TS Inter 2nd Year

Telangana TSBIE TS Inter 2nd Year Zoology Study Material Lesson 4(a) Endocrine System and Chemical Coordination Textbook Questions and Answers.

TS Inter 2nd Year Zoology Study Material Lesson 4(a) Endocrine System and Chemical Coordination

Very Short Answer Type Questions

Question 1.
What is acromegaly? Name the hormone responsible for this disorder. [March 2015 (A.P.)]
Answer:
Hyper secretion of growth stimulating hormone (somatotropin) in adults results in an abnormality called acromegaly. It is characterized by enlargement of bones of Jaw, hand and feet, thickened nose, lips and eyelids and gorilla like appearance of the person affected.

Question 2.
Which hormone is called anti-diuretic hormone ? Write the name of the gland that secretes it. [May/ June 2014]
Answer:
Vasopressin is called anti – diuretic hormone (ADH). It is secreted by posterior lobe of pituitary gland.

Question 3.
Name the gland that increases in size during childhood and decreases in size during adulthood. What important role does it play in case of infection? [March 2019]
Answer:
Thymus gland increases in size during childhood and decreases in size during adulthood. The secretion is thymosin plays a major role in the differentiation of T – lymphocytes, which provide cell mediated immunity. It also promote production of antibodies.

Question 4.
Distinguish between diabetes insipidus and diabetes mellitus. [March 2020, 2018 (A.P.); March 2014]
Answer:
a) Diabetes insipidus: Deficiency of vasopressin causes diabetes insipidus in which the patient excretes large volumes of urine, resulting in dehydration and thirst.
b) Diabetes mellitus: A condition resulting from lack of insulin as a result of which, the body cannot store or oxidise sugar efficiently (and sugar is lost through urine).

Question 5.
What are Islets of Langerhans?
Answer:
The endocrine portion of pancreas is just 1 to 2% and consists of 1 to 2 millions of Islets of Langerhans which are having α – cells and β – cells, α – cells produce hormone glucagon. β cells produce insulin.

Question 6.
What is ‘insulin shock’? [March 2018, ’15 (A.P.)]
Answer:
Hyper secretion of insulin leads to decreased level of glucose in the blood (hypoglycemia) resulting in insulin shock.

Question 7.
Which hormone is commonly known as fight and flight hormone? [March 2015 (T.S.)]
Answer:
Fight or flight hormone is the common name adrenaline or epinephrine. This hormone enhances alertness, dilation of pupils, piloerection, sweating. Increase heart beat to face emergency situation.

Question 8.
What are androgens? Which cells secrete them?
Answer:
Androgens are secreted by leiding cells or interstitial cells of testes in males. Androgens stimulate the secondary sexual characters in males and also enhance the process of spermatogenesis.

Question 9.
What is erythropoietin? What is its function? [Mar. 2019, ’14; May/June ’14]
Answer:
Kidneys produce a hormone called erythropoietin. It stimulates erythropoiesis (formation of RBC). The role of this hormone is to control the formation of red blood cells by regulating the differentiation and proliferation of erythroid progenitor cells in bone marrow.

Short Answer Questions

Question 1.
List out the names of endocrine glands present in human beings and mention the hormones they secrete.
Answer:
Endocrine glands of Man – Their Secretion

1. Hypothalamus – Growth hormone releasing hormone (GHRH)
2. Pituitary. – a) Anterior pituitary : Growth hormone, Prolactin, Thyroid stimulating hormone. (TSH)
   Andreno corticotropic hormone (ACTH), Follicle stimulating hormone (FSH), Lutenizing hormone (LH).
   b) Pars intermedia: Melanocyte stimulating Hormone (MSH)
   c) Posterior pituitary: Vasopressin, oxytocin.
3. Pineal gland – Melatonin
4. Thyroid gland – Thyroxine, calcitonin.
5. Parathyroid glands – Parathormone
6. Thymus gland – Thymosins
7. Adrenal gland – Cortex – Gluco corticoids, mineral corticoids.
   Medulla – Adrenaline, Noradrenaline.
8. Pancreas – Islets of langerhans – Insulin, Glucagon.
9. Testes – Androgens
10. Ovaries – Estrogen, Progesterone.

Question 2.
Describe the role of hypothalamus as a neuroendocrine organ.
Answer:
The hypothalamus is located below the thalamus, constituting the floor of the diencephalon, a part of the fore brain. It connects the neural and endocrine systems as it is closely tied to the pituitary gland. It responds to the sensory impulses received from different receptors by sending out appropriate neural or endocrine signals.

It regulates a wide range of body functions. It contains several groups of neurosecretory cells called ‘nuclei’ which produce hormones called neurohormones. They are transported to the neurohypophysis through the axons of the hypothalamo – hypophysial tract. The two types of hormones produced by the hypothalamus are 1) the releasing hormones (which stimulate secretion of pituitary hormones), and 2) the inhibiting hormones (which inhibit secretions of pituitary hormones.

Question 3.
Give an account of the secretions of pituitary gland.
Answer:
Previously, pituitary gland was called the “master” endocrine gland, because it controls several endocrine glands. Release of hormones by adenohypophysis is stimulated by releasing hormones and suppressed by inhibiting hormones of the hypothalamus.

Growth Hormone or Somatotropin: In response to human growth hormone, cells in the liver, skeletal muscle, cartilage, bone, and other tissues secrete insulin – like growth factors that cause cells to grow and multiply. These factors accelerate protein synthesis and decrease catabolism of proteins.

Thyroid – Stimulating Hormone :
It stimulates the synthesis and secretion of thyroid hormones by the thyroid gland.

Adrenocorticotropic Hormone (AGTH) :
It controls the secretion of glucocorticoids by the adrenal cortex.

Follicle – Stimulating Hormone :
In females FSH initiates the development of ovarian follicles. In males FSH stimulates spermatogenesis.

Luteinizing Hormone :
In females, LH stimulates ovulation, formation of the corpus luteum and the secretion of progesterone by the corpus luteum. In males, this hormone is called interstitial cell stimulating hormone. It stimulates leydig cells in the testis to secrete testosterone. FSH and LH are termed gonadotropins because their target organs are gonads.

Prolactin :
Prolactin, together with other hormones, initiates and maintains milk secretion by the mammary glands. The function of prolactin is not known in males.

Melanocyte – Stimulating Hormone (MSH) :
MSH increases skin pigmentation in lower vertebrates by stimulating the dispersion of melanin granules in melanocytes.

Neurohypophysis :
It does not synthesize hormones. It stores and releases oxytocin and vasopressin.

Oxytocin :
During delivery, oxytein enhances contraction of smooth muscle cells in the wall of uterus. After delivery, it stimulates milk ejection.

Vasopressin or Antidiuritic Hormone :
ADH causes the kidneys to absorb more water into the blood. In the absence of ADH, urine output increases from the normal 1 to 2 litres about 20 litres a day. ADH causes constriction of arterioles, which increases blood pressure. The amount of ADH secreted is regulated.

Question 4.
Compare a ‘pituitary dwarf’ and a ‘thyroid dwarf’ in respect of similarities and dissimilarities they possess.
Answer:
1. Pituitary dwarf :
Hypo secretion of growth harmone (STH) during childhood retards growth, resulting in a pituitary dwarf / midget. The pituitary dwarf is sexually and intellectually a normal individual.

2. Thyroid dwarf :
During pregnancy, due to hypothyroidism, defective development of the growing baby leads to a disorder called cretinism. Physical and mental growth get severely stunted and is called thyroid dwarf. This is due to untreated congenital hypothyroidism. Stunted growth, mental retardation, low intelligent quotient, abnormal skin, deafness and mutism are some of the characters of this disease.

Question 5.
Explain how hypothyroidism and hyperthyroidism can affect the body. [March 2017, May ’17 (A.P.)]
Answer:
Over activity of the thyroid, cancer of the gland or development of nodules of thyroid lead to hyperthyroidism. In adults, abnormal growth causes a disease called exophthalamic goiter, with characteristically protruded eyeballs. Hyperthyroidism also affects the physiology of the body (increased metabolic rate). Inadequate supply of iodine in the diet results in hypothyroidism and enlargement of the thyroid gland. This condition is called simple goiter.

During pregnancy, due to hypothyroidism, defective development of the growing baby leads to a disorder called cretinism. Physical and mental growth gets severely stunted (Thyroid dwarf) due to untreated ‘congenital hypothyroidism’ (deficiency of thyroid hormones by birth). Stunted growth, mental retardation, low intelligence quotient, abnormal skin, deafness and mutism are some of the characteristic features of the this disease. In adult women, hypothyroidism may cause irregular menstrual cycles. In adults the hypothyroidism results in a condition called myxedema, Lethargy, mental impairment, intolerance to cold, puffiness of face and dry skin and some of the symptoms of myxedema.

Question 6.
Write a note on Addison’s disease and Cushing’s syndrome.
Answer:
Addison’s disease is caused due to hyposecretion of glucocorticoids by the adrenal cortex. This disease is characterised by loss of weight, muscle weakness, fatigue and reduced blood pressure. Sometimes darkening of the skin in both exposed and nonexposed parts of the body occurs in this disorder. This disorder does not allow an individual to respond to stress.

Cushing’s syndrome :
It results due to over production of glucocorticoids. This condition is characterized by breakdown of muscle proteins and redistribution of body fat resulting in spindly arms and legs accompanied by a round moon face, buffalo hump on the back and pendulous abdomen. Wound healing is poor. The elevated level of cortisols causes hyperglycemia, over deposition of glycogen in liver and rapid gain of weight.

Question 7.
Why does sugar appear in the urine of a diabetic?
Answer:
Insulin secreted by a – cells of Islets of langerhans promotes conversion of glucose into glycogen in the target cells. Both glucagon and insulin maintain the homeostasis of glucose in the blood. Persistent hyperglycemia leads to a complex disorder called diabetes mellitus prolonged hyperglycemia leads to diabetes mellitus associated with loss of glucose through urine. It is called glycosuria and formation of harmful compounds called ‘Ketone bodies’. Insulin therapy is used to treat diabetic patients.

Question 8.
Describe the male and female sex hormones and their actions.
Answer:
Male sex hormones or Androgens are required for the development, maturation and functioning of the male accessory sex organs such as epididymis, vas deferens, seminal vesicles, prostate gland, urethra etc. These hormones control muscular growth, growth of facial and axillary hair, aggressiveness, low pitch voice (masculine voice) etc. Androgens stimulate the process of spermatogenesis. Androgens affect tbe central neural system, controlling the male sexual behavior (libido / sex drive / sexual urge) and also have an effect on protein and carbohydrate anabolism.

Ovaries act as endocrine glands too producing the female hormones chiefly: estrogen and progesterone. Ovarian follicles and stromal tissues are present in the ovary. The hormone estrogen is produced by the growing follicles of the ovary. After ovulation, the ruptured follicle becomes a ‘yellow body’ called corpus Juteum (which acts as a temporary endocrine gland) and secretes progesterone. After a few days, in the absence of pregnancy, the corpus luteum stops functioning and becomes the’corpus albicans’.

Estrogen is responsible for the development and the activity of the female secondary sex organs, development of the growing ovarian follicles, high pitch of voice etc., and the development of the mammary glands. Estrogen also controls the female sexual behaviour.

Progesterone has an important role in preparing the uterus for the implantation of the blastocyst in the wall of the uterus. It inhibits contraction of the uterus. Thus it supports pregnancy. In case of deficiency of this hormone, pregnancy fails to maintain. It stimulates the formation of alveoli (sac like structures which store milk) in the mammary glands and secretion of milk.

Question 9.
Write a note on the mechanism of action of hormones.
Answer:
Hormones stimulate or inhibit the target cells’ activities. Hence they are called regulators. Hormones play a vital role in regulating the functions of the body.

Hormones produce their effects on target tissue by binding to specific proteins called hormone receptors located in the target tissues only. Hormone receptors present on the cells membranes of the target cells are called membrane bound – receptors and the hormone receptors present inside the target cells are called intracellular receptors. Intracellular receptors are mostly nuclear receptors (present in the nucleus). Hormone receptors are specific, as each receptor is specific to a certain hormone only. A hormone and its receptor protein together form a hormone – receptor complex.

This hormone – receptor complex generates biochemical changes in the target cells. Hormones interacting with membrane bond receptors do not enter the target cell, but they generate ‘second messengers’ (e.g. Cyclic AMP produced from ATP by the action of the enzyme adenylate cyclase / Adenyl cyclase, IP₃, Ca⁺⁺ etc). These second messengers regulate cellular metabolism in the target cells in a cascading action amplifying the final effect. In this way even a very small quantity of the hormone can cause a series of enzymatic actions, each step having a multiplication effect, bringing a powerful cascading effect.

We can take an example to understand the action of hydrophilic hormone, such as Epinephrine, which cannot enter a cell. In the liver cells 1) Epinephrine attaches to cell membrane receptor 2) G protein of cell membrane binds to GTP and activates adenylate cyclase, a membrane enzyme 3) activated Adenylate cyclase forms cAMP from ATP 4) cAMP activates Protein Kinase – A, which activates the enzyme ‘phosphorylase’ 5) Phosphorylase ‘ phosphorylates” Glycogen to Glucose – 6 – phosphate and it, in turn produces glucose. Thus the liver cell is able to produce several molecules of glucose needed to the cell under the action of epinephrine (one of the fight and flight responses of the body).

(a) Membrane bound – receptor mechanism (b) Intracellular receptor mechanism
Hormones which interact with intracellular receptors (e.g. steroid hormones, iodothyronines, etc.) are lipid soluble and they diffuse through the plasma membrane into the cytoplasm. They bind to certain internal receptors, enter the nucleus and regulate gene expression. The hormonal mechanism of steroid hormones is called mobile -receptor mechanism (as the receptors are not fixed in the cell membrane). Cumulative biochemical actions result in physiological and developmental effects.

Mechanism of action of lipid soluble hormone : Aldosterone is a lipid soluble hormone which can easily diffuse through the cell membrane. It binds to a specific receptor in the cytoplasm forming an aldosterone – receptor complex molecule. This complex molecule enters the nucleus and binds to the DNA and stimulates the production of a specific mRNA molecule. The mRNA passes into the cytoplasm and attaches to ribosomes making them produce the specific protein. These proteins are produced by the cell as a response to aldosterone.

Thus hormones play a major role in maintaining homeostasis by their integrated actions and feedback mechanisms.

Telangana TSBIE TS Inter 2nd Year Zoology Study Material Lesson 3(b) Neural Control and Coordination Textbook Questions and Answers.

TS Inter 2nd Year Zoology Study Material Lesson 3(b) Neural Control and Coordination

Very Short Answer Type Questions

Question 1.
Name the cranial meninges covering the brain of man.
Answer:
The cranial meninges covering the brain of man are dura mater (outer) arachnoid mater (middle) and pia mater (innermost).

Question 2.
What is corpus callosum? [Mar. ’17 (A.P.); Mar. ’15 (T.S.)]
Answer:
In brain of man the two cerebral hemispheres are internally connected by a transverse, wide and flat bundle of myelinated fibres beneath the cortex called corpus callosum.

Question 3.
What do you know about arbor vitae? [March 2020]
Answer:
In the cerebellum, each hemisphere consists of 3 lobes namely anterior, posterior and floccular lobes. It has a branching tree-like core of white matter called arbor vitae surrounded by a sheath of grey matter.

Question 4.
Why the sympathetic division is called thoraco – lumbar division?
Answer:
In the sympathetic division, pre ganglionic neurons arise from the thoracic and lumbar regions of the spinal cord, hence called “Thoraco – lumbar division”.

Question 5.
Why the parasympathetic division is called craniosacral division?
Answer:
The cell bodies of the preganglionic neurons of the parasympathetic division are located in the brain and in the sacral region of the spinal cord. Hence, the parasympathetic division is also known as the cranio – sacral division.

Question 6.
Distinguish between the absolute and relative refractory periods.
Answer:

1. During the absolute refractory period, even a very strong stimulus cannot initiate a second action potential. This period coincides with the periods of depolarization and repolarization.
2. The relative refractory period is the time during which a second action potential can be initiated by a larger-than-normal stimulus. It coincides with the period of hyper polarization.

Question 7.
What is all -or- none principle?
Answer:
The action potential occurs in response to a threshold stimulus or suprathreshold stimulus but does not occur at sub threshold stimuli. It means the nerve impulse is either conducted totally or not conducted at all and this is called all or non principle.

Question 8.
How do rods and cones of human eye differ from each other chemically and functionally?
Answer:

1. Rods contain a purplish red protein called rhodopsin or visual purple, which contains a derivative of vitamin A and they are important in twilight.
2. Cones contain a visual pigment iodopsin made of a protein called photopsin and they are important in day light and colour vision.

Question 9.
Distinguish between the blind spot and the yellow spot
Answer:

1. The centre of the posterior portion of the retina is called macula lutea or yellow spot.
2. The site of the retina where the optic nerve exits the eye ball is called optic disc or blind spot.

Question 10.
What is organ of Corti? [Mar. 2019, ’17, ’15 (A.P.); May/June ’14]
Answer:
The internal ear consists of 3 parts cochlea, vestibula and semicircular canals. The cochlear epithelium forms a sensory ridge called organ of corti on basilar membrane. The organ of corti contains hair cells that act as “auditory receptors”.

Short Answer Type Questions

Question 1.
Draw a labelled diagram of the T.S. of the spinal cord of man. [May 2017 (A.P.); Mar. 15 (A.P. & T.S.)]
Answer:

Question 2.
Distinguish between somatic and autonomic neural systems.
Answer:
Somatic Neural System (SNS) :
The somatic neural system includes both sensory and motor neurons. The sensory neurons conduct sensory impulses from the different somatic receptors to the CNS. All these sensations normally are consciously perceived. Somatic motor neurons innervate the skeletal muscles and produce voluntary movements. The axon of a single myelinated somatic motor neuron extends from the CNS all the way to the muscle fibres. In the SNS, the effect of a somatic motor neuron always is excitation.

Autonomic Neural System (ANS) :
The ANS usually operates without conscious control. The autonomic neurons are associated with interoceptors (located in the viscera and sense internal stimuli), such as chemoreceptors. These sensory signals are generally not consciously perceived. Autonomic motor neurons regulate the involuntary activities of the cardiac muscle, smooth muscle and glands. The ANS has two divisions : 1. Sympathetic and 2. Parasympathetic divisions.

Question 3.
Give an account of the retina of the human eye.
Answer:
Retina (Nervous tunic) :
This is the third and inner coat of the eye. It consists of a pigmented epithelium (non – visual portion) and a neural portion (visual portion). The pigmented epithelium is a sheet of melanin – containing epithelial cells that lie between the choroid and the neural portion of the retina. The neural portion of the retina has three layers of retinal neurons namely : photoreceptor layer (the layer closest to the choroid coat), bipolar cell layer and ganglion cell layer.

Photoreceptor layer consists of two types of photoreceptor cells called rods and cones. Rods contain a purplish – red protein called the rhodopsin or visual purple, which contains a derivative of vitamin – A and they are important in twilight (scotopic vision – the vision of the eye under low light conditions).

Cones contain a visual pigment (iodopsin, made of a protein called photopsin) and they are important in daylight (photopic) vision and colour vision. There are three types of cones, each having different sensitivity and they provide ‘optimal response’ to red, green and blue colours.

The centre of the posterior portion of the retina is called the macula lutea or yellow spot. A small depression present in the centre of the yellow spot is called fovea centralis, and it contains only cones. Fobea is responsible for sharp, central vision, which is useful while walking, reading, driving etc. The axons of the ganglion cells extend posteriorly and exit the eye ball as the optic nerve. The site of the retina where the optic nerve exits the eye ball is called optic disc or blind spot which is devoid of photoreceptor cells (no image is formed at that spot).

Question 4.
Give an account of Synaptic transmission. [March 2018 (A.P.)]
Answer:
The functional junction formed between two neurons is called synapse. In a chemical synapse, the presynaptic neuron synthesizes the neurotransmitter and stores in the synaptic vesicles of synaptic terminals.

When an action potential reaches a synaptic terminal, it depolarizes the membrane. By this voltage gated calcium channels open. The rise is the Ca²⁺ concentration leads to the release of neurotransmitters by exocytosis. The neuro transmitter diffuses across the synaptic cleft.

The postsynaptic membrane has ligandgated in channels. Binding of the neurotransmitter to a receptor of the channel opens the channel and allows specific ions to diffuse across the postsynaptic membrane. It results in a postsynoptic membrane potential. Excitatory neurotransmitters depolarize the postsynaptic membrane while inhibitory neurotransmitters cause hyperpolarization of post synaptic membrane.

Acetylcholine is the most common neurotransmitter. It may be excitatory or inhibitory. Gama aminobutyric acid (GABA) and glycine are the inhibitory neurotransmitters.

Question 5.
List out the differences between sympathetic and parasympathestic neural systems in man.
Answer:
Differences between Sympathetic and Parasympathetic neural systems.

Long Answer Type Questions

Question 1.
Give a brief account of the structure and functions of the brain of man.
Answer:
Brain (‘the living super computer’) :
It is the site of information processing and control. It is protected in the cranial cavity and covered by three connective tissue membranes called ‘cranial meninges’ namely, dura mater, arachnoid mater and pia matter. Dura mater is the outer most, thick, double layered membrane which lines the inner surface of the cranial cavity. Arachnoid mater is a thin, webby middle membrane covering the brain. It is separated from the dura mater by a narrow subdural space. Pia matter is a thin, innermost meninx which closely adheres to the brain. Pia mater is separated from the arachnoid membrane by the subarachnoid space. The brain can be divided into three major parts called
i) Forebrain, ii) Midbrain and iii) Hindbrain.

I) Forebrain (Prosencephalon) The forebrain consists of i) Olfactory bulb, ii) Cerebrum and iii) Diencephalon.
i) Olfactory Bulb :
Olfactory bulbs receive impulses pertaining to smell from the olfactory epithelium.

ii) Cerebrum :
Cerebrum forms the major part of the brain and is longitudinally divided into the left and the right cerebral hemispheres by a deep cleft called ‘longitudinal fissure’. The two hemispheres are internally connected by a transverse, wide and flat bundle of myelinated fibres beneath the cortex, called ‘corpus callosum’ (colossal commissure).lt brings ‘coordination’ between the right and left sides of the cerebral hemispheres. The surface of the cerebrum is composed of grey matter and is called ‘cerebral cortex’. The neuronal cell bodies are concentrated int he cerebral cortex.

The surrace of the cerebral cortex shows many convolutions or folds and grooves. The folds are called gyri (singular: gyrus), the deepest and shallower grooves between the folds are called fissures and sulci, respectively. Gyri and sulci increase the surface area of the cerebral cortex (which is an indication of the higher level of evolution of the human being).

Cerebral cortex has three functioinal areas called a) sensory areas, that receive and interpret the sensory impulses b) motor areas. Which control voluntary muscular movements c) association areas, which are neither clearly sensory nor motor in function and they deal with more complex ‘integrative functions’ such as memory and communications. The cerebral medulla consists of mostly myelinated axons (white matter). Each cerebral hemisphere of the cerebrum is divided into four lobes namely frotnal, parietal, temporal and occipital lobes.

iii) Diencephalon (Thalamencephalon) :
The main parts of the diencephalon are the epithalamus, thalamus and hypothalamus.

i) Epithalamus :
It is the roof of the diencephalon. It is a non – nervous part which is fused with the pia mater to form the anterior choroid plexus. Just behind the anterior choroid plexus, the epithelium of the epithalamus forms a pineal stalk, which ends in a rounded structure called pineal body.

ii) Thalamus :
It lies superior to the mid brain. It is the major coordinating centre for sensory and motor signalling.

iii) Hypothalamus (the thermostat of the body):
It lies at the base of the thalamus. The hypothalamus forms a funnel – shaped downward extension called ‘infundibulum’, connecting the hypothalamus with the pituitary gland. It also contains several groups of neurosecretory cells, which secrete hormones called hypothalamic hormones. Hypothalamus controls and integrates the activities of the autonomous nervous system (ANS) and it has osmoregulatory, thermoregulatory, thirst, feeding (hunger) and satiety centres.

I) Limbic system :
The inner parts of the cerebral hemispheres and a group of associated deep structures like amygdala or amygdale, hippocampus etc., form the limbic system. The limbic system along with hypothalamus is involved in the regulation of sexual behaviour and expression of emotional reactions.

II) Midbrain (Mesencephalon) :
The midbrain is located between the thalamus / hypothalamus of the forebrain and the pons Varolii of the hindbrain. The ventral portion of the mid brain consists of a pair of longitudinal bands of nervous tissue called cerebral peduncles or crura cerebri (sing: crus cerebrum) (which connect the cerebral hemispheres with the pons). The dorsal portion of the midbrain consists of four rounded lobes called copora quadrigemina (Four optic lobes). The two larger anterior optic lobes are called superior colliculi and the smaller x posterior lobes are called inferior colliculi. The superior colliculi and the inferior colliculi are concerned with visual and auditory functions, respectively.

III) Hindbrain (Rhombencephalon) :
The hind brain comprises cerebellum, pons Varolii and medulla oblongata.

Cerebellum (‘the little brain’) :
It is the second largest part of the brain. It consists of two cerebellar hemispheres and a central vermis. Each cerebellar hemisphere consists of three lobes namely anterior, posterior and floccular lobes. It has a branching tree – like core of white mater called arbor vitae (the tree of life) ” surrounded by a sheath of grey matter (cerebellar cortex).

Pons Varolii :
It lies front of the cerebellum below the mid brain and above the medulla oblongata. It consists of nerve fibres which form a bridge between the two cerebellar hemispheres. It is a relay station between the cerebellum, spinal > cord and the rest of the brain. Pons has the pneumotaxic centre (involved in the control of the respiratory muscles as it regulates the amount of air a person can take in, each time).

Medulla oblongata :
It is the posterior most part of the brain. It extends from the pons Varolii above and continuous with the spinal cord below. It has a very . thin, vascular folded structure called posterior choroid plexus. Medulla includes cardiovascular and respiratory centers, the centers for swallowing, vomiting, coughing, sneezing and hiccupping. The midbrain, pons and the medulla oblongata . are together referred to as the’brain stem’ .The medulla oblongata passes out of the cranium through the foramen magnum and joins the spinal cord.

Human brain consists of four ventricles. The first and second ventricles (lateral ventricles or paracoels) are present in the right and left cerebral hemispheres respectively. The third ventricle (diocoel) occurs in the diencephalon. The two paracoels are connected to the median diocoel individually by the two ‘foramina of Monro’ (interventricular foramina). The fourth ventricle (myelocoel) is present in the medulla. The myelocoel and the diocoel are connected by a narrow canal called iter or aqueduct of Sylvius/cerebral aqueduct. The metacoel is continuous with the central canal of the spinal cord.

The ventricles of the brain, and the subarachnoid space are filled with Cerebro – spinal fluid (CSF). CSF is an alkaline, colourless fluid which is filtered from the choroid plexuses into the ventricles of the brain.

Question 2.
Explain the transmission of nerve impulse through a nerve fibre with the help of suitable diagrams.
Answer:
Generation and Conduction of Nerve Impulse :
Nerve cells exhibit a special property called electrical excitability. The signal that travels along the length of a nerve fiber and ends in the release of neurotransmitters is called a nerve impulse. Neurons can respond to external and internal stimuli and conduct nerve impulses (action potentials) because in a neuron a membrane potential is established across the neuronal membrane. It means there is an ‘unequal distribution of ions’ (charged atoms) on the two sides of a nerve cell membrane with the cell’s interior more negative with respect to that of the exterior. Ions keep moving in and out of an axon through several ‘ion channels’. The axolemma of a neuron has the following three different types of ion channels.

1) Leakage channels :
They are K⁺ and Na⁺ leakage channels. K⁺ leakage channels are more than those of Na⁺ leakage channels. Hence axolemma has greater permeability to K⁺ ions than Na⁺ ions.

2) Ligand – gated channels :
They are located in the post synaptic membrane (dendrites and cells bodies) and open or close in response to chemical stimuli.

3) Voltage gated channels :
These channels open in response to a change in membrane potential. There are sodium voltage gated and potassium voltage gated channels across the axolemma. Sodium voltage gated channels are of two types. They are sodium activation and inactivation voltage gated channels. For K⁺ only potassium activation voltage gated channel is present.

Resting membrane potential :
The resting membrane potential exists because of a small buildup of negative ions in the axoplasm along the inside of the membrane and an equal buildup of positive ions in the extra cellular fluid along the outer surface of the membrane. Such a separation of positive and negative electrical charges is a form of potential energy. In neurons, the resting membrane potential ranges from -40 to -99 mV. A typical value is -70 mV. The minus sign indicates that the inside of the cell is negative relative to the outside.

At resting phase, the axolemma is polarized. The membrane potential can change from its resting value when the membrane’s permeability to particular ions changes. If the inner side becomes less negative, it is said to be depolarized. If the inner side becomes more negative, it is said to be hyperpolarized. During the resting phase the activation gates of sodium are closed, the inactivation gates of sodium are open and the activation gates of potassium are closed.

Sodium – potassium pump :
Sodium and potassium ions diffuse inwards and outwards, respectively, down their concentration gradients through leakage channels. Such a movement of ions, if unchecked, would eventually disturb the resting membrane potential. These flows of ions are offset by sodium – potassium pumps (Na⁺/ K⁺ ATPases) present in the axonal walls. These pumps expel three Na+ ions for each two K⁺ ions imported. As these pumps remove more positive charges from the axoplasm than they bring into it, they contribute to the negativity of the resting membrane potential i.e., -70 mv.

Depolarization (Rising phase) :
When a nerve fibre is stimulated, the plasma membrane becomes more permeable to Na⁺ ions than to K⁺ ions as the activation and inactivation voltage gates of sodium open and activation voltage gates of potassium close. As a result the rate of flow of Na⁺ into the axoplasm exceeds the rate of flow of K⁺ to the ECF. Hence, the axolemma is positively charged inside and negatively charged outside. This reversal of electrical charge is called “depolarization”.

Outer face of the point which is adjacent to the site of depolarization remains positively charged. The electrical potential difference between these two areas is called “action potential”. An action potential occurs in the membrane of the axon of a neuron when depolarization reaches a certain level called ‘threshold potential’ (-55 mV). The particular stimulus which is able to bring the membrane potential to threshold is called ‘threshold stimulus’. The action potential occurs in response to a threshold stimulus or suprathreshold stimulus but does not occur at subthreshold stimuli. It means the nerve impulse is either conducted totally or not conducted at all and this is called ‘all – or – none principle’. Due to the rapid influx of Na⁺ ions, the membrane potential shoots rapidly up to +45mV (spike potential).

Repolarization (Falling phase) :
As the wave of depolarization passes away from its site of origin to the adjacent point, the activation gates of sodium remain open, inactivation gates of sodium close and activation gates of potassium open at the site of origin of depolarization. As a result the influx of Na⁺ ions into the axoplasm from the ECF is checked and ‘efflux’ of K⁺ ions occurs, which leads to the returning of axolemma to the resting state (exit of potassium ions causes a reversal of membrane potential to negative inside). This is called ‘repolarization’.

Hyperpolarization (Undershoot) :
The repolarization typically goes more negative than the resting potential to about – 90 mV. This is called ‘hyperpolarization’. This occurs because of the increased K⁺ permeability that exists while voltage – gated K⁺ channels are open (however they close rather slowly as K voltage gates are said to be ‘lazy’ gates), activation and inactivation gates of Na⁺ channels remain closed. The membrane potential returns to its original resting state as the K⁺ channels close completely. As the voltage falls below the – 70mV level of the resting state, it is called ‘undershoot’.

Telangana TSBIE TS Inter 2nd Year Zoology Study Material Lesson 3(a) Musculo-Skeletal System Textbook Questions and Answers.

TS Inter 2nd Year Zoology Study Material Lesson 3(a) Musculo-Skeletal System

Very Short Answer Type Questions

Question 1.
What is a ‘motor unit’ with reference to muscle and nerve?
Answer:
A motor neuron and the set of muscle fibres innervated by all the telodendrites constitute a motor unit.

Question 2.
What is triad system? [Mar. ’14; May/June ’14; Mar. ’15 (T.S.)]
Answer:
T- tubule and the two terminal cisternae at its sides form the triad system in a skeletal muscle fibre.

Question 3.
Write the difference between actin and myosin. [Mar. 2019, ’15, May ’17 (A.P.)]
Answer:
a) The light band in a myofibril contains actin and two regulatory proteins called troponin and tropomyosin. Actin filaments are thinner compared to myosin filaments.
b) The dark band in a myofibril contains (A band) myosin. Myosin filaments are thick and non-contractile.

Question 4.
Distinguish between red muscle fibers and white muscle fibers. [March 2018 (A.P.)]
Answer:
a) Red muscle fibres :
Myoglobin of these fibres is high which give a reddish appearance. Such muscle fibres are called red fibres. They also contain plenty of mitochondria.

b) White muscle fibres :
Some of the muscle fibres possess very less quantity of myoglobin in their muscle fibers and therefore appear pale or whitish. Hence called white muscle fibers.

Question 5.
Name two cranial sutures and their locations.
Answer:
1) Coronal suture :
Parietal bones that form major portion of sides and roof of cranial cavity are joined to the frontal bone by coronal suture.

2) lambdoid suture :
Parietal bones are posteriorly joined to the occiput by lambdoid suture.

Question 6.
Name the keystone bone of the cranium. Where is it located? [May 2017 (A.P.) March 2014]
Answer:
Sphenoid bone is called the keystone bone. It is present at the middle part of the base of the skull. It is named keystone bone because it articulates with all other cranial bones.

Question 7.
Human skull is described as dicondylic skull. Give the reason.
Answer:
In human skull two occipital condyles are present one on each side of the foramen magnum hence called dicondylic skull.

Question 8.
Name the ear ossicles and their evolutionary origin in human beings.
Answer:
Each middle ear contains 3 tiny bones – Malleus (modification of articular), Incus (modified quadrate) and Stapes (modified hyomandibula) collectively called as ear ossicles.

Question 9.
Name the type of joint between a) atlas/ axis b) carpal / metacarpal of the human thumb.
Answer:
1) Type of joint between atlas and axis is Pivot joint.
2) Type of joint between carpal / metacarpal is Condyloid joint.

Question 10.
Name the type of joint between a) atlanto – axial joint b) Femur-acetabulum joint.
Answer:
1) Type of joint between atlanto – axial joint is Pivot joint.
2) Type of joint between Femur – acetabulum is Ball and socket joint.

Question 11.
Name the type of joint between a) Cranial bones b) Inter – tarsal joint.
Answer:

1. Joint between cranial bones is fibrous joint called Suture.
2. Inter tarsal joint is Gliding joint.

Short Answer Type Questions

Question 1.
Write a short note on sliding filament theory of muscle contraction.
Answer:
Sliding Filament Hypothesis :
The sliding filament hypothesis of muscle contraction was put forth by Hugh Huxley and Hanson. According to their hypothesis, the contraction of sarcomere depends on the presence of two sets of filaments. During muscle contraction, thin filament slide over the thick filaments resulting in shortening of sarcomere. A series of events takes place during this process like, a) Stimulation of muscle b) Contraction of muscle c) Relaxation of muscle.

Question 2.
Describe the important steps in muscle contraction.
Answer:
Important steps in muscle contraction are
i) Excitation or stimulation of muscle :
Muscle contraction is initiated by a signal sent by the central nervous system (CNS) via a motor neuron. A neural signal reaching the neuromuscular junction releases a neurotransmitter (acetylcholine) which generates an ‘action potential’ in the sarcolemma. When the action potential spreads to the triad system through the T tubules, the cisternae of the sarcoplasmic reticulum release calcium ions into the sarcoplasm.

ii) Formation of Cross bridges :
Increase in the Ca²⁺ level leads to the binding of calcium ions to the subunit Tn – C of the troponin of the thin filaments. This makes troponin and tropomyosin complex to move away from the active sites of action molecules. Now, the active sites are exposed to the heads of the myosin. Utilizing the energy released from hydrolysis of ATP, the myosin head now binds to the exposed ‘active sites’ on the actin molecules to form a cross bridge and P₁ is released.

iii) Power Stroke :
The cross bridge pulls the attached actin filaments towards the centre of the ‘A’ band. The ‘Z’ lines attached to these actin filaments are also pulled inwards from both the sides, thereby causing shortening of the sarcomere, i.e., contraction. During the shortening of the muscle, the T bands get reduced in size / length (Z membranes of the sarcomere are brought closer) whereas the’A’ bands retain their size / length. It is important to note that myofilaments do not actually shorten. As the thin filaments are pulled deep in to the A bands making the H bands narrow, the muscle shows the effect contraction.

Question 3.
Describe the structure of a skeletal muscle.
Answer:
Structure of a skeletal Muscle :
Let us examine a skeletal muscle in detail to understand the structure and mechanism of contraction. Each organised skeletal muscle in our body is made of a number of muscle bundles or fascicles. Each fascicle contains a number of cylindrical muscle fibers. The fascicles are held together by a common collagenous connective tissue layer called fascia. Each muscle fibre is lined by the plasma membrane called sarcolemma enclosing the sarcoplasm. Skeletal muscle fibre is a ‘syncytium’, as each fibre is formed by fusion of embryonic, mononucleate ‘myoblasts’.

Hence, the skeletal muscle cells are multinucleate, with characteristically peripheral nuclei (just below the sarcolemma). The endoplasmic reticulum, also called sarcoplasmic reticulum of the muscle fibers is the store house of calcium ions. A characteristic feature of the muscle fiber is the presence of a large number of parallel filaments called myofilaments or myofibrils, in the sarcoplasm.

Question 4.
Write short notes on contractile proteins.
Answer:
Structure of Contractile Proteins :
Each actin (thin) filament is made of two ‘F’ (filamentous) actin molecules helically wound around each other. Each ‘F’ actin is a polymer of monomeric ‘G’ (globular) actin molecules. Two filaments of another protein, called tropomyosin also run close to the ‘F’ actin molecules, throughout their length. A complex protein called ‘troponin’ is distributed at regular intervals on the tropomyosin.

Troponin is made of three polypeptide units named Tn – T, Tn – I, and Tn – C. Tn – T binds to tropomyosin. Troponin – I (Tn – I), inhibits the myosin binding site on the actin. Tn – C can bind to Ca²⁺. When calcium ions are not bound to troponin (Tn – C), it stabilizes tropomyosin in its blocking position over the active sites of actin. When Calcium ions attach to the Tn – C of the troponin, the tropomyosin moves away/ is pulled away from the ‘active sites’ allowing the myosin heads to bind to the active sites of actin. Troponin and tropomyosin are often called ‘regulatory proteins’, because of their role in masking and unmasking the active sites.

Question 5.
Draw a neat labelled diagram of the Ultrastructure of muscle fibre.
Answer:

Question 6.
Draw the diagram of a Sarcomere of skeletal muscle showing different regions.
Answer:

Question 7.
What is Cori’s cycle – explain the process.
Answer:
Cori cycle :
The lactic acid produced during rapid contractions of skeletal muscles under low availability of oxygen is partly oxidized and a major part of it is carried to the liver by the blood, where it is converted into pyruvic acid (pyruvate) and then to glucose through gluconeogenesis. The glucose can enter the blood and be carried to muscles and immediately used. If, by this time the muscles have stopped contraction, the glucose can be used to rebuild reserve of glycogen through glycogenesis. This two way traffic between skeletal muscle and liver is called the Cori cycle.

Question 8.
List out the bones of the human cranium. .
Answer:
Cranium, the brain box, is formed by eight flattened bones. They are a) frontal bone (1), b) Parietals (2), c) Temporal bones (2), d) occipital bone (1), e) spenoid bone (1) f) Ethmoid bone (1)

i) Frontal bone :
It forms the forehead, anterior part of the cranial floor, and the roof of the orbits.

ii) Parietal bones :
They form the major portion of the sides and roof of the cranial cavity. They are joined to the frontal bone by a coronal suture and posteriorly to the occiput by lambdoid suture.

iii) Temporal bones :
They form the lateral parts and the floor of the cranium.

iv) Occipital bone :
It forms the posterior part and most of the base of the cranium. It has a large opening, the foramen magnum. Medulla oblongata passes out through this foramen and joins the spinal cord. Two occipital condyles are present one on each side of the foramen magnum (dicondylic skull).

v) Sphenoid bone :
It is present at the middle part of the base of the skull. It is the keystone bone of the cranium because it articulates with all the other cranial bones.

vi) Ethmoid bone :
It is present on the midline of the anterior part of the cranial floor.

Question 9.
Write short notes on the ribs of human being. [March 2020]
Answer:

Ribs :
Twelve pairs of ribs are present in the human chest. Each rib is a thin flat bone connected dorsally to the vertebral column and ventrally to the sternum. It has two articulation surfaces on its dorsal end, hence called bicephalic. The first seven pairs of ribs are called true ribs (vertebro – sternal ribs). Dorsally they are attached to the thoracic vertebrae and ventrally connected to the sternum with the help of hyaline cartilages. The remaining ribs are called ‘false ribs’. The 8th, 9th and 10th pairs of ribs do not articulate directly with the sternum but join the cartilaginous (hyaline cartilage) parts of the seventh rib. These are called vertebro – chondral (false ribs’) ribs. Last 2 pairs (11th and 12th) of ribs are not connected ventrally either to the sternum or the anterior ribs, hence called floating ribs. The thoracic vertebrae, ribs and sternum together form the rib cage.

Question 10.
List the bones of human fore limb.
Answer:
Bones of the Fore limb :
Each limb is made of 30 bones. The bones of the hand (forelimb) are humerus, radius and ulna, carpals (wrist bones – 8), metacarpals (palm bones – 5) and phalanges (digits -14).

Question 11.
List the bones of the human leg.
Answer:
Bones of the human leg : The bones of the leg are Femur (1) (thing bone – the longest bone), tibia (1) and fibula (1), tarsals (ankle bones – 7), metatarsals (5) and phalanges (digits – 14) are the bones of the legs (hind limbs). A cup shaped bone called patella (knee cap) (1) covers the knee joint ventrally.

Question 12.
Draw a neat labelled diagram of the skeleton of the fore limb of man.
Answer:

Question 13.
Draw a neat labeled diagram of pelvic girdle.
Answer:

Question 14.
Describe the structure of synovial joint with the help of a neat labelled diagram. [Mar. 2019, ’17 (A.P.); May/June; Mar.’14]
Answer:

Synovial joint is covered by a double layered synovial capsule. The outer layer consists of dense fibrous irregular connective tissue with more collagen fibres. This layer is continuous with the periosteum and resists stretching and prevents the dislocation of joints. Some fibres of these membranes are arranged in bundles called ligaments. The inner layer of synovial capsule is formed of areolar tissue and elastic fibers. It secretes a viscous synovial fluid which contains hyaluronic acid, phagocytes etc., and acts as a ‘lubricant1 for the free movement of the joints. Synovial joints include Ball and socket joint. Hinge joint, Pivot joint, Gliding joint, Condyloid joint, Saddle joint.

Long Answer Type Questions

Question 1.
Explain the mechanism of Muscle contraction.
Answer:
Mechanism of Muscle contraction :
Mechanism of muscle contraction is best explained by the ‘Sliding Filament Theory’. It states that contraction of a muscle fibre takes piace by the sliding of the thin filaments over / in between the thick filaments. It was proposed by Jean Hanson and Hugh Huxley. The process of muscle contraction can be studied under the following heads.

i) Excitation of muscle :
Muscle contraction is initiated by a signal sent by the central nervous system (CNS) via a motor neuron. A neural signal reaching the neuromuscular junction releases a neurotransmitter (acetylcholine) which generates an ‘action potential’ in the sarcolemma. When the action potential spreads to the triad system through the T tubules, the cisternae of the sarcoplasmic reticulum release calcium ions into the sarcoplasm.

ii) Formation of Cross bridges :
Increase in the Ca²⁺ level leads to the binding of calcium ions to the subunit Tn – C of the troponin of the thin filaments. This makes troponin and tropomyosin complex to move away from the active sites of actin molecules. Now, the active sites are exposed to the heads of the myosin. Utilizing the energy released from hydrolysis of ATP, the myosin head now binds to the exposed ‘active sites’ on the actin molecules to form a cross bridge and P₁ is released.

iii) Power Stroke :
The cross bridge pulls the attached actin filaments towards the centre of the ‘A’ band. The ‘Z’ lines attached to these actin, filaments are also pulls inwards from both the sides, thereby causing shortening of the sarcomere, i. e., contraction. During the shortening of the muscle, the T bands get reduced in size / length (Z membranes of the sarcomere are brought closer), whereas the ‘A’ bands retain their size / length. It is important to note that myofilaments do not actually shorten. As the thin filaments are pulled deep in to the A bands making the H bands narrow, the muscle shows the effect contraction.

iv) Recovery Stroke :
The myosin, goes back to its relaxed state and releases ADP. A new ATP moelcule binds to the head of myosin and the cross bridge is broken. The new ATP is again hydrolysed by the ATPase of the myosin head and the cycle of cross bridge formation, and breakage is repeated causing further sliding.

v) Relaxation of Muscle :
When motor impulses stop the Ca²⁺ ions are pumped back into the sarcoplasmic cisternae. It results in the masking of the active sites of the actin filaments. The myosin heads fail to bind with the active sites ofactin. These changes cause the return of ‘Z’ lines backtotheiroriginal position, i.e., relaxation.

Question 2.
List, in sequence, the events that take place during muscle contraction.
Answer:
[Refer the answer of above question and also add the following matter.]
For the contraction of muscle, continuous supply of energy is needed. ATP is the immediate source of energy for muscle contraction. As the ATP content is very low, it is actively replenished, continuously, by an energy rich muscle phosphagen.
ATP → ADP + Pi

The high energy phosphates of muscles that donate energy phosphate group to ADP are known as phosphagen. In vertebrate muscles creatine phsophate (CP) is the phsophagen, which is an immediate backup source. In invertebrate muscles, it is in the arginine phosphate. This reaction is catalysed by creatine kinase (CK). Creatine Phosphate is the immediate additional source of energy in the muscle.
Creatine phosphate + ADP → Creatine

When creatine phosphate gets exhausted, the next source of reserve energy is utilised, which includes the oxidation of glucose and fatty acids. The energy liberated in this process is transferred to ADP and creatine. Thus ATP and creatine phosphate are formed and which in turn supply energy for muscle contraction.

During rapid activity of a muscle, the respiratory system is unable to supply sufficient oxygen needed by it, which leads to oxygen debt. It is defined as the amount of extra oxygen required by a muscle during recovery from vigorous exercise. Thus the pyruvic acid produced by glycolysis is transformed into lactic acid in the absence of oxygen. Accumulation of lactic acid in the muscle leads to muscle fatigue.

Lactic acid, formed in the anaerobic degradation in the muscle, reaches the liver through blood circulation. In liver, during rest, 80% of lactic acid is utilised in the resynthesis of glycogen, which is transported back to muscle. This is known as Cori cycle. 20% of lactic acid is oxidised as CO₂ and H₂O.

Question 3.
Describe the structure of human skull.
Answer:
The skull :
It is composed of two sets of bones – cranial and facial bones (22 bones in all). Cranium, the brain box, is formed by eight flattened bones. They are a) frontal bone (1), b) Parietals (2), c) Temporal bones (2), d) Occipital bone (1), e) Sphenoid bone (1) and f) Ethmoid bone (1).
i) Frontal bone :
It forms the forehead, anterior part of the cranial floor and the roof of the orbits.

ii) Parietal bones :
They form the major portion of the sides and roof of the cranial cavity. They are joined to the frontal bone by a coronal suture and posteriorly to the occiput by lambdoid suture.

iii) Temporal bones :
They form the lateral parts and the floor of the cranium.

iv) Occipital bone :
It forms the posterior part and most of the base of the cranium. It has a large opening, the foramen magnum. Medulla oblongata passes out through this foramen and joins the spinal cord. Two occipital condyles are present one on each side of the foramen magnum (dicondylic skull).

v) Sphenoid bone :
It is present at the middle part of the base of the skull. It is the keystone bone of the cranium because it articulates with all the other cranial bones.

vi) Ethmoid bone :
It is present on the midline of the anterior part of the cranial floor.

A) The facial region is made up of 14 skeletal elements which form the front part of the skull. The bones of the facial skeleton are the nasals (2), the maxillae (2), the zygomatic bones (2) the mondible (1) the lacrimal bones (2), the palatine bones (2), inferior nasal conchae (2), and the vomer (1)
i) Nasal bones :
These are paired bones that from the bridge of the nose.

ii) Maxilla :
Two maxillae join together and form the upper jaw. The maxilla bears sockets (alveoli) for lodging the maxillary teeth. The palatine process is involved in the formation of the hard palate.

iii) Zygomatic bones :
These are known as a cheek bones.

iv) Lacrimal bones :
These are the smallest bones of the face.

v) Palatine bones :
They form the posterior portion of the hard palate.

vi) Nasal conchae :
These are scroll like bones that form a part of lateral wall of the nasal cavity. Nasal conchae are 3 pairs, namely superior, middle and inferior.

vii) Vomers :
It is a triangular bone present on the floor of nasal cavity.

viii) Mandible (Lower jaw) :
It is ‘U’ shaped and is the longest and strongest of all the facial bones. It is the only movable skull bone (except the ear ossicles).

B) Skeletal structures associated with sense organs:
i) The nasal cavity is divided into left and right cavities by a vertical partition called the nasal septum.
ii) Orbits : Orbits are bony depressions which accommodate the eyeballs and associated structures.
iii) Ear Ossicles : Each middle ear contains three tiny bones – Malleus (modification of articular), Incus (modified quadrate) and Stapes (modified hyomandibula), collectively called ear ossicles.

C) Hyoid Bone :
It is a single U shaped bone present at the base of the buccal cavity between the larynx and the mandible. The hyoid bone keeps the larynx open.

Telangana TSBIE TS Inter 2nd Year Zoology Study Material Lesson 2(b) Excretory Products and their Elimination Textbook Questions and Answers.

TS Inter 2nd Year Zoology Study Material Lesson 2(b) Excretory Products and their Elimination

Very Short Answer Type Questions

Question 1.
Name the blood vessels that enter and exit the kidney. [May 2017 (A.P.)]
Answer:
Renal artery enters the kidney and renal vein comes out of kidney.

Question 2.
What are renal pyramids and renal papillae?
Answer:
The medulla of kidney is divided into multiple cone shaped masses of tissue called renal pyramids. The base of each pyramid originates at the border between cortex and medulla and terminates in the renal papilla.

Question 3.
What are the columns of Bertin? [March 2019, 2015 (A.P.)]
Answer:
The renal pyramids are separated by the projections of the cortex called columns of Bertin (renal column).

Question 4.
Name the structural and functional unit of kidney. What are the two main types of structural units in it?
Answer:
Structural and functional unit of kidney is nephron. Each nephron has two types of structural units a) Bowman’s capsule and 2) the renal tubcile.

Question 5.
Distinguish between cortical and juxta medullary nephrons.
Answer:

1. In a majority of nephrons, the loops of Henle is too short and extends only very little into the medulla. Such nephrons are called Cortical nephrons.
2. In some of the nephrons, the loops of Henle are very long and run deep into the medulla. They are called Juxta medullary nephrons.

Question 6.
Define glomerular filtration. [Mar. ’14; May/June ’14; Mar. ’17 (A.P.)]
Answer:
Filtration of the blood from the glomerulus into the lumen of the Bowman’s capsule and this passive process is called glomerular filtration.

Question 7.
Define Glomerular Filtration Rate (GFR).
Answer:
The amount of filtrate formed by both the kidneys per minute is called Glomerular Filtration Rate (GFR).

Question 8.
What is meant by mandatory reabsorption? In which parts of nephron does it occur?
Answer:
About 85% of the filtrate formed is reabsorbed in a constant, unregulated fashion by the proximal convoluted tubule and descending limb of Henle’s loop. This is called obligatory or Mandatory reabsorption.

Question 9.
Distinguish between juxta glomerular cells and macula densa.
Answer:

1. Juxta glomerular cells are present in Juxta glomerular apparatus where the afferent arteriole comes into contact with DCT. These are modified smooth muscle cells of the afferent arteriole.
2. A group of modified epithelial cells of DCT which comes in contact with afferent arteriole are crowded in this region constitute macula densa.
   Macula densa together with JG cells form the JGA.

Question 10.
What is juxta glomerular apparatus? [March 2018 (A.P.)]
Answer:
The juxta Glomerular Aoparatus plays a complex regulating role. The JGA is the region in each nephron where the afferent arteriole comes into contact with DCT. Macula densa together with JG cells from the JGA.

Question 11.
Distinguish between the enzymes renin and rennin.
Answer:

1. A fall in glomerular blood flow / glomerular blood pressure / GFR can activate the JG cells to release an enzyme called renin into the blood. This enzyme catalyses the conversion of angiotensinogen.
2. Rennin is a digestive enzyme produced by gastric glands that converts milk into curd in infants. ‘

Question 12.
What is meant by the term osmoregulation?
Answer:
The process of maintaining the quantity of water and dissolved solutes in balance is referred to as osmoregulation.

Question 13.
What is the role of atrial natriuretic peptide in the regulation of urine formation?
Answer:
An increase in the flow of blood to the right atrium of the heart stretches its wall. It causes the release of atrial natriuretic peptide (ANP). It causes vasodilation and there by decrease the blood pressure. ANP mechanism therefore, acts as a counter check on the RAAS.

Short Answer Type Questions

Question 1.
Terrestrial animals are generally either ureotelic or uricotelic and not ammonotelic. Why?
Answer:
Aquatic animals excrete ammonia (ammonotelic) through body surface, gill surface etc. by diffuse. They can send ammonia through dilute urine as they can afford to lose much water. But that is not the case with terrestrial animals which may be excreting urea (ureotelic) or uric acid (uricotelic) water is the most important constituent of protoplasm, the living substance. Dehydration kills a person much faster than lack of food. Terrestrial adaptation necessitated the production of lesser toxic nitrogenous wastes such as urea and uric acid, for the conservation of water. Hence terrestrial animals are generally either ureotelic or uricotelic and not ammonotelic.

Question 2.
Differentiate vertebrates on the basis of the nitrogenous waste products they excrete, giving examples.
Answer:
Based on the nitrogenous waste products vertebrates are differentiated into 3 categories.
1. Ammonotelic animals :
Excrete nitrogenous Wastes in the form of ammonia, e.g.: Hydra, some bony fishes.

2. Ureotelic animals :
Excrete nitrogenous wastes in the form of urea, e.g.: Cartilaginous fishes, amphibians and mammals.

3. Uricotelic animals :
Excrete nitrogenous wastes in the form of uric acid, e.g.: Insects, reptiles and birds.

Question 3.
Draw a labelled diagram of the V.S. of Kidney.
Answer:

Question 4.
Describe the internal structure of kidney of man.
Answer:
Internal structure of Kidney :
A longitudinal section of the human kidney shows two distinct regions, the outer cortex and the inner medulla. The medulla is divided into multiple cone shaped masses of tissue called renal pyramids. The renal pyramids are separated by the projections of the cortex called columns of Bertin (renal column). The base of each pyramid originates at the border between the cortex and the medulla and terminates in the renal papilla. Renal papillae project into cup like calyces, formed by the funnel shaped pelvis, which continues out as the ureter.

Question 5.
Explain micturition.
Answer:
Micturition :
Urine formed by the nephrons is ultimately carried to the urinary bladder where it is stored till a voluntary signal is given by the central nervous system (CNS). This signal is initiated by the stretching of the urinary bladder as it gets filled with urine. In response, the stretch receptors on the walls of the bladder send signals to the CNS. The CNS passes on motor messages to initiate the contraction of smooth muscles of the bladder and simultaneous relaxation of the urethral sphincter, causing the release of urine. The process of passing out urine is called micturition and the neural mechanism involved is called ‘micturition reflex’.

Question 6.
What is the significance ofJuxta Glomerular Apparatus (JGA) in Kidney function?
Answer:
The Juxta Glomerular Apparatus plays a complex regulating role. The JGA is the region in each nephron where the afferent arteriole comes into contact with the DCT. A group of modified epithelial cells of the DCT are crowded in this region, constituting the macula densa. The wall of the afferent renal arteriole has JG cells (they are modified smooth muscle cells of the afferent arteriole). Macula densa together with JG cells form the JGA.

A fall in glomerular blood flow / glomerular blood pressure / GFR can activate the JG cells to release an enzyme called renin into the blood. This enzyme catalyses the conversion of angiotensinogen (a protein produced by the liver) into angiotensin I, which is converted into angiotensin II, by angiotensin converting enzyme (ACE). This conversion occurs primarily as blood passes through the capillaries of the lungs, where most of the converting enzyme is present. Angiotensin II stimulates the adrenal cortex to secrete aldosterone. Aldosterone causes reabsorption of Na⁺ and water from the DCT and CD to reduce loss through urine, and also promotes secretion of K⁺ ions into the DCT and CD. It leads to an increase in the blood pressure and GFR. This complex mechanism is generally known as renin – angiotensin – aldosterone system (RAAS).

An increase in the flow of blood to the right atrium of the heart stretches its wall. It causes the release of atrial natriuretic factor (ANF) / atrial natriuretic peptide (ANP). ANP can cause vasodilation (dilation of blood vessels) and there by decrease the blood pressure. ‘ANP’ mechanism therefore, acts as a counter check on the ‘RAAS’.

Question 7.
Give a brief account of the counter current mechanism.
Answer:
Mammals have the ability to produce concentrated urine. The Henle’s loop and vasa recta play a significant role in this. The flow of the renal filtrate in the two limbs of Henle’s loop is in opposite directions and thus forms a counter current. The flow of blood through the two limbs of vasa recta is also in a counter current pattern. The proximity between the Henle’s loop and vasa recta, as well as the counter currents of renal fluid and blood in them help in maintaining an increasing osmolarity towards the inner medullary interstitium. i.e., from 300 mOsml/ L in the cortex to about 120 mOsml/L in the inner medulla.

This gradient is mainly caused by NaCI and urea. NaCI passes out of the ascending limb of Henle’s loop, and it enters the blood of the descending limb of vasa recta. NaCI is returned to the interstitium from the ascending portion of the vasa recta. Similarly, small amounts of urea enter the thin segment of the ascending limb of Henle’s loop which is transported back to the interstitium, from the collecting duct.

Diagrammatic representation of a nephron and vasa recta showing counter current mechanisms

The above described transport of substances facilitated by the special arrangement of Henle’s loop and vasa recta is called the countercurrent mechanism (the two limbs of the loop of Henle constitute a counter current multiplier system). This mechanism helps to maintain a concentration gradient in the medullary interstitium. Presence of such interstitial gradient helps easy passage of water from the collecting duet, thereby concentrating the filtrate (urine). Human kidneys can produce urine nearly four times concentrated than the initial filtrate formed.

Question 8.
Explain the auto regulatory mechanism of GFR.
Answer:
The tubular epithelial cells in different segments of a nephron reabsorb certain substances of the glomerular filtrate either by active or passive mechanisms. About 85% of the filtrate formed is reabsorbed in a constant, unregulated fashion by the PCT and descending limb of Henle’s loop (obligatory or mandatory reabsorption) and the reabsorption of the rest of the fluid is “regulated”. Based on the necessity of re- absorption, the substances of glomerular filtrate can be categorized into ‘high threshold substances’ (essential and are efficiently reabsorbed e.g. glucose, amino acids, vitamins, some salts etc.,) ‘low threshold substances’ (absorbed in very little amounts e.g. urea, uric acid etc.) or ‘athreshold substances’ (actual excretory products and are not reabsorbed at all e.g. creatinine).

During the formation of urine, the tubular cells secrete substances such as H⁺, K⁺ and NH₃ into the filtrate. Tubular secretion is also an important step in the formation of urine as it helps in the maintenance of ionic and acid – base balance of the body fluids.

Question 9.
Describe the role of liver, lungs and skin in excretion.
Answer:
In addition to the kidneys, lungs, liver and skin also help in the elimination of excretory wastes.
a) Lungs :
Lungs regularly eliminat about 18L of C02 and also significant amount of water per day in the form of water vapour in normal resting condition. The quantity of water loss increases in dry climates. Various volatile materials are also eliminated through the lungs.

b) Liver :
Liver is the largest gland in our body. It changes the decomposed haemoglobin of the worn-out RBCs into bile pigments, namely, bilirubin and biliverdin. These pigments pass into the alimentary canal along with the bile for elimination. The liver also excretes cholesterol, degraded steroid hormones, certain vitamins and drugs via bile.

c) Skin :
Human skin possesses two types of glands for the elimination of certain substances through their secretion.

1. Sweat glands secrete a watery fluid called sweat. Primary function of sweat is to facilitate a cooling effect on the body surface. It also helps in the removal of some of the wastes like NaCI, small amounts of urea, lactic acid etc.
2. Sebaceous glands eliminate certain substances like sterols, hydrocarbons, waxes through sebum. This secretion provides a protective ‘oily covering’ to the skin.

Question 10.
Name the following:
a) A chordate animal having protonephridial type excretory structure.
b) Cortical portions projecting between the medullary pyramids in the human kidney.
c) Capillary network paralleling the loop of Henle.
d) A non-chordate animal having green lands as excretory structure.
Answer:
a) A chordate animal having protonephridial type excretory structures is Lancelet (with solenocytes) or Amphioxus.

b) Cortical portions projecting between the medullary pyramids in the human kidney are columns of Bertin.

c) Capillary network paralleling the loop of Henle is vaserecta.

d) A non-chordate animal having green glands as excretory structures is crustaceans like prawn & crab.

Long Answer Type Questions

Question 1.
Describe the excretory system of man, giving the structure of a nephron. [March 2015 (T.S.)]
Answer:
Human Excretory System :
In humans, the excretory system consists of a pair of kidneys, a pair of ureters, a urinary bladder and urethra.

Kidneys :
Kidneys are reddish brown, bean shaped structures, situated on either side of the vertebral column between the levels of the last thoracic and third lumbar vertebrae, in a ‘retroperitoneal position’. The right kidney is slightly lower than the left one due to the presence of liver.

The outer surface of the kidney is convex and the inner surface has a deep notch called hilum, the point at which the renal artery and nerves enter and the renal vein and ureter leave. Each kidney is surrounded by a tough, fibrous capsule that protects its delicate inner surface.

Internal structure:
A longitudinal section of the human kidney shows two distinct regions, the outer cortex and the inner medulla. The medulla is divided into multiple cone shaped masses of tissue called renal pyramids. The renal pyramids are separated by the projections of the cortex called columns of Bertin (renal column). The base of each pyramid originates at the border between the cortex and the medulla and terminates in the renal papilla. Renal papillae project into cup like calyces, formed by the funnel shaped pelvis, which continues out as the ureter.

Ureters :
These are slender whitish tubes which emerge from the pelvis of the kidneys. Their walls are lined by ‘transitional epithelium’. The ureters run down wards and open into the urinary bladder.

Urinary bladder :
It is a median storage sac, situated in the lower abdominal cavity. It has thick, muscular, distensible wall lined by ‘transitional epithelium’. The neck of the bladder leads into the urethra, which has an internal urethral sphincter (made of smooth muscles) and external urethral sphincter (made of striped muscles). Urethra opens near the vaginal orifice in the female and through penis in the males.

Structure of a Nephron :
Each kidney has nearly one million nephrons which are the ‘structural’ and ‘functional’ units. Each nephron has two parts – the ‘Bowman’s capsule’ and the renal tubule. The Bowmans’ capsule encloses a tuft of capillaries called glomerulus, formed by the afferent renal arteriole – a fine branch of the renal artery. Blood from the glomerulus is carried away by an efferent renal arteriole of a lesser diameter. The blind end of the tubule forms a double walled cup called Bowman’s capsule, which surrounds the glomerulus.

The inner wall of the Bowman’s capsule has certain unique cells called podocytes which wrap around each capillary. The podocytes are arranged in an intricate manner so as to leave some minute spaces called ‘filtration slits’ or ‘slit pores’. The endothelial cells of the capillaries have numerous pores or ‘fenestrations’. The glomerulus along with the Bowman’s capsule constitutes the Malpighian body or renal corpuscle.

The tubule continues further and forms a highly coiled proximal convoluted tubule (PCT). A hairpin shaped Henle’s loop, which has descending and ascending limbs, is the next part of the tubule. The proximal part of the ascending limb is thin and the distal part is thick . The thick ascending limb continues into the distal convoluted tubule (DCT). The DCT continues as the ‘initial collecting duct’ in the cortex. Some initial collecting ducts unite to form a straight collecting duct, which passes through the medullary pyramid. In the medulla, the tubes of each pyramid join and form the duct of Bellini, which finally opens on the tip of the renal papilla. The contents of the duct of Bellini are discharged into the renal pelvis through the renal calyx.

A diagrammatic representation of a nephron showing blood vessels, collecting duct and tubule

The Malpighian corpuscle, PCT and DCT of a nephron are situated in the cortical region of the kidney, whereas the loop of Henle is in the medulla. In a majority of nephrons, the loop of Henle is too short and extends only very little into the medulla. Such nephrons are called cortical nephrons. In some of the nephrons, the loops of Henle are very long and run deep into the medulla. These nephrons are called juxtamedullary nephrons.

The efferent arteriole emerging from the glomerulus forms a fine capillary network called the peritubular capillaries, around the renal tubule. The portion of the peritubular capillaries that surrounds the loop of Henle is called the vasa recta. The vasa recta is absent or highly reduced in the cortical nephrons. The juxtamedullary nephrons possess well developed vasa recta.

Question 2.
Explain the physiology of urine formation. [March 2020]
Answer:
Urine Formation :
The formation of urine involves three main processes namely, glomerular filtration, selective reabsorption and tubular secretion.

a) Glomerular filtration :
The first step in the formation of urine is the ‘filtration’ of the blood from the glomerulus into the lumen of the Bowman’s capsule and this ‘passive’ (non -energy consuming process) process is called glomerular filtration. The hydrostatic pressure of the blood while flowing in the glomerulus is 60 mm Hg. It is opposed by ‘glomerular colloidal osmotic pressure’ of 32 mm Hg (which is exerted by the non-filtered plasma proteins of the blood in the glomerular capillaries) and Bowman’s capsular hydrostatic pressure of 18mm Hg. The net filtration pressure is 10mm Hg (60 – 32 + 18 = 10). This causes the filtration of blood through the 3 layered filtrate membrane formed by the endothelial cells of glomerular capillary together with the basement membrane and podocytes of the Bowman’s cup.

Blood is filtered through the fine slit pores and fenestrations due to the NFP. Therefore, this process is called ‘ultrafiltration’. The filtrate contains almost all the constituents of the plasma, except the proteins. The filtrate thus formed is called ultra – filtrate or ‘glomerular filtrate’ or ‘primary urine’, which is hypotonic to the cortical fluid. It passes into the next part of .the renal tubule.

b) Selective reabsorption and secretion :
The tubular epithelial cells in different segments of a nephron reabsorb certain substances of the glomerular filtrate either by active or passive mechanisms. About 85% of the filtrate formed is reabsorbed in a constant, unregulated fashion by the PCT and descending limb of Henle’s loop (obligatory or mandatory reabsorption) and the reabsorption of the rest of the fluid is ‘regulated’. Based on the necessity of re-absorption, the substances of glomerular filtrate can be categorized into ‘high threshold substances’ (essential and are efficiently reabsorbed e.g. glucose, amino acids, vitamins, some salts etc.), ‘low threshold substances’ (absorbed in very little amounts e.g. urea, uric acid etc), or ‘athreshold substances’ (actual excretory products and are not reabsorbed at all e.g. creatinine).

During the formation of urine, the tubular cells secrete substances such as H⁺, K⁺ and NH₃ into the filtrate. Tubular secretion is also an important step in the formation of urine as it helps in the maintenance of ionic and acid-base balance of the body fluids. Mechanism of selective reabsorption and secretion in different parts of a nephron takes place as follows.

i) In the proximal convoluted tubule :
PCT is lined by simple cuboidal epithelium with ‘brush border’, which increases the surface area of absorption. Nearly all the essential nutrients and 70-80% of electrolytes and water are reabsorbed by this segment. Na⁺ is actively transported into the cortical interstitial fluid. This transfer of positive charge drives the passive transport of Cl^–. Glucose, amino acids, and other essential substances are also ‘actively’ transported. Movement of water occurs by ‘osmosis’.

PCT also helps to maintain the pH and ionic balance of the body fluids by selective secretion of hydrogen ions, and ammonia into the filtrate and by the absorption of HCO^–₃ from it.

ii) In the Henle’s loop :
Reabsorption in this segment is minimum. However, this region plays a significant role in the maintenance of high osmolarity of the medullary interstitial fluid.

The descending limb of loop of Henle is permeable to water and almost impermeable to electrolytes, hence reabsorption of water continues as the filtrate moves along the descending limb (passive transport). As a result, the filtrate concentration gradually increases as it moves towards the inner medulla. The ascending limb has two specialized regions, a proximal thin segment, in which NaCI diffuses out into the interstitial fluid passively, and a distal thick segment, in which NaCI is actively pumped out. The ascending limb is impermeable to water. Thus the filtrate becomes progressively more dilute as it moves up to the cortex (towards the DCT).

Reabsorption and secretion of major substances at different parts of the nephron (arrows indicate direction of movement of materials)

iii) In the distal convoluted tubule (DCT) :
The cells here are shorter than those in the proximal tubule and lack ‘microvilli’, indicating that they are not involved much in reabsorption ‘conditional reabsorption’ /’facultative reabsorption’ of Na⁺ and water takes place in this segment. The reabsorption of water is variable depending on several conditions and is regulated by ADH. DCT is also capable of reabsorption of HCO^–₃ and selective secretion of H⁺ and K⁺ ions and NH₃ into the DCT from the peritubular network, to maintain the pH and sodium – potassium balance in the blood.

iv) In the collecting duct (CD) :
This long duct carries the filtrate through the medulla to the renal pelvis. Considerable amount of water could be reabsorbed from the region to produce concentrated urine. This segment allows passage of small amount of urea to the medullary interstitum to keep up its osmolarity. It also plays a role in the maintenance of pH and ionic balance of blood by the selective secretion of H⁺ and K⁺ ions. The renal fluid after the process of facultative reabsorption in the CD, influenced by ADH, constitutes the ‘urine’, that is sent out. Urine in the CD is hypertonic to the plasma of blood.

Telangana TSBIE TS Inter 2nd Year Zoology Study Material Lesson 2(a) Body Fluids and Circulation Textbook Questions and Answers.

TS Inter 2nd Year Zoology Study Material Lesson 2(a) Body Fluids and Circulation

Very Short Answer Type Questions

Question 1.
Write the differences between ‘open’ and ‘closed’ systems of circulation.
Answer:
a) open type :
In this type, blood flows from the heart into the arteries. The arteries open into large spaces called sinuses. From sinuses, blood is carried by the veins to the heart. There are no inter connecting vessels, capillaries between arteries and veins. It is found in leeches, arthropods, molluscs, and echinoderms.

b) Closed type :
In this type blood flows through blood vessels. Blood flows from arteries to the veins through capillaries. Closed type of blood vascular system is found in annelids, cephalopods among nonchordates and all vertebrates.

Question 2.
Sino-atrial node is called the pacemaker of our heart. Why?
Answer:
Sino atrial node consists of specialized cardiomyocytes. It has the ability to generate action potentials without any external stimuli, hence called pace maker of our heart.

Question 3.
What is the significance of artrioventricular node and antrioventricular bundle in the functioning of the heart?
Answer:
Atrio ventricular node (AV node) is a relay point that relays the action potentials received from the SA node to the ventricular musculature. A bundle of nodal fibres called atrioventricular bundle (His bundle / AV bundle) continues from the AV node into the interventricular septum. The action potentials received by AVN are conducted through atrioventricular bundle causing simultaneous ventricular systole.

Question 4.
Name the valves that guard the left and right atrioventricular apertures in man. [March 2015 (T.S.)]
Answer:
The left atrio ventricular aperture is guarded by bicuspid or mitral valve. The right atrio ventricular aperture is guarded by tricuspid valve.

Question 5.
Where is the valve of Thebesius in the heart of man?
Answer:
The valve of the besius is situated where the coronary sinus opens into the right atrium of heart.

Question 6.
Name the aortic arches arising from the ventricles of the heart of man.
Answer:
The aortic arches arising from the ventricles of the heart of man are.

1. Pulmonary arch whose opening is guarded by pulmonary valve.
2. Systemic arch whose opening is guarded by aortic valve.
   Both valves are made up of 3 semilunar flaps each.

Question 7.
Name the heart sounds. When are they produced?
Answer:
Heart sounds are named as LUB and DUP.

1. The first heart sound LUB is produced when atrioventricular valves (AV valves) close preventing the back flow of blood.
2. The second heart sound DUP is produced when the semilunar valves close preventing the back flow of blood.

Question 8.
Define cardiac cycle and cardiac output. [March 2020]
Answer:

1. The cardiac events that occur from the beginning of one heart beat to the beginning of the next constitute a cardiac cycle.
2. The volume of blood pumped out by the heart from each ventricle per minute is termed cardiac output.

Question 9.
What is meant by double circulation? What is its significance?
Answer:
In this type of circulation blood circulate twice (two times) through the heart to complete circuit. There are two circuits.

1. Lesser circulation i.e., pulmonary circulation.
2. Greater circulation i.e., systemic circulation. Oxygenated and deoxygenated bloods never mix.

Question 10.
Why the arteries are more elastic than the veins?
Answer:
Arteries are most elastic than the veins because arteries and arterioles have two elastic laminae one on either side of the muscle layer. Veins have one elastic lamina inner to the muscle layer. The muscle layer is much thicker in the arteries than in the veins.

Short Answer Type Questions

Question 1.
Describe the evolutionary change in the structural pattern of the heart among the vertebrates.
Answer:
In the vertebrates the principal differences in the blood vascular system involve the gradual differentiation of the heart into two separate pumps as they evolved from the gill breathing aquatic life to the lung breathing complete terrestrial life.

Fishes have a 2 – chambered heart with an atrium and a ventricle. It pumps out deoxygenated blood to gills for oxygenation, hence the name ‘branchial heart’. Blood passes through the heart only once in a complete circuit, hence called single circulation.

Amphibians have a 3 – chambered heart with two atria and one ventricle. Reptile have two atria and an incompletely divided ventricle (except in the crocodiles in which the ventricle is divided into two chambers). The left atrium receives oxygenated blood from the gills/ lungs /skin and the right atrium receives deoxygenated blood from the other parts of the body through the venae cavae. However, the two types of blood get mixed up in the single ventricle, which pumps out mixed type of blood. Thus these animals (amphibians and reptiles) show an incomplete double circulation.

Birds and mammals possess a 4 – chambered heart with two atria and two ventricles. In these animals the oxygenated and the deoxygenated types of blood received by the left and right atria passes on to the left and right ventricles, respectively. The ventricles pump the blood out without any mixing of the oxygenated and deoxygenated types of blood i.e., there are two completely separate circulatory pathways namely systemic and pulmonary circulations. Hence, these animals are said to be showing ‘double circulation’.

Question 2.
Describe atria of the heart of man.
Answer:
Atria of the heart of man :
Atria are thin walled ‘receiving chambers’ (upper chambers). The right one is larger than the left. The two atria are separated by thin inter -atrial septum. In the fetal heart, the atrial septum has a small pore called foramen ovale. Normally the foramen ovale closes at birth, when lungs become functional. It is represented by a depression in the septum between the right and left atria, called fossa ovalis (that marks the position of the foramen ovale in the fetus). If, the foramen ovale does not close properly, it is called a patent foramen ovale.

The right atrium receives deoxygenated blood from different parts of the body (except the lungs) through three caval veins viz. the two precavals (right and left) and a post caval vein. It also receives blood from the myocardium (wall of the heart) through the coronary sinus, whose opening into the right atrium is guarded by the valve of Thebesius. Opening of the postcaval vein is guarded by the valve of the inferior vena cava or Eustachian valve. It directs the blood to the left atrium through the foramen ovale, in the foetal stage, but in the adult it becomes rudimentary and non – functional. The openings of the precaval veins into the right atrium have no valves. The left atrium receives blood from each lung through two pulmonary veins, which open into the left atrium. The two left pulmonary veins open by a common aperture in some.

Atria and ventricles are separated by a membranous atrio – ventricular septum, which possesses left and right atrioventricular apertures. The left and right apertures are guarded by bicuspid (mitral valve) and tricuspid valves respectively.

Question 3.
Describe the ventricles of the heart of man.
Answer:
Ventricles :
These are the thick walled blood pumping chambers (lower chambers), separated by an interventricular septum. The wall of the left ventricle is thicker than that of the right ventricle. The inner surface of the ventricles is raised into muscular ridges or columns called columnae carneae / trabeculae carneae projecting from the inner walls of the ventricles. Some of these ridges are large and conical, and are called papillary muscles, whose apices are connected to the chordae tendineae, or ‘heart strings’. They are cord – like collagenous processes that connect the papillary muscles to the tricuspid valve and the mitral valve in the heart. They prevent the cusps of the atrioventricular valves from bulging too far into atria during ventricular systole.

Question 4.
Draw a labelled diagram of the L.S of the heart of man.
Answer:

Question 5.
Describe the events in a cardiac cycle; briefly.
Answer:
Cardiac cycle :
The cardiac cycle consists of three phases, namely atrial systole, ventricular systole and cardiac diastole.

To begin with, all the four chambers of the heart are in a relaxed state / joint diastole stage. Blood from the pulmonary veins and venae cavae flows into the respective atria. As the A – V valves are in open condition, blood flows into the left and right ventricles, through the left and right atrioventricular apertures. The semilunar valves of the pulmonary and aortic arches are closed at this stage.

Atrial systole :
The SAN now generates an action potential which stimulates both the atria to contract simultaneously causing the ‘atrial systole’. It lasts about 0. 1 sec. This increases the flow of blood into the ventricles by about 30%. It means atrial systole accounts for about 30% of the filling of the ventricles, the remaining blood flows into the ventricles before the atrial systole.

Ventricular systole :
The action potentials from the SAN reach the AVN from where they are conducted through the bundle of His, its branches and the Purkinje fibres to the entire ventricular musculature. This causes the simultaneous ventricular systole. It lasts for about 0.3 sec. The atria undergo relaxation coinciding with the ventricular systole. Ventricular systole increases the pressure causing the closure of the AV valves preventing the ‘backflow’ of blood. It results in the production of the first heart sound known as ‘Lub’. As the ventricular pressure increases further, the semilunar valves guarding the pulmonary artery and the aorta are forced open. This allows the blood in the ventricles to flow into the aortic arches and enter the circulatory pathway.

Cardiac diastole :
The ventricles now relax and the ventricular pressure falls causing the closure of the semilunar valves which prevents the back flow of blood. This results in the production of the second heart sound known as ‘Dup’. As the ventricular pressure declines further, the AV valves are pushed open by the pressure in the atria exerted by the blood, which flowed into them through the larger veins.

Question 6.
Explain the mechanism of clotting of blood.
Answer:
Mechanism of blood clotting:
Clotting takes place in three essential steps.
i) Step – 1 :
It involves the formation of a complex of activated substances collectively called, prothrombin activator. It is formed by a complex cascade of chemical reactions that occur in the blood by the involvement of clotting factors in two pathways.

a) Intrinsic pathway :
It occurs when the blood is exposed to collagen of injured wall of blood vessel. This activates Factor XII, and in turn it activates another clotting factor, which activates yet another reaction (cascade fashion) which results in the formation of the prothrombin activator.

b) Extrinsic pathway :
It occurs when the damaged vascular waH or extra vascular tissue comes into contact with blood. This activates the release of tissue thromboplastin, from the damaged tissue. It activates the Factor VII . As a result of these cascade reactions, the final product formed is the prothrombin activator.

ii) Step – 2 :
The prothrombin activator, in the presence of sufficient amounts of ionic Ca++, causes the conversion of inactive prothrombin to active thrombin (activation of prothrombin).

iii) Step – 3 :
Thrombin converts the soluble protein fibrinogen into soluble fibrin monomers, which are held together by weak hydrogen bonds. The fibrin stabilizing factor (Factor XIII, released from platelets) replaces hydrogen bonds with covalent bonds and cross links the fibres to form a ‘mesh work . The insoluble mesh work of fibrin fibers spreading in all directions adhere to the damaged surfaces and trap the blood cells and platelets.

Question 7.
Distinguish between SAN and AVN.
Answer:
A specialized cardiac musculature called the nodal tissue is also distributed in the heart. A patch of this tissue called the sinoatrial node (SAN) is present in the right upper corner of the right atrium near the openings of the superior venae cavae. Another mass of this tissue, called the atrioventricular node (AVN), is seen in the lower left corner of the right atrium close to the atrioventricular septum. A bundle of nodal fibres, called atrioventricular bundle (AV Bundle / ‘His’ bundle) continues from the AVN into the inter-ventricular septum. It divides into right and left bundle branches. These branches give rise to minute fibres called purkinje fibres that extend throughout the ventricular musculature / walls of the respective sides. SAN consists of specialized cardiomyocytes. It has the ability to generate action potentials without any external stimuli, hence called pacemaker. AV mode is a relay point.

Question 8.
Distinguish between arteries and veins.
Answer:
Differences between arteries and veins.

Long Answer Type Questions

Question 1.
Describe the structure of the heart of man with the help of neat labelled diagram. [March 2018 (A.P.); March 2014; May/June ’14]
Answer:
The heart is mesodermal in origin. It is a thick walled, muscular and pulsating organ, situated in the mediastinum (the region in the thorax between the two lungs), and with its apex slighty turned to the left. It is the size of a clinched fist.

The heart is covered by a double walled pericardium which consists of the outer fibrous pericardium and the inner serous pericardium is double – layered, formed of an outer parietal layer and an inner visceral layer. The parietal layer is fused with the fibrous pericardium, whereas the visceral layer adheres to the surface of the heart and forms its outer layer, the epicardium. The two layers are separated by a narrow pericardial space, which is filled with the pericardial fluid. This fluid reduces friction between the two membranes and allows free movement of the heart.

The wall of the heart consists of three layers. They are the outer epicardium, the middle myocardium (a thick layer of cardiac muscles), and the inner most endocardium (a thin layer of endothelium). The endothelium covers the heart valves also and is continuous with the endothelial lining of the large blood vessels connected to the heart.

External structure:
Human heart has four chambers, with two relatively smaller upper chambers, called atria and two larger lower chambers called ventricles. Atria and ventricles are separated by a deep transverse groove called coronary sulcus (atrio – ventricular groove). The muscular pouch like projection from each atrium is called auricular appendix (auricular appendage). The ventricles are separated by two inter ventricular grooves (anterior and posterior), in which the coronary arteries and their branches are lodged.

Internal structure:
i) Atria :
Atria are thin walled ‘receiving chambers’ (upper chambers). The right one is larger than the left. The two atria are separated by thin inter-atrial septum. In the fetal heart, the atrial septum has a small pore called foramen ovale. Normally the foramen ovale closes at birth, when lungs become functional. It is represented by a depression in the septum between the right and left atria, called fossa ovalis (that marks the position of the foramen ovale in the fetus). If, the foramen ovale does not close properly, it is called a patent foramen ovale.

The right atrium receives deoxygenated blood from different parts of the body (except the lungs) through three caval veins viz. the two precavals (right and left) and a post caval vein. It also receives blood from the myocardium (wall of the heart) through the coronary sinus, whose opening into the right atrium is guarded by the valve of Thebesius. Opening of the postcaval vein is guarded by the valve of the inferior vena cava or Eustachian valve. It directs the blood to the left atrium through the foramen ovale, in the foetal stage, but in the adult it becomes rudimentary and non – functional. The openings of the precaval veins into the right atrium have no valves. The left atrium receives blood from each lung through two pulmonary veins, which open into the left atrium. The two left pulmonary veins open by a common aperture in some.

Atria and ventricles are separated by a membranous atrio – ventricular septum, which possesses left and right atrioventricular apertures. The left and right apertures are guarded by bicuspid (mitral valve) and tricuspid valves respectively.

ii) Ventricles :
These are the thick walled blood pumping chambers (lower chambers), separated by an interventricular septum. The wall of the left ventricle is thicker than that of the right ventricle. The inner surface of the ventricles is raised into muscular ridges or columns called columnae carneae / trabeculae carneae projecting from the inner walls of the ventricles. Some of these ridges are large and conical, and are called papillary muscles, whose apices are connected to the chordae tendineae, or ‘heart strings’. They are cord – like collagenous processes that connect the papillary muscles to the tricuspid valve and the mitral valve in the heart. They prevent the cusps of the atrioventricular valves from bulging too far into atria during ventricular systole.

Question 2.
Write notes on the working of the heart of man.
Answer:
The cardiac events that occur from the beginning of one heart beat to the beginning of the next constitute a cardiac cycle. This cardiac cycle consists of three phases, namely atrial systole, ventricular systole and cardiac diastole.

To begin with, all the fpur chambers of the heart are in a relaxed state / joint diastole stage. Blood from the pulmonary veins and venae cavae flows into the respective atria. As the A – V valves are in open condition, blood flows into the left and right ventricles, through the left and right atrioventricular apertures. The semilunar valves of the pulmonary and aortic arches are closed at this stage.

Atrial systole :
The SAN now generates an action potential which stimulates both the atria to contract simultaneously causing the ‘atrial systole’. It lasts about 0. 1 sec. This increases the flow of blood into the ventricles by about 30%. It means atrial systole accounts for about 30% of the filling of the ventricles, the remaining blood flows into the ventricles before the atrial systole.

Ventricular systole :
The action potentials from the SAN reach the AVN from where they are conducted through the bundle of His, its branches and the Purkinje fibres to the entire ventricular musculature. This causes the simultaneous ventricular systole. It lasts for about 0.3 sec. The atria undergo relaxation coinciding with the ventricular systole. Ventricular systole increases the pressure causing the closure of the AV valves preventing the ‘backflow’ of blood. It results in the production of the first heart sound known as ‘Lub’. As the ventricular pressure increases further, the semilunar valves guarding the pulmonary artery and the aorta are forced open. This allows the blood in the ventricles to flow into the aortic arches and enter the circulatory pathway.

Cardiac diastole :
The ventricles now relax and the ventricular pressure falls causing the closure of the semilunar valves which prevents the back flow of blood. This result in the production of the second heart sound known as ‘Dup’. As the ventricular pressure declines further, the AV valves are pushed open by the pressure in the atria exerted by the blood, which flowed into them through the larger veins. The blood now once again flows freely into the ventricles. All the heart chambers are now again in a relaxed state (joint diastolic phase). Soon, another cardiac cycle sets in.

Cardiac output :
The volume of blood pumped out by each ventricle, for each heart beat, is known as the stroke volume. The volume of blood pumped out by the heart from each ventricle per minute is termed cardiac output.

Cardiac output = Stroke volume × No. of beats per minute = 70 ml / beat × 72 beats / minute = 5040 ml/ min. or approximately 5 litres.

Telangana TSBIE TS Inter 2nd Year Zoology Study Material Lesson 1(b) Breathing and Exchange of Gases Textbook Questions and Answers.

TS Inter 2nd Year Zoology Study Material Lesson 1(b) Breathing and Exchange of Gases

Very Short Answer Type Questions

Question 1.
Define vital capacity. What is its significance?
Answer:
Vital Capacity (VC) :
The maximum volume of air a person can breathe in after forced expiration. This includes ERV, TV and IRV or the maximum volume of air a person can breathe out after forced inspiration.
VC = TV + IRV + ERV

Question 2.
What is the volume of air remaining in lungs after a normal expiration?
Answer:
The volume of air that remains in the lungs after normal expiration is called Functional Residual Capacity (FRC).
FRC = ERV + RV

Question 3.
Diffusion of oxygen occurs in the alveolar region only and not in the other parts of respiratory system. How do you justify the statement?
Answer:
Alveoli are primary sites of exchange of gases in the lungs. The diffusion membrane of Alveoli is made up of 3 major layers like thin squamous epithelium of Alveolar wall, the endothelium of the alveolar capillaries and the basement material in between them. As it is a very thin border, it is favourable for diffusion of gases.

Question 4.
What is the effect of pCO₂ on oxygen transport?
Answer:
The effect of pCO₂ and H⁺ concentration on the oxygen affinity of haemoglobin is called Bohr effect. (A rise in pCO₂ and fall in pH decreases the affinity of haemoglobin for oxygen. On other hand a fall in pCO₂ and rise in pH increases affinity of haemoglobin for oxygen).

Question 5.
What happens to the respiratory process in a man going up a hill?
Answer:
At a height of about 6000 m the pO₂ becomes almost half of what it is at the mean sea level, hence the mountain sickness in people ascending mountains.

Question 6.
What is Tidal volume? Find out the Tidal volume (approximate value) in a healthy human, in an hour?
Answer:
Volume of air inspired or expired during normal inspiration or expiration. It is approximately 500 ml. A healthy man can inhale or exhale approximately per hour is 3,60,000 -4,80,000 ml.

Question 7.
Define oxyhaemoglobin dissociation curve. Can you suggest any reason for its sigmoidal pattern?
Answer:
A sigmoid curve is obtained when percentage saturation of haemoglobin with O₂ is plotted against the pO₂. This curve is called oxyhaemoglobin dissociation curve. At normal condition that is on pCO₂ of 40mm Hg concentration, this curve is sigmoid and normal. By increasing concentration of CO₂ curve shifted towards rightside. By decreasing, concentration of CO₂ curve shifted towards left side.

Question 8.
What are conchae?
Answer:
Nasal chamber of human being is divided into 3 parts namely vestibularpart, respiratory part and olfactory part. In respiratory part, it has three thin twisted bony plates called turbinals or conchae.

Question 9.
What is meant by chloride shift? [March 2014]
Answer:
The exchange of chloride and bicarbonate ions between RBC and plasma at the tissues is called chloride shift or Hamburger’s phenomenon or Hamburger’s shift.

Question 10.
Mention any two occupational respiratory disorders and their causes in human beings. [March 2018(A.P)]
Answer:
Two occupational respiratory disorders are
a) Asbestosis :
It occurs due to chronic exposure to asbestos dust in the people working in asbestos industry.

b) Black lung disease :
It is a lung disease that develops from inhalation of coal dust. It is common in long time coal mine workers.

Question 11.
Name the muscles that help in normal breathing movements.
Answer:
Normal breathing movements are aided by

1. Phrenic muscles of diaphragm
2. External and internal intercostal muscles of ribs.

Question 12.
Draw a diagram of oxyhaemoglobin dissociation curve.
Answer:

Short Answer Type Questions

Question 1.
Explain the process of inspiration and expiration under normal conditions? [March 2015 (T.S.)]
Answer:
Inspiration :
Intake of atmospheric air into the lungs is called inspiration. It is an active process, as it takes place by the contraction of the muscles of the diaphragm and the external inter-costal muscles, which extend in between the ribs. The contraction of the diaphragm (phrenic muscles) increases the volume of the thoracic chamber in the antero-posterior axis. The contraction of external intercostal muscles lifts up the ribs and sternum causing an increase in the volume of the thoracic chamber in the dorso – ventral axis. The overall increase in the thoracic volume causes a similar increase in the ‘pulmonary volume’. An increase in the pulmonary volume decreases the intra – pulmonary pressure to less than that of the atmosphere, which forces the air from the outside to move into the lungs, i.e., inspiration.

Expiration :
Release of alveolar air to the exterior is called expiration. It is a passive process. Relaxation of the diaphragm and the external inter – costal muscles returns the diaphragm and sternum to their normal positions, and reduces the thoracic volume and thereby the pulmonary volume. This leads to an increase in the intra – pulmonary pressure to slightly above that of the atmospheric pressure, causing the expulsion of air from the lungs, i.e., expiration.

Question 2.
What are the major transport mechanisms for CO₂? Explain. [March 20191 May 2017 (A.P.)]
Answer:
Transport of Carbon Dioxide : CO₂ is transported in three ways.
i) In dissolved state :
7 percent of CO₂ is carried in a dissolved state (physical solution) through plasma.
CO₂ + H₂O → H₂CO₃

ii) As carbamino compounds :
About 20-25 percent of CO₂ combines directly with free amino group of the haemoglobin and forms carbamino-haemoglobin in a reversible manner.
Hb – NH₂ + CO₂ → Hb – NHCOO^– + H⁺

This binding of CO₂ is related to the partial pressure of CO₂. pO₂ is a major factor which could affect this binding. When pCO₂ is high and pO₂ is low as in the tissues, binding of more carbon dioxide occurs. When pCO₂ is low and pO₂ is high as in the alveoli, dissociation of CO₂ from carbamino – haemoglobin takes place, i.e., CO₂ which is bound to haemoglobin from the tissues is delivered at the alveoli. Carbamino compounds are also formed by the union of CO₂ with plasma proteins.

iii) As Bicarbonates: About 70 percent of CO₂ is transported as bicarbonate. RBCs contain a very high concentration of the enzyme, carbonic anhydrase and a minute quantity of the same is present in the plasma too. This enzyme facilitates the following reaction in both the directions.

At the tissues where partial pressure of CO₂ is high due to catabolism, CO₂ diffuses into the blood (RBC and Plasma) and forms carbonic acid which dissociates into HCO^–₃ and H⁺. At the alveolar site where pCO₂ is low, the reaction proceeds in the opposite direction leading to the formation of CO₂and water. Thus CO₂ is mostly trapped as bicarbonate at the tissues and transported to the alveoli where it is released out as CO₂.

Question 3.
How is respiratory movements regulated in Man? [March 2018 (A.P); March 2014]
Answer:
Respiratory movements are regulated in Man by neural system

1. A special centre present in the medulla region of brain called “Respiratory rhythm centre” is primarily responsible for this regulation.
2. Another centre present in the pons of the brain stem called ‘Pneumotaxic Centre’ can moderate the functions of the ‘respiratory rhythm centre1. Neural signal from this centre can reduce the duration of inspiration and there by alter the respiratory rate.
3. A chemo – sensitive area is situated adjacent to the respiratory rhythm centre which is highly sensitive to CO₂ and hydrogen ions. Increase in these substances can activate this centre, which inturn can send signals to the respiratory rhythm centre to make necessary adjustments in the respiratory process by which these substances can be eliminated.
4. Receptors associated with aortic arch and carotid artery also recognize changes in CO₂ and H⁺ concentration and send necessary signals to the respiratory rhythm centre for necessary actions (increase in the rate and depth of breathing when their concentration is high). The role of oxygen in the regulation of the respiratory rhythm is quite insignificant.

4. Distinguish between
a) IRV and ERV
b) Inspiratory Capacity and Expiratory Capacity.
c) Vital capacity and Total lung capacity.
Answer:
a) IRV and EftV :
IRV is the maximum volume of air that can be inhaled during forced breathing in addition to the tidal volume. This is about 2500 ml 3000 ml. ERV is the maximum volume of air that can be exhaled during forced breathing in addition to the tidal volume. This is about 1000 ml to 1100 ml.

b) Inspiratory Capacity and Expiratory Capacity (1C):
Inspiratory Capacity :
The total volume of air a person can inhale after normal expiration. This includes the tidal volume and inspiratory reserve volume i.e., Ic = TV + IRV. It is about 3000 ml to 3500 ml.

Expiratory Capacity :
The total volume of air a person can exhale after normal inspiration. This includes tidal volume and expiratory reserve volume. TV + ERV. It is about 1500 ml to 1600 ml.

c) Vital capacity and Total lung capacity :
The maximum volume of air a person can breathe in after forced expiration is called vital capacity. This includes ERV, TV and IRV.

Total lung Capacity :
The total volume of air accommodated in the lungs at the end of forced inspiration. This includes RV, ERV, TV and IRV.

Question 5.
Describe disorders of respiratory system. [Mar. ’20, ’17, ’15 (A.P.); May ’14]
Answer:
Disorders of the Respiratory System :
i) Asthma is a difficulty in breathing caused due to inflammation of bronchi and bronchioles. It is characterized by the spasm of smooth muscles present in the walls of the bronchi and bronchioles. Symptoms include coughing, difficulty in breathing and wheezing. In the case of asthma, the allergen causes release of histamine and other inflammatory substances which cause constriction of the bronchi.

ii) Emphysema is a chronic disorder in which alveolar walls are damaged and their walls coalesce due to which respiratory surface area of exchange of gases is decreased. The lung shows larger but fewer alveoli and more fibrous and less elastic. One of the major causes of this is ‘smoking’ of tobacco.

iii) Bronchitis is the inflammation of the bronchi, resulting in the swelling of mucous lining of bronchi, increased mucus production and decrease in the diameter of bronchi. Symptoms include chronic cough with thick mucus/ sputum (phlegm).

iv) Pneumonia is infection of lungs caused by bacteria such as Streptococcus pneumoniae and also by certain viruses, fungi, protozoans and mycoplasmas. Symptoms include inflammation of lungs, accumulation of mucus in alveoli, and impaired exchange of gases, leading to death if untreated.

v) Emphysema, chronic bronchitis and asthma come under Chronic Obstructive Pulmonary Diseases (COPDs).

Occupational Respiratory disorders :
These are caused by exposure of the body to the harmful substances from certain industries, especially those involving grinding or stone breaking. Long term exposure of the body to such substances can give rise to inflammation of respiratory passage and lungs leading to several disorders.

i) Asbestosis :
It occurs due to chronic exposure to asbestos dust in the people working in asbestos industry.

ii) Silicosis :
It occurs because of long term exposure to ‘silica dust’ in the people working in mining industries, quarries etc.

iii) Siderosis :
It occurs due to deposition of iron particles in tissues. It can cause different types of siderosis such as pneumoconiosis due to inhalation of iron particles, hyperferremia and hemosiderosis (which causes recurrent alveolar hemorrhage).

iv) Black-lung disease :
It is a lung disease that develops from inhalation of coal dust. It is common in long time coal mine workers.

Long Answer Type Questions

Question 1.
Describe the respiratory system in man.
Answer:
Human Respiratory System : Respiratory system of man includes the following :

I) External nostrils (External Nares) :
A pair of external nostrils opens out above the upper lip. They lead into nasal chambers through the nasal passages.

II) Nasal Chambers :
They lie above the palate and are separated from each other by a nasal septum. Each nasal chamber can be differentiated into three parts namely, (i) vestibular part (which has hair and sebaceous glands to prevent the entry of dust particles), (ii) respiratory part (which is involved in the conditioning the temperature of inhaled air, it has three thin, twisted bony plates called turbinals / conchae) and (iii) olfactory part (which is lined by an olfactory epithelium).

III) Naso-pharynx :
Nasal chambers lead into nasopharynx through a pair of internal nostrils, located above the soft palate. Nasopharynx is a portion of the pharynx, the common chamber for the passage of food and air. Nasopharynx leads into oropharynx and opens through glottis of larynx into the trachea.

IV) Larynx :
Larynx is a cartilaginous box which helps in sound production, hence called the voice box. Wall of larynx is supported by nine cartilages. Thyroid, cricoid and epiglottis are the unpaired cartilages, whereas corniculate cartilages (cartilages of Santorini – two small conical nodules of elastic cartilage articulating with the arytenoid cartilages), arytenoids, and cuneiform cartilages are the paired cartilages. Epiglottis is a thin leaf like elastic cartilaginous flap attached to the thyroid cartilage to prevent the entry of food into the larynx through the glottis. The yellow elastic fibres which connect the thyroid and arytenoid cartilages are called vocal cords / vocal folds. The space between the true vocal cords and the arytenoids cartilages is called rima glottidis.

The mid ventral part of the thyroid cartilage forms the laryngeal prominence called Adam’s apple.
In males, the vocal cords are thicker, longer, and produce low pitch voice, where as in women and children the vocal cords are usually short and produce high pitch voice.

V) Trachea :
Trachea, the wind pipe is a straight tube extending up to the mid – thoracic cavity. The wall of the trachea is supported by ‘C’ shaped rings of hyaline cartilage. These rings are incomplete dorsally and keep the trachea always open preventing collapse. Internally the trachea is lined by pseudostratified ciliated epithelium.

VI) Bronchi and Bronchioles :
On entering the mid thoracic cavity, trachea divides at the level of the fifth thoracic vertebra into right and left primary bronchi. Each primary bronchus enters the corresponding lung and divides into secondary bronchi that further divide into tertiary bronchi. Each tertiary bronchus divides and re – divides to form primary, secondary, tertiary terminal and respiratory bronchioles sequentially. Each respiratory bronchiole terminates in a cluster of alveolar ducts which end alveolar sacs. Bronchi and initial bronchioles re supported by incomplete cartilaginous rings. The branching network of trachea, bronchi and bronchioles constitute the’pulmonary tree’ (an upside down tree).

VII) Lungs :
Lungs occupy the greater part of the thoracic cavity. Lungs are covered by a double layered pleura, with pleural fluid between them. It reduces friction on the lung surface. The outer pleural membrane is in close contact with the thoracic lining whereas the inner pleural membrane is in contact with lung’s surface. The part starting with external nostrils up to the terminal bronchioles constitute the conducting part, whereas the alveoli and their ducts form the respiratory or exchange part of the respiratory system. The conducting part transports the atmospheric air to the alveoli, clears it from foreign particles, humidifies and also brings the inhaled air to the body temperature. Exchange part is the site of actual diffusion of and between blood and atmospheric air.

The lungs are situated in the thoracic chamber which is anatomically an air – tight chamber. It is formed dorsally by the vertebral column, ventrally the sternum, laterally by ribs and on the lower side by the dome – shaped diaphragm. The anatomical setup of lungs in the thorax is such that any change in the volume of thoracic cavity will be reflected in the lung cavity. Such an arrangement is essential for breathing, as the pulmonary volume cannot be directly altered.

Diagrammatic view of human respiratory system
(Sectional view of the left lung is also shown)

Question 2.
Write an essay on the transport of oxygen and carbon dioxide by blood.
Answer:
Transport of gases :
Blood is the medium of transport for O₂ and CO₂.

I. Transport of Oxygen :
Oxygen is transported from the lungs to the tissues through the plasma and RBC of the blood. 100 ml of oxygenated blood can deliver 5 ml of o₂ to the tissues under normal conditions.
i) Transport of oxygen through plasma :
About 3% of 02 is carried through the blood plasma in a dissolved state.

ii) Transport of oxygen by RBC :
About 97% of O₂ is transported by the RBCs in the blood. Haemoglobin is a red coloured iron containing pigment present in the RBCs. Each haemoglobin molecule can carry a maximum of four molecules of oxygen. Binding of oxygen with haemoglobin is primarily related to the partial pressure of O₂. At lungs, where the partial pressure of O₂ (oxygen tension) is high, pO₂ (mmHg) oxygen binds to haemoglobin (purplish – bluish-red in colour) in a reversible manner to form oxyhaemoglobin (bright red in colour). This is called oxygenation of haemoglobin.
Hb + 4O₂ → Hb (O₂)₄

At the tissues, where the partial pressure of 02 is low, oxyhaemoglobin dissociates into haemoglobin and oxygen. The other factors that influence binding of oxygen with haemoglobin are the partial pressure of C02, the hydrogen ion concentration (pH) and the temperature.

iii) Oxygen – haemoglobin dissociation curve :
It explains the relation between percentage saturation of haemoglobin and partial pressure of oxygen. A sigmoid curve is obtained when percentage saturation of haemoglobin with O₂ is plotted against the pO₂. This curve is called ‘oxyhaemoglobin dissociation curve’ and is highly useful in studying the effect of factors such as pCO₂, H⁺ concentration, temperature, etc., on the binding of O₂ with haemoglobin. In the alveoli, where there is a high pO₂, low pCO₂, lesser H⁺ concentration (high pH) and lower temperature, the factors are all favourable for the formation of oxyhaemoglobin.

In the tissues where low pO₂, high pCO₂, high H⁺ concentration (low pH) and higher temperature exist, the conditions are favourable for dissociation of oxygen from oxyhaemoglobin. Under these conditions, oxygen dissociation curve shifts away from the Y – axis (to the right). The effect of pCO₂ and H⁺ concentration on the oxygen affinity of haemoglobin is called Bohr Effect (increase of carbondioxide in the blood and decrease in pH results in the reduction of the affinity of hemoglobin for oxygen).

II. Transport of Carbon Dioxide :
CO₂ is transported in three ways.

i) In dissolved state :
7 percent of CO₂ is carried in a dissolved state (physical solution) through plasma.
CO₂ + H₂O → H₂CO₃

ii) As carbamino compounds :
About 20 – 25 percent of CO₂ combines directly with free amino group of the haemoglobin and forms carbamino-haemoglobin in a reversible manner.
Hb – NH₂ + CO₂ → Hb – NHCOO^– + H⁺

This binding of CO₂ is related to the partial pressure of CO₂. pO₂ is a major factor which could affect this binding. When pCO₂ is high and pO₂ is low as in the tissues, binding of more carbon dioxide occurs. When pCO₂ is low and pO₂ is high as in the alveoli, dissociation of CO₂ from carbamino – haemoglobin takes place, i.e., CO₂ which is bound to haemoglobin from the tissues is delivered at the alveoli. Carbamino compounds are also formed by the union of CO₂ with plasma proteins.

iii) As Bicarbonates :
About 70 percent of CO₂ is transported as bicarbonate. RBCs contain a very high concentration of the enzyme, carbonic anhydrase and a minute quantity of the same is present in the plasma too. This enzyme facilitates the following reaction in both the directions.

At the tissues where partial pressure of CO₂ is high due to catabolism, CO₂ diffuses into the blood (RBC and Plasma) and forms carbonic acid which dissociates into HCO^–₃ and H⁺. At the alveolar site where pCO₂ is low, the reaction proceeds in the opposite direction leading to the formation of CO₂ and water. Thus CO₂ is mostly trapped as bicarbonate at the tissues and transported to the alveoli where it is released out as CO₂. Every 100 mL of deoxygenated blood delivers approximately 4mL of CO₂ to the alveolar air.

Telangana TSBIE TS Inter 2nd Year Zoology Study Material Lesson 1(a) Digestion and Absorption Textbook Questions and Answers.

TS Inter 2nd Year Zoology Study Material Lesson 1(a) Digestion and Absorption

Very Short Answer Type Questions

Question 1.
Give the dental formula of adult human [March 2015 (T.S.)]
Answer:
Dental formula of adult human being is = \(\frac{2123}{2123}\)

Question 2.
Bile juice contains no digestive enzymes, yet it is important for digestion. How?
Answer:
Bile salts of the bile help in the emulsification of fats (breakdown of fats into very small micelles). Bile also activates lipases of pancreatic juice (Steapsin) and intestinal lipases.

Question 3.
Describe the role of chymotrypsin. Name two other digestive enzymes of the same category and secreted by the same gland.
Answer:
Chymotrypsin acts upon proteins, proteoses and peptones to convert them into tripeptides and dipeptides.

Two other digestive enzymes of the same category and secreted by the same gland pancreas are a) trypsin b) carboxypeptidase.

Question 4.
What would happen if, HCl were not secreted in the stomach?
Answer:
HCI secreted by gastric glands in stomach provides the acidic pH (1.8) which is optimal for the action of pepsin. It also kills the microorganisms ingested along with food. Pepsinogen and prorenin are not activated if HCI were not secreted in the stomach.

Question 5.
Explain the terms thecodont and diphyodont dentitions.
Answer:

1. Teeth of human beings are embedded in the sockets of the Jaw bones, hence called thecodont.
2. Human beings form two sets or teeth during their life time, milk set and permanent adult teeth. This type of dentition is called diphyodont dentition.

Question 6.
What is auto catalysis? Give two examples.
Answer:
Trypsinogen is activated by the enzyme enterokinase into active trypsin which inturn can activate trypsinogen into trypsin it is called auto catalysis. Another such example is Pepsin.

Question 7.
What is chyme? [March 20191 May 2017 (A.P.)]
Answer:
The food is mixed thoroughly with the acidic gastric Juice of the stomach by the churning movements of its muscular wall and the product is called Chyme. (The semi digested paste like acidic food of stomach is called chyme).

Question 8.
Name the different types of salivary glands of man, and their locations in the human body. [March 2020]
Answer:
There are 3 pairs of salivary glands of Man.

1. Parotid glands – present below the pinna and inner surface of cheeks.
2. Sub maxillary glands-located at the angles of lower jaw.
3. Sub lingual glands – present below the tongue.

Question 9.
Name different types of papillae present on the tongue of man. [March 2019, May 2017 (A.P.)]
Answer:
The different types of papillae present on the tongue of Man are a) fungi form b) filiform c) circumvallate papillae.

Question 10.
What is the hardest substance in the human body? What is its origin?
Answer:
The hardest substance in the body is enamel which is secreted by ameloblasts of ectodermal origin.

Question 11.
Name the structure of gut which is vestigial in human beings, but well developed in the herbivores and mention the type of tissue with which it is mostly formed.
Answer:
Structure of gut which is vestigeal in human beings is Vermiform appendix arises from the caecum, it is mostly formed from epithelial tissue.

Question 12.
Distinguish between deglutition and mastication.
Answer:

1. Deglutition is swallowing of food facilitated by buccal cavity.
2. Mastication is chewing. Teeth and tongue with the help of saliva masticate and mix up the food thoroughly.

Question 13.
Distinguish between diarrhoea and constipation.
Answer:
a) Diarrhoea :
The abnormal frequency of bowl movement and increased liquidity of the faecal discharge is known as diarrhoea. It results in loss of water (dehydration).

b) Constipation :
In constipation, the faeces are retained with in the rectum as it is hard due to low content of water and the movement of the bowel occurs irregularly.

Question 14.
Name two hormones secreted by the duodenal mucosa.
Answer:
Two hormones secreted by duodenal mucosa.

1. Enterocrinin
2. Secretin
3. Cholecystokinin.

Question 15.
Distinguish between absorption and assimilation.
Answer:

1. Absorption is the process by which the end products of digestion pass through the intestinal mucosa into blood or lymph.
2. The absorbed substances finally reach the tissues where food materials become integral components of the living protoplasm and are used for the production of energy, growth and repair. This process is called assimilation.

Short Answer Type Questions

Question 1.
Draw a neat labelled diagram of L.S. of tooth. [Mar. ’20, 19, 18,17,14 (A.P.); May/June ’14]
Answer:

Question 2.
Describe the process of digestion of proteins in the stomach. [March 2015 (A.P.)]
Answer:
The gastric glands of stomach secrete acidic gastric juice. Gastric juice contains HCl, prorennin, pepsinogen and bicarbonates.

Proenzymes pepsinogen and prorennin, on exposure to HCl, are converted into the active enzymes, pepsin and rennin respectively.

Pepsin converts proteins into proteoses and peptones. Rennin is found in gastric juice of infants. It acts on milk protein casein in the presence of calcium ions and converts it into calcium paracaseinate (curd) and proteoses. Pepsin again acts on calcium para- caseinate and converts it into peptones. Certain other cells in the wall of stomach produce bicarbonate, a base to buffer the acidic contents of the stomach. Bicarbonates and mucus produced by stomach wall forms a physical barrier to prevent HCl from damaging the wall of stomach.

Question 3.
Explain the role of Pancreatic Juice in the digestion of proteins.
Answer:
Pancreatic juice contains sodium bicarbonate, trypsinogen, chymotrypsinogen, carboxypeptidase, pancreatic lipase (steapsin), pancreatic amylase and nucleases such as DNA ase and RNA ase.

Trypsinogen is activated by an enzyme enterokinase into active trypsin which in turn activates the other enzymes in the pancreatic juice. Trypsin itself can similarly activate trypsinogen into trypsin (autocatalysis).

Proteolytic enzymes of the pancreatic juice act upon proteins, proteoses and peptones when they reach intestine.

Thus the end products of digestion of proteins namely amino acids are formed in the small intestine.

Question 4.
How are polysaccharides and disaccharides digested?
Answer:
Polysaccharides are nothing but carbohydrates. Carbohydrates in the chyme are hydrolysed by the pancreatic amylase into disaccharides. Intestinal disaccharidases act on the disaccharides and convert them into monosaccharides like Glucose etc.

Question 5.
If, you take butter in your food, how does it get digested and absorbed in the body? Explain.
Answer:

1. Butter is a content of macromolecules and is fat or lipid.
2. Bile salts of the bile help in breaking down of large fat molecules into very small micelles. This process is called Emulsification. They move into intestinal mucosal cells.
3. These micelles are reformed into very small protein coated fat globules called chylomicrons which are transported into the lacteals in the villi by exocytosis.
4. Bile also activates lipases of pancreatic juice (steapsin) and intestinal lipases. These lipases act on emulsified fats and convert them into fatty acids and glycerols.
   

Question 6.
What are the functions of liver? [May 2017 (A.P.); March 2015 (T.S.)]
Answer:
Functions of the liver :
Liver performs a variety of functions such as synthesis, storage and secretion of various substances. There are as follows :

1. Liver secretes bile juice. It does not contain enzymes, but it contains bile salts such as glycocholates and taurocholates of sodium and potassium and ‘bile pigments’ the bilirubin and biliverdin.
2. Liver plays the ‘key role’ in carbohydrate metabolism (glycogenesis, glycogenolysis, gluconeogenesis and lipogenesis).
3. Liver also plays a role in lipid metabolism (synthesis of cholesterol and production of triglycerides).
4. Deamination of proteins (removal of NH2 group from the amino acids) and conversion of ammonia into to urea – via the ornithine cycle).
5. The lactic acid formed during anaerobic muscle contraction is converted into glycogen (gluconeogenesis) in the liver by Cori cycle.
6. Liver is the chief organ of detoxification of toxic substances that enter the gut along with food.
7. Liver acts as thermoregulatory organ (like skeletal muscle, liver too takes part in thermogenesjs as it has high glucose at its disposal).
8. Liver acts as a haemopoietic organ in the foetus and erythroclastic organ in the adult.
9. The liver synthesizes the plasma proteins such as albumins, globulins, blood clotting factors such as fibrinogen, prothrombin, etc., and the anticoagulant, called heparin.
10. Kupffer’s cells/ Kupffer cells are the large phagocytic cells which remove unwanted substances and microbes that attack the liver by phagocytosis. They are present in the sinusoids that lie in between hepatic cords and they are also called hepatic macrophages.

Long Answer Type Questions

Question 1.
Describe the physiology of digestion of various types of food in the human digestive system.
Answer:
Physiology of digestion :
Digestion is the process of conversion of complex non – diffusible food substances into simple diffusible forms. The process of digestion is accomplished by mechanical (cutting and chewing of food by teeth and churning of food by peristalsis) and chemical (enzymatic reactions by hydrolysing enzymes) processes.

i) Digestion in the buccal cavity :
Buccal cavity performs two major functions, mastication of food and facilitation of swallowing (deglutition). Teeth and tongue with the help of saliva masticate and mix up the food thoroughly. Mucus in saliva helps in lubricating and adhering the masticated food particles into a bolus. The saliva secreted into oral cavity contains electrolytes such as Na⁺, K⁺, Cl^–, HCO^–₃ and enzymes, such as salivary amylase (ptyalin) and lysozyme. The chemical process of digestion is initiated in the oral cavity (buccal cavity) by the hydrolytic action of carbohydrate (starch) splitting enzyme, the salivary amylase. About 30% of starch is hydrolyzed here into a disaccharide called maltose by the enzyme ptyalin. Lysozyme present in the saliva acts as an antibacterial agent that prevents infections.

ii) Digestion in the stomach :
The stomach stores food for 4-5 hours. The food is mixed thoroughly with the acidic gastric juice of the stomach by the churning movements of its muscular wall and the product is called chyme. The mucus and bicarbonates present in the gastric juice play an important role in the lubrication and protection of the mucosal epithelium from ‘excoriation’ by the highly concentrated hydrochloric acid. HCl provides the acidic pH (1.8) which is optimal for the action of pepsin. It also kills the microorganisms ingested along with food. The proenzymes of gastric juice, the pepsinogen and prorennin, on exposure to hydrochloric acid are converted into the active enzymes pepsin and rennin respectively. Pepsin converts proteins into proteoses and peptones. Rennin is a proteolytic enzyme found in the gastric juice of infants.

It acts on the milk protein, the casein in the presence of calcium ions and converts it into calcium paracaseinate (curdling of milk) and proteoses. Pepsin acts on calcium paracaseinate and converts it into peptones. Proteoses are a group of compounds formed during protein digestion and they are more complex than peptones. Certain other cells in the wall of the stomach produce bicarbonate, a base, to buffer the acidic contents of the stomach. They also prevent too much acidity in the stomach. The mucus produced by the wall of the stomach forms a physical barrier to prevent HCl from damaging the wall of the stomach.

iii) Digestion in the small intestine :
Various types of movements are generated by the muscularis external layer of the small intestine. These movements help in thorough mixing up of the food with bile, pancreatic juice and intestinal juice in the intestine and thereby facilitate digestion. The mucus along with the bicarbonates from pancreas protects the intestinal mucosa from the acidic medium and provides an alkaline medium (pH 7.8) for enzymatic activities. The duodenal cells of the proximal part also produce large amounts of bicarbonate to completely neutralize any gastric acid that passes further down into the digestive tract. All the enzymes of the pancreatic juice and succus entericus (mentioned above) act only in alkaline medium.

i) Digestion of proteins :
Trypsinogen, chymotrypsinogen and procarboxy peptidases are inactive enzymes. Trypsinogen is activated by the enzyme, enterokinase, secreted by the intestinal mucosa into active trypsin, which in turn activates the other enzymes in the pancreatic juice. Trypsin itself can similarly activate trypsinogen into trypsin (autocatalysis).

Proteins, proteoses and peptones (partly hydrolysed proteins) in the chyme, reaching the intestine are acted upon by the proteolytic enzymes of the pancreatic juice and intestinal juice as shown below :

Thus the end products of digestion of proteins namely amino acids are formed in the small intestine.

ii) Digestion of fats:
Bile salts of the bile help in the emulsification of fats i.e., break down of fats into very small micelles. Bile also activates lipases of pancreatic juice (steapsin) and intestinal lipases. These lipases act on emulsified fats and convert them into fatty acids and glycerols.

iii) Digestion of Carbohydrates :
Carbohydrates in the chyme are hydrolysed by the pancreatic amylase into disaccharides. Intestinal disaccharidases act on the ‘disaccharides’ and convert them into monosaccharides.

iv) Digestion of nucleic acids :
Nucleases (DNAase, RNAase) of the pancreatic juice act on the nucleic acids to form nucleotides and nucleosides. Nucleotidases and nucleosidases of the intestinal juice convert the nucleotides and nucleosides into pentose sugars and nitrogen bases.

Question 2.
Explain the digestive system of man with neat labelled diagram.
Answer:
Digestive System :
Human digestive system consists of the alimentary canal and the associated glands.

Alimentary canal / digestive tract :
The alimentary canal of man begins with the anterior opening, the mouth and ends with the posterior opening, the anus. Parts of the alimentary canal between the mouth and the anus include buccal cavity, pharynx, oesophagus, stomach, small intestine and the large intestine in the given order.

I. Mouth and Buccal (oral) cavity :
The mouth, bordered by the movable upper and lower lips (labia), leads into the buccal or oral cavity. The palate separates the ventral buccal cavity from the dorsal nasal chamber and facilitates chewing and breathing simultaneously. The anterior bony hard palate is lined by palatine rugae. The posterior soft palate that hangs down into the pharynx is called uvula. The jawbones bear four kinds of teeth and a tongue occurs at the base of the buccal cavity.

i) Teeth :
These are ecto-mesodermal in origin. Teeth of human beings are embedded in the sockets of the jaw bones – hence called thecodont. Majority of mammals including human beings form two sets of teeth during their life time, a set of temporary/milk or deciduous teeth replaced by a set of permanent or adult teeth. This type of dentition is called diphyodont dentition. An adult human has 32 permanent teeth, which are of four different types namely, incisors (I), canines (C), premolars (PM), and molars (M), and such a type of dentition is called heterodont dentition. The arrangement of different types of teeth in each half of both the jaws in the order I, C, PM, M is represented by the dental formula. In adult humans it is \(\frac{2123}{2123}\) = 32. The dental formula of the ‘milk dentition’ of a baby is \(\frac{2102}{2102}\) = 20 teeth. The third molar teeth appear very late (usually at about 21 years of age) and are called wisdom teeth. Incisors, the ‘chisel shaped’ teeth are useful in cutting, canines, the dagger like teeth help in tearing, premolars and molars, the cheekteeth, help in grinding the food.

ii) Tongue :
It is a freely movable, muscular sense organ, attached to the floor of the oral cavity by a fold of tissue called frenulum. The upper surface of the tongue has small projections called papillae, some of which bear taste buds. In humans the tongue bears 3 types of papillae namely 1. fungi form (at the anterior margin and tip of tongue), 2. filiform (on the surface of the tongue) and 3. circumvallate papillae (on the posterior surface / base of the tongue). The tongue acts as ‘universal tooth brush’, and helps in mixing saliva with food, taste detection, deglutition and speaking.

II. Pharynx :
The oral cavity leads into a short pharynx which serves as a common passage for food and air. It is divided into nasopharynx (lies above the soft palate), oropharynx (the middle part) and laryngopharynx (the lower part) by the soft palate. Oesophagus and trachea open into the laryngopharynx. The trachea opens into the laryngopharynx through the glottis. A cartilaginous flap called epiglottis prevents the entry of food into glottis during swallowing. Pharynx possesses voluntary muscles to assist in swallowing. Tonsils (lymphoid tissues) present in the pharynx, include i) pharyngeal tonsil or adenoids, ii) a pair of palatine tonsils and iii) a pair of lingual tonsils.

III. Oesophagus :
The oesophagus is a thin long tube which extends posteriorly, passing through the neck, thorax and diaphragm and it finally leads into the stomach. A muscular sphincter (gastro – esophageal / cardiac sphincter) regulates the opening of the oesophagus into the stomach.

IV. Stomach :
The stomach is a wide, J – shaped, distensible muscular bag like structure, located in the upper left portion of the abdominal cavity just below the diaphragm. It has three major parts, an anterior cardiac portion into which the oesophagus opens, a middle large fundic region (main body) and a posterior pyloric portion which opens into the first part of the small intestine through the pyloric aperture which is guarded by the pyloric sphincter.

V. Small intestine :
The small intestine is the longest part of the alimentary canal. It is distinguished serially into three regions namely proximal duodenum, middle long coiled jejunum and distal highly coiled ileum. Duodenum receives the hepato – pancreatic duct. Ileum opens into the large intestine.

VI. Large Intestine :
It consists of caecum/colon and rectum. Caecum is a small blind sac which hosts some symbiotic microorganisms. A narrow finger – like tubular projection, the vermiform appendix (abdominal tonsil) which is a vestigial organ, arises from the caecum. The caecum opens into the colon which is divided into – an ascending, a transverse, a descending parts and a sigmoid colon that continues behind into the rectum. Colon shows the external bulged out pouches called haustra. Rectum is a small dilated sac which leads into anal canal that opens out through the anus.

It is guarded by an internal anal sphincter formed by ‘smooth muscle’ and external anal sphincter formed by a ring of voluntary, striped muscle. There is no significant digestive activity in the large intestine. It is concerned with the absorption of some water, minerals and certain drugs. It secretes mucus which helps in keeping the undigested particles together and lubricating its passage to the exterior.

Digestive glands :
The digestive glands present in the wall of the alimentary canal are gastric glands, Brunner’s glands, and crypts of Lieberkuhn. The salivary glands, liver and pancreas are the digestive glands associated with the gut (extra alimentary canal glands).

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