biodiversity in an area. For example : The use of Bt toxin to decrease the amount of pesticide is posing a threat for beneficial insect pollinators such as honey bee. If the gene expressed for Bt toxin gets expressed in the pollen, then the honey bee might be affected. As a result, the process of pollination by honey bees would be affected. Also genetically modified crops are affecting human health. They supply allergens and certain antibiotic resistance markers in the body. Also they can cause genetic pollution in the wild relatives of the crop plants. Hence it is affecting our natural environment.

Question 4.
What are Cry proteins? Name an organism that produces it. How has man exploited this protein to his benefit?
Answer:
Cry proteins are encoded by cry genes. These proteins are toxins which are produced by Bacillus thuringiensis bacteria. This bacterium contains these proteins in their inactive form. When the inactive toxin protein is ingested by the insect it gets activated by the alkaline pH of the gut. This results in the lysis of epithelial cell and eventually the death of the insect. Therefore man has exploited this protein to develop certain transgenic crops with insect resistance such as Bt Cotton, Bt Corn, etc.

Question 5.
List the advantages of recombinant insulin.
Answer:

1. Its molecular structure is absolutely identical to that of the natural molecule.
2. It helps to have continuous supply of insulin and stabilization of its market price etc.

Question 6.
What is meant by the term bio-pesticide? Name and explain the mode of action of a popular bio-pesticide.
Answer:
Bio pesticides are the plants which are having resistance to insects.
Example: Bt cotton.

1. Bt cotton is created by using some strains of a bacterium, Bacillus thuringiensis (Bt is short form).
2. Bt toxin protein exist as inactive protoxins, but once an insect ingets the inactive toxin, it is converted into an active form of toxin due to the alkaline pH of the gut which solubilise the crystals.
3. The activated toxin binds to the surface of midgut epithelial cells and creates pores that cause cell swelling and lysis leading to death of an insect.

Telangana TSBIE TS Inter 2nd Year Botany Study Material 11th Lesson Biotechnology: Principles and Processes Textbook Questions and Answers.

TS Inter 2nd Year Botany Study Material 11th Lesson Biotechnology: Principles and Processes

Very Short Answer Type Questions

Question 1.
Define biotechnology.
Answer:

1. The European Federation of Biotechnology (EFB) defines Biotechnology as the intergration of natural science and organisms, cells, parts thereof, and molecular analogues for products and services.
2. Biotechnology is the science of utilising the properties and uses of microorganisms or to exploit cells and the cell constituents at industrial level for generating useful products essential to life and human welfare.

Question 2.
What are molecular scissors? Where are they obtained from?
Answer:

1. Molecular scissors are the restrition endonucleases that recognize and cut specific nucleotide sequences of DNA.
2. They are usually obtained from Bacteria. For the first time, Herbert Boyer (1969) isolated two restriction enzymes from E.coli.

Question 3.
Name any two artificially restructured plasmids. [May 2014]
Answer:

1. pBR 322 (named after Boliver and Rodriguez)
2. pUC 19,101 (named after University of California)

Question 4.
What is EcoRI? How does it function?
Answer:

1. EcoRI is a restriction enzyme obtained from a bacterium, Escherichia coli.
2. This enzyme specifically recognises GAA sites on the DNA and cuts it between G and A.

Question 5.
What are cloning vectors? Give an example.
Answer:

1. The DNA used for transforming and multiplying the foreign DNA sequences in a suitable host is called cloning vector.
2. Plasmids, Bacteriophages, Cosmids, and artificial chromosomes are commonly used cloning vectors.

Question 6.
What is recombinant DNA?
Answer:

1. The hybrid (chimeric)DNA formed by the intergration of a gene of interest within a suitable vector.
2. Both source DNA and vector DNA are cut with same restriction enzyme and are joined with DNA ligase to make rDNA.

Question 7.
What is palindromic sequence?
Answer:

1. Palindrome sequence: A sequence of base pairs that reads same on the two strands when orientation of reading is kept the same.
2. Eg : 5′ – GAATTC – 3′
   3′ – CTTAAG – 5′

Question 8.
What is the full form of PCR? How is it useful in biotechnology? [March 2018]
Answer:

1. PCR stands for Polymerase Chain Reaction. In this process, multiple copies of a gene are synthesized using a computerized machine called Thermocycler.
2. Multiple copies (1 billion) of the gene of interest are synthesized in vitro using two sets of primers and a DNA polymerase (Taq polymerase).

Question 9.
What is downstream processing? [March 2019, May ’17, Mar. ’14]
Answer:

1. Downstream processing : Separation and purification of a product that carried out after completion of the biosynthetic stage and before it is ready for marketing as a finished product.
2. This includes formulation with preservatives, clinical trials (for drugs) and quality control testing etc.

Question 10.
How does one visualize DNA on an agar gel? [March 2020]
Answer:

1. The separated DNA fragments can be visualised only after staining the DNA with a compound known as ethidium bromide followed by exposure to UV radiation.
2. Bands of DNA in an ethidium bromide stained gel appear in bright in orange colour under UV light, in an instrument called transilluminator.

Question 11.
How can you differentiate between exonucleases and endonucleases?
Answer:
1. Exonucleases :
Nucleases that cut DNA and remove nucleotides from the ends.

2. Endonucleases :
Nucleases that cut specific positions within the DNA.

Short Answer Type Questions

Question 1.
Write short notes on restriction enzymes.
Answer:
Restriction enzymes or molecular scissors belong to a class of enzymes called nucleases. It always cut DNA molecules at a particular point by recognizing a specific sequence of six base pairs known as recognition sequence.
They are of two types.

1. Exonucleases, which remove nucleotides from the ends of DNA.
2. Endonucleases, which cut the DNA at specific portions anywhere within its length.

Each restriction endonuclease recognizes a specific palindromic nucleotide sequence in the DNA Palindrome is a group of letters that forms the same words when read both forward and backward, eg: MALAYALAM. It is a sequence in DNA of base pairs that reads same on the two strands. When orientation of reading is kept same.

For example, the following sequence reads the same on the two strands in 5′ → 3′ direction as well as 3′ → 5′ direction
5′ – GAATTC – 3′
3′ – CTTAAG – 5′

The restriction enzymes are named as follows.

• The first letter of the name comes from the genus and the next two letters from the name of the species of the prokaryotic cell from which it is isolated.
• The next letter comes from the strain of the prokaryote.
• The Roman numbers following these four letters indicate the order in which the enzymes were isolated from that strain of the bacterium.
  eg : EcoRI from Escherichia coli RY13.
  Hind 11 from Haemophilus influenza.
  Bam H1 from Bacillus amyloliquefaciens.

Question 2.
Give an account of amplification of gene of interest using PCR.
Answer:
PCR stands for Polymerase Chain Reaction.

In this reaction, multiple copies of the gene (or DNA) of interest are synthesised in vitro using two sets of primers and the enzyme DNA polymerase. The enzyme extends the primers, using the nucleotides provided in the reaction and the genomic DNA as template.

If the replication of DNA is repeated many times, the segment of DNA can be amplified to approximately billion times i.e., 1 billion copies are made. Such repeated amplification is achieved by the use of a thermostable DNA polymerase such as Taq polymerase (isolated from a bacterium Thermus aquaticus) which remain active even during the high temperature induced denaturation of double stranded DNA. The amplified fragment, if desired, can now be used to ligate with a vector for further cloning.

Question 3.
What is a bio-reactor? Describe briefly the stirring type of bio-reactor.
Answer:
Bioreactors are large volume (100 – 1000L) vessels in which raw materials are biologically converted into specific products, individual enzymes etc. using microbial plant, animal or human ceils.

1. It provides all the optimal conditions for achieving the desired product by providing optimal growth conditions like temperature, pH, substrate, salt, vitamins and oxygen.
2. The most commonly used bioreactors are of stirring type as shown in figure.
3. The stirred-tank reactor is usually cylindrical or with a curved base to facilitate the mixing of the reactor contents.
4. The stirrer facilitates even mixing and oxygen availability throughout the bioreactor.
5. The components of a bioreactor are
   a) An agitator system
   b) An oxygen delivery system
   c) A foam control system
   d) A temperature control system
   e) pH control system
   f) Sampling ports to withdraw cultures periodically

(a) Simple stirred – tank bioreactor;
(b) Sparged stirred-tank bioreactor through which sterile air bubbles are sparged

Question 4.
What are the different methods of insertion of recombinant DNA into the host cell?
Answer:
Methods of insertion of rDNA into the host cell:

• The rDNA can be forced into the competent cells by incubating the cells with recombinant DNA on ice followed by placing them at 42°C and then putting them back on ice.
• Microinjection is a method in which the recombinant DNA is directly injected into the nucleus of the animal cell with the help of microneedles or micropipettes.
• Gene gun or biolistics is a method suitable for plant cells, where cells are bombarded with high velocity microparticles of gold or tungsten coated with DNA.
• Disarmed pathogens are used as vectors, when they are allowed to infect the cell, they transfer the recombinant DNA into the host.

Long Answer Type Questions

Question 1.
Explain briefly the various processes of recombinant DNA technology. [Mar. 18 17, 14; May 14]
Answer:
Recombinant DNA technology involves several steps in specific sequence such as
a) Isolation of DNA
b) Fragmentation of DNA by restriction endonucleases
c) Isolation of a desired DNA fragment
d) Ligation of the DNA fragment into a vector
e) Transferring the recombinant DNA into the host
f) Culturing the host cells in a medium at large scale
g) Extraction of the desired product.

a) Isolation of DNA :

1. DNA is enclosed within the membranes. To release DNA along with other macromolecules such as RNA, proteins, polysaccharides and lipids, bacterial cells / – plants or animal tissue are treated with enzymes such as lysozyme (bacteria) cellulose (plant cells), chitinase (fungus) to digest cell wall.
2. RNA can be removed by treatment with ribonuclease.
3. Proteins can be removed by treatment with protease.
4. Other molecules can be removed by appropriate treatments and purified DNA ultimately precipitates out after the addition of chilled ethanol.

b) Fragmentation of DNA by restriction endonucleases :
Cutting of DNA at specific locations can be done by using restriction enzymes. The purified DNA is incubated with the specific restriction-enzymes at conditions optimum for the enzyme to act.

c) Isolation of a desired DNA fragment :
Is carried out using agarose gel electrophoresis the DNA is negatively charged, it moves towards the positive electrode or anode and in the process, DNA separates out. The desired DNA fragment is eluted out. Amplification of the gene of interest: Using Polymerase Chain Reaction (PCR) is a reaction in which amplification of specific DNA sequences is carried out in vitro.

i) PCR technique requires :

1. A DNA template which is a double-stranded DNA that needs to be amplified.
2. Primers are small chemically made oligonucleotides of about 10-18 nucleotides that are complementary to a region of template DNA.
3. Enzymes used are Taq polymerase (from Thermus aquaticus) and the vent polymerase (from Thermococcus litoralis).

ii) Steps in PCR :

1. Denaturation of double-stranded DNA is carried out by applying high temperature of 95°C for 15 seconds. Each separated single-stranded strand acts as a template for DNA synthesis.
2. Annaling is carried out in two sets of primers. Which are added and anneal to the 3′ end of each separated strand. Primers act as initiator of replication.
3. Extension is done by DNA polymerase of primers by adding nucleotides complementary to the template provided in the reaction.
4. A thermostable DNA polymerase (taq polymerase) is used in the reaction which can tolerate the high temperature of the reaction.
5. These steps are repeated many times to obtain several copies of desired DNA.

d) Ligation of the DNA fragment into a vector requires a vector DNA and source DNA.

1. These are cut with the same endonuclease to obtain sticky ends.
2. Both are then ligated by mixing vector DNA, gene of interest and enzyme DNA ligase to form recombinant DNA.

e) Insertion of recombinant DNA into the host cell :
Organism occurs by several methods, before which the recipient cells are made competent to receive the DNA.

1. If recombinant DNA carrying antibiotic resistance (eg., ampjcillin) is transferred into E.coli cells, the host cell is transformed into ampicillin resistant cells
2. The ampicillin resistant gene can be called as selectable marker.
3. When transformed cells are grown on agar plates containing ampicillin, only transformants will grow and other will die.

f) Culturing the host cells in a medium at a large scale :
It is carried out in appropriate medium at optimal conditions. The DNA gets multiplied and express itself to form desired products.

g) Extraction of desired gene products :
It is carried out by following steps.

1. A protein encoded gene expressed in a heterologous host is called recombinant protein.
2. Cells having genes of interest can be grown on a small scale or on a large scale.
3. In small scale cells are grown on cultures and then expressed protein is extracted and purified by various separation methods.
4. In large scale cells are grown in a continuous culture system in which fresh medium is added from one side to maintain cells growth phase and the desired protein is collected from the other side.

Question 2.
Give a brief account of the tools of recombinant DNA technology. [March 2020, March 2019, May 2017]
Answer:
The tools of recombinant DNA technology
i) Restriction enzymes
ii) Polymerase enzymes
iii) Ligases
iv) Vectors
v) Host organism

i) Restriction enzymes :
Belong to a larger class or enzymes called Nucleases. These are two kinds.
a) Exonucleases :
Exonucleases remove nucleotides from the ends of the DNA.

b) Endonucleases :
Endonucleases make cuts at specific positions with in the DNA. Restriction enzymes cut the strand of DNA a little away from the centre of the palindrome sites, but between the same two bases on the opposite strands.
Eg : EcoRI recognises 5′ GAATTC 3′ sites on the DNA and cuts in between G and A.

Naming of Restriction Enzymes :

• The first letter of the name comes from the genus and the next two letters from the name of the species of the prokaryotic cell from which it is isolated.
• The next letter comes from the strain of the prokaryote.
• The Roman numbers following these four letters indicate the order in which the enzymes were isolated from that strain of the bacterium.
  Eg: EcoRI from Escherichia coli RY13
  Hind II is from Haemophilus influenzae
  Bam HI is from Bacillus amyloliquefaciens

ii) Polymerase enzymes :
Thermas aquaticus a bacterium yields DNA polymerase used in biotechnology.

1. This enzyme remains active during the high temperature applied during denaturation of double-stranded DNA.
2. It extends the primers using the nucleotides provided in the reaction and the genomic DNA as template.
3. Repeated amplification is achieved by this enzyme. The amplified fragments, if desired can be used to ligate with a vector for further cloning.

iii) Ligases :
The enzyme DNA ligase joins the complementary ends of the plasmid DNA with that of desired gene by covalent bonding to regenerate a circular hybrid called Recombinant (r) DNA or chimeric DNA.

iv) Vectors :
The DNA used as a carrier for transferring a fragment of foreign DNA into a suitable host called vector.

Vectors used for multiplying the foreign DNA sequence are called cloning vectors. Commonly used vectors are plasmids, bacteriophages, cosmids.

Properties of cloning vector

1. It must have low molecular weight.
2. It must have unique cleavage site for the activity of restriction enzymes at single point.
3. It must be able to replicate inside the host cell after its introduction (through ori gene – origin of replication).
4. The vectors requires a selectable marker, which helps in identifying and eliminating non transformants and selectively permitting the growth of transformants.

v) Host organisms :
Competent host for transformation with Recombinant DNA is required as tool.

Intext Question Answers

Question 1.
Do eukaryotic cells have restriction endonucleases? Justify your answer.
Answer:
No, eukaryotic cells do not have restriction endonucleases. This is because the DNA of eukaryotes is highly methylated by a modification enzyme called m&thylase. Methylation protects the DNA from the activity of restriction enzymes. These enzymes are present in prokaryotic cells were they help prevent the invasion of DNA by virus.

Question 2.
Besides better aeration and mixing properties, what other advantages do stirred tank bioreactors have over shake flasks?
Answer:
The shake flask method is used for a small scale production of biotechnological products in a laboratory. On the other hand, stirred tank bioreactors are used for a large scale production of the biotechnology products stirred tank bioreactors have several advantages over shake flasks.

1. Small volumes of culture can be taken out from the reactor for sampling or testing.
2. It has a foam breaker for regulating the foam.
3. It has a control system that regulates the temperature and pH.

Question 3.
Can you recall meiosis and indicate at what stage a recombinant DNA is made?
Answer:
Meiosis is a process that involves the reduction in the amount of genetic material. It is two types, namely meiosis I and meiosis II. During the pachytene stage of prophase I, crossing over of chromosomes takes place where the exchange of segments between non-sister chromatids of homologous chromosomes takes place. This results in the formation of recombinant DNA.

Question 4.
Describe briefly the following :
a) Origin of replication
b) Bioreactors
c) Downstream processing
Answer:
a) Origin of replication :
Origin of replication is defined as the DNA sequence in a genome from where replication initiates. The initiation of replication can be either uni-directional or bi-directional. A protein complex recognizes the ‘on’ site, unwinds the two strands and initiates the copying of the DNA.

b) Bioreactors :
Bioreactors are large vessels used for the large scale production of biotechnology products from raw materials. They provide optimal conditions to obtain the desired product by providing the optimum temperature, pH, vitamin, oxygen, etc. Bioreactors have an oxygen delivery system, a foam control system, a pH, a temperature control system, and a sampling port to obtain a small volume of culture for sampling.

c) Downstream processing :
Downstream processing is a method of separation and purification of foreign gene products after the completion of the biosynthetic stage. The product is subjected to various processes in order to separate and purify the product. After downstream processing, the product is formulated and is passed through various clinical trails for quantity control and other test.

Question 5.
Explain briefly
a) PCR
b) Restriction enzymes and DNA
c) Chitinase
Answer:
a) PCR :
PCR stands for Polymerase Chain Reaction. In this reaction, multiple copies of the gene (or DNA) of interest are synthesised in vitro using two sets of primers and the enzyme DNA polymerase.

b) Restriction enzymes and DNA :
It is also called as molecular scissors. It belongs to class of enzyme called nucleases. It always cut DNA molecules at a particular point by recognizing a specific sequence of six base pairs known as recognition sequence. DNA : DNA stands for Deoxyribo Nucleic Acid. It is a double stranded molecule. It contains deoxyribose sugar. It has the ability to replicate. It is a component of the chromosome.

c) Chitinase :
Chitinase is a class of enzymes used for the degradation of chitin, which forms a major component of the fungal cell wall. Therefore, to isolate the DNA enclosed within the cell membrane of the fungus, enzyme chitinase, is used to break the cell for releasing its genetic material.

Question 6.
Discuss with your teacher and find out how to distinguish between
a) Plasmid DNA and Chromosomal DNA
b) RNA and DNA
c) Exonuclease and Endonuclease
Answer:
a) Plasmid DNA and Chromosomal DNA :

b) RNA and DNA :

c) Exonuclease and Endonuclease :

Question 7.
What does ‘H’ in’d’ and ‘III refer to in. the enzyme Hind III?
Answer:
In naming restriction enzymes, the first letter comes from the genus Haemophilus (H)
The next two letters comes from the name of the species influenzae (in).
The next letter comes from the strain of the prokaryote (d).
The Roman numbers following these four letters indicate the order in which the enzymes were isolated from that strain of the bacterium.

Question 8.
Restriction enzymes should not have more than one site of action in the cloning site of a vector. Comment.
Answer:
Cloning sites are required to link alien DNA with vector.

1. For this, the vector requires single recognition sites forthe commonly used restriction enzymes.
2. Presence of more than one restriction sites within the vector will generate several fragments leading to complication to gene cloning.

Question 9.
What does ‘competent’ refer to in ‘competent cells’ used in transformation experiments?
Answer:
Since DNA is a hydrophillic molecule, it cannot pass through cell membranes. In order to force bacteria to take up the plasmid, the bacterial cells must first be made competent to take up DNA. This is done by treating them with a specific concentration of a divalent cation, such as calcium, which increases the efficiency with which DNA enters the bacterium through pores in its cell wall.

Question 10.
What is the significance of adding proteases at the time of isolation of genetic material (DNA).
Answer:
Proteins can be removed by treating them with proteases at the time of isolation of genetic material (DNA).

Question 11.
While doing a PCR, ‘denaturation’ step is missed. What will be its effect on the process?
Answer:
Annealing and Extensions and amplifing does not takes place.

Question 12.
What modification is done on the Ti plasmid of Agrobacterium tumefaciens to convert it into a cloning vector?
Answer:
Agrobacterium tumefaciens, a pathogen of several dicot plants is able to deliver a piece of DNA known as T-DNA to transform normal plant cells into tumor and directs these tumor cells to produce the chemicals required by the pathogens. The tumor inducing (Ti) plasmids of Agrobacterium tumifaciens has now been modified into a cloning vector such that it is no longer pathogenic to plants but it still able to use the mechanisms to deliver genes of our interest into a variety of plants.

Question 13.
What is meant by gene cloning?
Answer:
The selected fragments of DNA are inserted into a suitable vector to produce a large number of copies of genes. This is called gene cloning.

Question 14.
Decide the ratio between ester bonds and hydrogen bonds that are broken in each palindromic sequence of a DNA when treated with EcoRI during the formation of sticky ends.
Answer:
Equal ratio 2 : 8, i.e 1: 4.

Telangana TSBIE TS Inter 2nd Year Botany Study Material 10th Lesson Molecular Basis of Inheritance Textbook Questions and Answers.

TS Inter 2nd Year Botany Study Material 10th Lesson Molecular Basis of Inheritance

Very Short Answer Type Questions

Question 1.
What is the function of histones in DNA packaging?
Answer:

1. Histones are positively charged basic proteins. They are organized to form a unit of eight molecules called histone octamer.
2. The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called nucleOsome.

Question 2.
Distinguish between heterochromatin and euchromatin. Which of the two is transcriptionally active? [Mar. 2020]
Answer:
1. Euchromation :
Regions of chromatin that are loosely packed and stained lightly. It is transcriptionally active.

2. Heterochromation :
Regions of chromatin that are densely packed and stained dark. It is transcriptionally inactive.

Question 3.
Who proved that DNA is genetic material? What is the organism they worked on? [May 2017]
Answer:

1. Alfred Hershey and Martha Chase proved that DNA is genetic material.
2. They worked with viruses that infect bacteria, bacteriophages.

Question 4.
What is the function of DNA polymerase?
Answer:

1. DNA polymerase uses a DNA template to catalyze polymerization of deoxynucleotides.
2. It is highly efficient and catalyses polymerization in only one direction (5′ → 3′) with high degree of accuracy.

Question 5.
What are the components of a nucleotide? [Mar. ’18, ’17; May ’14]
Answer:

1. A pentose sugar, a phosphate group and a nitrogen base are the 3 components of a nucleotide.
2. The pentose sugar is ribose in nucleotides of RNA and it is deoxyribose in nucleotides of DNA.

Question 6.
Write the structure of chromatin.
Answer:

1. Nucleosomes are formed due to wrapping of negatively charged DNA around the positively charged Histone octamers. These repeating units in the nucleus form chromatin.
2. The nucleosomes in chromatin are seen as ‘beads – on – string’.

Question 7.
What is the cause of discontinuous synthesis of DNA on one of its parental strands? What happens to these short stretches of synthesised DNA?
Answer:

1. The DNA – dependent DNA polymerase can catalyze polymerization in only one direction, 5′ → 3′ and hence DNA synthesis is discontinuous on lagging of strand (with 5′ → 3′ polarity) of parental DNA.
2. The discontinuously synthesized Okazaki fragments on lagging strand are later joined by DNA ligase to form a continuous strand.

Question 8.
Given below is the sequence of coding strand of DNA in a transcription unit 3′ – AATGCAGCTATTAGG – 5′
Write the sequence of
a) its complementary strand
b) the mRNA
Answer:
a) Its complementary strand
5′ – TTACGTCGATAATCC – 3′

b) The mRNA
5′- UUACGUCGAUAAUCC – 3′

Question 9.
In a nucleus, the number of ribonucleoside triphosphates is 10 times the number of deoxy ribonucleoside triphosphates, but only deoxyribonucleotides are addded during the DNA replication. Suggest a mechanism.
Answer:

1. DNA is a polymer made of deoxyribonucleotides.
2. Absence of 2 – OH group is nuclosides confers stability to DNA molecules.

Question 10.
Name any three viruses which have RNA as the genetic material.
Answer:

1. Tobacco Mosaic virus
2. Polio virus
3. HIV
4. QB Bacteriophage

Question 11.
Write the sequence of nucleotides after single base insertion and deletion in the given gene:
Gene: THE CAT ATE THE FAT RAT
Answer:
1. If single base T is inserted between ‘THE’ and ‘CAT’, then
Gene : THE TCA TAT ETH EFA TRA T

2. If single base C is deleted from ‘CAT’, then
Gene: THE ATA TET HEF ATR AT

Question 12.
Why was DNA chosen over RNA as genetic material in the majority of the organisms?
Answer:

1. RNA consists of 2 – OH’ group at every nucleotide which is reactive group and makes RNA liable and eassily degradable. It is single stranded, catalyst, reactive and hence unstable.
2. DNA lacks 2 – OH’ group, double stranded, consists of thymine and resists changes by evolving a process of repair. Hence it is a stable genetic material in majority of the organisms.

Question 13.
What are the components of a transcription unit? [March 2019]
Answer:
The three major components of a transcription unit are (1) A Promoter (2) The structural gene (3) A terminator.

Question 14.
What is the difference between exons and introns?
Answer:
1. Exons :
The coding (expressed sequences in split genes of eukaryotes and will appear in processed (matured) RNA.

2. Introns :
The non coding sequences in split genes of eukaryotes that interrupt introns and they do not appear in processed RNA.

Question 15.
What is meant by capping and tailing? [May 2017]
Answer:
1. Capping :
It is a process in which, an unusual nucleotide (methyl guanosine triphosphate) is added to the 5′ end of hn RNA.

2. Tailing :
In this process, adenylate residues (200 – 300) are added at 3′ – end in a template independent manner.

Question 16.
What is meant by point mutation? Give an example. [May 2014]
Answer:

1. The mutation that occurs in a single base pair of a gene is called point (gene) mutation.
2. A point mutation in the gene for Beta globin chain (in human haemoglobin) results a disease, sickle cell anaemia.

Question 17.
Define peptide bond. Why are proteins referred to as polypeptide chains?
Answer:

1. The bond between two adjacent amino acids in a protein is known as peptide bond.
2. Proteins are the macromolecules and polymers. Amino acids, are joined by many peptide bonds to form proteins and hence are referred as polypeptide chains.

Question 18.
What is meant by charging of tRNA?
Answer:

1. Charging of tRNA (amino acylation of tRNA): Amino acids are activated in the presence of ATP and linked to their cognate tRNAs.
2. This favours the formation of peptide bond by providing energy.

Question 19.
What is a regulator and a promoter?
Answer:

1. Regulator : This gene directs the activity of the operator gene by producing an inhibitor protein called repressor.
2. Promoter is a region of DNA where RNA polymerase binds and initiates transcription.

Question 20.
During DNA replication, why is it that the entire molecule does not open in one go? Explain replication fork.
Answer:

1. For long DNA molecules, the entire molecule does not open in one go due to very high energy requirement.
2. The replication occur within a small opening of the DNA helix referred to as replication fork. This is the Y-shaped structure formed when the double-stranded DNA is unwounded upto a point during its replication.

Question 21.
What is the function of the codon-AUG? [Mar. ’20, ’14]
Answer:

1. AUG acts as the initiation codon (start codon) during formation of mRNA.
2. It also codes for an aminoacid, Methionine.

Question 22.
Define stop codon. Write the codons. [March 2019]
Answer:

1. The codons that do not code for any amino acid and responsible for termination of protein synthesis / translation process are called as stop codons.
2. There are 3 stop codons ie., UAA, UAG and UGA.

Question 23.
What is the difference between the template strand and a coding strand in a DNA molecule? [May 2014]
Answer:

Question 24
Write any two chemical differences between DNA and RNA. [March 2017]
Answer:

Question 25.
In a typical DNA molecule, the proportion of Thymine is 30% of the N bases. Find out the percentages of other N bases.
Answer:

1. According to Erwin Chargaff in double stranded DNA the ratio between A and T and between G and C are constant and each equals one.
2. Adenine – 30%, Guanine – 20%, Cytocine – 20%.

Question 26.
The proportion of nucleotides in a given nucleic acid are: Adenine 18%, Guanine 30%, Cytosine 42%, and Uracil 10%. Name the nucleic acid and mention the number of strands in it. [March 2018]
Answer:

1. As there is Uracil (pyramidine) is the given problem the nucleic acid is RNA.
2. The proportion of A ≠ U and G ≠ C, so it is single stranded RNA.

Question 27.
If the base sequence of a codon in mRNA is 5′ AUG-3′. What is the sequence of tRNA pairing with it?
Answer:
3′ – UAC – 5′

Short Answer Type Questions

Question 1.
Define transformation in Griffith’s experiment. Discuss how it helps in the identification of DNA as genetic material.
Answer:
Transformation is defined as the uptake of a naked DNA molecule of the fragments of a bacterial cell and the incorporation of this DNA molecule into the recepient chromosome in a heritable form.

In 1928 Frederick Griffith performed the experiments on Bacterial transformation with Streptococcus pneumonia the bacterium causing pneumonia. During the course of his experiment he found that a living organism (bacteria) could change in physical form.

• He observed two strains of this bacterium, one forming smooth colonies with capsule (S-type) and the other forming rough colonies without capsule (R-type).
• The S-type cells are virulent while R-type cells are non-virulent.
• When live S-type cells are injected into the mice, they suffered from pneumonia and died.
• When live R-type cells are injected into the mice, the disease did not appear and the mice survived.
• When heat killed S-type cells were injected, the disease did not appear.
• When heat killed S-type were mixed with live R-type cells and injected into the mice, the mice died of pneumonia and live S-type cells were isolated from the body of mice.
• He concluded that the R-strain had some how been transformed by the heat killed S- strain bacteria, which must be due to the transfer of genetic material the transforming principle.

Question 2.
Who revealed the biochemical nature of the transforming principle? How was it . done? Oswald Avery, Colin Mac Lead, and Madyn Me carty.
Answer:
Avery Mac Lead and Me carty revealed the biochemical nature of the transforming principle in a Griffith experiment.

• They purified biochemicals like proteins, DNA and RNA from the heat killed S-cells to see which one could transform live R cells into S cells.
• When their fraction were added to the culture of live R-cells, DNA was able to cause transformation of R-cells into S-cells.
• They also found that protein digesting enzymes and RNA digesting enzymes did not affect transformation indicating that transforming substance is not a protein or RNA.
• Digestion with DNase did inhibit transformation, this suggests that the DNA cause transformation.

Question 3.
Discuss the significance of heavy isotope of nitrogen in Meselson and Stahl’s experiment.
Answer:
a) Matchew Meselson and Franklin Stahl performed an experiment using Escherischia coli to prove that DNA replication is semi conservative.

b) They grew E.coli in a medium containing ¹⁵NH₄Cl (¹⁵N is the heavy isotope of nitrogen) as the only nitrogen source for many generations. This heavy DNA can be separated from the normal DNA by centrifugation in Cesium Chloride (CsCl) density gradient (Note that ¹⁵N is not a radioactive isotope and it can be separated from ¹⁴N based on densities only).

c) Then they transferred the cells into a medium with normal ¹⁴NH₄Cl and took out samples at various time intervals and extracted DNA and centrifuged them to measure their densities.

d) Thus the DNA that was extracted from the culture one generation after transfer from ¹⁵N to ¹⁴N medium (that is after 20 minutes: E.coli divides in 20 minutes) had a hybrid or intermediate density.

e) The DNA extracted after two generations (i.e., after 40 minutes) consisted of equal amounts of light DNA and hyrbid DNA.

Question 4.
Define a cistron. Differentiate between monocistronic and polycistronic transcription unit with suitable examples. [May 2014]
Answer:
Cistron is defined as a segment of DNA coding for a polypeptide.

Monocistronic transcription :
Mostly in eukaryotes. The structural gene only one transcription unit can translate only one protein chain (or) one polypeptide chain. So it be said to be monocistronic transcription.

Polycistronic transcription:
Mostly in prokaryotes. The structural gene in a transcription unit can translate many polypeptides be said to be polycistronic transcription.

Question 5.
Recall the experiments done by Frederick Griffith in which DNA was speculated to be the genetic material. If RNA, instead of DNA, was the genetic material, would the heat killed strain of Pneumococcus have transformed the R-strain into virulent strain? Explain.
Answer:

1. Stability as one of the properties of genetic material was clearly evident in Griffith’s transforming principle.
2. Heat which killed bacteria, at least did not destroy some properties of genetic material.
3. This can be explained by DNA. The two strands, being complementary, if separated by heating, come together when appropriate conditions are provided.
4. RNA is has 2′ – OH group present at every nucleotide. It make RNA labile and easily degradable, RNA is reactive.
5. RNA cannot be genetic material as RNA is degradable to heat, it cannot transform R strain into virulent strain.

Question 6.
There is only one possible sequence of amino acids when deduced from a given nucleotides. But a multiple nucleotide sequence can be deduced from a single amino acid sequence. Explain this phenomenon.
Answer:
AUG UUU UUC UUC UUU UUU UUC the only possible sequence of amino acid is
Met Phe Phe Phe Phe Phe Phe

Similarly from the sequence of following amino acids coded by an mRNA. Met-Phe-Phe-Phe-Phe-Phe-Phe the nucleotide sequence may be any one of UUU or UUC because these two codon code for Phenylalanine (Phe). As there are 64 codons for 20 amino acid and in which 3 codon do not code for any amino acids. Remaining codons must code for 20 amino acids. Therefore there will be more nucleotide for a single amino acid.

Question 7.
A single base mutation in a gene may not always result in loss or gain of function. Do you think the statement is correct? Defend your answer.
Answer:
No, the statement is not correct.

A classical example of point mutation is a change of single base pair in a gene for beta globin chain (in human haemoglobin) that results in the change of amino acid residue glutamate to valine. It results in a diseased condition called sickle cell anemia.

Thus a change in a single nucleotide in a codon alters the amino acid in a polypeptide chain.

For example :
Consider a statement that is made up of following words each having three letters like a genetic code.
RAM HAS RED CAP

If we insert a letter B in between HAS and RED and rearrange the statement it would read as follows.
RAM HAS BRE DCA P

The same can be repeated by deleting the one letter R then it will be the statement to make a triplet word.
RAM HAS EDC AP

Insertion or deletion of one or two bases changes the reading frame from the point of insertion or deletion. It can disrupt the protein structure and affect the functioning.

Question 8.
A low level of expression of lac operon occurs all the time. Can you explain the logic behind this phenomenon.
Answer:

• In lac operon (here lac refers to lactose), a polycistronic structural gene is regulated by a common promoter and regulatory genes. Such an arrangement is very common in bacteria and is referred to as operon.
• Lactose is the substrate for the enzyme beta-galactosidase and it regulates switching on and off of the operon. Hence it is termed as inducer.
• In the absence of a preferred carbon source such as glucose, if lactose is provided in the growth medium of the bacteria. It is transported into the cells through the action of permease.
• A very low level of expression of lac operon has to be present in the cell all the time, otherwise, lactose cannot enter the cells.

Question 9.
What background information did Watson and Crick have for developing a model of DNA? What was their contribution?
Answer:
Background information for developing a model of DNA was : DNA as an acidic substance present in the nucleus was first identified by Friedrich Meischer in 1869. He named it “Nuclein”. However, due to technical limitation in isolating such a long polymer intact, the elucidation of structure of DNA remained elusive for a long period of time.

It was in 1953 that James Watson and Francis Crick, based on the X-ray diffraction data produced by Maurice Wilkins and Rosalind Franklin, proposed a very simple and famous Double Helix model for the structure of DNA.

One of the hallmarks of their proposition was base pairing between the two strands of polynucleotide chains. However, this proposition was also based on the observation of Erwin Chargaff that for a double stranded DNA, the ratio between Adenine and Thymine and that between Guanine and Cytosine are constant and each equal ones.

Question 10.
What are the functions of
i) methylated guanosine cap,
ii) poly-A “tail” in a mature on RNA?
Answer:i) In capping and unusual nucleotide (methylated guanosine cap) is added to the 5′ end of hn RNA.

ii) In tailing, adenylate (Poly-A-tail) residues are added at 3’ end in the template independent manner. It is a fully processed hn RNA, now called mRNA.

Question 11.
How many types of RNA polymerases exist in cells? Write their names and functions.
Answer:
There are 3 types of RNA polymerases in the nucleus (in addition to the RNA polymerase found in the organelles).
They are

1. RNA polymerase I transcribes rRNAs (28s, 18s & 5.8s)
2. RNA polymerase II transcribes the precursor of mRNA, the heterogenous nuclear RNA (hn RNA).
3. RNA polymerase III. It is responsible for transcription of tRNA < 5sr RNA and sn RNAs (Small nuclear RNAs).

Question 12.
Write briefly about DNA polymerase.
Answer:
DNA polymerase is the main enzyme in the replication of DNA.

It uses a DNA template to catalyse the polymerisation of deoxynucleotides only, in one direction i.e., 5′ → 3′ leading to one strand replication continuous and the other one as discontinuous.

The DNA polymerase can not initiate the process of replication on their own. A small stretch of RNA (called a primer) is required for initiation.

Question 13.
What are the contributions of George Gamow, H.G. Khorana, Marshall Nirenberg in deciphering the genetic code?
Answer:
George Gamow, a physicist argued that since there are only 4 bases and if they have to code for 20 amino acids, the code should constitute a combination of bases. He suggested that in order to code for all the 20 amino acids, the code should be made up of three nucleotides. This was a bold proposition, because a permutation and combination of 43 (4 × 4 × 4) would generate 64 codons, generating more codons than required.

Har Gobind Khorana developed a chemical method in synthesising RNA molecules with defined combinations of bases (homopolymers such as UUU and copolymers such as UUC, CCA).

Marshall Nirenberg’s made cell free system for protein synthesis. It finally helped the code to be deciphered.

Question 14.
On the diagram of the secondary structure of tRNA shown below indicate the location of the following features :
a) Anticodon b) Acceptor stem c) Anticodon stem d) 5′ end e) 3′ end

Answer:

Question 15.
If there are 2.9 × 10⁹ complete turns in a DNA molecule estimate the length of the molecule.
(1 angstrom = 10^-8 cm).
Answer:
Distance between two consecutive base pair is 0.34 nm (0.34 × 10^-9 m)
Complete turns in DNA = 2.9 × 10⁹
Length of the DNA = Total no. of multiply with distance
between two consecutive bp = 2.9 × 10⁹ × 0.34 × 10^-9 = 0.986 m

Question 16.
Draw the schematic / diagrammatic presentation of the lac operon.
Answer:

Question 17.
What are the differences between DNA and RNA? [March ’20, ’14]
Answer:

Question 18.
Write the important features of Genetic code. [Mar. 18, 17]
Answer:
The important features of Genetic code are

1. The codon is Triplet. 61 codons code for amino acids and 3 codons do not code for any amino acids, hence they function as stop codons.
2. One codon codes for only one amino acid, hence it is unambiguous and specific.
3. Some amino acids are coded by more than one codon, hence the code is degenerate.
4. The codon is read in mRNA in a contiguous fashion. There are no punctuations.
5. The code is nearly universal. For example, from bacteria to human, UUU would code for pheylalaline (Phe). Some exceptions to this rule have been found in mitochondrial codons, and in some protozoans.
6. AUG has dual functions. It codes for Methionine (Met) and also acts as the initiotor codon.

Question 19.
Describe the sequential steps in the replication of a DNA molecule.
Answer:

• The process of replication involves a number of enzymes / catalysts of which DNA- dependent DNA-polymerase, is the major enzyme. It catalyses the polymerisation of the deoxy-ribonucleotides approximately at a rate of 2000 bs/second.
• The interwined DNA strand start separating from a particular point called origin of replication (which is single is prokaryotes and many in eukaryotes).
• This unwinding is catalysed by enzymes called helicases.
• Enzymes called topoisomerases break and reseal one of the strands of DNA, so that the unwind strands will not wind back.
• It is easy to add the base onto an already existing chain called primer.
• Primer is a short strech of RNA formed on the DNA template catalysed by enzyme primase.
• When a double stranded DNA is unwind upto a point, it shows a Y-shaped structure called replication fork.
• As new strands grow from the fork, it appears as if the point of divergence at the fork is moving.
• Enzyme DNA polymerase catalyses the joining of nucleotides (A, T, G, C) in the 5′ → 3′ direction.
• The enzymes forms one new strand in a continuous stretch (leading strand) in the 5′ → 3′ direction, on one of the template strands.
• On the other template strand, the enzyme forms short stretches of DNA (Okazaki fragments) also in the 5′ → 3′ direction.
• The Okazaki fragments are later joined by DNA – ligase to form the lagging strand.
• In prokaryotes, the DNA-polymerase does proof reading by removing any wrong base added, before proceeding to add new bases.
• The two strands are held together by hydrogen bonds between nucleotides.

Question 20.
Give a diagrammatic presentation of the process of transcription in a bacterial cell.
Answer:

Question 21.
Write briefly on nudeosomes. [Mar. 2019, May ’17]
Answer:

In DNA of eukaryotes, there are a set of positively charged basic proteins called histones.

• Histones are organised to form a unit of eight molecules called histone octamer.
• The negative charged DNA is wrapped around the positively charged histone octamer to form a structure called nudeosome.
• A typical nucleosome contains 200 bp of DNA helix.
• Nucleosome constitute the repeating unit of a structure in nucleus called chromatin.
• The nudeosomes in chromatin are seen as “beads-on-string” when viewed under electron microscope.

Long Answer Type Questions

Question 1.
Give an account of the Hershey and Chase experiment. What did it conclusively prove? If both DNA and proteins contained phosphorus and sulphur do you think the result would have been the same.
Answer:
Harshey and Chase Experiment:

• Their experiment is to prove that DNA is the genetic material and not the protein.
  
• They worked with bacteriophage T₂ which attacks bacterium E.coli.
• The grew some viruses on the medium that contained radioactive phosphorus and some others on a medium that contained radioactive sulphur.
• Virus grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus but protein does not.
• Similarly, virus, grown on radioactive sulphur contained radioactive protein but not radioactive DNA because DNA does not contain sulphur.
• Radioactive phages were allowed to attach to E.coli bacteria.
• The infection proceeded, the viral coats were removed from the bacteria by agitating them in a blender. The virus particles were separated from the bacteria by spinning them in a centrifuge.
• Bacteria which were infected with viruses that had radioactive DNA were radioactive, indicating that DNA was the material that passed from the virus to the bacteria.
• Bacteria that were infected with viruses that had radioactive proteins were not radioactive.
• This indicates that proteins did not enter the bacteria from the viruses.
• Thus it proves that DNA is the genetic material that is passed from virus to bacteria.
• If both DNA and proteins contained phosphorus and sulphur the result would not be same.

Question 2.
Give an account of post transcriptional modifications of a eukaryotic mRNA.
Answer:

• The primary transcripts contain both exons and introns and they are non-functional. Hence they are subjected to a process called splicing where the introns are removed and exons are joined in a defined order.
  
• hn RNA undergoes additional processing called capping and tailing, ft In capping an unusual nucleotide (methyl guanosine triphosphate) is added to the 5′ end of hn RNA.
• In tailing, adenylate residues (200-300) are added at 3′ end in a template independent manner.
• It is the fully processed hn RNA, now called mRNA, that is transported out of the nucleus for translation.

Question 3.
Discuss the process of translation in detail.
Answer:
Translation is the process of polymerization of amino acids to form a polypeptide.
i) The amino acids are joined by a bond which is known as a peptide bond. This process requires energy.
ii) The different phases of translation are
a) Activation of amino acids
b) Initiation of polypeptide synthesis
c) Elongation of polypeptide chain
d) Termination of polypeptide chain

a) Activation of amino acids :
It occur in presence of ATP and linked to their cognate tRNA i.e., charging of tRNA or aminoacylation of tRNA. If two such charged tRNA, are brought close, the formation of peptide bond between them would occur energitically in presence of a catalyst.

b) Initiation of polypeptide synthesis occurs in ribosomes (cellular factory for protein synthesis):

1. Ribosome consists of structural RNAs and about 80 different proteins.
2. In its inactive state, it exists as two subunits – a large and a small subunit.
3. When the small subunit encounters an mRNA, the process of translation of the mRNA to protein begins P-site and A-site.
4. There are two sites in the large subunits – the P-site and A-site for subsequent aminoacids to bind to and thus close enough to each other for the formation of a peptide bond.
5. The small subunit attaches to the large subunit in such a way that the initiation (AUG) comes to the P-site.

Elongation of polypeptide chain :
When a second tRNA charged with an appropriate amino acid binds to the A-site of the ribosome.
i) The peptide bond (CO – NH) forms between the carboxyl group of methionine and the amino group of the second amino acid. The reaction is catalysed by the enzyme peptidyl transferase.

ii) The complexes composed of an amino acid linked to tRNA, sequentially bind to the appropriate codon in mRNA by forming complementary base pairs with the tRNA anticodon.

iii) The ribosome moves from codon to codon along with the mRNA. Amino acids are added one by one, translated into polypeptide sequence dictated by DNA and represented by mRNA.

Termination of polypeptide synthesis occur when a released factor binds to the stop codon. As a result, the polypeptide synthesis or elongation stops releasing the complete polypeptide from the ribosome.

Question 4.
Write briefly about Griffith’s experiments on S. pneumoniae bacteria. What was his conclusion?
Answer:
Frederick Griffith (1928) performed the experiments on Bacterial transformation with Streptococcus pneumoniae, the bacterium causing pneumonia.

• He observed two strains of this bacterium, one forming smooth colonies with capsule (S-type) and the other forming rough colonies without capsule (R-type).
• The S-type cells are virulent while the R-type cells are non-virulent.
• When live S-type cells were injected into the mice, they suffered from pneumonia and died.
• When live R-type cells were injected into the mice, the disease did not appear and the mice survived.
• When heat killed S-type cells were injected, the disease did not appear.
• When heat killed S-type cells were mixed with live R-type cells and injected into the mice, the mice died of pneumonia and live S-type cells were isolated from the body of the mice.
• He concluded that the R-strain had somehow been transformed by the heat killed S-strain bacteria, which must be due to transfer of genetic material, the transforming principle.

Question 5.
Define an operon, giving an example. Explain what is an Inducible operon.
Answer:
Operon is a group of closely packed structural genes and regulatory elements (DNA sequences) such as a regulator, a promoter and an operator, which function in a coordinated manner.
F. Jacob and J. Monod were first described a transcriptionally regulated system.

The lac operon (here lac refers to lactose) consists of one regulatory gene (i gene), the promoter (p) and operator (o) and the structural genes (z, y and a).
i) i – codes for repressor of the operon
z – codes for beta galactosidase (β-gal) y – codes for enzyme permease
a – codes for transacetylase enzyme ‘

ii) All the three gene products in lac operon are required for metabolism of lactose.

iii) Lactose is the substrate for the enzyme betagalactosidase and its regulates switching on and off of the operon. Hence it is termed as inducer.

iv) The lactose induces operon in the following way.
a) Repressor of the operon is synthesized from the i-gene.
b) Repressor protein binds to the operator region of the operon and prevents RNA polymerase from transcribing the operon.
c) In presence of an inducer, such as lactose or allolactose, the repressor is inactivated by the interaction with inducer. This allows RNA polymerase access to the promoter and transcription proceeds.

v) Regulation of lac operon by repressor is referred to as negative regulation.

Question 6.
Give the salient features of the Double helix structure of DNA.
Answer:
The salient features of a Double-Helix structure of DNA are as follows.

1. It is made of two polynucleotide chains, where the backbone is constitued by sugar- phosphate and the bases project inside.
2. The two chains have anti – parallel polarity. This means that if one chain has the polarity 5′ → 3′, the other has 3′ → 5′.
3. The bases in two strands are paired through hydrogen bonds (H-bonds) forming base pairs (bp). Adenine forms two hydrogen bonds with Thymine from the opposite strand and vice-versa. Similarly Guanine is bonded with Cytosine with three hydrogen bonds. As a result, a Purine always comes opposite to a Pyramidine. This generates approximately uniform distance (20 A) between the two strands of the helix.
   
4. The two chains are coiled in a right handed fashion. The pitch of the helix is 3.4 nm a nanometer is one billionth of a metre, that is 10-9 m) and there are roughly 10 bp in each turn. Consequently, the distance between two successive base pairs (bp) in a helix is approximately equal to 0.34 nm.
5. The plane of one base pair stacks over the other in a double helix. This, in addition to H-bonds, confers stability of the helical structure.
   

Question 7.
Replication was allowed to take place in the presence of radioactive Deoxy-nucleotide precursors in E.coli that was a mutant for DNA ligase. Explain how the newly synthesised radioactive DNA will be when purified.
Answer:

1. In the long DNA molecule, the replication occurs within a small opening of DNA Helix, called Replication fork.
2. On one strand, the template with polarity (3′ – 5′) the replication is continuous.
3. On the other strand, the template with polarity (5′ – 3′), it would be discontinuous.
4. The discontinuously synthesized fragments are joined by the DNA ligase.
5. When Replication was allowed to take place in the presence of radioactive Deoxy – nucleotide precusors in E.Coli that was a mutant for DNA ligase, their Okazaki fragments will not be joined.

Intext Question Answers

Question 1.
Group the following as nitrogenous bases and nucleosides: Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.
Answer:
Nitrogenous bases present in the list are adenine, thymine, uracil and cytocine. Nucleosides present in the list are cytidine and guanosine.

Question 2.
If a double stranded DNA has 20 per cent of cytosine, calculate the percent of adenine in the DNA.
Answer:
According to Chargaff’s rule, the DNA molecule should have an equal ratio of pyrimidine (cytocine and thymine) and purine (adenine and guanine). It means that the number of adenine molecules is equal to thymine molecules and the number of guanine molecules is equal to cytosine molecules.
% A = % T and % G = % C

If ds DNA has 20% of cytosine then according to the law, it would have 20% of guanine. Thus percentage of G + C content = 40%. The remaining 60% represents both A + T molecules. Since adenine and guanine are always present in equal number the percentage of adenine molecule is 30%.

Question 3.
If the sequence of one strand of ONA is written as follows :
Write down the sequence of complementary strand in 3′ → 5′ direction.
5′ – ATGCATGCATGCATGCATGCATGCATGC – 3′
Answer:
The DNA strand are complementary to each other with respect to base sequence. Hence if the sequence of one strand of DNA is
5′ ATGCATGCATGCATGCATGCATGCATGC -3′

Then the sequence of complementary strand in
3′ TACGTACGTACGTACGTACGTACGTACG – 5′ direction.

Therefore the sequence of nucleotides in DNA polypeptide direction in 5′ – GCATGCATGCATGCATGCATGCATGCAT 3′.

Question 4.
If the sequence of the coding strand in a transcription unit is written as follows : 5′ – ATGCATGCATGCATGCATGCATGCATGC – 3′. Write down the sequence of mRNA.
Answer:
If the coding strand in a transcription unit is 5′ ATGCATGCATGCATGCATGCATGC ATGC 3′ Then the template strand in 3′ to 5′ direction would be 3′ TACGTACGTACGTACGTACGTACGTACG 5′

It is known that the sequence of mRNA is same on the coding strand of DNA. However, in RNA, thymine is replaced by uracil. Hence the sequence of mRNA will be 5′ AUGCAUGCAUGC AUGC AUGC AUGC – 3′

Question 5.
Which property of DNA double helix led Watson and Crick to hypothesise semiconservative mode of DNA replication? Explain.
Answer:
Watson and Crick observed that the two strands of DNA are anti parallel and complementary to each other with respect to their base sequences. This type of arrangement in DNA molecule led to the hypothesis that DNA replication is semiconservative. It means that the double stranded DNA molecule separates and then each of the separated strand acts as a template for the synthesise of a new complementary strand. As a result, each DNA molecule would have one parental strand and a newly synthesized daughter strand. Since only one parental strand is conserved in each daughter molecule. It is semi-conservative mode of replication.

Question 6.
Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesized from it (DNA or RNA), list the types of nucleic acid polymerases.
Answer:
There are two different types of polymerases.

1. DNA – dependent DNA polymerases.
2. DNA dependent RNA polymerases.

The DNA dependent DNA polymerases use a DNA template for synthesizing a new strand of DNA. Where as DNA-dependent RNA polymerases use a DNA template strand for synthesizing DNA.

Question 7.
How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
Answer:
Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus but protein does not. Similarly viruses grown on radioactive sulphur contained radioactive protein but not radioactive DNA because DNA does not contain sulphur.

Question 8.
Differentiate between the followings :
a) mRNA and tRNA
b) Template strand and Coding strand.
Answer:
a) mRNA is a single stranded and unfolded polynucleotide molecule. It carries a genetic information required for the synthesis of a specific protein from DNA to ribosomes in the form of triplet codons.

tRNA is the smallest RNA also called soluble RNA (sRNA) or Adaptor RNA or Translator RNA. It is like clover leaf. It is single stranded which is folded forming three distinct loops and a 4th indistinct loop which is considered as accessory arm along with a tail.

b) DNA at promoter sites gives two single strands, one of which is called template strand which is transcribed where as the other strand is called coding strand which does not code for anything and is not transcribed.

Question 9.
List two essential roles of ribosome during translation.
Answer:
Ribosomes are responsible for protein synthesis. The ribosomes consists of structural RNAs and about 80 different proteins. In its inactive state, it exists as two subunits : a large subunit and a small subunit. When the small subunit encounter an mRNA, the process of translation of the mRNA to proteins begins. There afe two sites in the large subunit for subsequent amino acids to bind to and thus be close enough to each other for the formation of a peptide bond. The ribosome also acts as a catalyst for the formation of a peptide bond.

Question 10.
In the medium where E.coli was growing, lactose was added, which induced the lac operon. Then, why does lac operon shut down some time after addition of lactose in the medium?
Answer:

• Lactose is the inducer for lac operon.
• The active form of lactose binds to the repressor protein and brings about a conformational change in the repressor.
• As a result the repressor is inactive i.e., it cannot bind to the operator.
• This provides access of the RNA polymerase to structural genes and transcription and production of enzymes continue and metabolism of lactose takes place.
• In its absence, the repressor is active and binds to the operator. There by switching off the process.

Question 11.
Explain (in one or two lines) the function of the followings :
a) Promoter b) t RNA c) Exons t
Answer:
a) Promoter :
It is a region of DNA where RNA polymerase binds and initiates transcription.

b) t RNA :
Transfer RNA is an adaptor molecule, that is used by all living organisms to bridge the genetic code in messenger RNA with the twenty amino acids in proteins. tRNA carry amino acids to ribosomes, where they are linked into proteins.

c) Exons :
In eukaryotes, the monocistronic structural genes have interrupted coding sequences – the genes in eukaryotes are split. The coding sequences or expressed sequences are defined as Exons. Exons are interrupted by introns.

Question 12.
Briefly describe the following:
a) Transcription
b) Translation
Answer:
a) Transcription :
Transfer of the genetic information from the DNA blue print to the mRNA is called transcription.

b) Translation :
Arrangement of aminoacids in a linear, specific sequence and formation of polypeptide chain according to the specific sequence of the information written on the m RNA is called translation.

Question 13.
How the polymerization of nucleotides can be prevented in a DNA molecule.
Answer:
DNA polymerases on their own cannot initiate the process of replication. By removing primer DNA template and absence of DNA polymerase.

Question 14.
In an experiment, DNA is treated with a compound which tends to place itself amongst the stacks of nitrogenous base pairs. As a result of this, the distance between two consecutive base pairs increases from 0.34 nm to 0.44 nm calculate the length of DNA double helix (which has 2 × 10⁹ bp) in the presence of saturating amount of this compound.
Answer:
Distance between 2 consecutive base pair in 0.44 nm (0.44 × 10^-9 m)
The length of DNA double helix = 6.6 × 2 × 10⁹ bp × 0.44 × 10^-9 m / bp = 5.8 metres.

Question 15.
Recall the experiments done by Frederick Griffith. Where DNA was speculated to be the genetic material. If RNA, instead of DNA was the genetic material, would the heat killed strain of Pneumococcus have transformed the R-strain into virulent strain? Explain.
Answer:
Heat, which killed the bacteria, at least did not destroy some of the properties of the genetic material.

Moreover 2′ – OH group present at every nucleotide in RNA is a reactive group and makes RNA liable and easily degradable.

RNA is known to be catalyst hence reactive.

So, If RNA instead of DNA was the genetic material, the heat killed strain of pseudococcus could not transform the R-strain into virulent strain.

Question 16.
You are repeating the Hershey-Chase experiment and are provided with two isotopes: ³²P and ¹⁵N (in place of ³⁵S in the original experiment). How do you expect your results to be different?
Answer:
Harshey and Chase worked to discover whether it was protein or DNA from the virus that entered the bacteria.

If we repeat the same experiment by using two isotopes ³²P and ¹⁵N (in place of ³⁵S in the original experiment) we cannot distinguish the difference between the protein or DNA.

³²P and ¹⁵N both are present in DNA in the Bacteria which was infected with virus. But we can not distinguish that protein did not enter the bacteria.

Question 17.
Do you think that the alternate splicing of exons may enable a structural gene to code for serveral isoproteins from one and the same gene? If yes, how? If not, why so?
Answer:
Yes. A gene is split into several sections separated by non-coding segments of DNA. Such genes are called discontinuous genes or split genes.

In split genes there are two sections – introns and exons.

Introns are non-coding sections of DNA segment. Before translation, the introns have to be removed and the exons reattached to produce a final copy of mRNA with continuous codons. This is called gene splicing.

According to the one gene one protein hypothesis of Beadle’arnd Tatum, each gene will separately code for one protein and that genes do not overlap.

So alternative splicing of exons may enable a structural gene to code for several isoproteins from one and the same gene.

Question 18.
Can you recall what centrifugal force is, and think why a molecule with higher mass / density would sediment faster?
Answer:
Centrifugal force is the force that acts away from the centre of the circle. Higher mass / density will be thrown towards outside because heavier particles experience more centrifugal force. So they would sediment faster.

Question 19.
Do Retroviruses follow central Dogma? Give one example.
Answer:
Francis Crick proposed the central dogma in molecular biology, which states that genetic information flows from

Retroviruses do not follow central dogma.

Retroviruses contain RNA as genetic material flow information in the reverse direction, that is from RNA to DNA.
Example: HIV

Telangana TSBIE TS Inter 2nd Year Botany Study Material 9th Lesson Principles of Inheritance and Variation Textbook Questions and Answers.

TS Inter 2nd Year Botany Study Material 9th Lesson Principles of Inheritance and Variation

Very Short Answer Type Questions

Question 1.
What is the cross between the Fa progeny and the homozygous recessive parent called? How is it useful?
Answer:

1. The cross between the Ft progeny -having dominant phenotype and the homozygous recessive parent is called Test cross.
2. The progenies of such a cross can be easily analysed and the genotype of the test organism can be determined.

Question 2.
Do you think Mendel’s laws of inheritance would have been different if the characters that he chose were located on the same chromosome?
Answer:

1. Yes, Mendel’s law of independent assortment would not be true for the genes that are located on the same chromosome.
2. Linkage refers to the physical association of genes on a chromosome and are called linked genes.

Question 3.
Who proposed the Chromosome theory of Inheritance? [Mar. 2019, 17]
Answer:

1. Walter Sutton and Theodore Boveri.
2. They used chromosomal segregation during meiosis to explain Mendel’s laws.

Question 4.
Define true breeding. Mention its significance.
Answer:

1. A true breeding line is one that has undergone continuous self pollination.
2. It shows the stable trait inheritance and expression for several generations.

Question 5.
Explain the terms phenotype and genotype. [May ’17, Mar. ’14]
Answer:

1. The physical or external appearance of a character (trait) is called phenotype.
2. The genetic makeup of an individual is called genotype.

Question 6.
What is point mutation? Give an example. [Mar. ’18, May ’14]
Answer:

1. The mutation that occurs in a single base pair of DNA molecule is called gene mutation or point mutation.
2. A classical example for point mutation is sickle cell anemia.

Question 7.
A person has to perform crosses for the purpose of studying inheritance of a few traits / characters. What should be the criteria for selecting the organisms?
Answer:

1. Organism must have well defined characteristics, can be grown and crossed easily.
2. It possess bisexual flowers and can be self fertilized conveniently.
3. It must have short life cycle and produce large number of offsprings.

Question 8.
In order to obtain the F₁ generation, Mendel pollinated a pure-breeding tall plant with a pure breeding dwarf plant. But, to get the F₂ generation, he simply self-pollinated the tall F₁ plants. Why?
Answer:

1. Pure breedingtall and dwarf plants are homozygous and give the same on self pollination. So he bred pure tall with pure dwarf to produce a hybrid.
2. Mendel selfed the F₁ to understand the inheritance of tall and dwarf characters and to know fate of suppressed trait in F₂.

Question 9.
How are alleles of a particular gene differ from each other? Explain its significance.
Answer:

1. Alleles are slightly different forms of the same gene ie., they differ by a single base pair.
2. They are significant in understanding inheritance and behaviour of the genetic polymorphs.

Question 10.
In a monohybrid cross between red and white flowered plants, a geneticist got only red flowered plants. On self-pollinating these Ft plants, he got both red and white flow¬ered plants in 3 :1 ratio. Explain the basis of using RR and rr symbols to represent the genotype of plants of parental generation.
Answer:

1. Red flowered plant crossed with white flowered plant gave red flowered F₁ plant and hence red is dominant over white.
2. In general, first letter of the dominant allele (Red) is used as symbol (R) to denote the genotype and its respective recesssive allele (white) is denoted by small letter (r).

Question 11.
What is the genetic nature of wrinkled phenotype of pea seeds? [Mar. 2020]
Answer:

1. Wrinkled character of seed in pea is a recessive tract.
2. Hence genotype of wrinkled phenotype for pea seeds is ‘rr’

Short Answer Type Questions

Question 1.
In a Mendelian monohybrid cross, the F₂ generation shows identical genotypic and phenotypic ratios. What does it tell us about the nature of alleles involved? Justify your answer.
Answer:
Incomplete dominance :
It is the phenomenon in which neither of the genes is completely dominant or completely recessive. As a result, the hybrid shows intermediate character. For example, the inheritance of flower colour in the dog flower. (Snapdragon or Antirrhinum sp). The cross between true breeding (homozygous) red flower (RR) and true breeding (homozygous) white flower plant (rr) F₁ (R r) was pink.

The phenotypic ratio deviates from Mendelian monohybrid ratio of 3 : 1 to 1 : 2 : 1 (Red flower – 1, Pink flowers – 2, White flower -1) since the heterozygous / hybrid shows a different phenotype genotype ratio remains the same as Mendelian ratio 1 : 2 : 1.

Question 2.
Mention the advantages of selecting pea plant for experiment by Mendel. [Mar. 2018, ’17 ’14]
Answer:

1. It is an annual plant that has well defined characteristics.
2. It can be grown and crossed easily.
3. It has bisexual flowers containing both female and male parts.
4. It can be self fertilized coveniently.
5. It has a short life cycle and produces large number of offsprings.

Question 3.
Differentiate between the following :
a) Dominant and Recessive [Mar. 2020]
b) Homozygous and Heterozygous [Mar. 2020]
c) Monohybrid and Dihybrid.
Answer:
a) The character which is expressed in F₁ generation is called dominant and that which is unexpressed is called recessive.

b) Parents carrying similar genes such as TT or tt are called homozygous and the parents with unlike genes like Tt are called heterozygous.

c) Cross involving two parents differing in only one character is called Monohybrid cross. For example, cross between tall (TT) and dwarf (tt) parents. Cross involving two parents differing in two characters is called a Dihybrid cross. For example, cross between pea plant having yellow round seeds (YYRR) with homozygous pea plant having green wrinkled seeds (yyrr).

Question 4.
Explain the Law of Dominance using a monohybrid cross. [May 2017]
Answer:
Mendel observed that in F₁ hybrids the character of tallness dominated or supress the dwarf character. The character which is expressed in F₁ generation is called dominant trait and that which remained unexpressed is called recessive trait.

Based on his observation on monohybrid crosses, Mendel proposed Law of Dominance.

Law of Dominance :

1. Characters are controlled by discrete units called factors.
2. Factors occur in pairs.
3. In a dissimilar pair of factors pertaining to a character one member of the pair dominates (dominant) the other (recessive).

The Law of Dominance is used to explain the expression of only one of the parental characters in a monohybrid cross in the F₁ and the expression of both in the F₂. It also explains the proportion of 3 : 1 obtained at the F₂.

The phenotypic ratio is 3 Tall: 1 Dwarf.
The genotypic ratio is 1 TT, 2Tt, 1 tt.

Question 5.
Define and design a test -cross.
Answer:

1. When the F₁ individuals are crossed with the recessive parent or organism similar in phenotype and genotype to the recessive parent, it is called test cross.
2. Test cross is used to test whether an individual is homozygous (pure) or heterozygous (hybrid).
3. A monohybrid test cross gives a phenotypic ratio of 1 : 1

Monohybrid Test cross

Question 6.
Using a punnet square, work out the distribution of phenotypic features in the first filial generation after a cross between a homozygous female and a heterozygous male for a single locus.
Answer:
Punnett square is a graphical representation to calculate the probability of the offsprings.

1. No recessive individuals are obtained in the progeny.
2. All the plants show phenotypically dominant character.
3. They show a genotypic ratio 1 : 1.
4. This is a test cross.

Question 7.
When a cross is made between tall plant with yellow seeds (It Yy) and a tall plant with green seeds (Tt yy). What proportion of phenotype in the offspring is expected to be
a) tall and green?
b) dwarf and green?
Answer:

a) Tall and green – 3/S
b) Dwarf and green – 1/8

Question 8.
Explain the following terms with examples. [May 2014]
a) Co-dominance
b) Incomplete dominance
Answer:
a) Co-dominance :
It is the phenomenon In which both the genes are equally dominant and so the character of both genes is well expressed in next generation. So in F1 generation regeneration resemble both parents.

Examples are different types of red blood cells that determine ABO blood grouping in human beings and seed coat pattern and size in lentil plants.

Lentil is a major grain legume (pulse) crop in N. America, A cross between pure-breed- ing spotted (having a few big irregular pathes) lentils and pure breeding dotted (having several small circular dots) lentils produce heterozygotes that are both spotted and dotted. Thus it shows the phenotypic features of both parents, which means that neither the spotted nor the dotted allele is dominant or recessive to the other.

b) Incomplete dominance :
It is the phenomenon in which neither of the genes is completely dominant or completely recessive. As a result the hybrid shows intermediate character. For example, the inheritance of flower colour in the dog flower. (Snapdragon or Antirrhinum sp). The cross between true breeding (homozygous) red flower (RR) and true breeding (homozygous) white flower plant (rr) F₁ (R r) was pink.

The phenotypic ratio deviates from Mendelian monohybrid ratio of 3 : 1 to 1 : 2 : 1 (Red flower -1, Pink flowers – 2, White flower -1) since the heterozygous / hybrid shows a different phenotype genotype ratio remains the same as Mendelian ratio 1 : 2 : 1.

Question 9.
Write a brief note on chromosomal mutations and gene mutations.
Answer:
Chromosomal Mutation :
Any change in structure and No. of chromosome is called chromosomal mutation.

1. Loss or gain of a segment of DNA, results in alternation in chromosomes. Because genes are located on chromosomes, alteration in chromosomes results in abnormalities.
2. Chromosomal alternations are common in cancer cells.

Gene Mutation :
Mutation that occurs due to change in a single base pair of DNA is called Gene Mutation or point mutation. E.g.: Sickle cell anaemia. Deletion and insertion of base pairs of DNA, Cancer from shift mutation.

Question 10.
How was it concluded that genes are located on chromosomes?
Answer:
Morgar hybridised yellow bodied and white eyed females to brown bodied, red eyed males and intercrossed their F₁ progeny. He observed that the two genes did not segregate independently with each other and the F₂ ratio deviated very significantly from the 9 : 3 : 3 : 1 ratio. He saw quickly that when the two genes in a dihybrid cross were situated on the same chromosome, the proportion of parental combinations was much higher than the non – parental type. He attributed this due to the physical association or linkage of the two genes and coined the term linkage and the term recombination to describe the generation of non parental gene combinations. He also found that, even when genes were grouped on the same chromosome, some genes were tightly linked and some genes were loosely linked.

Alfred Sturtevant used the frequency of recombination between gene pairs on the same chromosome as a measure of distance between genes and mapped their position on the chromosome.

Question 11.
A plant with red flowers was crossed with one having yellow flowers. If F₁ showed all flowers in orange colour, explain the inheritance.
Answer:

The cross between red flower and yellow flower, showed all orange colour flowers in F₁ generation. This shows that neither of the genes is completely dominant or completely recessive. As a result, the hybrid shows intermediate character. This is incomplete dominance.

Question 12.
Define Law of Segregation and Law of Independent Assortment.
Answer:
Law of Segregation or law of purity of gametes :
Based on the results obtained from the monohybrid cross Mendel postulated the first law “the law of segregation or law of purity of gametes.” It states that the two alleles of a gene when present together in a heterozygous state, do not fuse or blend in any way, but remain distinct and segregate during meiosis or in the formation of gametes so that each meiotic product or gamete will carry only one of them.

Law of Independent Assortment :
Based on the results of the dihybrid crosses in pea plant, Mendel postulated his second law known as “the law of Independent Assortment.” Which states that “when the plants differ from each other in two or more pairs of contrasting characters or factors, then the inheritance of one pair of factors is independent to that of the other pair of factors.

Question 13.
In peas, tallness is dominant over dwarfness and violet colour of flowers is dominant over the white colour. When a tall plant bearing violet flowers was pollinated with a dwarf plant bearing white flowers, different phenotypic groups were obtained in the progeny in numbers mentioned against them.
Tall, violet = 138 Dwarf, violet = 136
Tall, white = 132 Dwarf, white = 128
Mention the genotypes of the two parents and of the four offspring types.
Answer:
When a tall plant bearing violet flowers was pollinated with a dwarf plant bearing white flowers, the different phenotypic groups were in ratio nearly 1: 1:1: 1. This is Dihybrid test cross ratio. The genotype of two parents Tt Vv (Tall violet) and tt vv (Dwarf white) 

Question 14.
How do genes and chromosomes share similarity from the view point of genetical studies?
Answer:
Chromosomes are made up of DNA and proteins DNA is a special kind of molecules made up of nucleotides. There are four types of nucleotides. A DNA molecule consists of a long string of nucleotide pairs linked together. The sequence of nucleotide forms a code gene refer to a specific sequence of this DNA code that actually means characters all genes are DNA but not all the DNA are the genes.

Chromosomes are DNA molecule made up of a sequence of nucleotides. Genes are stretches of DNA that contain code to make protein. Thus both chromosome and genes are similar from the point of genetic studies.

Question 15.
With the help of an example differentiate between incomplete dominance and com-plete dominance.
Answer:
Incomplete dominance :
It is the phenomenon in which neither of the two alleles of a gene is completely dominant over the other.

1. The inheritance of flower colour in dog flower / snap dragon (Antirrhinum majus) and that in Mirabilis jalapa (4 o’ clock plant) are examples of this phenomenon.
2. When a cross is made between a red flowered plant with a white flowered plant of snap dragon the F₁ hybrid has pink flowers.
3. When the F₁ individual was self pollinated and an F₂ raised the F₂ contained individuals bearing red, pink and white flowers.

The phenotypic and genotypic ratio are same i.e., 1 Red : 2 Pink : 1 White.

Complete dominance is a phenomenon in which one allele of a gene expresses itself and suppresses the expression of the other allele of the same, when they are present to-gether in a hybrid.

1. When a cross was made between two individuals, one with tall stem (homozygous) and the other with dwarf stem, F₁ individual had tall stem.
2. When a F₁ individual is self pollinated the F₂ produced tall and dwarf individuals in the ratio of 3 : 1

The phenotypic ratio is 3 Tall : 1 Dwarf.
The genotypic ratio is 1 TT : 2 T t: 111.

Long Answer Type Questions

Question 1.
In a plant, tallness is dominant over dwarfness and red flower is dominant over white. Starting with the parents workout a dihybrid cross. What is standard dihybrid ratio? Do you think the values would deviate if the two genes in question are interacting with each other?
Answer:
In a plant tallness is dominant over dwarfness and red flower is dominant over the white cross between two parents with one parent, Tall with red flowers (TT RR) and other parent, Dwarf with white flowers (tt rr). Such a cross involving two parental plants differing in two character is called dihybrid cross.

When the F₁ hybrids were allowed to undergo self pollination, the F₂ generation progeny was showing four kinds of phenotypes in a definite pattern.

The phenotypic ratio is 9 : 3 : 3 : 1 and the genotypic ratio is 1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1.

The F₂ progeny showed not only the parental combination i.e., Tall with red flowers and Dwarf with white flowers but also the new combination i.e., Tall white and Dwarf red.

These new combination are produced due to genetic recombination.

The F₂ progeny are Tall purple 9 ; Tall white 3 ; Dwarf purple : 3; Dwarf white 1.
The dihybrid ratio is 9 : 3 : 3 : 1.
Yes, the values would deviate if the two genes interact with each other.

Intext Question Answers

Question 1.
What will be the phenotypic ratio in the offsprings obtained from the following crosses,
a) Aa × aa b) AA × aa c) Aa × Aa d) Aa × AA
Note : Gene ‘A’ is dominant over gene ‘a’.
Answer:
a) Aa × aa – P generation

The progeny shows 1 : 1 ratio.

b) AA × aa – P generation

The progeny shows Aa hybrid.

c) Aa x Aa – P generation

The progeny shows 1 : 2 : 1, 1 AA : 2 Aa : 1 aa genotype, 3 : 1 phenotype.
d) Aa × AA – generation

All the progeny shows dominant character.

Question 2.
In garden pea, the gene T for tall is dominant over its allele for dwarf. Give the geno-types of the parents in the following crosses.
a) tall × dwarf producing all tall plants.
b) tall × tall producing 3 tall and 1 dwarf plants.
c) tall × dwarf producing half tall and half dwarf number of plants.
Answer:
a) TT × tt → Tt all tall plants.
b) Tt × Tt → 3 tall and 1 dwarf plant.
c) Tt × tt → 1 tall (T t) and 1 dwarf (t t)

Question 3.
Mendel crossed pea plants producing round seeds with those producing wrinkled seeds. From a total of 7324 F₂ seeds, 5474 were round and 1850 were wrinkled. Using the symbols R and r for genes, predict the :
a) the parental (p) genotypes
b) the gametes
c) F₂ progeny
d) the cross between F hybrids
e) genotypes, phenotypes, genotypic frequency, phenotypic ratio of F₂ progeny.
Answer:
a) The parental genotypes are RR, rr (homozygous).
b) The gametes are (R) and (r).
c) F₁ progeny gene R r.
d) Cross between F₁ hybrids are R r × R r.
e) Genotypes are 3
Phenotypes are 2
Genotypic frequency 1 : 2 : 1
Phenotypic ratio is 3 : 1

Question 4.
The following data was obtained from an experiment on peas. The grey coloured seed is dominant over white coloured seed. Use the letter G for grey and g for white traits. Predict genotypes of the parents in each of the following crosses.

Answer:
a) Genotypes of the
Grey × white
Gg × gg

b) Grey × grey
Gg × Gg

c) White × white
gg × gg

d) Grey × grey
Gg × GG

Question 5.
In tomatoes, red fruit colour (R) is dominant to yellow (r). Suppose a tomato plant homozygous for red is crossed with one homozygous for yellow. Determine the appearance of the following.
a) the F₁ b) F₂, c) the offspring of a cross of the F₁ back to the red parent
d) the offspring of a cross of the F₁ back to the yellow parent.
Answer:
a) The appearance in F₁ are all red (Rr).

b) The appearance in F₂ are 3 : 1 (3 red; 1 yellow).

c) The appearance of the offspring of a cross of the F₂ back to the red parent are all red (homozygous red and heterozygous red). It is back cross.

d) The appearance of the offspirng of a cross of the F₁ back to the yellow parent are red and yellow in 1 : 1 ratio. It is test cross.

Question 6.
In pea, axillary position of flowers (T) is dominant over its terminal position (t). Coloured flowers (C) are dominant to white flowers (c). A true breeding plant with coloured flowers in axils is crossed to one with white terminal flowers. Give the phenotypes, genotypes and expected ratios of F₂, F₂ back cross and test cross progenies. What genotypic ratio is expected in the F₂ progeny?
Answer:
Axillary coloured flowers × Terminal white flowers – P

All the F₁ hybrids phenotypes are found to have Axillary coloured flowers.
All the F₁ hybrids genotypes are T t Cc.
The phenotype ratio of F₂ are 9 : 3 : 3 : 1.
The genotype ratio of F₂ are 1 : 2 : 2 :4 : 1 : 2 : 1 : 2 : 1.
Dihybrid test cross ratio is 1 : 1 : 1 : 1

In back cross when F₂ individuals are crossed with the parent having dominant traits no recessive individuals are obtained. All are Axillary coloured flowers.

Question 7.
In summer squash, a plant with white flowers and disc-shaped fruits is crossed to a plant with yellow flowers and sphere shaped fruits. The F₁ hybrids had white flowers and discshaped fruits. Which phenotypes are dominant? Give the genotypes of the parents and the hybrids. If these hybrids are selfed and 256 progeny are obtained. What would be the frequencies of the various phenotypes?
Answer:
The dominant phenotypes are white flowers with disc shaped fruits.
The dominant white flower (W) and recessive yellow flower (w).
The dominant disc shaped fruit (D) and recessive sphere shaped fruit (d).
The genotype of parents WWDD (white flowered disc shaped fruits) and wwdd (yellow flower and sphere shaped fruits).
The genotype of hybrid yellow flower and sphere shaped fruits is WwDd.
Phenotype frequency will be 9 : B : 3 : 1 ratio.

Question 8.
Give the ratios of the following:
a) Monohybrid test cross
b) Dihybrid test cross
c) F₂ phenotypic ratio of monohybrid cross
d) F₂ phenotypic ratio of dihybrid cross
e) F₂ Genotypic ratio of monohybrid cross and
f) F₂ genotypic ratio of dihybrid cross.
Answer:
a) Monohybrid test cross ratio 1 : 1
b) Dihybrid test cross ratio 1 : 1 : 1 : 1
c) F₂ phenotypic ratio of monohybrid cross 3 : 1
d) F₂ phenotypic ratio of dihybrid cross 9 : 3 : 3 : 1
e) F₂ genotypic ratio of monohybrid cross 1 : 2 : 1
f) F₂ genotypic ratio of dihybrid cross 1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1

Question 9.
A diploid organism is heterozygous for 4 loci. How many types of gametes can it produce?
Answer:
A diploid organism is heterozygous for 4 loci. It can produce sixteen types of gametes.

Question 10.
What is crossing over? In which stage of cell division crossing over occurs? What is its significance?
Answer:
The exchange of chromatids between non-sister chromatids leading to genetic recombination is known as crossing over. Crossing over occurs during pachytene sub stage of meiosis. Crossing over significance is

1. It is the cause of variation and genetic recombination.
2. It is essential for natural selection and evolution.

Question 11.
“Genes contain the information that is required to express a particular trait”. Explain.
Answer:
Mendel conducted experiments like monohyrid cross. He crossed tall and dwarf plants to study the inheritance of gene. He observed that only one of the parent traits was expressed in the F₁ generation while at the F₂ stage both the traits were expressed in the proportion 3 : 1. The constracting characters traits did not show any blending at.either F₁ or F₂ stage.

Based on these observations, Mendel proposed that something was being stably passed down, unchanged from parents to offspring through the gametes over successive generation. He called them as factors. Now a days we call them genes. Genes therefore are the units of inheritance. They contain the information that is required to express a particular trait in an organism.

Question 12.
For the expression of traits genes provide only the potentiality and the environment provides the opportunity. Comment on the veracity of the statement.
Answer:

1. Two individuals with the same genotype may become different in phenotype when they come into contact with different environmental conditions like temperature, light, humidity are referred to as environmental variation or modifications.
2. The phenotype of any organism is necessarily a result of the interaction of a genotype with an environment, both are absolutely necessary.
3. Inbred lines are obtained in which genotype is uniform. One can measure the effect of environmental factors, such as amount and kind of fertilizer and type of soil in crop plants by growing members of an inbred line under different environmental conditions. They are different phenotypically.
4. For example, sweet potatoes grown in soil rich in potassium are round and fleshy but when this element is scare they became long and spindling.
5. Other example is the effect of temperature a hair colour in the Himalayan rabbit. The white hair on a small area of the back of this rabbit was pulled out and the animal was put in a cold room. The hair that grew in under cold condition was black. Hair is this region which grows under warm condition is white.

Then it proves that for the expression of traits gene provide only the potentiality and the environment provides the opportunity.

Question 13.
Two heterozygous parents are crossed. If the two loci are linked what would be the distribution of phenotypic features in the F₁ generation for a dihybrid cross?
Answer:
Linkage is defined as the coexistence of two or more genes are situated on the same chromosome and lie close to each other, then they are inherited together and are said to be linked genes. For example, a cross between yellow body and white eyes and wild type parent in a Drosophila will produce wild type and yellow white prggenies. It is because yellow bodied and white eyed genes are linked. Therefore, they are inherited together in progenies.

Telangana TSBIE TS Inter 2nd Year Botany Study Material 8th Lesson Viruses Textbook Questions and Answers.

TS Inter 2nd Year Botany Study Material 8th Lesson Viruses

Very Short Answer Type Questions

Question 1.
Mention the living and non-living characters of viruses.
Answer:
Living characters:

1. They contain nucleic acid.
2. They maintain genetic continuity through multiplication and undergo mutations.
3. The live as obligate intracellular parasites.

Non-living characters:

1. They do not exhibit most of the life processes like growth, irritability.
2. They are inert life less molecules.
3. They are acellular.
4. They do not have metabolic system.

Question 2.
What is the shape of T₄ phage? What is its genetic material?
Answer:

1. T₄ phage is tadpole shaped with a large head and a tail.
2. Genetic material is double stranded DNA.

Question 3.
What are virulent phages? Give an example.
Answer:

1. The viruses that attack the bacterium E.coli, causes lysis of the cells are called virulent phages.
2. Eg: T – even phages.

Question 4.
What is lysozyme and what is its function?
Answer:

1. The viral enzyme which dissolves the plasma membrane of the host cell (bacterial cell) is called lysozyme.
2. Lysozyme is synthesized within the cell and the bacterial cell wall breaks releasing the newly produced phage particles / virions.

Question 5.
Define ‘lysis’ and ‘burst size’ with reference to viruses and their effects on host cells.
Answer:
1. Lysis :
It is the final stage of lytic cycle, the host cell wall bursts during this phase and all the newly produced virions are released. This is known as lysis.

2. Burst size :
The number of newly synthesized phage particles released from a single host cell.

Question 6.
What is a prophage?
Answer:

1. Prophase : It is the phage DNA that is incorporated into bacterial DNA and remains latent during lysogenic cycle.
2. It is found in temperate phages and the prophase also undergoes replication.

Question 7.
What are temperate phages? Give one example.
Answer:

1. Temperate phage: A bacteriophage whose DNA is incorporated into the host DNA to form prophase during lysogenic cycle.
2. It does not cause immediate lysis and death of host, when they multiply Eg: Coliphase-λ.

Question 8.
Mention the differences between virulent phages and temperate phages.
Answer:
1. Virulent phages :
Bacteriophages that cause lysis of host at the end of replication (lytic cycle) Eg: T-even phages.

2. Temperate phages :
Bacteriophages, whose DNA is incorporated into host DNA to form prophases (lysogenic cycle) and do not cause immediate lysis of host. Eg: Coliphase-λ,.

Short Answer Type Questions

Question 1.
What is ICTV? How are viruses named?
Answer:
ICTV means International Committee on Taxonomy of Viruses. It regulates the norms of classification and nomenclature of viruses.

The ICTV scheme has only three hierarchial levels
– Family (including some sub – families)
– Genus
– Species

The family names end with the suffix ‘viridae’ while the genus names with ‘virus’ and the species names are common English expressions describing their nature.

Viruses are named after the disease they cause. Eg : Polio virus

Using the ICTV system, the virus that causes Acquired Immune Deficiency Syndrome (AIDS) in human beings is classified as :
Family : Retroviridae Genus : Lentivirus
Species : Human Immune Deficiency Virus (HIV).

Question 2.
Explain the chemical structure of viruses. [Mar. 2020]
Answer:

1. All viruses consists of two basic components 1) core 2) capsid.
2. Core is the nucleic acid that forms the genome. Capsid is the surrounding protein coat.
3. Capsid gives shape and protection. It is made up of protein subunits called capsomeres. The no. of capsomeres is characteristic for each type of virus.
4. Virus contains its genetic information in either double stranded (ds) DNA or single stranded (ss) DNA.
5. Generally, viruses that infect plants have single stranded RNA and viruses that infect animals have double stranded DNA.
6. Bacteriophages are usually ds DNA.
7. Viral nucleic acid molecules are either circular or linear.
8. Most viruses have a single nucleic acid molecule, but a few have more than one (Eg : HIV which has two identical molecules of RNA).

Question 3.
Write briefly about the symmetry of viruses.
Answer:
Helical virus are helical symmetry. They resemble long rods. Eg: RSbies virus and Tobacco mosaic virus.

Polyhedral symmetry :
Many plants and animals has polyhedral symmetry (many sides). Eg : Herpes simplex and polio viruses.

Binal symmetry :
Bacteriophage have both polyhedral symmetry in the head and helical symmetry in the tail sheath.
Spherical virus are enveloped virus. Eg : Influenza virus.

Question 4.
Explain the structure of TMV. [Mar. ’18 ’14; May ’17]
Answer:

1. TMV was the first virus to be crystallized by Stanley.
2. Fraenkel described the structure of TMV.
3. Tobacco Mosaic Virus is a rod shaped virus. It is about 300 nm long and 18 nm in diameter, with a molecular weight of 39 x 106 Daltons.
4. The capsid is made up of 2,130 subunits called capsomeres.
5. Capsomeres are arranged in a helical manner around a central core of 4 nm.
6. Each protein subunit is made up of a single polypeptide chain with 158 amino acids.
7. Single stranded RNA is present inside the capsid and is also spirally coiled.
8. RNA of TMV consists of 6,500 nucleotides.

Question 5.
Explain the structure of T – even bacteriophages. [Mar. 17, May’14]
Answer:

1. The body of T-even bacteriophage can be distinguished into head and tail regions joined by a collar.
2. The tail region includes a tail sheath, a base plate, pins and tail fibres which help the virus attach to host cell.
3. The tail sheath aids in injecting viral DNA into the host cell.

Question 6.
Explain the lytic cycle with reference to certain viruses. [Mar. 2019]
Answer:
T – even phages that attack the bacterium E.coli cause lysis of the cells and are called virulent phages. They show lytic cycle. It involves 5 step process. They are 1. attachment 2. penetration 3. biosynthesis 4. maturation and 5. release.

1. Attachment:

1. Contact of the virion to the surface of host bacterium is called attachment or adsorption.
2. The phages use tail fibres for attachment to the complementary receptor sites on the bacterial cell wall.

2. Penetration :

1. The injection of phage nucleic acid into the host cell is called penetration.
2. The phage DNA is injected into the bacterium through the tail core like a hypodermal syringe.
3. The capsid remains outside the bacterial cell and is referred to as ghost.

3. Biosynthesis :

1. Once the phage DNA reaches the cytoplasm of the host cell, many copies of phage DNA, enzymes and capsid proteins are synthesized, using the cellular machinery of the host cell.
2. Host cell do not contain any complete infective viruses.

4. Maturation :

1. In this process bacteriophage DNA and capsids are assembled into complete virions.
2. This period of time between the infection by a virus and the appearance of the mature virus within the cell is called the eclipse period.

5. Release :

1. The final stage of viral multiplication is the lysis phase of the host cell and the release of virions from the host cell.
2. The plasma membrane of the host cell gets dissolved or lysed due to viral enzyme called lysozyme.
3. The bacterial cell wall breaks releasing the newly produced phage particles /virions.

Question 7.
Explain how temperate phages play a role in transduction.
Answer:

1. In lysogenic cycle, some bacteriophages such as X (Lambda) phages do not cause lysis
   during multiplication.
2. Instead, the phage DNA upon penetration into the E.coli gets integrated in to the circular bacterial DNA, becomes part of it and remains latent (inactive). Such phages are called temperate phages. This inserted phage DNA is now called prophage.
3. Every time the bacterial genetic material replicates, the prophage also gets replication. The prophage remains latent in the progeny cells.
4. However, rarely prophage gets disintegrated when they are exposed to UV light or some chemicals and enter into lytic cycle.
5. Thus temperate phase plays a role of transduction by the transfer of genetic material from one bacterium to another through bacteriophage.

Question 8.
Mention the differences between lytic and lysogenic cycles.
Answer:

Long Answer Type Questions

Question 1.
Write about the discovery and structural organization of viruses.
Answer:

1. Viruses have been causing diseases in humans, animals and plants from the ancient time.
2. ‘Germ theory of disease’ was put forth by Louis Pasteur but the agent responsible for the diseases are not known.
3. In 1892 for the first time, the Russian pathologist Dmitri Iwanowski, while studying tobacco mosaic disease, filtered the “sap of diseased tobacco leaf” through filter which was designed to retain bacteria. However the infectious agent passed through the pores of the filter. After injecting the filtered sap into the healthy plant, he found the development of symptoms of mosaic disease in it. Unable to see any microorganism in a sap, he reported that the filterable agent was responsible for the disease.
4. Martinus Beijerinck repeated Iwanowski’s experiments and concluded that the disease causing agent was a contagious living fluid (contagium vivum fluidum).
5. W.M. Stanley (1935) purified the sap and announced that the virus causing mosaic disease in tobacco could be crystallized. It was named as Tobacco Mosaic Virus (TMV).
6. Fraenkel Conrat (1956) confirmed that the genetic material of the TMV is RNA.

Question 2.
Describe the process of multiplication of viruses.
Answer:
The process of multiplication of viruses is done by two alternative mechanisms.
a) Lytic cycle b) Lysogenic cycle

a) Lytic cycle :
T – even phages that attack the bacterium E.coli cause lysis of the cells and are called virulent phages. They show lytic cycle. It involves 5 step process. They are 1. attachment 2. penetration 3. biosynthesis 4. maturation and 5. release.

1. Attachment:

1. Contact of the virion to the surface of host bacterium is called attachment or adsorption.
2. The phages use tail fibres for attachment to the complementary receptor sites on the bacterial cell wall.

2. Penetration :

1. The injection of phage nucleic acid into the host cell is called penetration.
2. The phage DNA is injected into the bacterium through the tail core like a hypodermal syringe.
3. The capsid remains outside the bacterial cell and is referred to as ghost.

3. Biosynthesis :

1. Once the phage DNA reaches the cytoplasm of the host cell, many copies of phage DNA, enzymes and capsid proteins are synthesized, using the cellular machinery of the host cell.
2. Host cell do not contain any complete infective viruses.

4. Maturation :

1. In this process bacteriophage DNA and capsids are assembled into complete virions.
2. This period of time between the infection by a virus and the appearance of the mature virus within the cell is called the eclipse period.

5. Release :

1. The final stage of viral multiplication is the lysis phase of the host cell and the release of virions from the host cell.
2. The plasma membrane of the host cell gets dissolved or lysed due to the viral enzyme called lysozyme.
3. The bacterial cell wall breaks releasing the newly produced phage particles / virions.

b) Lysogenic cycle :

1. In lysogenic cycle, some bacteriophages such as X (Lambda) phages do not cause lysis during multiplication.
2. Instead the phage DNA upon penetration into the E.coli gets integrated in to the circular bacterial DNA, becomes part of it and remains latent (inactive). Such phages are called temperate phages. This inserted phage DNA is now called prophage.
3. Every time the bacterial genetic material replicates, the prophase also gets replication. The prophage remains latent in the progeny cells.
4. However, rarely prophage gets disintegrated when they are exposed to UV light or some chemicals and enter into lytic cycle.
5. Thus temperate phase plays a role of transduction by the transfer of genetic material from one bacterium to another through bacteriophage.

Intext Question Answers

Question 1.
When discussing the multiplication of viruses, Virologists prefer to call the process as replication, rather than reproduction. Why?
Answer:

1. Viral reproduction requires a living cell to takes place. Virus do not go through mitosis
   or cytogenesis nor do they have any mitotic machinery to produce new virus.
2. Virus cannot reproduce without a host cell or infect whereas a reproductive cell are show independent replication.
3. The cell infected by virus die immediately (lytic cycle) or after sometime (lysogenic cycle). Hence virologist prefer to call the multiplication of viruses as replication rather than reproduction.

Question 2.
In dealing with public health, the approach to deal with bacterial diseases is treatment. Can you guess the nature of the general public health approach to viral diseases ? What example do you cite to support your answer?
Answer:
The most effective medical approaches to viral diseases are vaccinations to provide immunity to infection and antiviral therapy to overcome drug resistance. Antibiotics have no effect on viruses. Ex : AIDS, Viral hepatitis and many more.

Telangana TSBIE TS Inter 2nd Year Botany Study Material 7th Lesson Bacteria Textbook Questions and Answers.

TS Inter 2nd Year Botany Study Material 7th Lesson Bacteria

Very Short Answer Type Questions

Question 1.
Write briefly on the occurrence of micro-organisms.
Answer:

1. Microorganisms are ubiquitous. They are found in soil, water, air, and inside living beings.
2. They occur in a variety of foods and can withstand extreme cold, heat, and drought conditions.

Question 2.
Define Microbiology.
Answer:

1. Microbiology is a branch of biological science that deals with the study of microorganisms like protozoa, microscopic algae, fungi (yeasts and molds) bacteria and viruses.
2. It is concerned with structure, function, classification, ways to control and using the activities of microorganisms.

Question 3.
Name the bacteria which is a common inhabitant of human intestine. How is it used in biotechnology? [May 2014]
Answer:

1. Escherichia coli (E.coli) is a common inhabitant of human intestine.
2. It is used in Recombinant DNA technology for the production of insulin harmone.

Question 4.
What are pleomorphic bacteria? Give an example. [May ’17, Mar. ’14]
Answer:

1. The bacteria that keep on changing their shape depending upon the type of environment and nutrients available are called pleomorphic bacteria.
2. Eg : Acetobacter.

Question 5.
What is sex pilus? What is its function?
Answer:

1. The process of conjugation requires a special conjugation apparatus called the conjugation tube or pilus or sex pilus.
2. Its function is to transfer F plasmid from F⁺ donor to F^– recipient.

Question 6.
What is a genophore? [Mar. 2019]
Answer:

1. Bacterial chromosome is also called as genophore.
2. It is the main genetic material of bacteria.

Question 7.
What is a plasmid? What is its significance?
Answer:

1. Plasmids : Small circular, double stranded DNA molecules present is addition to the bacterial chromosome (genophore) in Bacteria.
2. Plasmids contain genes, for resistance to drugs, production of toxins and enzymes. These are used as tools (vectors) in modern genetic engineering technique.

Question 8.
What is conjugation? Who discovered it and in which organism? [Mar. 2017]
Answer:

1. Conjugation is a process, in which two live bacteria come together and the donor cell directly transfers DNA to the recipient cell.
2. This process was first observed by Lederberg and Tatum (1946) in Escherichia coli.

Question 9.
What is transformation? Who discovered it and in which organism? [Mar. 2020]
Answer:

1. Transformation is uptake of naked DNA fragments from the surrounding environment and the expression of that genetic information in the recipient cell.
2. Frederick Griffith (1928) discovered it in Streptococcus pneumoniae.

Question 10.
What is transduction? Who discovered it and in which organism? [March 2010]
Answer:

1. The transfer of genetic material from one bacterium to anotherthrough bacteriophage is known as transduction,
2. It was discovered by Lederberg and Zinder (1951) in Salmonella typhimurium.

Short Answer Type Questions

Question 1.
Explain the importance of Microbiology.
Answer:
Microbiology is a branch of biological science that deals with the study of micro-organisms. Micro-organisms are useful in several ways for the welfare of human society.
1. Soil fertility :
Micro-organisms decompose dead plants and animals thereby enrich the soil with nutrients which are utilized by plants. Microbes also play vital role in recycling of elements like C, N, O, S, and P.

2. Antibiotics :
Alexander Fleming isolated an antibiotic Penicillin from a fungus, Penicillium notatum. Waksman isolated an antibiotic Streptomycin from a bacterium, Streptomyces griseus.

3. Industrial products :
Industrial products like enzymes, amino acids, vitamins, organic acids and alcohols are commercially produced using microbes.

4. Dairy products :
Lactobacillus, commonly known as Lactic Acid Bacteria (LAB) grows in milk and convert it to curd, which also improves its natural quality by increasing vitamin B12, food stuff, like cheese, yogurt are the byproducts of microbial growth.

5. Mining :
Metals like Uranium can be extracted with 50% reduced cost by using microbes.

6. Tools in Genetic Engineering :
Microbes are used as tools in altering the genetic make up of organisms.

7. Biocontrol agents :
Micro-organisms like Bacillus thuringiensis are used to control plant diseases and pests.,

8. Production of Biogas :
Biogas, a mixture of gases (predominantly methane) produced by the microbial activity. It may be used as fuel.

9. Exomicrobiology :
Microbes are used for the exploration of life in the outer space.

10. Sewage disposal :
Bacteria and fungi are also used in sewage disposal.

Question 2.
How are bacteria classified on the basis of morphology?
Answer:
Based on their shape bacteria are classified into following types.
1. Cocci :
Spherical bacteria are called cocci. They are 6 types.
a) Monococcus : A single spherical bacterium.
b) Diplococcus : A pair of spherical bacterium.
c) Tetracocci : A group of four spherical bacteria.
d) Streptococcus : A linear chain of spherical bacteria arranged in a single row.
e) Sarcinae : Cocci arranged in cubes of eight.
f) Staphylococci : A group of cocci bacteria forming irregular shapes producing bunches.

2. Bacillus :
Elongated rod shaped bacteria are called bacillus. They are 3 types.
a) Monobacillus : A single elongated rod shaped bacterium.
b) Diplobacillus : Paired cells of bacilli.
c) Streptobacillus : Chains of bacilli appearing like straws.

3. Spiral forms :
Vibrioid : Cells having less than one complete twist.
Spirillum : Cells that have more than one complete twist – a distinct helical shape. Spirochete : Cells are slender, long and cork-screw shaped.

4. Pleomorphic :
These bacteria change their shape depending upon the type of environment and nutrients available. This phenomenon is called pleomorphism.
Eg : Acetobacter.

Question 3.
How are bacteria classified on the basis of number and distribution of flagella?
Answer:
Based on the number and distribution of flagella bacteria are classified into 4 types.

1. Monotrichous : A single flagellum is present on one side of the cell.
2. Lophotrichous : A tuft of flagella are present on one side of the cell.
3. Amphitrichous : Tufts of flagella or a single flagellum on either end of the cell.
4. Peritrichous : Many flagella are distributed all over the cell surface.

Question 4.
What are the nutritional groups of bacteria based on their source of energy and carbon?
Answer:
Four major nutritional groups of bacteria based on the source of energy and carbon are described below.

Question 5.
Write briefly about chemoheterotrophs and their significance.
Answer:
Chemoheterotrophs derive both carbon and energy from organic compounds. Processing these organic molecules by respiration or fermentation releases energy in the form of ATP.

These bacteria are divided into saprophytes and parasites basing on how they obtain their organic compounds.

Saprophytes :
These are free living bacteria which grow on dead, organic matter are called Saprophytes. Eg : Bacillus

Parasites :
The bacteria which grow on or in living host and cause diseases are called parasites.
Eg : Xanthomonas, Salmonella.

Significance :
Most microbes of biomedical importance belong to these two categories. Bdellovibrio bacteriovorous grows as a parasite on some harmful bacteria and their abundance is supposed to be responsible for the microbial purity of Ganges waters.

Question 6.
Explain the conjugation in bacteria.
Answer:
Conjugation :

1. The transfer of genetic material (DNA) through direct cell to cell contact is known as conjugation.
2. It was first reported by Lederberg and Tatum in 1946 in Escherichia coli.
3. In E.coli, in addition to the bacterial chromosome or genophore which is the main genetic material, bacteria contain small circular, double stranded DNA molecules called plasmids or ‘F’ factor.
4. E.coli having F factor are called F⁺ cells or donor cells and the cells without F factor are called F^– cells or acceptor cells.
5. F⁺ cells have pilus or sex pilus. During conjugation the F⁺ and F^– strains come close together. Once contact is established, the pilus shortens to bring the two bacteria close together.
6. A conjugation.tube is established. The plasmid in F⁺ replicates and forms a copy of it, which moves to the acceptor cell (F). Thus donor bacterium generally retain a copy of genetic material that is being transferred.

Long Answer Type Questions

Question 1.
Explain different methods of sexual reproduction in Bacteria.
Answer:
Sexual reproduction :
True sexual reproduction is absent in bacteria. However the exchange of genetic material is reported through other methods.

Three types of genetic recombinations are reported in different species of bacteria. They are

1. Conjugation
2. Transformation and
3. Transduction.

1. Conjugation :

1. The transfer of genetic material (DNA) through direct cell to cell contact is known as conjugation.
2. It was first reported by Lederberg and Tatum in 1946 in Escherichia coli.
3. In E.coli, in addition to the bacterial chromosome or genophore which is the main genetic material, bacteria contain small circular, double stranded DNA molecules called plasmids or ‘F’ factor.
4. E.coli having F factor are called F⁺ cells or donor cells and the cells without F factor are called F^– cells or acceptor cells.
5. F⁺ cells have pilus or sex pilus. During conjugation the F⁺ and F strains come close together. Once contact is established, the pilus shortens to bring the two bacteria close together.
6. A conjugation tube is established, The plasmid in F⁺ replicates and forms a copy of it, which moves to the acceptor cell (F). Thus donor bacterium generally retain a copy of genetic material that is being transferred.

2. Transformation :
Transformation is uptake of naked DNA fragments from the surrounding environment and the expression of that genetic information in the
recipient cell. That is, the recipient cell has now acquired a characteristic that is previously lacked. This mode of bacterial genetic recombination was discovered by Frederick Griffith in streptococcus pneumoniae.

3. Transduction :
The transfer of genetic material from one bacterium to another through bacteriophage is known as transduction.

Question 2.
“Bacteria are friends and foes of man” – discuss.
Answer:
Bacteria are known to be the casual agents of plant, animal and human diseases. At the same time there are many bacteria which are directly or indirectly beneficial to man. Thus, these organisms can be considered both as ‘friends and foes of man’.

Beneficial activities:

1. Microbes are now used in extracting valuable metals like uranium from rocks. The process is known as Bio-mining. The use of microbes in mining reduces the cost of production by more than 50%.
2. DNA components from bacteria are used as Biosensors that can detect biologically active toxic pollutants.
3. Microbes also find application in medical diagnostics, food and fermentation operations.
4. The most important development in Biotechnology depends on the possibility of altering the genetic makeup of bacteria through genetic engineering.
5. Microbes in household products :
   a) A common example is the production of curd from milk micro-organisms such as Lactobacillus and others, commonly called lactic acid bacteria (LAB), grow in milk and convert into curd. It also improves its nutritional quality by increasing vitamin B12.
   b) Some of our food stuffs like cheese, yogurt are also actually the by- products of microbial growth.
6. Microbes are used as biocontrol agents. Biocontrol refers to the use of biological methods for controlling plant diseases and pests.
7. Many industrial products like enzymes, amino acids, vitamins, organic acid and alcohols are commercially produced by micro organisms.
8. Microorganism decompose dead plants and animals and enrich the soil nutrients which can be used by plants. They play an important role in recycling of elements.
9. Microbes cause diseases in plants and human beings. On the other hand, they help in creating disease free world by producing antibiotics and vaccinations. Penicillin was the antibiotic discovered by Alexander Fleming from a fungus, penicillin notatum. The antibiotic obtained from bacteria streptomyces griseus is known as streptomycin.
10. Biogas is a mixture of gases (containing predominantly methane) produced by the microbial activity and which may be used as fuel.

Harmful activities:
Some bacteria that cause human diseases are

Bacteria causes plant diseases :

Bacteria also cause animal diseases :

Intext Question Answers

Question 1.
Many people believe that bacteria do little more than cause human illness and infectious diseases. How does the information in this chapter help you correct that misconception?
Answer:
Most of the bacteria are beneficial activities than harmful activities.

Bacteria are known to be the casual agents of plant, animal and human diseases. At the same time there are many bacteria which are directly or indirectly beneficial to humans. Thus, these organisms can be considered both as ‘friends and foes of man’.”

Beneficial activities:

1. Microbes are now used in extracting valuable metals like uranium from rocks. The process is known as Bio-mining. The use of microbes in mining reduces the cost of production by more than 50%.
2. DNA components from bacteria are used as Biosensors that can detect biologically active toxic pollutants.
3. Microbes also find application in medical diagnostics, food and fermentation operations.
4. The most important development in Biotechnology depends on the possibility of altering the genetic makeup of bacteria through genetic engineering.
5. Microbes in household products :
   a) A common example is the production of curd from milk micro-organisms such as Lactobacillus and others, commonly called lactic acid bacteria (LAB), grow in milk and convert into curd. It also improves its nutritional quality by increasing vitamin Bir
   b) Some of our food stuffs like cheese, yogurt are also actually the by-products of microbial growth.
6. Microbes are used as biocontrol agents. Biocontrol refers to the use of biological methods for controlling plant diseases and pests.
7. Many industrial products like enzymes, amino acids, vitamins, organic acid and alcohols are commercially produced by microorganisms.
8. Microorganism decompose dead plants and animals and enrich the soil nutrients which can be used by plants. They play an important role in recycling of elements.
9. Microbes cause diseases in plants and human being. On the other hand, they help in creating disease free world by producing antibiotics and vaccinations. Penicillin was antibiotic obtained from bacteria streptomyces griseus is known as streptomycin.
10. Biogas is a mixture of gases (containing predominantly methane) produced by the microbial activity and which may be used as fuel.

Harmful activities:
Some bacteria that cause human diseases are

Bacteria causes plant diseases ;

Bacteria also cause animal diseases :

Question 2.
Humans produce about 50 grams of feces per day. Scientists estimated in broad terms about one -third of human feces is composed of bacteria. If one E.coli cell weighs 1 × 10^-12 g, how many bacteria are there in a day’s feces? How can this be possible?
Answer:
Feces per day = 50 gram
Bacteria present = \(\frac{1}{3}\) × 50 gm
Each bacteria weight = 1 × 10^-12 gm
Bacteria in a day’s feces = \(\frac{1}{3}\) × 50 × 1 × 10^-12 gm
i.e., 16.66 × 10^-12(approx)

Question 3.
An organism is described as a peritrichous bacillus. How might you translate this bacteriological language into a description of the organism?
Answer:
Bacillus indicates that bacteria are rod elongated in shape. Peritrichous indicates that bacteria is having many flagella distributed all over the cell surface flagella are locomotory in function.

Telangana TSBIE TS Inter 2nd Year Botany Study Material 6th Lesson Plant Growth and Development Textbook Questions and Answers.

TS Inter 2nd Year Botany Study Material 6th Lesson Plant Growth and Development

Very Short Answer Type Questions

Question 1.
Define plasticity. Give an example.
Answer:

1. Plasticity is the ability of plants to follow different pathways in response to the environment or phases of life to form different kinds of structures.
2. Heterophylly is an example of plasticity.

Question 2.
What is the disease that formed the basis for the identification of gibberellins in plants? Name the causative fungus of the disease.
Answer:

1. “Bakane” (foolish seedling) disease of rice seedlings.
2. It is caused by a fungal pathogen Gibberella fujikuroi.

Question 3.
What is apical dominance? Name the growth hormone that causes it.
Answer:

1. Apical dominance : The growing apical bud inhibits the growth of lateral (axillary) buds.
2. It is caused by Auxins.

Question 4.
What is meant by bolting ? Which hormone causes bolting?
Answer:

1. Bolting : Elongation of internodes just prior of flowering in plants with rosette habit (Eg : Beat, Cabbage).
2. Gibberellins are responsible for bolting.

Question 5.
Define respiratory climactic. Name the PGR associated with it.
Answer:

1. The rise in the rate of respiration during the ripening of the fruits is known as respiratory climatic.
2. Ethylene is responsible for it.

Question 6.
What is ethephon? Write its role in agricultural practices.
Answer:

1. Ethephon in an aqueous solution. It is readily absorbed and transported within the plant and releases ethylene slowly.
2. Ethephon hastens fruit ripening in tomatoes and apples and accelerates abscission in flowers and fruits. It also promotes female flowers in cucumbers and thereby increasing the yield.

Question 7.
Which of the PGR (Plant Growth Regulator) is called stress hormone and why?
Answer:

1. Abscisic acid (ABA) is called as the stress hormone.
2. It stimulates the closure of stomata in the epidermis and increases the tolerance of plants to various kinds of stresses.

Question 8.
What do you understand by vernalisation? Write its significance.
Answer:

1. Vernalisation is the method of inducing early flowering in plants by pre-chilling treatment of their seeds or young shoots.
2. Vernalization of winter varieties of wheat, barley and rye useful to early harvesting of crop.

Question 9.
Define the terms quiescence and dormancy.
Answer:

1. Quiescence is the condition of a seed which is unable to germinate only because favourable external conditions normally required for growth are not present.
2. Dormancy is the condition of a seed when it fails to germinate because of internal conditions, even though external conditions are suitable.

Short Answer Type Questions

Question 1.
Write a note on agricultural/horticultural applications of auxins. [Mar. ’17,’14; May ’14]
Answer:

1. Auxins help to initiate rooting in stem cuttings, an application widely used for plant propagation in horticulture.
2. Auxins promote flowering e.g. in pineapples.
3. Auxins help to prevent fruit and leaf drop at early stages but promote the abscission of older mature leaves and fruits.
4. Removal of shoot tips usually results in the growth of lateral buds. This is widely applied in tea plantations.
5. Auxins also induce parthenocarpy. e.g. in tomatoes.
6. Auxins (2, 4D) are widely used as herbicides.
7. Auxins control xylem differentiation and help in growth.

Question 2.
Write the physiological responses of gibberellins in plants. [Mar. 2019]
Answer:

1. Gibberellins promote bolting (internode elongation just before flowering) in beet, cabbages and many plants with rosette habit.
2. The ability to cause an increase in the length of axis is used to increase the length of grapes stalks.
3. Gibberellins cause fruits like apple to elongate and improve their shape.
4. Gibberellins delay senescence. Thus, fruits can be left on the tree longer so as to extend the market period.
5. GA₃ is used to speed up the malting process in brewing industry.
6. Spraying Gibberellins on sugarcane crop increases the length of the stem, thus increasing the yield by as much as 20 tonnes per acre.
7. Spraying juvenile conifers with Gibberellins fastens the maturity period, thus leading to early seed production.

Question 3.
Write any four physiological effects of cytokinins in plants. [Mar. ’18, May ’17]
Answer:

1. Cytokinins have specific effects on cytokinesis (that means division of cytoplasm during cell division). It helps in occuring rapid cell division for example: root apices, developing shoot buds, young fruits etc.
2. Cytokinins help to produce new leaves, chloroplasts in leaves, lateral shoot growth and adventitious shoot formation.
3. Cytokinins help to overcome the apical dominance.
4. Cytokinins promote nutrient mobilisation which helps in the delay of leaf senescence.

Question 4.
What are the physiological processes that are regulated by ethylene in plants? [Mar. 2020]
Answer:

1. Ethylene accelerates the ripening of fruits. It is regarded as fruit ripening hormone.
2. Effects of ethylene on plants include horizontal growth of seedlings, swelling of the axis and apical hook formation in dicot seedlings.
3. Ethylene promotes senescence and abscission of plant organs, especially of leaves and flowers.
4. Ethylene breaks seed and bud dormancy. It initiates germination in peanut seeds and sprouting of potato tubers.
5. Ethylene promotes rapid internode/petiole elongation in deep water rice plants.
6. Ethylene helps leaves/upper parts of the shoot to remain above water.
7. Ethylene promotes root growth and root hair formation, thus helping plants to increase their absorption surface.
8. Ethylene is used to initiate flowering and for synchronising fruit – set in pineapples. It also induces flowering in mango.

Question 5.
Write short notes on seed dormancy.
Answer:

1. Dormancy is the condition of seed when it fails to germinate because of internal conditions, even though external conditions (e.g. temperature, moisture) are suitable.
2. The dormancy may be caused by hard seed coats that prevents uptake of oxygen or water (e.g. fabaceae).
3. Dormancy caused by hard seed coats can be broken by scarification.
4. Seeds of certain plants (e.g.: tomato) contain chemical compounds, which inhibit their germination.
5. Many seeds (e.g.: Polygonum) will not germinate untill they have been exposed to low temperatures in moist conditions in the presence of oxygen for weeks to months.
6. Motet seeds respond to high temperatures and several seeds respond best when daily temperatures alternate between high and low.

Question 6.
Which one of the plant growth regulators would you use if you are asked to
a) Induce rooting in a twig
b) QCiickly ripen a fruit
c) Delay leaf senescence
d) Induce growth in axillary buds
e) ‘Bolt’ a rosette plant
f) Induce immediate stomatal closure in leaves
g) Overcome apical dominance
h) Kill dicotyledonous weeds.
Answer:
a) Auxins like IBA, NAA
b) Ethylene
c) Cytokinin
d) Cytokinins
e) Gibberellins
f) Abscisic acid (ABA)
g) Cytokinins
h) Auxins (2-4 D)

Long Answer Type Questions

Question 1.
Define growth, differentiation, development, dedifferentiation, redifferentiation, determinate growth, meristem and growth rate.
Answer:
1) Growth :
Growth is defined as a permanent or irreversible increase in dry weight, size, mass or volume of a cell, organ or organism.

2) Differentiation :
The cells derived from root apical and shoot-apical meristems and cambium differentiate and mature to perform specific functions. This act leading to maturation is termed as differentiation.

3) Dedifferentiation :
The living differentiated cells, that have lost the capacity to divide, can regain the capacity of division under certain conditions. This phenomenon is called dedifferentiation, e.g.: Formation of meristems – interfascicular cambium and cork cambium from parenchyma cells.

4) Redifferentiation :
In the process of redifferentiation meristems/tissues are able to divide and produce cells that once again lose the capacity to divide but mature to perform specific functions. That means they get redifferentiation.

5) Determinate growth :
Leaves are specialised organs characterised by defined developmental destiny and determinate growth.

6) Meristem :
It is located at root and shoot tips. The meristem refers to the cells that remain dividing.

7) Growth rate :
Increased growth per unit time is termed as growth rate.

Question 2.
Describe briefly
a) Arithmetic growth
b) Geometric growth
c) Sigmoid growth curve
d) Absolute and relative growth rates
Answer:
a) Arithmetic growth :
In arithmetic growth, following mitotic cell division, only one daughter cell continues to divide while the other differentiates and matures.

The simplest expression of arithmetic growth is exemplified by a root elongating at a constant rate. On plotting the length of the organ against time, a linear curve is obtained. Mathematically, it is expressed as

Constant linear growth, a plot of length L against time t
L₁ = L₀ + rt
L₁ = Length at time’t’
L₀ = Length at time ‘zero’
r = growth rate/elongation per unit time.

b) Geometric growth :
In several systems, the initial growth is slow (lag phase). Growth increases rapidly thereafter at an exponential rate (log or exponential phase) S – curve is observed

Diagrammatic representation of: (a) Arithmetic (b) Geometric growth and (c) Stages during embryo development showing geometric and arithmetic phases

c) Sigmoid curve growth :
A sigmoid curve is a characteristic of living organism growing in a natural environment. It is typical for all cells, tissues and organs of a plant.

An idealised sigmoid growth curve typical of cells in culture, and many higher plants and plant organs

It consists of 3 phases namely
(a) Lag phase
(b) log phase
(c) Stationary phase

d) Absolute and relative growth rate :
Measurement and comparison of the total growth per unit time is called the absolute growth rate.

The growth of the given system per unit time expressed on a common basis e.g. per unit initial parameter is called the relative growth rate.

Question 3.
List five natural plant growth regulators. Write a note on discovery, physiological functions an agricultural/horticultural applications of any one of them.
Answer:

Note : Write any one of the above.

Intext Question Answers

Question 1.
Fill in the blanks with appropriate word/words.
a) The phase in which growth is most rapid is …………… .
b) Apical dominance as expressed in dicotyledonous plants is due to the presence of more …………… in the apical bud than in the lateral ones.
c) In addition to auxin, a ………….. must be supplied to the culture medium to obtain a good callus in plant tissue culture.
d) …………… of vegetative plants are the sites of photoperiodic perception.
Answer:
a) Log phase
b) Auxins
c) Cytokinins
d) Shoot apex of plant

Question 2.
A primary root grows from 5 cm to 19 cm in a week. Calculate the growth rate and relative growth rate over the period.
Answer:
Growth rate is = \(\frac{19-5}{7}=\frac{14}{7}\) = 2 cm

Question 3.
Gibberellins promote the formation of ………….. flowers on genetically …………… plants in Cannabis whereas ethylene promotes formation of …………. flowers on genetically …………… plants.
Answer:
a) male
b) dwarf
c) female
d) dwarf

Question 4.
Classify the following plants into Long day plants (LDP), Short day plants (SDP) and Day neutral plants (DNP)
Xanthium, Spinach, Henbane (Hyoscyamus niger), Rice, Strawberry, Bryophyllum, Sunflower, Tomato, Maize.
Answer:
Xanthium – Short day plant
Spinach – Long day plant
Henbane – Short day plant
Rice – Short day plant
Strawberry – Short day plant
Bryophyllum – Short day plant
Sunflower – Day neutral plant
Tomato – Day neutral plant
Maize – Long day plant

Question 5.
A farmer grows cucumber plants in his field. He wants to increase the number of female flowers. Which plant growth regulator can be applied to achieve this?
Answer:
Auxins.

Question 6.
Where are the following hormones synthesized in plants?
a) IAA
b) Gibberellins
c) Cytokinins
Answer:
a) Shoot apex and to some extent root apex also contribute to the synthesis of auxins.

b) Gibberellins are synthesised from Acetyl co enzyme A. They are mostly synthesized in apical tissues. They also synthesised in developing seeds, fruits, young leaves of developing apical bud, and elongation shoots.

c) Natural cytokinin are synthesised in the regions where rapid cell division occurs for example root apices, developing shoot buds, young fruits, etc.

Question 7.
Light plays an important role in the life of all organisms. Name any three physiological processes in plants which are influenced by light.
Answer:

1. Photosynthesis
2. Photoperiodism
3. Growth

Question 8.
Growth is one of the characteristics of all living organisms. Do unicellular organisms also grow? If so, what are the parameters?
Answer:
Yes. If we plot the parameter of growth against time, we get a typical sigmoid or . S-curve.

Question 9.
Rice seedlings infected with fungus Gibberella fujikuroi are called foolish seedlings. What is the reason?
Answer:
The reason is the rice plants grew excessively tall and very little grain production. The plants were taller, thinner, and paler with little tillering as compared to healthy plants.

Question 10.
Why isn’t any one parameter good enough to demonstrate growth throughout the life of a flowering plant?
Answer:
Because growth depends upon cell division, nutrient supply, time, etc.

Question 11.
‘Both growth and differentiation in higher plants are open.’ Comment.
Answer:
Plants retain the capacity for unlimited growth throughout the life due to the presence of meristems at certain locations of their body. The cells of such meristems have the capacity to divide and self perpetite. The product, however, soon loses the capacity to divide and such cells make up plant body.

This form of growth where in new cells are always being added to the plant body by the activity of the meristem is called the open form of growth.

The cells divided from root apical and shoot apical meristems and cambium differentiate the nature to perform specific functions. It is called differentiation.

Differentiation in plants is open because cells/tissues arising out of the same meristems have different structures at maturity. The final structure at maturity of cell tissue is also determined by the location of the cell with in.

Question 12.
‘Both a short day plant and a long day plant can produce flowers simultaneously in a given place.’ Explain.
Answer:
A long day plant required the exposure to light for a period exceeding well defined critical duration while short day plant must be exposed to light for a period less than this critical duration before the flowering is initiated in them. It means that not only the duration of light but also the duration of dark period is of equal importance. Hence flowering depends not only a combination of light and dark exposure but also their relative durations.

That is why, both a short day and a long day plant produce flowering simultaneously in a given place when they are exposed to necessary inductive photoperiod required by them in artificial condition.

Question 13.
Would a defoliated plant respond to photoperiodic cycle? Why?
Answer:
No, a defoliated plant would not respond to photoperiodic cycle. Because the site of perception of light/dark duration is the leaves. In defoliated plants leaves are absent.

It has been hypothesised that there is a hormonal substances that is responsible for flowering. This hormonal substance migrates from leaves to shoot apices for inducing flowering only when the plants are exposed to the necessary inductive photoperiod.

Question 14.
What would be expected to happen if
(1) GA₃ is applied to rice seedlings.
(2) Dividing cells stop differentiating.
(3) A rotten fruit gets mixed with unripe fruits
(4) You forget to add cytokinin to the culture medium
Answer:
(1) Elongation of coleoptile occurs quickly.
(2) Undifferentiated mass of cells are formed.
(3) All the fruits will be mature.
(4) Shoot growth is inhibited.

Telangana TSBIE TS Inter 2nd Year Botany Study Material 5th Lesson Respiration in Plants Textbook Questions and Answers.

TS Inter 2nd Year Botany Study Material 5th Lesson Respiration in Plants

Very Short Answer Type Questions

Question 1.
Energy is released during the oxidation of compounds in respiration. How is this energy stored and released as and when it is needed?
Answer:

1. Energy contained in respiratory substrates is released in a series of slow step wise reactions controlled by enzymes and stored as ATP.
2. ATP is broken down whenever and wherever energy needs to be utilised.

Question 2.
Explain the term ‘Energy currency’. Which substance acts as energy currency in plants and animals?
Answer:

1. ATP is broken down whenever and wherever energy needs to be utilised is cells of living organisms.
2. ATP acts as energy currency in plants and animals.

Question 3.
Different substrates get oxidised during respiration. How does respiratory quotient (RQ) indicate which type of substrate i.e., carbohydrate, fat or protein is getting oxidised?
RQ = A/B
What do A and B stand for?
What type of substrates have RQ of 1, <1, >1?
Answer:
1. RQ is an index for type of substrate being used in respiration RQ = A/B, in which A stands for volume of CO₂ evolved and B stands for volume of O₂ consumed. Thus RQ = volume of CO₂ evolved/Volume of O₂ consumed.

2. RQ is 1, when carbohydrates are used as substrate. It is less than 1, when fats or proteins are used as substrate.

Question 4.
What is the specific role of F₀ – F₁ particles in respiration?
Answer:

1. F₀ is the integral membrane protein complex that forms the channel through which protons cross enter into matrix by crossing the inner membrane.
2. F₁ head piece is the peripheral membrane protein complex and contains the site for synthesis of ATP from ADP and inorganic phosphate.

Question 5.
When does anaerobic respiration occur in man and yeast?
Answer:

1. Anaerobic respiration occurs during exercise in the muscles of man when oxygen is inadequate for cellular respiration.
2. In yeast, anaerobic respiration occurs during fermentation.

Question 6.
Distinguish between obligate anaerobes and facultative anaerobes.
Answer:
1) Obligate anaerobe :
An organism that cannot grow in the presence of oxygen. It neither requires O₂ to grow nor does it tolerate it.

2) Facultative anaerobe :
An organism can grow in either the presence or absence of oxygen. It does not require oxygen to grow but it does tolerate its presence.

Question 7.
Explain the economic importance of fermentation.
Answer:

1. Fermentation is useful in making bread.
2. It is useful for industrial production of organic acids and alcohol.

Question 8.
What is the common pathway for aerobic and anaerobic respirations? Where does it take place?
Answer:

1. Glycolysis (EMP pathway) is the common pathway for both aerobic and anaerobic respirations.
2. It occurs in cytoplasm of the cell and takes place is all living organisms.

Question 9.
Why are mitochondria termed as the power houses of the cell?
Answer:

1. Mitochondria are found in eukaryotic cells and are seat of Krebs cycle and oxidative phosphorylation.
2. They produce energy currency of the cell (ATP).

Question 10.
What is the reason for describing ATP synthesis in FQ – Fx particles of mitochondria as oxidative phosphorylation?
Answer:

1. Oxygen acts as the final Hydrogen acceptor is ETS of aerobic respiration.
2. Production of proton gradient and synthesis of ATP by ATP synthase in respiration are oxygen driven, hence it is called oxidative phosphorylation.

Question 11.
Which substance is known as the connecting link between glycolysis and Krebs cycle? How many carbons does it have?
Answer:

1. Acetyl CoA is the connecting link between glycolysis and Krebs cycle.
2. It consists of 2 carbon atoms.

Question 12.
What cellular organic substances are never used as respiratory substrates?
Answer:
Pure proteins or fats are never used as respiratory substrates.

Question 13.
Why is the RQ of fats less than that of carbohydrates?
Answer:
1. It requires more oxygen for complete oxidation, since C : O ratio in fats in very high.

Question 14.
What is meant by ‘Amphibolic pathway’?
Answer:

1. The pathway involved both in catabolism and anabolism is termed as Amphibolic pathway.
2. Eg. Respiratory pathway.

Question 15.
Name the mobile electron carriers of the respiratory electron transport chain in the inner mitochondrial membrane.
Answer:

1. Cytochrome C that transfer electrons between complex III and IV of ETS.
2. U biquinone that acts as electron carries in between complex II and III of ETS.

Question 16.
What is the final acceptor of electrons in aerobic respiration? From which complex does it receive electrons?
Answer:

1. Oxygen is the final acceptor of electrons in aerobic respiration.
2. It receives electrons from complex IV (cytochrome ‘c’ oxidase).

Short Answer Type Questions

Question 1.
What is meant by the statement ‘Aerobic respiration is more efficient’?
Answer:

1. Aerobic respiration is the process that leads to a complete oxidation of organic substance in the presence of oxygen.
2. It releases CO₂, water and a large amount of energy present in the substrate.
3. 686 k.cal of energy is released where as in anaerobic respiration only 56 k.Cal. is released.
4. In aerobic respiration 36 ATP molecules are formed. Where as in anaerobic respiration only 2 ATP molecules are formed.

Question 2.
Pyruvic acid is the end product of glycolysis. What are the three metabolic fates of pyruvic acid under aerobic and anaerobic conditions?
Answer:
The fate of pyruvate depends upon the availability of O₂ and the type of organism. The three metabolic fates of pyruvic acid are

1. lactic acid fermentation
2. alcoholic fermentation under anaerobic condition and
3. aerobic respiration under aerobic condition.

Question 3.
The energy yield in terms of ATP is higher in aerobic respiration than during anaerobic respiration. Why is there anaerobic respiration even in organisms that live in aerobic condition like human beings and angiosperms?
Answer:

1. Under the conditions of oxygen scarcity anaerobic respiration takes place in organism that live in aerobic conditions.
2. For example because Muscle tissue under intense use muscle demands too much energy (ATP) and consume much more Oxygen to produce that energy. This high consumption leads to oxygen scarcity and the muscle cells begin to make lactic acid by anaerobic respiration to fulfill their energy needs.
3. The first cells on this planet lived in an atmosphere that lacked O₂. Even today all living organisms retain the enzymatic machinery to partially oxidise glucose without the help of oxygen during glycolysis.

Question 4.
Oxygen is an essential requirement for aerobic respiration but it enters the respiratory process at the end. Discuss.
Answer:

1. Although the aerobic process of respiration takes place only in the presence of oxygen, the role of oxygen is limited to the terminal stage of the process.
2. The presence of oxygen is vital, since it drives the whole process by removing hydrogen from the system.
3. Oxygen acts as the final hydrogen acceptor.
4. In respiration, it is the energy of oxidation-reduction utilised for Phosphorylation. It is for this reason that the process is called oxidative phosphorylation.

Question 5.
Respiration is an energy releasing and enzymatically controlled catabolic process which involves a stepwise oxidative breakdown of organic substances? Inside living cells.
In this statement about respiration, explain the meaning of i) step wise oxidative breakdown ii) organic substance (used as substrates).
Answer:
i) a) Respiration involves a stepwise oxidative breakdown of organic substances.
b) In the process of aerobic respiration it is divided into 4 phases. They are Glycolysis, TCA cycle, Electron Transport System and Oxidative Phosphorylation.
c) It is generally assumed that the process of respiration and production of ATP in each phase takes place in a stepwise manner.
d) The product of one pathway forms the substrate of the other pathway.
e) Various molecules produced during respiration are involved in other biochemical processes.
f) The respiratory substrate enters and withdraws from pathway.
g) ATP gets utilised whenever required and enzymatic rates are generally controlled.
h) Stepwise breakdown of organic substances makes the system more efficient in extracting and storing energy.

ii) Usually carbohydrates, glucose are oxidised to release energy but proteins, fats and even organic acids can be used as respiratory substrates in some plants under certain conditions.

Question 6.
Comment on the statement – Respiration is an energy producing process but ATP is used in some steps of the process.
Answer:

1. Though respiration is an energy producing process ATP is utilised in two steps during glycolysis.
2. First in the conversion of glucose into glucose 6 phosphate.
   Glucose + ATP → Glucose 6 phosphate + ADP.
3. Second in the conversion of fructose 6 phosphate into Fructose 1-6 bisphosphate.
   Fructose 6 phosphate + ATP → fructose 1, 6 bisphosphate + ADP.

Question 7.
Explain briefly the process of glycolysis.
Answer:
Glycolysis occurs in the cytoplasm of all living organisms. In this process, glucose undergoes partial oxidation to form two molecules of pyruvic acid. The scheme of glycolysis was given by Gustav Embden, Meyerhof and J. Parnas and is often referred to as the EMP pathway.

In glycolysis, a chain of ten reactions, under the control of different enzymes takes place to produce pyruvate from glucose. They are

1) Phosphorylation :
Glucose is phosphorylated to give rise to glucose 6-phosphate one ATP is utilised.

2) Isomerisation :
Glucose 6 phosphate converts into isomer fructose 6 phosphate in the presence of phosphohexose isomerase.

3) Phosphorylation :
The enzyme phosphofructokinase uses another ATP molecule to transfer a phosphate group to fructose 1, 6 biphosphate.

4) Cleavage :
The enzyme aldolase splits fructose 1,6 bisphosphate into two sugars that are isomers to each other. They are Glyceraldehyde 3 phosphate & Dihydroxy acetone phosphate (DHAP).

5) Isomerisation :
Enzyme Triose phosphate isomerase rapidly inter-converts the molecules of DHAP and G3P.

6) Oxidation :
Glyceraldehyde-3-phosphate undergoes oxidation and phosphorylation in the presence of Glyceraldehyde dehydrogenase resulting in the dehydrogenase formation of 1, 3 bisphosphoglyceric acid and NADH.

7) Dephosphorylation :
Enzyme phosphoglycerokinase catalyses the dephosphorylation of 1, 3 bisphosphoglyceric add resulting 3 PGA. ATP is formed.

8) Intramolecular shift (Isomerisation) :
Enzyme phosphoglyceromutase transfers phosphate group from 3-carbon position to 2-carbon position resulting 2 PGA.

9) Dehydration :
The enzyme enolase catalyses the removal of one water molecule from 2 PGA resulting in PEP.

10) Dephosphorylation :
Phosphoenol pyruvic acid undergoes dephosphorylation in the presence of enzyme pyruvic kinase and results in the formation of pyruvic acid ATP is formed.

Question 8.
Why is the respiratory pathway referred to as an amphibolic pathway? Explain.
Answer:
a) Fatty acids would be broken down to acetyl CoA before entering the respiratory pathway when it is used as a substrate. But when the organism needs to synthesise fatty acids, acetyi CoA would be withdrawn from the respiratory pathway for it. Hence the respiratory pathway acts as both breakdown and the synthesis of fatty acids.

b) Similarly during the breakdown and the synthesis of protein too, respiratory intermediates forms the link.

c) The breaking down process within the living organism constitute catabolism, which synthesis is anabolism.

d) Because the respiratory pathway is involved in both anabolism and catabolism, it would be better to consider the respiratory pathway as an amphibolic pathway rather than as a catabolic one.

Question 9.
We commonly call ATP the energy currency of the cell. Can you think of some other energy carriers present in the cell? Name any two.
Answer:

1. NADH and FADH₂
2. Oxidation of one molecule of NADH gives rise to 3 molecules of ATP.
3. Oxidation of one molecule of FADH₂ give rise to 2 molecules of ATP.
4. This takes place in electron transport system for mitochondria.

Question 10.
ATP produced during glycolysis is a result of substrate level phosphorylation. Explain.
Answer:
Substrate level phosphorylation is a type of metabolic reaction that results in the formation of ATP by the direct transfer and donation of phosphate group to ADP. This occurs twice in Glycolysis.

1. 1, 3 bis phospho glyceric acid form 3 phosphoglyceric acid. The enzyme phosphoglycero kinase, catalyses the phosphorylation. ADP molecule accepts the phosphate released in this reaction and gets converted into ATP.
2. Phosphoenol pyruvic acid undergo dephosphorylation in the presence of the enzyme pyruvic kinase and results in the formation of pyruvic acid. ADP is converted to ATP.

Question 11.
Do you know of any step in Krebs cycle where there is a substrate level phosphorylation? Explain.
Answer:
In Krebs cycle, Succinyl coenzyme A splits into succinic acid and co-enzyme A by the catalytic activity of ‘Succinic acid thiokinase’. The energy released in this reaction is utilized to form ATP from ADP and inorganic phosphate (Pi). Because ATP formation is linked directly to conversion of substrate, this reaction is an example of substrate level phosphorylation.

Question 12.
When a substrate is being metabolised. Why doesn’t all the energy that is produced get released in one step ? Instead it is released in multiple steps. What is the advantage of step wise release of energy?
Answer:

1. The stepwise release of chemical bond energy enables the cell to utilize that energy in ATP synthesis.
2. It minimises the wastage of energy.
3. It keeps the temperature of the cell low to prevent its burning.
4. The activities of enzymes for different steps of respiration can be controlled.
5. The intermediates of respiratory pathway can be used for the synthesis of other biomolecules.
   Thus stepwise release of energy makes the system more efficient in extracting and storing energy.

Question 13.
Respiration requires O₂. How did the first cells on earth manage to survive in an atmosphere that lacked oxygen?
Answer:
The first cells on earth manage to survive in an atmosphere that lack oxygen are anaerobic bacteria. The facultative anaerobic organism is an organism usually bacterium, that makes ATP by aerobic respiration if oxygen is present but is also capable of switching to fermentation under anaerobic conditions. Thus facultative anaerobes is an organism that can grow in either the presence or absence of oxygen. It does not require O₂ to grow but it does tolerate its presence. An obligate anaerobe is an organism that cannot grow in the presence of organism. It neither require O2 to grow nor does it tolerate it.

In any case, all living organisms remain the enzymatic machinery to partially oxidise glucose without the help of oxygen.

Question 14.
The energy yield in terms of ATP is higher in aerobic respiration than during anaerobic respiration. Explain.
Answer:

1. The energy yield in terms of ATP during aerobic respiration is 38 ATP molecules.
2. Glucose molecule gets completely oxidised in the presence of Oxygen.
   C₆H₁₂O₆ + 6O₂ → 6 CO₂ + 6 H₂O + energy
3. Aerobic respiration involves Glycolysis, Krebs cycle and electron transport. 2 ATP molecules are utilised during Glycolysis. Hence net gain in aerobic respiration is 36 ATP molecules.
4. The energy yield in terms of ATP during an aerobic respiration is only 2 ATP molecules. It involves Glycolysis and Fermentation Glucose molecule gets incomplete oxidation in the absence of oxygen.
   C₆H₁₂O₆ → CO₂ + Ethyl alcohol + Energy

Question 15.
RUBP carboxylase, PEPase, pyruvate dehydrogenase ATPase, cytochrome oxidase, Hexokinase, Lactic dehydrogenase.
Select the enzymes from the list above which are involved in
a) Photosynthesis
b) Respiration
c) Both photosynthesis and respiration.
Answer:
a) Enzymes in Photosynthesis : RUBP caboxylase, PEPase.

b) Enzymes in Respiration : Lactic dehydrogenase, Pyruvic dehydrogenase, Hexokinase, Cytochrome oxidase.

c) Enzymes in both photosynthesis and respiration : ATPase.

Question 16.
How does a tree trunk exchange gases with the environment although it lacks stomata?
Answer:
In stems, the living cells are organised in thin layers inside and beneath the bark. They also have openings called lenticels. The cells in the interior are dead and provide only mechanical support. Thus, most cells of a plant have at least a part of their surface in contact with air. This is also facilitated by the loose packing of parenchyma cells in leaves, stems and roots, which provide an interconnected network of air spaces.

Question 17.
Write about two energy yielding reactions of glycolysis.
Answer:
1) Phosphoglycerokinase enzyme catalyses the 1, 3 – bisphosphoglyceric acid resulting in the formation of 3-phosphoglyceric acid. ADP molecule accepts the phosphate released in this reaction and get converted into ATP. This process of ATP is known as substrate level phosphorylation.
phosphoglycerokinase

2) In the final step of glycolysis, phosphoenol pyruvic acid undergoes dephosphorylation in the presence of enzyme ‘pyruvic kinase’ and results in the formation of pyruvic acid. ADP is converted into ATP. This is also substrate level phosphorylation.

Question 18.
Name the site(s) of pyruvate synthesis. Also write the chemical reaction wherein pyruvic acid dehydrogenase acts as a catalyst.
Answer:
Sites of pyruvate are Cytosol
Pyruvate which is formed by the glycolytic catabolism of carbohydrates in the cytosol, after entering into mitochondrial matrix, undergoes oxidative decarboxylation by a complex set of reactions catalysed by pyruvic dehydrogenase. The reactions catalysed by pyruvate dehydrogenase require the participation of several coenzymes, including NAD+ and coenzyme A.

Question 19.
Mention the important series of events of aerobic respiration that occur in the matrix of the mitochondrion as well as the one that takes place in the inner membrane of the mitochondrion.
Answer:

1. The important events that occur in the matrix of mitochondrion is TCA cycle or Krebs cycle or citric acid cycle.
2. During citric acid cycle complete oxidation of pyruvate by the stepwise removal of all the hydrogen atoms leavings three molecules of CO₂, Eight molecules of NADH + H⁺ and two molecules of ATP are formed.
3. The important events that takes place in the inner membrane of the mitochondrion is Electron Transport System (ETS) and Oxidative Phosphorylation.
4. During electron transport system the passing one of the electrons removed as part of the hydrogen atoms to molecules O₂ with simultaneous synthesis of ATP takes place.

Question 20.
The respiratory pathway is believed to be a catabolic pathway. However, the nature of TCA cycle is amphibolic. Explain.
Answer

1. Respiration has been considered as catabolic process and the respiratory pathway as a catobolic pathway.
2. Fatty acids would be broken down to acetyl CoA before entering the respiratory pathway when it is used as a substrate. But when the organism needs fatty acids, acetyl CoA would be withdrawn from the respiratory pathway for it. Thus the respiratory pathway comes into the picture for both during the breakdown and synthesis of fatty acids.
3. Similarly during the breakdown and synthesis of proteins too, respiratory intermediates form the link.
4. The breakdown process within the living organism is called catabolism. The synthesis of his process is called anabolism.
5. Thus the respiratory pathway includes both anabolism and catabolism. Hence it is better to consider respiration as an amphibolic pathway rather than catabolic one.

Question 21.
The net gain of ATP for the complete aerobic oxidation of glucose is 36. Explain.
Answer:
ATP produced during complete aerobic oxidation of one molecule of glucose is as follows.

I) Glycolysis:
i) ATP produced by substrate level phosphorylation

ii) From NADH generated in glycolysis

II) Oxidative decarboxylation of Pyruvic acid

III) Krebs’s Cycle

Long Answer Type Questions

Question 1.
In the following flow chart, replace the symbols a, b, c and d with appropriate terms. Briefly explain the process and give any two applications of it.

Answer:

This process is Fermentation.
In fermentation, the pyruvic acid the incomplete oxidation of glucose is achieved under anaerobic conditions by sets of reactions, where pyruvic acid is converted to C02 and ethanol. In some bacteria lactic acid is formed from pyruvic acid.

Applications :

1. Fermentation is used for preparing alcohol by using yeast cell.
2. In animal cells, like muscles during exercise when oxygen is inadequate for cellular respiration, pyruvic acid.is reduced to lactic acid by lactate dehydrogenase.

Question 2.
Explain Mitchell’s chemiosmosis in relation to oxidative phosphorylation.
Answer:
The synthesis of ATP in respiration is associated with the consumption of oxygen, hence it is referred as oxidative phosphorylation. The mechanism of mitochondrial ATP synthesis can be explained by Mitchell’s chemiosmosis. The transfer of electrons from NADH or FADH to oxygen through electron transport system results in proton transfer from matrix to inner membrane space of mitochondria. Due to this, proton concentration gradient is established across the inner mitochondrial membrane (more number of H+ on inner membrane space side and less on the matrix side).

The inner membrane Of mitochondria is virtually impermeable to protons and thus prevents the return of proton into the matrix.

However the ATP synthase (complex V) consists of the major components F₁ and F₀ .

F₀ is the integral membrane protein complex that forms the channel through which protons cross the inner membrane.

F₁ head piece is a peripheral membrane protein complex and contains the site for the synthesis of ATP from ADP and inorganic phosphate.

When H⁺ move down the gradient, energy is released some amount of energy helps in combining ADP and iP leading to the form of ATP. The energy of 3H⁺ moving down the potential gradient is sufficient to form one ATP molecule.

Diagramatic presentation of ATP synthesis in mitochondria

Question 3.
Oxygen is critical for aerobic respiration. Explain its role with respect to ETS.
Answer:
The energy stored in NADH + H⁺ and FADH₂ are oxidised through electron transport system. The electrons are passed on to O₂ resulting in the formation of H₂O.

The metabolic pathway through which an electron passes from one carrier to another is called the Electron Transport System (ETS). It is present in the inner mitochondrial membrane.

1. Electrons from NADH produced in the mitochondrial matrix during the citric acid cycle are oxidised by an NADH dehydrogenase (complex I) and the electrons are transferred to ubiquinone located within the inner membrane.
2. Ubiquinone also receives reducing equivalents via FADH₂ (complex II) that is generated during oxidation of succinate in the citric acid cycle.
3. The reduced ubiquinone is then oxidised with the transfer of electrons to cytochrome c via cytochrome bcx complex (Complex III).
4. Cytochrome c is a small protein attached to the outer surface of the inner membrane and acts as a mobile carrier for the transfer of electrons between complex III and IV.
5. Complex IV refers to cytochrome c oxidase complex containing cytochromes a and a3 and two copper centres.

When the electrons pass from one carrier to another via complex I to IV in the electron transport chain, they are coupled to ATP synthase (complex V) for the production of ATP from ADP and inorganic phosphate.

The number of ATP molecules synthesised depends upon the nature of electron donor. Oxidation of one molecule of NADH gives rise to 3 molecules of ATP, while that of one molecule of FADH₂ produces 2 molecules of ATP. Although the aerobic process of respiration takes place only in the presence of oxygen, the role of oxygen is limited to the terminal stage of the process. The presence of oxygen is vital, since it drives the whole process by removing hydrogen from the system.

Oxygen acts as the final hydrogen acceptor. Unlike photophosphorylation where it is the light energy that is utilised for the production of proton gradient required for phosphorylation.

In respiration it is the energy of oxidation – reduction utilised for the same process. It is for this reason that the process is called oxidative phosphorylation.

Question 4.
Enumerate the assumptions that we undertake in making the respiratory balance sheet. Are these assumptions valid for a living system ? Compare fermentation and aerobic respiration in this context.
Answer:
It is possible to make calculations of the net gain of ATP for every glucose molecule oxidised; but in reality this can remain only a theoretical exercise. These calculations can be made only on certain assumptions. These assumptions are :

1. There is a sequential, orderly pathway functioning, with one substrate forming the next and with glycolysis, TCA cycle and ETS pathway following one after another.
2. The NADH synthesised in glycolysis is transferred into the mitochondria and undergoes oxidative phosphorylation.
3. None of the intermediates in the pathway are utilised to synthesise any other compound.
4. Only glucose is being respired – no other alternative substrates are entering in the pathway at any of the intermediary stages.

These kinds of assumptions are not really valid in a living system because (1) all pathway work simultaneously and do not take place one after another. (2) Substrates enter the pathways and are withdrawn from it as and when necessary. (3) ATP is utilised as and when needed. (4) Enzymatic rates are controlled by multiple means.

Comparison of fermentation and aerobic respiration :

1. Fermentation accounts for only a partial breakdown of glucose whereas in aerobic respiration glucose completely degraded to CO₂ and H₂O.
2. In fermentation there is a net gain of only two molecules of ATP for each molecule of glucose degraded to pyruvic acid whereas many more molecules of ATP are generated under aerobic conditions.
3. NADH is oxidised to NAD⁺ rather slowly in fermentation; however the reaction is very vigorous in the case of aerobic respiration.

Question 5.
Give an account of glycolysis. Where does it occur? What are the end products? Trace the fate of these products in both aerobic and anaerobic respiration. [Mar. ’20, 18, May ’17]
Answer:
Glycolysis occurs in the cytoplasm. The end product of glycolysis is pyruvic acid.

Glycolysis is the process in which glucose undergoes partial oxidation to form two molecules of pyruvic acid. It is as follows.
1) In the first of glycolysis, a phosphate group is added to glucose molecule in the formation of glucose-6-phosphate. ATP is utilised. This reaction is catalysed by hexokinase enzyme.

2) Phosphohexokinase enzyme catalyses the conversion of glucose-6-phosphate to its isomer fructose-6-phosphate.

3) Fructose-6-phosphate undergoes phosphorylation resulting Fructose-1, 6 bisphosphate. In this step ATP is utilised. This reaction is done by enzyme phosphofructokinase.

4) Fructose 1-6 bisphosphate with the help of enzyme aldolase splits into 2 molecules. They are glyceraldehyde 3 phosphate and dihydroxyacetone phosphate (DHAP).

5) Among the two trioses, only Glyceraldehyde 3 phosphate directly participates in the subsequent reactions.

On the other hand, dihydroxyacetone phosphate gets converted into glyceraldehyde- 3-phosphate in the presence of triose phosphate isomerase and can participate in the next reaction of glycolysis.

6) Glyceraidehyde-3-phosphate undergoes oxidation as well as phosphorylation in the presence of glyceraldehyde 3 phosphate dehychoganase resulting in the formation of 1-3-bisphospho glyceric acid and NADH.

7) Phosphoglycerokinase enzyme catalyses the dephosphorylation of 1,3 bisphosphoglyceric acid resulting in the formation of 3 phosphoglyceric acid. ADP molecule accepts the phosphate released in this reaction and gets converted into high energy compound ATP. This process of formation is known as substrate level phosphorylation.

8) Phosphoglyceromutase enzyme catalyses the transfer of phosphate group from 3-carbon position of 3 PGA to 2-carbon position leading to the formation of 2PGA.

9) The enzyme enolase catalyses the removal of one water molecule from 2-phospho- glyceric acid resulting in the formation of phosphoenol pyruvic acid.

10) Phosphoenol pyruvic acid undergoes dephosphorylation in the presence of the enzyme pyruvic kinase and results in the formation of pyruvic acid. ADP is converted to ATP. This is substrate level of

Fate of Pyruvic acid :
The ultimate fate of pyruvic acid which is an end product of glycolysis depends upon the availability of oxygen. In the presence of oxygen it is completely oxidized into C02 and H20. If oxygen is not available, the pyruvic acid is converted to ethyl alcohol or other organic substances by fermentation during anaerobic respiration.

Question 6.
Explain the reactions of Kreb’s cycle.
Answer:
The reaction sequence of Kreb’s cycle is as below.
1. Condensation :
Acetyl co-enzyme A condenses with oxaloacetic acid and results in the formation of citric acid and co-enzyme A. This condensation reaction is catalysed by the enzyme ‘citric synthetase.’

2. Dehydration :
The enzyme ‘aconitase’ catalyses the removal of one molecule of water from citric acid leading to the formation of cis-aconitic acid.

3. Hydration :
Addition of one molecule of water is done in the presence of aconitase. Cis-aconitic acid forms Isocitric acid.

4. Oxidation -1 :
Isocitric acid undergoes dehydrogenation (oxidation) in the presence of ‘isocitric dehydrogenase’ enzyme, leading to the formation of oxalosuccinic acid. In this reaction NAD is reduced to NADH + H⁺.

5. Decarboxylation :
Oxalosuccinic acid releases one molecule of CO₂ in the presence of oxalosuccinic decarboxylase enzyme and forms a – Ketoglutaric acid.

6. Oxidation – II :
α-ketoglutaric acid undergoes oxidation (dehydrogenase) decarboxylation and condensation with one molecule of CoA leading to the formation of succinyl coenzyme A.

7. Cleavage :
Succinyl coenzyme A splits into succinic acid and coenzyme A by the enzyme activity of succinic acid thiokinase. Energy released in this reaction is utilised to form ATP from ADP and inorganic phosphate.

8. Oxidation III :
Succinic acid then undergoes oxidation and forms Fumeric acid. Instead of NAD+, FAD serves as hydrogen acceptor in this reaction. Therefore FAD is reduced to FADH₂. The enzyme which catalyses this reaction is known as succinic dehydrogenase.

9. Hydration :
The enzyme Fumerase mediates the addition of one water molecule to fumaric acid and leads to the formation of malic acid.

10. Oxidation IV :
In the presence of Malic dehydrogenase, Malic acid releases two hydrogen atoms and gives rise to oxaloacetic acid. In this step NAD+ acts as hydrogen acceptor and is converted into NADH + H⁺.

Intext Question Answers

Question 1.
Differentiate between
a) Respiration and Combustion
b) Glycolysis and Krebs cycle
c) Aerobic respiration and Fermentation
Answer:

a) Differences between Respiration and combustion :

b) Differences between Glycolysis and Kreb’s cycle :

c) Differences between Aerobic respiration and Fermentation.

Question 2.
What are respiratory substrates? Name the most common respiratory substrate.
Answer:
The compounds that are oxidised during respiration are known as respiratory substrate. The most common respiratory substrate is glucose. Usually carbohydrates are oxidised to release energy but proteins, fats and even organic acids can be used as respiratory substrate.

Question 3.
Give the schematic representation of glycolysis.
Answer:

Question 4.
What are the main steps in aerobic respiration ? Where does it take place?
Answer:
The main steps in aerobic respiration are glycolysis, Krebs cycle, electron transport system and oxidative phosphorylation.

Glycolysis occurs in the cytosol of the cell, Krebs cycle occurs in the matrix of mitochondrion. Electron transport system and oxidative phosphorylation takes palce in the inner membrane of mitochondria.

Question 5.
Give the schematic representation of an overall view of Krebs cycle.
Answer:

Question 6.
Explain ETS.
Answer:
The metabolic pathway through which an electron passes from one carrier to another is called the electron transport system. It is present in the inner mitochondrial membrane.

1. Electrons from NADH produced in the mitochondrial matrix during the citric acid cycle are oxidised by an NADH dehydrogenase (Complex – I)
2. The electrons are then transferred to ubiquinone located within the inner membrane. Ubiquinone also receives reducing equivalents via FADH₂ (complex II) that is generated during oxidation of succinate in the citric acid cycle.
3. The reduced ubiquinone (ubiquinol) is then oxidised with the transfer of electrons to cytochrome c via cytochrome bc₁ complex (complex III).
4. Cytochrome c is a small protein attached to the outer surface of the inner membrane and acts as a mobile carrier for the transfer of electrons between complex III and IV.
5. Complex IV refers to cytochrome c oxidase complex containing cytochrome a and a₃ and two copper centres.

When the electrons passes from one carrier to another via complex I to IV in the electron transport chain, they are coupled to ATP synthase (Complex V) for the production of ATP from ADP and inorganic phosphate.

The number of ATP molecules synthesised depends on the nature of the electron donor. Oxidation of one molecule of NADH gives rise to 3 molecules of ATP, while the molecule of FADH₂ produces 2 molecules of ATP.

Aerobic respiration takes place in the presence of oxygen. Oxygen drives the whole process by removing hydrogen from the system. Oxygen acts as a final hydrogen acceptor. Synthesis of ATP from ADP and ip coupled to electron transport from substrate to molecular oxygen is called oxidative phosphorylation.

Question 7.
Distinguish between the following :
a) Aerobic respiration and Anaerobic respiration.
b) Glycolysis and Fermentation.
c) Glycolysis and Citric acid cycle.
Answer:
a) Differences between Aerobic respiration and Anaerobic respiration :

b) Differences between Glycolysis and Fermentation.

c) Differences between Glycolysis and C trie acid cycle:

Question 8.
What are the assumptions made during the calculation of net gain of ATP?
Answer:
The assumptions are

1. There is a sequential, orderly pathway functioning with one substrate forming the next and with glycolysis, TCA cycle and ETS pathway following one after another.
2. The NADH synthesised in glycolysis is transferred into the mitochondria and undergoes oxidative phosphorylation.
3. None of the intermediates in the pathway are utilised to synthesise any other compound.
4. Only glucose is being respired – no other alternative substrates are entering in the pathway at any of the intermediary stages.

Question 9.
Discuss “The respiratory pathway is an amphibolic pathway.”
Answer:
The respiratory pathway is involved in both anabolism and catabolism. So it is considered as amphibolic pathway rather than as a catabolic one.
e.g. : Fatty acids would be broken down to acetyl CoA before entering the respiratory pathway when it is used as a substrate. But when the organism needs to synthesise fatty acids, acetyl CoA would be withdrawn from the respiratory pathway for it. Hence the respiratory pathway shows break down and synthesis of fatty acids.
e.g.: Similarly during the breakdown and the synthesis of proteins too, respiratory intermediate form the link. Thus the breaking down process with in the living organisms constitute catabolism while synthesis is anabolism.

Question 10.
Define RQ. What is its value for fats?
Answer:
The ratio of the volume of CO₂ evolved to the volume of O₂ consumed in respiration is called the Respiratory Quotient (RQ) or respiratory ratio.

Question 11.
What is oxidative phosphorylation?
Answer:
Synthesis of ATP from ADP and iP coupled to electron transport from substrate to molecular oxygen is called oxidative phosphorylation.

Question 12.
What is the significance of stepwise release of energy in respiration?
Answer:

Question 13.
Find the correct ascending sequence of the following, on the basis of energy released in respiratory oxidation.
a) 1 gm of fat
b) 1 gm of protein
c) 1 gm of glucose
d) 0.5 gm of protein + 0.5 gm of glucose
Answer:
Ascending order
a) 1 gm of fat
b) 1 gm of protein
c) 0.5 gm of protein + 0.5 gm of glucose
d) 1 gm of glucose

Question 14.
Name the products, respectively, in aerobic glycolysis in skeletal muscle and anaerobic fermentation in yeast.
Answer:
The product of glycolysis in aerobic skeleton muscle is — The product of anaerobic fermentation in yeast is C02 and ethanol.

Question 15.
If a person is feeling dizzy, glucose or fruit juice is given immediately but not a cheese, sandwich, which might have more energy. Why?
Answer:
Energy is released quickly in case of glucose or fruit juice, it is a quick process.

Question 16.
In a way green plants and cyanobacteria have synthesised ail the food on earth. Comment.
Answer:
Green plants and cyanobacteria prepare food materials by photosynthesis. They are producers. Consumers depend on them directly or indirectly. Thus green plants and cyanobacteria prepare food materials but not all the food on earth.

Question 17.
It is known that red muscle fibres in animals can work for longer periods of time continuously. How is this possible?
Answer:
Red muscle fibres are striated muscle. They contain large number of mitochondria which release energy. They work very speed and active for longer period but with little force. As they work, they are well developed according to Lamark’s theory. Smooth muscle fibres works more longer periods than red muscle fibres.

Telangana TSBIE TS Inter 2nd Year Botany Study Material 4th Lesson Photosynthesis in Higher Plants Textbook Questions and Answers.

TS Inter 2nd Year Botany Study Material 4th Lesson Photosynthesis in Higher Plants

Very Short Answer Type Questions

Question 1.
Name the processes which take place in the grana and stroma regions of chloroplasts.
Answer:
1. Grana :
Light reactions – trapping the light energy by membrane systems and synthesis of ATP and NADPH.

2. Stroma :
Dark reactions – carbon fixation leading to synthesis of sugars.

Question 2.
Can chloroplasts be passed on to progeny? How?
Answer:

1. Yes, nearly equal number of chloroplasts among 2 daughter cells formed from a mother cell during mitosis.
2. Maternal inheritance (cytoplasm of egg) in a sexually reproducing plant.

Question 3.
Where does the photolysis of H₂O occur? What is its significance? [Mar. ’20, 17; May ’14]
Answer:

1. Membrane of grana thylakoid and is associated with PSII.
2. Oxygen liberated by splitting of H₂Ois the main source of atmospheric O₂, electrons released during photolysis replace those removed for PSI.

Question 4.
Where is the enzyme NADP reductase located? What is released if the proton gradient breaksdown?
Answer:

1. The enzyme NADP reductase is located on the stroma side of the grana thylakoid membrane.
2. Energy is released if the proton gradient breaks down and is stored in ATP.

Question 5.
Which tissue transports photosynthates ? What experiments prove this?
Answer:

1. Phloem
2. Ringing experiment

Question 6.
How many molecules of ATP and NADPH are needed to fix a molecule of CO₂ in C₃ plants? Where does this process occur?
Answer:

1. In C₃ plants, 3 molecules of ATP and 2 molecules of NADPH are required to fix one CO₂ molecule.
2. This process occurs by Calvin cycle in the stroma of chloroplast.

Question 7.
Explain the terms:
a) Hatch-Slack pathway
b) Calvin cycle
c) PEP carboxylase
d) Bundle sheath cells
Answer:
a) Hatch-Slack pathway :
It is also called as C₄ cycle because the first stable compound in this cycle is a four carbon compound OAA (Oxalo Acetic Acid).

b) Calvin cycle :
It is also called as C₃ pathway, because the first stable compound is this cycle is a three carbon compound PGA (phosphoglyceric acid)

c) PEP carboxylase :
The enzyme responsible for primary fixation of CO₂ in C₄ plants. It mediates the formation of C₄ acids by reaction of CO₂ with phosphoenolpyruvate.

d) Bundle sheath cells :
The large cells with thick walls impervious to gaseous exchange and no intercelliilo spaces present around the vascular bundles of the C₄ pathway plants. These cells are characterised by a large number of agranular chloroplasts.

Question 8.
What is the role of NADP reductase in the development of proton gradient?
Answer:

1. The NADP reductase enzyme is located on the stroma side of the membrane.
2. It mediates the reduction of NADP⁺ to NADH + H⁺, creates protos gradient across the granathylakoid membrane.

Question 9.
Mention the components of ATPase enzyme. What is their location? Which part of the enzyme shows conformational change?
Answer:

1. The ATPase enzyme consists of two parts : a) F₀ (stalk) is embedded in thylakoid membrane and b) F₁ (head) that protrudes into the stroma.
2. F₁ particle of the ATPase undergoes conformational changes to synthesize ATP molecules.

Question 10.
What products drive Calvin Cycle? What process regenerates them?
Answer:

1. ATP and NADPH
2. Light reaction – phostophosphorylation.

Question 11.
What is the basis for designating. C₃ and C₄ pathway of photosynthesis?
Answer:
1. C₃ Pathway :
In plants with this pathway, the first product of CO₂ fixation is a 3 Carbon containing acid (3 – phosphoglyceric acid / 3 – PGA).

2. C₄ Pathway :
In plants with this pathway, the first product of CO₂ fixation is a 4 Carbon containing acid (oxaloacetic acid / OAA).

Question 12.
Distinguish between action spectrum and absorption spectrum.
Answer:

1. Action spectrum is the graph showing rate of photosynthesis as a function of different wavelengths of light.
2. Absorption spectrum is the graph shows the absorption of different wavelengths of light by photosynthetic pigments.

Question 13.
Of the basic raw materials of photosynthesis, what is reduced? What is oxidised?
Answer:

1. CO₂ and H₂O are the two raw materials of photosynthesis.
2. CO₂ is reduced and H₂O is oxidized during photosynthesis.

Question 14.
Define the law of limiting factors proposed by Blackman. [Mar. 2019, May 2017]
Answer:
If a process (like photosynthesis) is conditioned as to its rapidity by a number of separate factors, the rate of the process is limited by the factor that is present in a relative minimal value.

Question 15.
What is Joseph Priestley’s contribution to the study of photosynthesis?
Answer:

1. Plants restore to the air whatever breathing animals and burning candles remove, i.e., O₂ is evolved during photosynthesis.
2. His experiments revealed the essential role of air in the growth of green plants.

Question 16.
Comment on the contribution of Van Niel to the understanding of photosynthesis.
Answer:

1. Experiments by Van Niel, on purple and green bacteria demonstrated that photosynthesis is essentially a light – dependent reaction.
2. Niel inferred that the O₂ evolved by the green plant comes from H₂O, not from CO₂.
   

Question 17.
With reference to photosystem, bring out the meaning of the terms a) antennae b) reaction centre.
Answer:
a) Antennae :
All the pigments of a photosynten that form a light harvesting system. Those pigments absorb different wavelengths of light,

b) Reaction centre :
A special chlorophyll a in a photosystem forms. It converts light energy into chemical energy. In PS I, the reaction centre is P700 that has an absorption peak at 700 nm. In PS II, the reaction centre is P680 that has an absorption maximum at 680nm.

Question 18.
Why is photosynthetic electron transport from H₂O to NADP⁺, named as Z-scheme?
Answer:

1. The electrons released by photolysis are transported to NADP⁺ via PS II, electron carriers, PS I and electron carriers in non cyclic manner.
2. In this, when all the carriers are placed in a sequence on a redox potential scale it appears like ‘Z’ and hence called as z – scheme.

Question 19.
What is the primary acceptor of CO₂ in C₃ plants? What is the first stable compound formed in a Calvin cycle?
Answer:

1. The primary acceptor of CO₂ in C₃ plants is RuBP (Rubilose bi phosphate).
2. The first stable compound formed in a Calvin cycle is PGA (phosphoglyceric acid).

Question 20.
What is the primary acceptor of CO₂ in C₄ plants? What is the first compound formed as a result of primary carboxylation in the C₄ pathway? [Mar. 2018]
Answer:

1. The primary acceptor of CO₂ in C₄ plants is PEP (phosphoenol pyruvate).
2. The first compound formed due to primary carboxylation in the C₄ pathway is OAA (oxaloacetic acid).

Short Answer Type Questions

Question 1.
Succulents are known to keep their stomata closed during the day to check transpiration. How do they meet their photosynthetic CO₂ requirements?
Answer:

1. Succulent plants have only one kind of photosynthetic cell in which CO₂ is fixed during night and used to make glucose during day.
2. In succulent plants there is an alternative pathway of CO₂ fixatibn called Crassulacean Acid Metabolism (CAM). E.g.: Cacti Crassulaceae is one family of succulent plants.
3. CAM pathway is similar to C₄ pathway in that CO₂ is trapped by highly efficient PEP carboxylase during night time.
4. During day time, the malic acid undergoes oxidative carboxylation to form Pyruvic acid & CO₂. CO₂ is used for photosynthesis.

Question 2.
Chlorophyll ‘a’ is the primary pigment for light reaction. What are accessory pigments? What is their role in photosynthesis?
Answer:

1. Chlorophyll ‘a’ is the primary pigment for light reaction other thylakoid pigments like chlorophyll b, xanthophylls and carotenoids are accessory pigments.
2. Accessory pigments also absorb light and transferthe energy to chlorophyll a.
3. They utilise wider range of wavelength of incoming light for photosynthesis.
4. They also protect chlorophyll a from photo oxidation.

Question 3.
Does ‘dark reaction’ of photosynthesis require light? Explain.
Answer:

1. No. Dark reaction does not depend on the presence of light but is dependent on the products of the light reaction i.e., ATP and NADPH besides H₂O and CO₂.
2. ATP and NADPH are used to drive the process leading to the synthesis of food, sugars. This is called biosynthetic phase of photosynthesis.
3. By convention, it is called dark reaction. However this should not be taken to mean that they occur in darkness.
4. This could be verified. It is simple. Immediately after light becomes unavailable, the biosynthetic process continues for sometime and then stops. If then, light is made available, the synthesis starts again.

Question 4.
How are photosynthesis and respiration related to each other?
Answer:

1. Photosynthesis is an anabolic process. Simple substances like carbondioxide and water combine and yield complex carbohydrates. E.g.: Glucose
   Respiration is an catabolic process. Complex carbohydrates (food materials) are broken into simpler substances like CO₂ and H₂O.
2. Photosynthesis is a reductive process where as respiration is a oxidative process.
3. Photosynthesis is an endogonic process where as respiration is endergonic process.
4. In photosynthesis O₂ is a by product. In Respiration O₂ is utilised.
5. Thus photosynthesis is opposite to respiration Photosynthesis equation is
   
   Respiration equation is
   C₆H₁₂0, + 6O₂ → 6CO₂ + 6H₂O + 686 k.Cal

Question 5.
What conditions enable “RuBisCO” to function as oxygenase? Explain the ensuing process.
Answer:
RuBisCO is characterised by the fact that its active site can bind tb both CO₂ and O₂. It is the relative concentration of O₂ and CO₂ that determines which of the two will bind to the enzyme. If O₂ concentration is more, RuBisCO function as Oxygenase, it binds with O₂ and instead of forming 2 molecules of PGA, it forms one molecule of phosphoglycerate and phosphoglycolate. This pathway is called photorespiration.

In photorespiration pathway, sugars or ATP are not formed, moreover, there is a release of CO₂ with the utilisation of ATP. Therefore photorespiration is a wasteful process.

Question 6.
Why does the rate of photosynthesis decrease at higher temperatures?
Answer:

1. The effect of temperature is linked with the optimum range of enzymatic activity.
2. The dark reactions, being enzymatic, they are temperature controlled. Though the light reactions are also temperature sensitive they are affected to a much lesser extent.
3. Tropical plants (C₄ plants) respond to higher temperatures than temperate plants (C₃ plants).
4. As enzymes gets denatured at higher temperatures the rate of photosynthesis decreases.

Question 7.
Explain how, during light reaction of photosynthesis, ATP synthesis is a chemiosmotic phenomenon.
Answer:

1. ATP synthesis in the chloroplast is a chemiosmotic phenomenon.
2. Like in respiration, in photosynthesis also ATP synthesis is linked to development of a proton gradient across the membrane.
3. During splitting of water and also during oxidation of PQ. (plastoquinol) protons are released into the lumen of thylakoid.
4. Hence the concentration of protons is many times greater than that of stroma, thus leading to proton concentration gradient.

Because of the concentration difference the protons try to move into the stroma, but thylakoids are impermeable to protons (H⁺).

However an enzyme complex ATPase enzyme is present in the membrane. It allows th’e movement of H⁺ through it into the stroma.

The ATPase enzyme consists of two parts.

1. F₀ is embeded in the membrane and forms a transmembrane channel that carries out facilitated diffusion of protons across the membrane.
2. F₁ that comes out of the stroma.
3. The breakdown of the gradient provides enough energy to cause a conformational change in the F₁ particle of the ATPase, which makes the enzyme synthesis several molecules of energy – packed ATP.

Question 8.
Explain how Calvin worked out the complete biosynthetic pathway for the synthesis of sugar.
Answer:

1. The process leading to the synthesis of sugars is called biosynthetic phase of photosynthesis.
2. ATP and NADPH are used in biosynthetic phase CO₂ combines with H₂O toTorm (CH₂O) or sugar.
3. The use of radioactive ¹⁴C by him in algal photosynthesis led to the discovery that the first CO₂ fixation product was a 3 – carbon organic acid.
4. Calvin also contributed to working out the complete biosynthetic pathway. Hence it is called Calvin cycle.
5. Calvin cycle occurs in all photosynthetic plants.
6. Calvin cycle can be described in three stages.

a) Carboxylation :
During which CO₂ combines with ribulose 1, 5 bisphosphate and form 3 phosphoglyceric acid.

b) Reduction :
During which carbohydrate is formed at the expense of the photochemically made ATP and NADPH.

c) Regeneration :
During which the CO₂ acceptor ribulose-1, 5 bisphosphate is formed again so that the cycle continues.

Question 9.
Six turns of Calvin cycle are required to generate one molecule of glucose. Explain.
Answer:

1. With each turn of Calvin cycle, one CO₂ molecule enters the system.
2. With three turns, uptake of 3 molecules of CO₂ and one triose (G – 3P) is available.
3. Hence six turns, results in the uptake of six molecules of CO₂ accounts for two triose (G – 3P), ultimately forms a glucose molecule.
4. Thus the fixation of six molecules of CO₂ and 6 turns of the cycle are required for the removal (net gain) of two molecules of triose (= one glucose molecule) from the pathway.
5. For every CO₂ molecule entering the Calvin cycle 3 molecules of ATP and 2 of NADPH are required. To make one molecule of glucose 6 turns of the cycle are required. Then for 6CO₂, 18 ATP and 12 NADPH are required.

Question 10.
With the help of diagram, explain briefly the process of cyclic photo – phosphorylation. [May 2014]
Answer:

1. In cyclic photo phosphorylation only PS I is functional, the electron is circulated within the photosystem and the phosphorylation occurs due to the cyclic flow of electrons.
2. This takes place in stroma lamellae.
3. Electrons in the reaction centre of PS I gets excited when they receive red light of wavelength 700 nm.
4. In this process, the electrons from the reaction centre of P700 are conveyed to e- acceptor.
5. The excited electron is cycled back to PS I complex through electron transport system.
6. The cyclic flow results in the synthesis of ATP.
7. Cyclic photo phosphorylation also occurs when only light of wavelength beyond 680 nm is available.
8. In green plants, cyclic photo phosphorylation is an additional source of ATP required for chloroplast activities over and above that is required in the Calvin cycle.

Question 11.
In what type of plants do you come across ‘Kranz’ anatomy? To which conditions are those plants better adapted? How are these plants better adapted than the plants, which lack this anatomy?
Answer:
a) C₄ plants.
b) C₄ plants are better adapted to dry tropical regions, they tolerate high temperatures.
c) C₄ plants are special. They have a special type of leaf anatomy.
They tolerate higher temperatures.
They show a response to high light intensities.
They lack a process called photorespiration.
They have greater productivity of biomass.

Question 12.
Explain the structure of the chloroplast. Draw a neat labelled diagram.
Answer:
Chloroplasts are double membrane bound. The space limited by the inner membrane of the chloroplast is called stroma. In the stroma, flattened membrane sacs called Thylakoids are present. Thylakoids are arranged in stacks like a piles of coins called grana. In addition there are flat membranous tubules called the stroma lamellae connecting the thylakoids of the different grana. The space in the thylakoids is called lumen. The stroma contains enzymes for the synthesis of carbohydrates and protein Dark reaction takes place in stroma. It also contain circular DNA and ribosomes. Thylakoids contain photosynthetic pigment. In it light reaction occurs.

Diagrammatic representation of an electron micrograph of a section of chloroplast

Question 13.
Explain why 12 molecules of water are used as substrate, instead of 6 molecules of water, in the following equation.

Answer:
By the middle of nineteenth century the key features of plant photosynthesis were know, namely, that plants could use light energy to make carbohydrates from CO₂ and water.

In the equation when 6 molecules of water is used it denotes that O₂ is evolved from CO₂.

But Cornelius Van Neil of based on his studies (1897 -1985) purple and green bacteria demonstrated that photosynthesis is – light essentially a dependent reaction in which hydrogen from a suitable oxidisable compound reduces CO₂ to carbohydrates. This is expressed as

In green plants H₂O is the hydrogen donor and is oxidised to O₂. Some organisms do not release O₂ during photosynthesis. When H₂S is the hydrogen donor for purple and green sulphur bacteria, the ‘oxidation’ product is sulphur or sulphate depending on the arganism and not O₂. Hence Niel inferred that the O₂ evolved by the green plant comes from H₂O, not from CO₂. This was later proved by using isotopic techniques. The correct equation is

Question 14.
Compare and contrast the absorption spectrum of chlorophylls and carotenoids.
Answer:
Absorption spectrum is a graph showing the absorption of light by pigments at different wavelengths. Chlorophyll a shows the maximum absorption of light and also shows higher rate of photosynthesis in the blue and red regions. Hence we can conclude that chlorophyll a is the chief pigment associated with photosynthesis.

Carotenoid is an accessory pigment. It also absorbs light and transfer the energy to chlorophyll a.

Question 15.
Which group of plants exhibits two types of photosynthetic cells? What is the first product of carboxylation? What carboxylating enzyme is present in bundle sheath cells and mesophyll cells?
Answer:

1. C₄ plants exhibit two types of photosynthetic cells.
2. OAA (oxaloacetic acid) is the first product of carboxylation.
3. PEP carboxylase is present in mesophyll cells Ribulose biphosphate carboxylase- oxygenase (RuBisCO) is present in bundle sheath cells.

Question 16.
A cyclic process is occurring in a C₃ plant, which is light dependent and needs O₂. This process does not produce energy rather it consumes energy.
a) Can you name the given process?
b) Is it essential for survival?
c) What are the end products of this process?
d) Where does it occur?
Answer:
a) Yes, it is photorespiration.
b) Yes, though it is wasteful process, it is essential, as it protects from photo oxidative damage.
c) CO₂ is the end product.
d) It occurs in chloroplast, peroxysomes and mitochondria.

Question 17.
Suppose Euphorbia and Maize are grown in the tropical area.
a) Which one of them do you think will be able to survive under such conditions?
b) Which one of them is more efficient in terms of photosynthetic activity?
c) What differences do you think are there in the leaf?
Answer:
a) Euphorbia (C₄ plants grown in tropical area)
b) Maize CO₂ fixation is done both in Mesophyll cell and Bundle sheath cell.
c) Kranz anatomy in Maize leaf.

Long Answer Type Questions

Question 1.
The entire process of photosynthesis consists of a number of reactions. Where in the cell do each of these take place?
a) Synthesis of ATP and NADPH
b) Photolysis of water
c) Fixation of CO₂
d) Synthesis of sugar molecule
e) Synthesis of starch
Answer:
a) This occur in Grana thylakoid.
b) PS II – Oxygen Evolving Complex (OEC) is associated with the PS II, which itself is physically located on the inner side of the membrane of the thylakoid.
c) Stroma of chloroplast.
d) Cytoplasm
e) Cytoplasm

Question 2.
Which property of pigments is responsible for its ability to initiate the process of photosynthesis? Why is the rate of photosynthesis higher in red and blue regions of the spectrum of light?
Answer:

1. Ability to absorb light at specific wavelength initiates the process of photosynthesis.
2. Absorption spectrum of chlorophyll a show maximum absorption of light at blue and red regions.
3. Action spectrum shows the rate of photosynthesis it is maximum at blue and red region.
4. Absorption spectrum and action spectrum clearly indicate that chlorophyll a is the chief pigment associated with photosynthesis. The rate of photosynthesis is higher in red and blue regions of the spectrum due to it absorption of light wavelengths.

Question 3.
Under what conditions are C₄ plants superior to C₃?
Answer:

Because of these above differences C₄ plants are considered to be superior than C₃ plants.

4. a) How can we plot an action spectrum? What does the action spectrum indicate? Explain with an example.
b) How can we derive an absorption spectrum for any substance?
c) If chlorophyll ‘a’ is responsible for light reaction of photosynthesis, why do the action spectrum and absorption spectrum not overlap?
Answer:
a) Rate of photosynthesis is measured by O₂ released. By drawing a graph between the rate of photosynthesis on x-axis and the wavelength of light on y – axis we can plot an action spectrum.

It indicates the wavelengths at which maximum photosynthesis occurs in plants. Graph shows that chlorophyll a show maximum rate of photosynthesis in the blue and the red regions. Hence chlorophyll ‘a’ is considered as chief pigment associated with photosynthesis.

b) The absorption spectrum for any substance is the ability of the substance to absorb lights at different wavelength.

c) The action spectrum and absorption spectrum does not overlap because the rate of photosynthesis at different wave length is not uniform. Accessory pigment absorbs light at different wavelengths.

Question 5.
What are the important events and end products of light reactions?
Answer:
In light reaction, important events are light absorption, water splitting, oxygen release and the end products are ATP and NADPH.

Light absorption :
The molecules that absorb light are called pigments. The pigments are organised into two discrete photochemical light harvesting complexes (LHC) within the photosystem I (PS I) and photosystem II (PS II). Each photosystem has all the pigments forming a light harvesting system called antennae. The antennae serves to absorb radiant energy. They supply this energy finally to reaction centre. Reaction centre is chlorophyll ‘a’ molecule. It converts light energy into chemical energy. The reaction centre in PS I is P700. The reaction centre in PS II is P680.

Cyclic and Non-cyclic Photo phosphorylation :
The process of forming ATP by cells is named phosphorylation. As it is done in the presence of light, it is called photo phosphorylation. It is of two types.

1. Cyclic photo phosphorylation
2. Non-cyclic photo phosphorylation.

1) Cyclic photo – phosphorylation :

1. In cyclic photo phosphorylation only PS I is functional. The electron is circulated within the photosystem and the phosphorylation occurs due to cyclic flow of electrons.
2. It takes place in stroma lamellae.
3. The electrons in the reaction centre of PS I gets excited when they receive red light of wave length 700 nm.
4. In this process, the electrons from the reaction centre of PS I are conveyed to e“ acceptor.
5. The excited electron is cycled back to PS I through electron transport system.
6. The cyclic flow results in the synthesis of ATP.

2) Non-cyclic photo phosphorylation :

1. When the two photosystems work in a series, first PS I! and then PS I a process called non-cyclic photo phosphorylation occurs.
2. In PS II, the reaction centre chlorophyll a absorbs 680 nm wavelength of red light causing electrons to become excited and jump.
3. The electrons are picked up by an electron acceptor.
4. From there it passes through electron transport system consists of cytochrome and reaches PS I.
5. Simultaneously electrons in the reaction centre PS I also gets excited when they receive red light of 700 nm wavelength.
6. The electrons from the PS i are conveyed to e acceptor.
7. These electrons then are moved downhill again, to a molecule of energy rich NADP⁺.
8. The addition of these electrons reduces NADP⁺ to NADPH + H⁺.
   Non-cyclic photo phosphorylation is also called Z-scheme.

Splitting of water:

1. The electrons that are moved from PS II are supplied with electrons continuously by splitting of water.
2. The electrons removed from PS I are provided by PS II.
3. Water splitting complex (oxygen evolving complex) is responsible for photolysis of water.
   2H₂O → 4H⁺ + O₂ + 4e^–

Thus oxygen released is one of the net products of photosynthesis.
The end products of light reaction are ATP, NADPH and O₂.

Question 6.
Explain various aspects of Mitchell chemiosmotic hypothesis, with the help of diagrams.
Answer:
Mitchell proposed chemiosmotic hypothesis to explain ATP synthesis.
ATP synthesis is linked to development of proton gradient across the membrane.

a) Since splitting of the water molecule takes place on the inner side of the membrane, the protons or hydrogen ions that are produced by the splitting of water accumulate within the lumen of the thylakoids.

b) As electrons move through the transport chain, protons are transported across the membrane. This happens because the primary acceptor of electrons which are located towards the outer side of the membrane transfers its electron not to electron carrier, but to (H⁺) proton carrier (PQ). Hence it removes protons, from the stroma which transporting an electron. When this molecule passes its electron, to the electron carrier on the inner side of the membrane, the proton is released into the inner side or lumen of the membrane. Then proton gradient across the membrane increases due to quinones.

c) The NADP reductase enzyme is located on the stroma side of the membrane. Along with the electrons that come from the acceptor of electrons of PS I, protons are necessary for the reduction of NADP⁺ to N ADPH + H⁺. These protons are also removed from the stroma. Thus protons in the stroma decrease in number. Where as in lumen protons are accumulated. This creates a proton gradient across the thylakoid membrane and the pH in the lumen decreases.

The proton gradient is broken down due to the movement of protons across the membrane to the stroma through the transmembrane channel of ATPase. ATPase consists of two parts.

1. One is F₀. F₀ is embeded in the membrane that forms transmembrane channel it carries facilitated diffusion of protons across the membrane.
2. Other one is F₁ It protrudes on the outer surface of the thylakoid membrane.

Diffusion of protons across the membrane releases enough energy to activate ATPase enzyme that catalyses the formation of ATP.

Question 7.
Comment on the dual role of RuBisCO. What is the basis for its oxygenation activity? Why is this activity absent or negligible in C₄ plants?
Answer:
RuBisCO is the most abundant enzyme in the world. It is characterised by the fact that its active site can bind to both CO₂ and O₂. Thus it shows dual role. It is written a Ribulose bisphosphate carboxylase – oxygenase. Generally, RuBisCo has greater affinity for CO₂ than O₂. This binding is competitive. It is the relative concentration of O₂ and CO₂ that determines which of the two will bind to the enzyme.

In C₃ plants some O₂ does bind to RuBisCO and hence CO₂ fixation is reduced. Here RuBP instead of forming 2 molecules of PGA forms one molecule of phosphoglycerate and phosphoglycelate. This pathway is called photorespiration. In this photorespiration sugar or ATP or NADP are not formed. Moreover, CO₂ is released with the utilisation of ATP molecule. Hence it is wasteful process.

In C₄ plants photorespiration is negligible or absent because mechanism that increases the concentration of CO₂ at the enzyme site. C₄ acid from the mesophyll is broken down in the bundle sheath cells to release CO₂. This results in increasing the intracellular concentration of CO₂. This makes RuBisCO function as carboxylase, minimising oxygenase activity.

Intext Question Answers

Question 1.
By looking at a plant externally, can you tell whether a plant is C₃ or C₄? Why and how?
Answer:
In C₃ plants photorespiration occurs. In C₃ plants transpiration is more. They occur in all types climates. They have much lower temperature optimum. C₃ plants respond to increased CO₂ concentration. C₄ plants are found in dry tropical and subtropical regions. E.g. : Sugarcane, maize, sorghum, amaranthus etc. They tolerate higher temperatures. They show a response to high light intensities. They lack photorespiration. They have greater productivity of biomass. They transpire less.

Question 2.
By looking at which internal structure of a plant can you tell whether a plant is C₃ or C₄? Explain.
Answer:
By looking at the vertical section of leaf we can tell whether a plant is C₃ or C₄. The C₄ plant leaves have Kranz anatomy. Large cells around the vascular bundle called bundle sheath cells are present. The bundle sheath cells have a large number of chloroplasts, thick walls impervious to gaseous exchange and no intercellular spaces.
E.g.: Maize, sorghum etc.

In C₃ plants, chloroplast dimorphism is not present, cells participating will have only one type of chloroplast. In C₄ plants, chloroplast dimorphism is present. Cells participating will have two types of chloroplasts (agranal and granal).

Question 3.
Even though a very few cells in a C₄ plant carry out the biosynthetic – Calvin pathway, yet they are highly productive. Can you discuss why?
Answer:
RuBisCO has an active site which can bind to both CO₂ and O2. RuBisCO has greater affinity for CO₂ than for O₂. It is the relative concentration of 02 and C02 that determine which of the two will bind to the enzyme. In C₄ cycle, RuBisCO is present in bundle sheath cells, C02 concentration is more in bundle sheath cells. Hence productivity is high. In C₃ cycle some O₂ does bind to RuBisCO, leading to photorespiration. So CO₂ fixation is decreased.

Question 4.
RuBisCO is an enzyme that acts both as a carboxylase and oxygenase. Why do you think RuBisCO carries out more carboxylation in C₄ plants?
Answer:
RuBisCO is an enzyme that acts both as a carboxylase and oxygenase. In C₄ plants RuBisCO is present only in bundle sheath cells. So oxygenation of RuBisCO is avoided. These plants have CO₂ concentrating mechanism. That is why RuBisCO carries more carboxylation in C₄ plants.

Question 5.
Suppose there were plants that had a high concentration of chlorophyll b, but lacked chlorophyll a, would it carry out photosynthesis? Then why do plants have chlorophyll b and other accessory pigments?
Answer:
Plants having high concentration of chlorophyll b, but lacked chlorophyll a would not carry photosynthesis. Chlorophyll a shows maximum absorption in the blue and red regions where maximum photosynthesis occurs. Hence chlorophyll a is the chief pigment of photosynthesis. Plants have chlorophyll b and other accessory pigments to absorb light and transfer the energy to chlorophyll a and protect chlorophyll a from photo oxidation.

Question 6.
Why is the colour of a leaf kept in the dark frequently yellow, or pale green? Which pigment do you think is more stable?
Answer:
Colour of the leaf is not due to single pigment but due to four pigments. We can separate these pigments by chromatographic separation.

1. Chlorophyll a – bright or blue green
2. Chlorophyll b – yellow green
3. Xanthophylls – yellow
4. Carotenoids – yellow to yellow orange
   Chlorophyll a is the most stable pigment.

Question 7.
Look at leaves of the same plant on the shady side and compare it with the leaves on the sunny side. Or, compare the potted plants kept in the sunlight with those in the shade. Which of them has leaves that are darker green? Why?
Answer:
The leaves of the plant potted in the sunlight will be more green as compared to the leaves of the plant potted in the shade. This is because the amount of chlorophyll in a leaf has a direct relationship with the rate of photosynthesis. In dark, the photosynthesis decreases. So it appears pale.

Question 8.
Figure shows the effect of light on the rate of photosynthesis. Based on the graph, answer the following questions :
a) At which point/s (A, B or C) in the curve is light a limiting factor?
b) What could be the limiting factor/s in region A?
c) What do C and D represent on the curve?

Answer:
a) At higher light intensities the rate of photosynthesis become limiting factor – A/B.

b) Light

c) Saturation of light (C).
The rate of photosynthesis is not increased by increasing intensity of light (D).

Question 9.
Give comparison between the following :
a) C₃ and C₄ pathways.
b) Cyclic and non-cyclic photophosphorylation,
c) Anatomy of leaf in C₃ and C₄ paints.
Answer:

a)

b)

c)

Question 10.
Cyanobacteria and some other photosynthetic bacteria do not have chioroplasts. How do they conduct photosynthesis?
Answer:
Due to presence of chlorophyll pigments bacteria xanthophyll, phycobillins. They conduct photosynthesis in cyanobacteria. In xanthophyll, photosynthetic bacteria have bacterial chlorophyll, chlorobial chlorophyll conducts photosynthesis.

Question 11.
Does moonlight support photosynthesis?
Answer:
Yes, at very low rate.

Question 12.
Why does photorespiration not occur in C₄ plants?
Answer:
In C₄ plants RuBisCO is present only in bundle sheath cells. So oxygenation of RuBisCO is avoided. So photorespiration does not occur.

Question 13.
Tomatoes, chillis and carrots are red in colour due to the presence of one pigment. Name the pigment. Is it a photosynthetic pigment?
Answer:

1. Carotenoids.
2. It absorbs light of different wavelength and also protect of photo-oxidation.

Question 14.
If a green plant is kept in dark with proper ventilation, can this plant carry out photosynthesis? Can anything be given as supplement to maintain its growth or survival?
Answer:
No, it cannot carry photosynthesis.
Yes, artificial light.

Question 15.
Photosynthetic organisms occur at different depths in the ocean. Do they receive qualitatively and quantitatively the same light? How do they adapt to carry out photosynthesis under these conditions?
Answer:
No, they don’t receive same light.
Diffuse light.

Question 16.
In tropical rain forests, the canopy is thick and shorter plants growing below it, receive filtered light. How are they able to carry out photosynthesis?
Answer:
They get modified into epiphytes.

Question 17.
Why do you believe chloroplast and mitochondria to be semi-autonomous organelle.
Answer:
Due to the presence of circular double stranded DNA.

Question 18.
Is it correct to say that photosynthesis occurs only in the leaves of a plant ? Besides leaves, what are the other parts that may be capable of carrying out photosynthesis ? Justify.
Answer:
Young plants which are green in colour Herbs.
Herbs contain green colour stem.
Photosynthetic roots. E.g.: Taeniophyllum.

Question 19.
What can we conclude from the statement that the action and absorption spectrum of photosynthesis overlap? At which wavelength do they show peaks?
Answer:
Maximum absorption by chlorophyll a and the rate of photosynthesis is highest in the blue and the red regions. It is at its peak in blue and red wavelengths.

Telangana TSBIE TS Inter 2nd Year Botany Study Material 3rd Lesson Enzymes Textbook Questions and Answers.

TS Inter 2nd Year Botany Study Material 3rd Lesson Enzymes

Very Short Answer Type Questions

Question 1.
How are prosthetic groups different from co-factors?
Answer:
1. Prosthetic group :
An organic co-factor that is tightly bound to the apoenzyme.
Eg. Hemee in peroxidase enzyme.

2. Cofactor :
Non protein part of a holo enzyme. It may be a metal ion or orgain compound.
Eg. Zinc in carboxy peptidase.

Question 2.
What is meant by ‘feedback inhibition’?
Answer:

1. Feedback inhibition: The end product of a chain of enzyme catalysed reactions inhibits the enzyme of the first reaction as part of homeostatic control of metabolism.
2. Eg. Pyruvic acid in Glycolysis.

Question 3.
Why are ‘oxido reductases’, so named?
Answer:
Oxido reductases are the enzymes which catalyse oxido reduction between two substrates S and S.
E.g.: S reduced + S’ oxidised → S oxidised + S’

Question 4.
Distinguish between apoenzyme and cofactor. [March 2014]
Answer:

1. The protein part of a holoenzyme is called apoenzyme.
2. The non-protein part of a holoenzyme is called cofactor, it may be a metal ion or an organic compound.

Question 5.
What are competitive enzyme inhibitors? Mention one example. [May 2014]
Answer:

1. Competetive inhibitor: An inhibitor closely resembles the substrate in its molecular structure and inhibits the activity of the enzyme
2. E.g.: Inhibition of succinic dehydrogenase by malonate which closely resembles the substrate succinate in structure.

Question 6.
What are non-competitive enzyme inhibitors? Mention one example.
Answer:

1. Non-competitive enzyme inhibitor has no structural similarity with the substrate and forms an enzyme inhibitor complex at a point other than its active site, So that the globular structure of the enzyme is changed and catalysis cannot take place.
2. E.g.: Metal ions of copper, mercury, silver, etc.

Question 7.
What do the four digits of an enzyme code indicate?
Answer:

1. The four digits of an enzyme code helps to identify individual enzyme.
2. For example : Glucose-6- phosphotransferase has the enzyme code (E.C.) 2.7.I.2. The first, second, third and fourth of the code indicate the major class, subclass, sub-sub class and serial number of the enzyme respectively.

Question 8.
Who proposed ‘Lock and Key hypothesis’ and Induced fit hypothesis?
Answer:

1. ‘Lock and Key’hypothesis was proposed by Emil frsher. (1884)
2. ‘Induced – Fit’ hypothesis was proposed by Daniel E. Koshland. (1973)

Short Answer Type Questions

Question 1.
Explain how pH affects enzyme activity with the help of a graphical representation.
Answer:

1. Generally enzymes function in a narrow range of pH value.
2. Each enzyme shows its highest activity at a particular pH called optimum pH.
3. Activity declines both below and above the optimum value.

Question 2.
Explain the importance of (ES) complex formation.
Answer:
The formation of (ES) complex is essential for catalysis.

1. First, the substrate binds to the active site of the enzyme, fitting into the active site.
2. The binding of the substrate induces the enzyme to alter its shape; fitting more tightly around the substrate.
3. The active site of the enzyme, now in close proximity to the substrate, breaks the chemical bonds of the substrate and the new enzyme product complex is formed.
4. The enzyme releases the products of the reaction and the free enzyme is ready to bind to another molecule of the substrate and runs through the catalytic cycle once again.

Question 3.
Write briefly about enzyme inhibitors. [Mar. 2019, ’18, ’17 ; May ’17]
Answer:
a) The chemicals that block the enzyme activity are called inhibitors. The process is called inhibition.
b) Inhibitors are three types. They are
i) Competitive inhibitors
ii) Non-competitive inhibitors
iii) Feedback inhibitors

i) Competitive inhibitors :
These are the substances which are structurally similar to substrate molecules and compete for the. active sites of an enzyme. E.g : Inhibition of succinic dehydrogenase by malonate which closely resembles the substrate succinate in structure.

ii) Non-competitive inhibitors :
These chemicals do not resemble the substrate in structure. They bind to an enzyme at locations other than the active sites and makes it inactive. So no new products are formed. E.g: Metal ions of copper, mercury, silver etc.

iii) Feedback inhibitors :
In many biochemical reactions, the accumulation of end products of reactions will inhibit the first step of reaction. It is a part of homeostatic control of metabolism.

Question 4.
Explain different types of cofactors.
Answer:
a) Enzymes having non-protein part along with protein part is called Holoenzyme. The non-protein part is called co-factor and protein part is called apoenzyme.
Co-factor + Apoenzyme = Holoenzyme
b) Co-factors are of three kinds. They are
i) Prosthetic groups
ii) Co-enzymes
iii) Metal ions

i) Prosthetic groups :
Prosthetic groups are the organic co-factors which are tightly bound to the apoenzyme. For example : In peroxidase and catalase, which catalyse the breakdown of hydrogen peroxide to water and oxygen, haem is the prosthetic group and it is the active part of enzyme.

ii) Co-enzymes :
Coenzymes are the organic molecules which are loosely associated with the apoenzyme. These co-enzymes are derived from wajer soluble vitamins. Eg: Coenzyme NAD and NADP contain vitamin niacin.

iii) Metal ions :
Mostly enzymes require metal ions for their activity which form coordinate bonds with side chains at the active site and at the same time form one or more coordination bonds with the substrate. E.g. : Zinc is a cofactor for the proteolytic enzyme carboxypeptidase.

Long Answer Type Questions

Question 1.
Write an account of the classification of enzyme.
Answer:
Enzymes have been classified into different groups based on the type of reactions they catalyse. Enzymes are divided into 6 classes. Each class is again divided into sub-class and sub-subclasses. They are

1) Oxidoreductases / dehydrogenases :
Enzymes which catalyse oxidoreduction between two substrates S and S’.
E.g.: S.reduced + S’ oxidised → S oxidised + S’ reduced

2) Transferases :
Enzymes catalysing a transfer of a group G (other than hydrogen) between a pair of substrate S and S’.
E.g.: S — G + S’ → S + S’ — G

3) Hydrolases :
Enzymes catalysing hydrolysis of ester, ether, peptide, glycosidic, C – C, C – halide or P – N bonds.

4) Lyases :
Enzymes that catalyse removal of groups from substrates by mechanisms other than hydrolysis leaving double bonds.

5) Isomerases :
Includes all enzymes catalysing inter – conversion of optical, geometric or positional isomers.

6) Ligases :
Enzymes catalysing the linking together of 2 compounds. E.g.: enzymes which catalyse joining of C – O, C – S, C – N, P – O etc., bonds.

The above classification provides for a four digit code to identify individual enzymes. For E.g.: Glucose – 6 – Phosphotransferase has the enzyme code (E.C.) 2,7.1.2
The first digit of code indicates the major class.
The second digit of code indicates the subclass.
The third digit of code indicates the sub-subclass.
The fourth digit indicates the serial number of the enzyme in a particular sub-subclass.

Question 2.
Explain the mechanism of enzyme action. [Mar. 2020]
Answer:

1. During the enzyme action the enzyme (E) combines with its specific substrates (S) to form a enzyme – substrate complex (E – S) which is short – lived.
2. Energy that is required for a substrate to react inorder to get converted into end product is called “activation energy”.
3. This activation energy is available in different forms like heat, ATP etc.
4. The activation energy of the formation of this E – S complex is low, hence many molecules can react and participate in the reaction, leading to the formation of products (P).
5. The E – S complex dissociates into its products P and the unchanged enzyme E with an intermediate formation of the enzyme – product complex (EP).
6. The formation of the ES complex is essential for catalysis. E + S → (ES) → (EP) → E + P
7. Formation of (ES) complex has been explained with ‘Lock and Key’ hypothesis by Emil Fisher and later with “Induced Fit” hypothesis by Daniel E. Koshland (1973).
8. According to this theory every enzyme possess “Active sites”.
9. The substrate (S) gets attached to the active site of the enzyme (E) and forms an enzyme substrate (ES) complex.
10. The enzyme remains unchanged while the substrate is broken into products (P).

Intext Question Answers

Question 1.
Enumerate the properties of enzymes.
Answer:
Enzymes show following properties :
a) Catalytic property :
An enzyme is organic catalyst. It does not undergo any change during a reaction, catalysed by it. It only speeds up a rate of a reaction.

b) Specificity :
Enzymes are specific and act only on specific substances. For example, sucrase acts only on sucrose.

c) Active in minute quantity :
Enzymes are active in small quantities. The number of substrate molecules to be converted into products by one molecule of enzyme per minute is called turn over number.

d) Reversibility :
Most of the enzymes are reversible in their action. They can speed up a particular reaction either in forward or in backward direction.

e) Thermolability :
Enzymes are heat sensitive. At high temperature, enzymes are denatured and at low temperature, they are inactive because enzymes are Chemically proteins.

f) Sensitivity of pH :
The enzyme activity is by pH controlled by pH concentration. Most of the enzymes work at neutral pH.

g) Proteinaceous nature :
All enzymes are chemically proteins, having high molecular weights, ranging from 10,000 to several million deltron. Based on the composition enzymes are of two types. 1) Simple enzymes 2) Conjugated or Holoenzymes.

Question 2.
What is Michaelis constant?
Answer:
a) The Michaelis constant Km. is very important in determining enzyme substrate interaction.
b) The value of enzyme range widely and often dependent on environmental conditions such as pH, temperature and ionic strength.
c) The Michaelis constant is able to detect two factors.

One is the concentration of the substrate when the reaction velocity is half that of the maximal velocity, thus Michaelis constant measures the concentration of substrate required for a significant catalysis to take place.

Second is the Michaelis constant is able to detect the strength of the enzyme-substrate complex (ES).

Question 3.
Distinguish between feedback inhibition and allosteric inhibition.
Answer:
In many biochemical reactions, the accumulation of end products of reactions will inhibit the first step of reaction. This is called Feedback inhibition.

Allosteric inhibition :
Some enzymes, possess allosteric (alios = other; steros = site) sites. The enzymes which possess allosteric sites are known as allosteric enzymes. The binding of substance to allosteric site may stimulate or inhibit enzyme action. The substances that reduce the activity of an enzyme by binding at allosteric site are known as allosteric inhibitors.

Question 4.
What are isoenzymes?
Answer:
Isoenzymes are proteins with different structure which catalyze the same reaction.

Question 5.
What is turnover number? What is the fastest acting enzyme?
Answer:
The number of substrate molecules converted into products by one molecule of an enzyme in one minute time is called turn over number (TON). The fastest acting enzyme is carbonic- anhydrase.