NCERT Solutions For Class 11 Biology Chapter 19-Chemical Coordination and Integration

Question 19.1. Define the following:
(a) Exocrine gland
(b) Endocrine gland
(c) Hormone

Solution:

(a) Exocrine gland is a gland that pours its secretion on the surface or into a particular region by means of ducts for performing a metabolic activity, e.g., sebaceous glands, sweat glands, salivary glands and intestinal glands.

(b) Endocrine gland is an isolated gland (separates even from epithelium forming it) which secretes informational molecules or hormones that are poured into venous blood or lymph for reaching the target organ because the gland is not connected with the target organ by any duct. Therefore endocrine gland is also called ductless gland e.g., thyroid gland.

(c) Hormone is a substance that is manufactured and secreted in very small quantities into the blood stream by an endocrine gland or a specialized nerve cell and regulates the growth or functioning of a specific tissue organ in a distant part of the body e.g., insulin.

Question 19.2. Diagrammatically indicate the location of the various endocrine glands in our body.

Solution:

Question 19.3. List the hormones secreted by the following:
(a) Hypothalamus
(b) Pituitary
(c) Thyroid
(d) Parathyroid
(e) Adrenal
(f) Pancreas
(g) Testis
(h) Ovary
(i) Thymus
(j) Atrium
(k) Kidney
(l) G.I. Tract

Solution:

(a) Two types of hormones are produced by hypothalamus: releasing hormones (that stimulate secretion of pituitary hormones) and inhibiting hormones (that inhibit secretion of pituitary hormones). These hormones are:
– Thyrotrophin-releasing hormone
– Adrenocorticotrophin-releasing hormone
– Follicle-stimulating hormone-releasing hormone
– Luteinizing hormone-releasing hormone
– Growth hormone-releasing hormone
– Growth inhibiting hormone
– Prolactin releasing hormone
– Prolactin inhibiting hormone
– Melanocyte stimulating hormone-releasing hormone
– Melanocyte stimulating hormone-inhibiting hormone.

(b) Different parts of pituitary secrete different hormones.
– Hormones secreted by anterior lobe of pituitary are: Follicle stimulating hormone ($\text{FSH}$), Luteinizing hormone ($\text{LH}$), Thyroid stimulating hormone ($\text{TSH}$), Adrenocorticotrophic hormone ($\text{ACTH}$), Somatotrophic or Growth hormone ($\text{GH}$), Prolactin hormone ($\text{PRL}$) or Luteotrophic hormone.
– Middle (intermediate) lobe of pituitary: Melanocyte stimulating hormone ($\text{MSH}$).
– Posterior lobe of pituitary: (i) Oxytocin (ii) Vasopressin or antidiuretic hormone ($\text{ADH}$).

(c) Thyroid secretes 3 hormones:
– Thyroxine or tetraiodothyronine ($\text{T}_4$)
– Triiodothyronine ($\text{T}_3$)
– Calcitonin.

(d) Parathyroid gland secretes a single hormone called parathormone ($\text{PTH}$) or Collip’s hormone.

(e) Adrenal glands have two regions, namely, outer adrenal cortex and inner adrenal medulla. Both these regions secrete different hormones.
– Hormones of adrenal cortex are grouped into three categories: Glucocorticoids (e.g., cortisol), Mineralocorticoids (e.g., aldosterone), Sexcorticoids (e.g., testosterone).
– Adrenal medulla secretes two hormones: (i) Epinephrine (adrenaline) (ii) Nor-epinephrine (nor-adrenaline).

(f) Pancreas secretes following hormones:
– Insulin
– Glucagon
– Somatostatin.

(g) Testis secretes androgens such as testosterone.

(h) Ovary secretes:
– Estrogens such as estradiol
– Progesterone
– Relaxin.

(i) Thymus secretes thymosin hormone.

(j) Atrium secretes atrial natriuretic factor ($\text{ANF}$).

(k) Kidney secretes: (i) Renin (ii) Erythropoietin.

(l) G.I. tract secretes: Gastrin, Secretin, Cholecystokinin ($\text{CCK}$), Enterocrinin, Duocrinin, Villikinin.

Question 19.4. Fill in the blanks:
Hormones                                               Target gland
(a) Hypothalamic hormones                  ………………..
(b) Thyrotrophin (TSH)                        ………………..
(c) Corticotrophin (ACTH)                    ………………..
(d) Gonadotrophins (LH, FSH)             ………………..
(e) Melanotrophin (MSH)                    ………………..

Solution:

(a) Pituitary
(b) Thyroid
(c) Adrenal cortex
(d) Gonads – Testes in male and ovaries in female
(e) Skin.

Question 19.5. Write short notes on the functions of the following hormones:
(a) Parathyroid hormones (PTH)
(b) Thyroid hormones
(c) Thymosin
(d) Androgens
(e) Estrogens
(f) Insulin and Glucagon

Solution:

(a) Parathyroid hormone increases the level of calcium and decreases the level of phosphate in the blood.

(b) Thyroid gland secretes three hormones: thyroxine, triiodothyronine and calcitonin. Thyroxine and triiodothyronine control the general metabolism of the body, promote growth of body tissues and stimulate tissue differentiation. Calcitonin regulates the concentration of calcium in the blood.

(c) Thymosin is secreted by thymus. It accelerates cell division, stimulates the development and differentiation of T-lymphocytes and also hastens attainment of sexual maturity.

(d) Androgens are secreted by testis. They stimulate the development of male reproductive system, formation of sperms, development of male accessory sex characters and also determine the male sexual behaviour and the sex urge.

(e) Estrogens are secreted by ovaries. They stimulate the female reproductive tract to grow to full size and become functional, differentiation of ova and development of accessory sex characters.

(f) Insulin is secreted by the $\beta$-cells of the pancreas. It lowers blood glucose level, and promotes synthesis of proteins and fats. Glucagon is secreted by the $\alpha$-cells of the pancreas. It increases the level of glucose in the blood.

Question 19.6. Give example(s) of
(a) Hyperglycemic hormone and hypoglycemic hormone
(b) Hypercalcemic hormone
(c) Gonadotrophic hormones
(d) Progestational hormone
(e) Blood pressure lowering hormone
(f) Androgens and estrogens

Solution:

(a) Glucagon, Insulin

(b) Parathormone ($\text{PTH}$)

(c) Follicle stimulating hormone ($\text{FSH}$) and Luteinizing hormone ($\text{LH}$)

(d) Progesterone

(e) Atrial natriuretic factor ($\text{ANF}$)

(f) Testosterone and Estradiol.

Question 19.7. Which hormonal deficiency is responsible for the following:
(a) Diabetes mellitus
(b) Goitre
(c) Cretinism

Solution:

(a) Insulin

(b) Thyroxine and Triiodothyronine (due to iodine deficiency leading to undersecretion)

(c) Thyroxine and Triiodothyronine (congenital deficiency).

Question 19.8. Briefly mention the mechanism of action of FSH.

Solution:

Follicle stimulating hormone being glycoprotein is insoluble in lipids, therefore, cannot enter the target cells. It binds to the specific receptor molecules located on the surface of the cell membrane to form hormone-receptor complex. This complex causes the release of an enzyme adenylate cyclase from the receptor site. This enzyme forms the cell cyclic adenosine monophosphate ($\text{cAMP}$) from $\text{ATP}$. The $\text{cAMP}$ activates the existing enzyme system of the cell. This accelerates the metabolic reactions in the cell. The hormone is called the first messenger and the $\text{cAMP}$ is termed the second messenger. The hormone-receptor complex changes the permeability of the cell membrane to facilitate the passage of materials through it. This increases the activities of the cell as it receives the desired materials.

Question 19.9. Match the following:

Solution:

(a) – (ii); (b) – (iv); (c) – (i); (d) – (iii)

Question 18.1. Briefly describe the structure of the following:
(a) Brain (b) Eye (c) Ear

Solution:

(a) Brain: The brain acts as control and command system of the body. It is protected by skull and is covered by three meninges. It is divisible into three main regions: forebrain, midbrain and hindbrain.
(i) Forebrain – It consists of three regions:
– Olfactory lobes: These are a pair of very small, solid club-shaped bodies which are widely separated from each other. They are fully covered by cerebral hemispheres.
– Cerebrum – It is the largest and most complex of all the parts of human brain. A deep cleft divides the cerebrum into right and left cerebral hemispheres, connected by myelinated fibres, the corpus callosum.
– Diencephalon – It encloses a slit-like cavity, the third ventricle. The thin roof of this cavity is known as the epithalamus, the thick right and left sides as the thalami, and floor as the hypothalamus.
(ii) Midbrain – It is located between thalamus/hypothalamus of forebrain and pons of hindbrain. Its upper surface has two pairs of rounded protrusions called corpora quadrigemina and two bundles of fibres called crura cerebri.
(iii) Hindbrain – It consists of:
– Cerebellum – The second largest part of the human brain is the cerebellum. It consists of two lateral cerebellar hemispheres and central worm-shaped part, the vermis.
– Pons varolii – An oval mass lying above the medulla oblongata, consisting mainly of nerve fibres which interconnect different regions of the brain.
– Medulla oblongata – It extends from the pons varolii above and is continuous with the spinal cord below. The midbrain, pons varolii and medulla oblongata are collectively called brain stem.

(b) Eye: Eye is a hollow spherical structure composed of three coats:
(i) Fibrous coat: It has two distinct regions – sclera and cornea. Sclera covers most of the eyeball and contains collagen fibres. Cornea is a transparent portion that forms the anterior one-sixth of the eyeball.
(ii) Vascular coat: Comprises choroid, iris, and ciliary body.
(iii) Nervous coat: Consists of retina which is the neural and sensory layer containing ganglion cells, bipolar cells and photoreceptor cells (rods and cones).

(c) Ear: There are three portions in an ear:
(i) External ear: Consists of pinna and external auditory canal or meatus.
(ii) Middle ear: Consists of 3 small bones called ear ossicles – malleus, incus and stapes.
(iii) Internal ear: Consists of bony and membranous labyrinth, including three semicircular ducts, utricle, saccule and cochlea.

Question 18.2. Compare the following:
(a) Central neural system (CNS) and Peripheral neural system (PNS).
(b) Resting potential and action potential.
(c) Choroid and retina.

Solution:

(a) CNS and PNS:
– CNS lies along the mid-dorsal axis of the body, comprising the brain and spinal cord, serving as the center of information processing and control.
– PNS consists of nerves arising from the CNS (cranial and spinal nerves) that carry information to and from the CNS.

(b) Resting potential and action potential:
– Resting potential is the potential difference across the plasma membrane of a resting nerve fibre, where the outside is positively charged relative to the inside (electrically polarized).
– Action potential is the reversal of polarity across the plasma membrane (depolarization) during the excited state when the inside becomes positively charged relative to the outside.

(c) Choroid and retina:
– Choroid lies adjacent to the sclera, contains numerous blood vessels to supply nutrients and oxygen, and is pigmented dark brown.
– Retina is the inner neural and sensory layer of the eyeball containing photoreceptor cells (rods and cones), bipolar cells, and ganglion cells.

Question 18.3. Explain the following processes:
(a) Polarisation of the membrane of a nerve fibre.
(b) Depolarisation of the membrane of a nerve fibre.
(c) Conduction of a nerve impulse along a nerve fibre.
(d) Transmission of a nerve impulse across a chemical synapse.

Solution:

(a) Polarisation: In resting nerve fibres, $\text{Na}^+$ ions predominate outside while $\text{K}^+$ ions predominate inside. The differential ion permeability and active transport create a positive charge outside and a negative charge inside the membrane, making it polarized.

(b) Depolarisation: When a stimulus is applied, $\text{Na}^+$ channels open, causing $\text{Na}^+$ ions to rush into the axon. This reverses the polarity (inside becomes positive relative to outside), resulting in depolarisation.

(c) Conduction of a nerve impulse: An action potential generated at one spot on the nerve fibre is self-generated and propagated as a wave along the entire length of the axon.

(d) Transmission across a chemical synapse: When an action potential reaches the axon terminal, synaptic vesicles fuse with the presynaptic membrane and release neurotransmitters into the synaptic cleft. These neurotransmitters bind to specific receptors on the postsynaptic membrane, opening ion channels and generating a new potential in the postsynaptic neuron.

Question 18.4. Draw labelled diagrams of the following:
(a) Neuron (b) Brain
(c) Eye (d) Ear

Solution:

a.

b.

c.

d.

Question 18.5. Write short notes on the following:
(a) Neural coordination (b) Forebrain
(c) Midbrain (d) Hindbrain
(e) Retina (f) Ear ossicles
(g) Cochlea (h) Organ of Corti
(i) Synapse

Solution:

(a) Neural coordination: The working together of various organs of a multicellular organism in a systematic manner to complement each other’s functions, achieved through sensory input, integration, and motor output.

(b) Forebrain: Consists of olfactory lobes, cerebrum (divided into hemispheres connected by corpus callosum), and hypothalamus/thalamus.

(c) Midbrain: Located between forebrain and hindbrain, contains corpora quadrigemina on its dorsal portion.

(d) Hindbrain: Comprises pons, cerebellum, and medulla oblongata, controlling vital reflexes and balance.

(e) Retina: Inner layer of the eye containing photoreceptor cells (rods for twilight vision, cones for daylight and colour vision).

(f) Ear ossicles: Chain of three bones in the middle ear – malleus, incus, and stapes – that amplify sound wave transmission.

(g) Cochlea: Spirally coiled structure of the inner ear responsible for hearing.

(h) Organ of Corti: Located on the basilar membrane inside the cochlea, containing hair cells that act as auditory receptors.

(i) Synapse: Junction between the axon of one neuron and the dendrite/cyton of another neuron for impulse transmission.

Question 18.6. Give a brief account of
(a) Mechanism of synaptic transmission.
(b) Mechanism of vision.
(c) Mechanism of hearing.

Solution:

(a) Refer to Solution 18.3(d).

(b) Mechanism of vision: Light rays focus on the retina, inducing dissociation of retinal from opsin, which alters membrane permeability and generates receptor potentials. This triggers action potentials in ganglion cells via bipolar cells, which are transmitted via optic nerves to the visual cortex.

(c) Mechanism of hearing: Sound waves strike the tympanum, producing vibrations transmitted via ear ossicles to the oval window, creating fluid waves in the cochlea. These waves stimulate hair cells in the organ of Corti against the tectorial membrane, generating nerve impulses transmitted via auditory nerves to the brain.

Question 18.7. Answer briefly.
(a) How do you perceive the colour of an object?
(b) Which part of our body helps us in maintaining the body balance?
(c) How does the eye regulate the amount of light that falls on the retina?

Solution:

(a) Through three types of cone cells in the retina containing photopigments sensitive to red, green, and blue light. Equal stimulation produces white light sensation, while combinations produce various colours.

(b) Ears (specifically cristae and maculae in the vestibular apparatus of the inner ear).

(c) By the iris, which contains circular sphincter muscles (making the pupil smaller in bright light) and radial dilator muscles (widening the pupil in dim light).

Question 18.8. Explain the following.
(a) Role of $\text{Na}^+$ in the generation of action potential.
(b) Mechanism of generation of light-induced impulse in the retina.
(c) Mechanism through which a sound produces a nerve impulse in the inner ear.

Solution:

(a) $\text{Na}^+$ ions rush into the axon upon stimulus, decreasing membrane potential, reversing polarity (depolarization), and peaking at $+40$ to $+50\text{ mV}$ to form the action potential.

(b) Refer to Solution 18.6(b).

(c) Refer to Solution 18.6(c).

Question 18.9. Differentiate between
(a) Myelinated and non-myelinated axons
(b) Dendrites and axons
(c) Rods and cones
(d) Thalamus and Hypothalamus
(e) Cerebrum and Cerebellum

Solution:

(a) Myelinated vs Non-myelinated axons:

(b) Dendrites vs Axons:

(c) Rods vs Cones:

(d) Thalamus vs Hypothalamus:

(e) Cerebrum vs Cerebellum:

Question 18.10. Answer the following.
(a) Which part of the ear determines the pitch of a sound?
(b) Which part of the human brain is the most developed?
(c) Which part of our central neural system acts as a master clock?

Solution:

(a) The receptor cells in the organ of Corti (internal ear).

(b) Cerebrum (cerebral hemispheres).

(c) Pineal gland (associated with diencephalon of forebrain) acts as a master clock maintaining biological rhythms.

Question 18.11. The region of the vertebrate eye, where the optic nerve passes out of the retina, is called the
(a) fovea (b) iris
(c) blind spot (d) optic chiasma

Solution:

(c) blind spot

Question 18.12. Distinguish between
(a) Afferent neurons and efferent neurons
(b) Impulse conduction in myelinated nerve fibre and unmyelinated nerve fibre
(c) Aqueous humour and vitreous humour
(d) Blind spot and yellow spot
(e) Cranial nerves and spinal nerves

Solution:

(a) Afferent vs Efferent neurons:

(b) Refer to Solution 18.9(a).

(c) Aqueous vs Vitreous humour:

(d) Blind spot vs Yellow spot (Fovea):

(e) Cranial vs Spinal nerves:

Question 17.1. Draw the diagram of a sarcomere of skeletal muscle showing different regions.

Solution:

Question 17.2. Define sliding filament theory of muscle contraction.

Solution:

According to sliding filament theory of muscle contraction, the actin and myosin filaments slide past each other with the help of cross-bridges to reduce the length of the sarcomeres.

Question 17.3. Describe the important steps in muscle contraction.

Solution:

Mechanism of muscle contraction is explained by sliding filament theory which states that contraction of a muscle fibre takes place by the sliding of the thin filaments over the thick filaments. As a nerve impulse reaches the terminal end of the axon, synaptic vesicles fuse with the axon membrane and release a chemical transmitter, acetylcholine and binds to receptor sites of the motor end plate. When depolarisation of the motor end plate reaches a certain level, it creates an action potential. An action potential (impulse) passes from the motor end plate over the sarcolemma and then into the T-tubules and sarcoplasmic reticulum and stimulates the sarcoplasmic reticulum to release calcium ions into the sarcoplasm. The calcium ions bind to troponin causing a change in its shape and position. This in turn alters shape and the position of tropomyosin, to which troponin binds. This shift exposes the active sites on the F-actin molecules. Myosin cross-bridges are then able to bind to these active sites. The heads of myosin molecules project laterally from thick myofilaments towards the surrounding thin myofilaments. These heads are called cross bridges. The head of each myosin molecule contains an enzyme myosin ATPase. In the presence of myosin ATPase, $\text{Ca}^{2+}$ and $\text{Mg}^{2+}$ ions, $\text{ATP}$ breaks down into $\text{ADP}$ and inorganic phosphate, releasing energy in the head.

Energy from $\text{ATP}$ causes energised myosin cross bridges to bind to actin.

The energised cross-bridges move, causing thin myofilaments to slide along the thick myofilaments.

Question 17.4. Write true or false. If false change the statement so that it is true.
(a) Actin is present in thin filament.
(b) H-zone of striated muscle fibre represents both thick and thin filaments.
(c) Human skeleton has 206 bones.
(d) There are 11 pairs of ribs in man.
(e) Sternum is present on the ventral side of the body.

Solution:

(a) True

(b) False – H-Zone of striated muscle fibres represents only thick filaments.

(c) True

(d) False – There are 12 pairs of ribs in man.

(e) True

Question 17.5. Write the differences between:
(a) Actin and Myosin
(b) Red and White muscles
(c) Pectoral and Pelvic girdle

Solution:

(a) Actin filaments and myosin filaments can be differentiated as follows:

(b) Differences between red muscle fibres and white muscle fibres:

(c) Differences between pectoral and pelvic girdles:

Question 17.6. Match Column I with Column II:

Solution:

(a) – (iv), (b) – (ii), (c) – (i), (d) – (iii)

Question 17.7. What are the different types of movements exhibited by the cells of human body?

Solution:

The cells of human body show three types of movements: amoeboid, ciliary and muscular.

Amoeboid movements: These are found in leucocytes of blood and phagocytes of certain body organs. In such cells, movements are brought with the help of temporary finger-like cytoplasmic projections, called pseudopodia or false feet. So it is also called pseudopodial movement. These pseudopodia are formed by flow of cytoplasm, called cyclosis, and cytoskeletal structures like microfilaments.

Ciliary movements: Large number of our internal tubular organs are lined by ciliated epithelium. For instance, the cilia of the cells lining the trachea, oviducts and vasa efferentia propel dust particles, eggs and sperms respectively by their coordinated movements in specific directions in these organs.

Muscular movements: These are brought about by the action of skeleton, joints and muscles. These are of two types: movements of body parts and locomotion.

Question 17.8. How do you distinguish between a skeletal muscle and a cardiac muscle?

Solution:

Question 17.9. Name the type of joint between the following:
(a) atlas/axis
(b) carpal/metacarpal of thumb
(c) between phalanges
(d) femur/acetabulum
(e) between cranial bones
(f) between pubic bones in the pelvic girdle

Solution:

(a) Pivot joint

(b) Saddle joint

(c) Hinge joint

(d) Ball and socket joint

(e) Fibrous joint (sutures)

(f) Cartilaginous joint (pubic symphysis)

Question 17.10. Fill in the blank spaces:
(a) All mammals (except a few) have ……. cervical vertebra.
(b) The number of phalanges in each limb of human is …….
(c) Thin filament of myofibril contains two ‘F’ actins and two other proteins namely ……. and …….
(d) In a muscle fibre $\text{Ca}^{2+}$ is stored in …….
(e) ……. and ……. pairs of ribs are called floating ribs.
(f) The human cranium is made of ……. bones.

Solution:

(a) 7

(b) 14

(c) tropomyosin, troponin

(d) sarcoplasmic reticulum

(e) $11^{\text{th}}$ and $12^{\text{th}}$

(f) 8

Question 16.1. Define Glomerular Filtration Rate (GFR).

Solution:

The amount of filtrate formed by the kidneys per minute is called glomerular filtration rate ($\text{GFR}$). It is approximately $125\text{ mL/min}$ in a healthy person.

Question 16.2. Explain the autoregulatory mechanism of GFR.

Solution:

The kidneys have built-in mechanisms for the regulation of glomerular filtration rate. One such efficient mechanism is carried out by juxta glomerular apparatus ($\text{JGA}$). $\text{JGA}$ is a special sensitive region formed by cellular modifications in the distal convoluted tubule and the afferent arteriole at the location of their contact. A fall in $\text{GFR}$ can activate the JG cells to release renin which can stimulate the glomerular blood flow and thereby the $\text{GFR}$ back to normal.

Question 16.3. Indicate whether the following statements are true or false.
(a) Micturition is carried out by a reflex.
(b) ADH helps in water elimination, making the urine hypotonic.
(c) Protein-free fluid is filtered from blood plasma into the Bowman’s capsule.
(d) Henle’s loop plays an important role in concentrating the urine.
(e) Glucose is actively reabsorbed in the proximal convoluted tubule.

Solution:

(a) True
(b) False
(c) True
(d) True
(e) True

Question 16.4. Give a brief account of the counter current mechanism.

Solution:

The kidneys have a special mechanism for concentrating the urine, it is called counter current mechanism. The mechanism is said to be a counter current mechanism because the outflow (in the ascending limb) of Henle’s loop runs parallel to and in the opposite direction of the inflow (in the descending limb) and vasa recta. As the mechanism begins to function, the ascending limb of loop of Henle actively transports chloride and sodium ions out into the vasa recta from where it is secreted into the interstitial fluid. As a result the interstitial fluid around the loop of Henle contains large quantities of $\text{NaCl}$. The filtrate passes from the ascending limb of loop of Henle and enters a collecting duct. The collecting duct passes adjacent to the loop of Henle where the interstitial fluid contains large amounts of $\text{NaCl}$. The high osmotic pressure created by $\text{NaCl}$ causes water to diffuse out of the collecting duct in the interstitial fluid and eventually to the blood of vasa recta. The filtrate becomes greatly concentrated and is now called urine. A similar counter current mechanism operates between the interstitial fluid and blood passing through the vasa recta. As the blood capillary runs along the ascending limb of loop of Henle, $\text{NaCl}$ diffuses out of the blood. The direction is reversed as the blood capillary passes along the descending limb of Henle. The blood flows in the vasa recta around the loop of Henle from ascending to the descending side while the fluid passing through the loop of Henle goes in the opposite direction. The arrangement helps to maintain the concentration gradient of $\text{NaCl}$.

The overall function of counter current mechanism is to concentrate sodium chloride in the interstitial fluid and thereby cause water to diffuse out of the collecting ducts and concentrate the urine.

Question 16.5. Describe the role of liver, lungs and skin in excretion.

Solution:

Other than the kidneys, lungs, liver and skin also help in the elimination of excretory wastes. Lungs remove large amounts of $\text{CO}_2$ ($18\text{ litres/day}$) and also significant quantities of water every day. Liver secretes bile which contains substances like bilirubin, biliverdin, cholesterol, degraded steroid hormones, vitamins and drugs. Most of these substances ultimately pass out along with digestive wastes. The sweat and sebaceous glands in the skin can eliminate certain substances through their secretions. Sweat produced by the sweat glands is a watery fluid containing $\text{NaCl}$, small amounts of urea, lactic acid etc. Sebaceous glands eliminate certain substances like sterols, hydrocarbons and waxes through sebum.

Question 16.6. Explain micturition.

Solution:

The process of passing out urine from the urinary bladder is called micturition. Urine formed by the nephrons is ultimately carried to the urinary bladder where it is stored. This causes stretching of the wall of bladder that leads to the stimulation of stretch receptors on the walls of the bladder. This sends signal 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.

Question 16.7. Match the items of column I with those of column II.

Solution:

(a) – (iii), (b) – (v), (c) – (iv), (d) – (i), (e) – (ii)

Question 16.8. What is meant by the term osmoregulation?

Solution:

The regulation of water and solute contents of the body fluids by the kidney is called osmoregulation.

Question 16.9. Terrestrial animals are generally either ureotelic or uricotelic, not ammonotelic, why?

Solution:

Ammonotelic animals are aquatic animals that excrete ammonia which is highly soluble in water, thus large amount of water is also excreted. Terrestrial animals cannot afford to lose such large quantities of water from their bodies as they live in environments having water scarcity. They, therefore, excrete either urea (ureotelic) or uric acid (uricotelic) as these are less soluble in water.

Question 16.10. What is the significance of juxta glomerular apparatus (JGA) in kidney function?

Solution:

Juxta glomerular apparatus ($\text{JGA}$) is a special sensitive region formed by cellular modifications in the distal convoluted tubule and the afferent arteriole at the location of their contact. The $\text{JGA}$ plays a complex regulatory role. A fall in glomerular blood flow / glomerular blood pressure / $\text{GFR}$ can activate the JG cells to release renin which converts angiotensinogen in blood to angiotensin I and further to angiotensin II. Angiotensin II, being a powerful vasoconstrictor, increases the glomerular blood pressure and thereby $\text{GFR}$. Angiotensin II also activates the adrenal cortex to release aldosterone. Aldosterone causes reabsorption of $\text{Na}^+$ and water from the distal parts of the tubule. This also leads to an increase in blood pressure and $\text{GFR}$.

Question 16.11. Name the following.
(a) A chordate animal having flame cells as excretory structures.
(b) Cortical portions projecting between the medullary pyramids in the human kidney.
(c) A loop of capillary running parallel to the Henle’s loop.

Solution:

(a) Cephalochordate – Amphioxus

(b) Columns of Bertini

(c) Vasa recta

Question 16.12. Fill in the gaps.
(a) Ascending limb of Henle’s loop is________to water whereas the descending limb is________to it.
(b) Reabsorption of water from distal parts of the tubules is facilitated by hormone________
(c) Dialysis fluid contains all the constituents as in plasma except________
(d) A healthy adult human excretes (on an average)________gm of urea/day.

Solution:

(a) Ascending limb of Henle’s loop is impermeable to water whereas the descending limb is permeable to it.

(b) Reabsorption of water from distal parts of the tubules is facilitated by hormone ADH.

(c) Dialysis fluid contains all the constituents as in plasma except nitrogenous wastes.

(d) A healthy adult human excretes (on an average) 25 – 30 gm of urea/day.

Question 15.1. Name the components of the formed elements in the blood and mention one major function of each of them.

Solution:

Blood corpuscles are the formed elements in the blood, they constitute $45\%$ of the blood. Formed elements are – erythrocytes ($\text{RBC}$s or red blood corpuscles), leucocytes ($\text{WBC}$s or white blood corpuscles) and thrombocytes or blood platelets. The major function of $\text{RBC}$s is to transport oxygen from lungs to body tissues and $\text{CO}_2$ from body tissues to the lungs. White blood cells provide immunity to the body. Blood platelets play important role in blood clotting.

Question 15.2. What is the importance of plasma proteins?

Solution:

Plasma proteins constitute about $7$ to $8\%$ of plasma. These mainly include albumin, globulin, prothrombin and fibrinogen. Prothrombin and fibrinogen are needed for blood clotting. Albumins and globulins retain water in blood plasma and helps in maintaining osmotic balance. Certain globulins

Question 15.3. Match Column I with Column II.

Solution:

(a) – (iii); (b) – (v); (c) – (ii); (d) – (i); (e) – (iv).

Question 15.4. Why do we consider blood as a connective tissue?

Solution:

A connective tissue connects different tissues or organs of the body. It consists of living cells and extracellular matrix. Blood is vascular connective tissue, it is a mobile tissue consisting of fluid matrix and free cells. Blood transports materials from one place to the other and thereby establishes connectivity between different body parts.

Question 15.5. What is the difference between lymph and blood?

Solution:

The differences between blood and lymph are given below:

Question 15.6. What is meant by double circulation? What is its significance?

Solution:

The type of blood circulation in which oxygenated blood and deoxygenated blood do not get mixed is termed double circulation. It includes systemic circulation and pulmonary circulation. The circulatory pathway of double circulation is given in the following flow chart.

Flow chart: Double blood circulation Double circulation or separation of systemic and pulmonary circulations provides a higher metabolic rate to the body and also allows the two circulations to have different blood pressures according to the need of the organs they supply.

Question 15.7. Write the differences between:
(a) Blood and lymph
(b) Open and closed system of circulation
(c) Systole and diastole
(d) P-wave and T-wave

Solution:

(a) Refer answer 5.

(b) The differences between open and closed circulatory system are given below:

(c) Systole is contraction of heart chambers in order to pump out blood while diastole is relaxation of heart chambers to receive blood. The contraction of a chamber or systole decreases its volume and forces the blood out of it, whereas its relaxation or diastole brings it back to its original size to receive more blood.

(d) P wave is a small upward wave of electrocardiograph that indicates the atrial depolarisation (contraction of atria). It is caused by the activation of SA node. T-wave is a dome shaped wave of electrocardiograph which represents ventricular repolarisation (ventricular relaxation).

Question 15.8. Describe the evolutionary change in the pattern of heart among the vertebrates.

Solution:

Vertebrates have a single heart. It is a hollow, muscular organ composed of cardiac muscle fibres. Two types of chambers in heart are atria and ventricles. The heart of lower vertebrates have additional chambers, namely sinus venosus and conus arteriosus or bulbus arteriosus or truncus arteriosus. During the course of development, in higher vertebrates, the persistent portions viz, auricles and ventricles are retained. However, these get complicated by incorporating several valves inside them and becoming compartmentalised.

In fishes, heart is two chambered ($1$ auricle and $1$ ventricle). Both the accessory chambers, sinus venosus and conus arteriosus are present. The heart pumps out deoxygenated blood which is oxygenated by the gills and sent to the body parts from where deoxygenated blood is carried to the heart. It is called single circulation and heart is called venous heart. Lung fish, amphibians and reptiles have three chambered heart, ($2$ auricles and $1$ ventricle). The left atrium gets oxygenated blood from the gills/lungs/skin/buccopharyngeal cavity and the right atrium receives the deoxygenated blood from other body parts. But both oxygenated and deoxygenated blood get mixed up in single ventricle which pumps out mixed blood. This is called incomplete double circulation.

Crocodiles, birds and mammals have a complete four chambered heart (right and left auricles; right and left ventricles). Oxygenated and deoxygenated blood never get mixed. Right parts of the heart receive deoxygenated blood from all other body parts and send it to lungs for oxygenation whereas left parts of heart receive oxygenated blood from lungs and send it to other body parts. This mode of circulation is termed as complete double circulation which includes systemic and pulmonary circulation. There are no accessory chambers in heart of birds and mammals.

Question 15.9. Why do we call our heart myogenic?

Solution:

The heart of molluscs and vertebrates including humans is myogenic. It means heart beat is initiated in heart itself by a patch of modified heart muscle called sino-atrial node or pacemaker which lies in the wall of the right atrium near the opening of the superior vena cava.

Question 15.10. Sino-atrial node is called the pacemaker of our heart. Why?

Solution:

Sino-atrial node ($\text{SAN}$) is a mass of neuromuscular tissue which lies in the wall of right atrium. It is responsible for initiating and maintaining the rhythmic contractile activity of the heart. Therefore, it is called the pacemaker.

Question 15.11. What is the significance of atrio-ventricular node and atrio-ventricular bundle in the functioning of heart?

Solution:

atrio-ventricular node ($\text{AVN}$) is a mass of neuromuscular tissue, which is situated in wall of right atrium, near the base of inter-atrial septum. AV node is the pacesetter of the heart, as it transmits the impulses initiated by SA node to all parts of ventricles. Atrio-ventricular bundle ($\text{A-V}$ bundle) or bundle of His is a mass of specialised fibres which originates from the AVN. Within the myocardium of the ventricles the branches of bundle of His divide into a network of fine fibres called Purkinje fibres. The bundle of His and the Purkinje fibres convey impulse of contraction from the AVN to the myocardium of the ventricles.

Question 15.12. Define a cardiac cycle and the cardiac output.

Solution:

The sequential events in the heart which are repeated cyclically is called cardiac cycle and it consists of systole (contraction) and diastole (relaxation) of both the atria and ventricles. The duration of a cardiac cycle is $0.8\text{ seconds}$. Periods of cardiac cycle are atrial systole ($0.1\text{ second}$), ventricular systole ($0.3\text{ second}$) and complete cardiac diastole ($0.4\text{ second}$).

The amount of blood pumped by heart per minute is called cardiac output. It is calculated by multiplying stroke volume (volume of blood pumped by each ventricle per minute) with heart rate (number of beats per minute). The heart of normal person beats $72$ times per minute and pumps out about $70\text{ mL}$ of blood per beat. Therefore, cardiac output averages $5000\text{ mL}$ or $5\text{ litres}$.

Question 15.13. Explain heart sounds.

Solution:

The beating of heart produces characteristic sounds which can be heard by using stethoscope. In a normal person, two sounds are produced per heart beat. The first heart sound ‘lubb’ is low pitched, not very loud and of long duration. It is caused partly by the closure of the bicuspid and tricuspid valves and partly by the contraction of muscles in the ventricles.

The second heart sound ‘dubb’ is high pitched, louder, sharper and shorter in duration. It is caused by the closure of the semilunar valves and marks the end of ventricular systole.

Question 15.14. Draw a standard ECG and explain the different segments in it.

Solution:

ECG is graphic record of the electric current produced by the excitation of the cardiac muscles. The instrument used to record the changes is an electrocardiograph. A normal electrogram ($\text{ECG}$) is composed of a P wave, a QRS wave (complex) and a T wave. The P Wave is a small upward wave that indicates the atrial depolarisation (contraction of atria). It is caused by the activation of SA node. The impulses of contraction start from the SAnode and spread throughout the artia.

The QRS Wave (complex) represents ventricular depolarisation (ventricular contraction). It is caused by the impulses of the contraction from AV node through the bundle of His and Purkinje fibres and the contraction of the ventricular muscles. Thus this wave is due to the spread of electrical impulse through the ventricles.

The T Wave represents ventricular repolarisation (ventricular relaxation). The potential generated by the recovery of the ventricle from the depolarisation state is called the repolarisation wave. The end of the T-wave marks the end of systole.

ECG gives accurate information about the heart. Therefore, ECG is of great diagnostic value in cardiac diseases.

Question 14.1. Define vital capacity. What is its significance?

Solution:

Vital capacity is defined as the maximum volume of air a person can breathe in after a forced expiration or the maximum volume of air a person can breathe out after a forced inspiration. It represents the maximum amount of air one can renew in the respiratory system in a single respiration. Thus, the greater the vital capacity the more is the energy available to the body.

Question 14.2. State the volume of air remaining in the lungs after normal breathing.

Solution:

When a person breathes normally, the amount which remains in the lung after normal expiration is called functional residual capacity. It is the sum of residual volume and the expiratory reserve volume ($\text{FRC} = \text{RV} + \text{ERV}$). It is about $2100\text{ – }2300\text{ mL}$ of air.

Question 14.3. Diffusion of gases occurs in the alveolar region only and not in the other parts of respiratory system. Why?

Solution:

For efficient exchange of gases, respiratory surface must have certain characteristics such as (i) it must be thin, moist and permeable to respiratory gases (ii) it must have large surface area, (iii) it must be highly vascular. Only alveolar region has these characteristics. Thus, diffusion of gases occurs in this region only.

Question 14.4. What are the major transport mechanisms for $\text{CO}_2$? Explain.

Solution:

Nearly $20\text{–}25$ percent of $\text{CO}_2$ is transported by haemoglobin of $\text{RBC}$s, $70$ percent of it is carried as bicarbonate ion in plasma and about $7$ percent of $\text{CO}_2$ is carried in a dissolved state through plasma. $\text{CO}_2$ is carried by haemoglobin as carbamino-haemoglobin. This binding is related to the partial pressure of $\text{CO}_2$.

Question 14.5. What will be the $p\text{O}_2$ and $p\text{CO}_2$ in the atmospheric air compared to those in the alveolar air?
(i) $p\text{O}_2$ lesser, $p\text{CO}_2$ higher
(ii) $p\text{O}_2$ higher, $p\text{CO}_2$ lesser
(iii) $p\text{O}_2$ higher, $p\text{CO}_2$ higher
(iv) $p\text{O}_2$ lesser, $p\text{CO}_2$ lesser

Solution:

(ii) Air that has entered the alveoli through the bronchioles is called alveolar air. It has a different partial pressure of $\text{CO}_2$ and $\text{O}_2$ compared to atmospheric air. Then, there occurs gaseous exchange between the adjacent blood capillaries and the alveoli. $\text{CO}_2$ diffuses from blood into the alveolar air and $\text{O}_2$ diffuses from alveolar air to the blood. As a result, alveolar air has higher $p\text{CO}_2$ and lesser $p\text{O}_2$ than the atmospheric air (meaning atmospheric air has higher $p\text{O}_2$ and lesser $p\text{CO}_2$).

Question 14.6. Explain the process of inspiration under normal conditions.

Solution:

Inspiration is a process by which fresh air enters the lungs. The diaphragm, intercostal muscles and abdominal muscles play an important role. The muscles of the diaphragm and external intercostal muscles are principal muscles of inspiration. Volume of thoracic cavity increases by contraction of diaphragm and external intercostal muscles. During inspiration, relaxation of abdominal muscles also occurs which allows compression of the abdominal organs by diaphragm. Thus, overall volume of the thoracic cavity increases and as a result, there is a decrease of the air pressure in the lungs. The greater pressure outside the body now causes air to flow rapidly into the lungs.

Question 14.7. How is respiration regulated?

Solution:

Respiration is under both nervous and chemical regulation.

The respiratory centre in the brain is composed of groups of neurons located in the medulla oblongata and pons varolii. The respiratory centre regulates the rate and depth of breathing.

– Dorsal respiratory group of neurons are located in the dorsal portion of the medulla oblongata. This group of neurons mainly causes inspiration.

– Ventral group of neurons are located in the ventrolateral part of the medulla oblongata. These can cause either inspiration or expiration.

– Pneumotaxic centre is located in the dorsal part of pons varolii. It sends signals to all the neurons of dorsal respiratory group and only to inspiratory neurons of ventral respiratory group. Its job is primarily to limit inspiration.

Chemically, respiration is regulated by the large numbers of chemoreceptors located in the carotid bodies and in the aortic bodies. Excess carbon dioxide or hydrogen ions mainly stimulate the respiratory centre of the brain and increases the inspiratory and expiratory signals to the respiratory muscles. Increased $\text{CO}_2$ lowers the pH resulting in acidosis. The role of oxygen in the regulation of respiratory rhythm is quite insignificant.

Question 14.8. What is the effect of $p\text{CO}_2$ on oxygen transport?

Solution:

Increase in $p\text{CO}_2$ tension in blood brings a rightward shift of the oxygen dissociation curve of haemoglobin thereby decreasing the affinity of haemoglobin for oxygen. This effect is called Bohr’s effect. It plays an important role in the release of oxygen in the tissues.

Question 14.9. What happens to the respiratory process in a man going up a hill?

Solution:

Rate of breathing will increase in order to supply sufficient oxygen to blood because air in mountainous regions is deficient in oxygen.

Question 14.10. What is the site of gaseous exchange in an insect?

Solution:

Tracheae (Tracheal respiration) is the site of gaseous exchange in an insect.

Question 14.11. Define oxygen dissociation curve. Can you suggest any reason for its sigmoidal pattern?

Solution:

The relationship between the partial pressure of oxygen ($p\text{O}_2$) and percentage saturation of the haemoglobin with oxygen ($\text{O}_2$) is graphically illustrated by a curve called oxygen haemoglobin dissociation curve (also called oxygen dissociation curve).

The sigmoidal pattern of oxygen haemoglobin dissociation curve is the result of two properties which play a significant role in the transport of oxygen. These two properties are:

(i) Minimal loss of oxygen from haemoglobin occurs above $p\text{O}_2$ of $70\text{–}80\text{ mm Hg}$ despite significant changes in tension of oxygen beyond this. This is depicted by the relatively flat portion of the curve.

(ii) Any further decline in $p\text{O}_2$ from $40\text{ mm Hg}$ causes a disproportionately greater release of oxygen from the haemoglobin. It results in the steeper portion of the curve and causes the curve to be sigmoid.

Question 14.12. Have you heard about hypoxia? Try to gather information about it, and discuss with your friends.

Solution:

Hypoxia is a condition of oxygen shortage in the tissues. It is of two types:

(i) Artificial hypoxia: It results from shortage of oxygen in the air as at high altitude. It causes mountain sickness characterised by breathlessness, headache, dizziness and bluish tinge on skin.

(ii) Anaemic hypoxia: It results from the reduced oxygen carrying capacity of the blood due to anaemia or carbon monoxide poisoning. In both cases, less haemoglobin is available for carrying $\text{O}_2$.

Question 14.13. Distinguish between
(a) IRV and ERV
(b) Inspiratory capacity and expiratory capacity.
(c) Vital capacity and total lung capacity.

Solution:

(a) Differences between IRV and ERV:

(b) Differences between inspiratory capacity and expiratory capacity:

(c) Differences between vital capacity and total lung capacity:

Question 14.14. What is tidal volume? Find out the tidal volume (approximate value) for a healthy human in an hour.

Solution:

Tidal volume is the volume of air inspired or expired with each normal breath. This is about $500\text{ mL}$ in an adult person. It is composed of about $350\text{ mL}$ of alveolar volume and about $150\text{ mL}$ of dead space volume. The alveolar volume consists of air that reaches the respiratory surfaces of the alveoli and engages in gas exchange. The dead space volume consists of air that does not reach the respiratory surfaces.

A healthy man can inspire or expire approximately $6000$ to $8000\text{ mL}$ of air per minute. Therefore, the volume of air respired by a healthy human in an hour is approximately $360\text{ L}$ to $480\text{ L}$ (or calculated via minute ventilation).

Question 13.1. Define growth, differentiation, development, dedifferentiation, redifferentiation, determinate growth, meristem and growth rate.

Solution:

Growth is defined as a vital process which brings about an irreversible and permanent change in the shape, size, form, weight and volume of a cell, organ or whole organism, accompanied with increase in dry matter.

Differentiation is a localised qualitative change in size, biochemistry, structure and function of cells, tissues or organs, e.g., fibre, vessel, tracheid, sieve tube, mesophyll, leaf etc. Thus it is a change in form and physiological activity. It results in specialisation for particular functions.

Development may be defined as a process which includes growth, differentiation and maturation in a regular sequence in the life history of a cell, organ or organism viz., seed germination, growth, differentiation, flowering, seed formation and senescence.

Dedifferentiation is the process by which the differentiated cells which have lost the ability to divide under certain circumstances, become meristematic and regain the divisibility.

Redifferentiation is defined as maturation or differentiation of dedifferentiated cells to form cells which are unable to divide e.g., secondary xylem elements, cork cells etc., are formed by redifferentiation of secondary cambial cells.

Determinate growth is the ability of a cell, tissue or the organism to grow for a limited period of time.

Meristem is a tissue consisting of unspecialised immature cells, possessing the power of continuous cell division and adding new cells to the body.

Growth rate is defined as the increased growth per unit time.

Question 13.2. Why is not any one parameter good enough to demonstrate growth throughout the life of a flowering plant?

Solution:

A flowering plant consists of a number of organs viz., roots, stem, leaves, flowers, fruits etc. growing differently under different stages of life cycle. These plant organs require different parameters to demonstrate their growth. In plant organs like fruits, bulbs, corms etc. fresh weight is used for measuring their growth. In case of fruits, increase in volume, diameter etc., are also used as other parameters for the measurement of their growth. For flat organs like leaves, increase in surface area is used as the parameter. Stem and roots primarily grow in length and then in girth, thus increase in length and diameter are used for measuring their growth. Consequently, the flowering plants exhibit several parameters to demonstrate growth.

Question 13.3. Describe briefly
(a) Arithmetic growth
(b) Geometric growth
(c) Sigmoid growth curve
(d) Absolute and relative growth rates

Solution:

(a) Arithmetic growth: If the length of a plant organ is plotted against time it shows a linear curve, the growth is called arithmetic growth. In this growth, the rate of growth is constant and increase in growth occurs in arithmetic progression e.g., length of a plant is measured as $2, 4, 6, 8, 10, 12\text{ cm}$ at a definite interval of $24\text{ hrs}$. It is found in root or shoot elongating at constant rate. Arithmetic growth is expressed as:

$$L_t = L_0 + rt$$

Here, $L_t = \text{length after time } t$, $L_0 = \text{length at the beginning}$, $r = \text{growth rate}$.

(b) Geometric growth: Geometric growth is the growth where both the progeny cells following mitosis retain the ability to divide and continue to do so. It occurs in many higher plants and in unicellular organisms when grown in nutrient-rich medium. Number of cells is initially small so that initial growth is slow which is called lag phase. Later on, there is rapid growth at exponential rate. It is called log or exponential phase.

(c) Sigmoid growth curve: Geometric growth cannot be sustained for long. Some cells die. Limited nutrient availability causes slowing down of growth. It leads to stationary phase. There may be actually a decline. Plotting the growth against time will give a typical sigmoid or S-curve.

S-curve of growth is typical of most living organisms in their natural environment. It also occurs in cells, tissues and organs of plants.

(d) Absolute growth rate is the measurement of total growth per unit time. Relative growth rate is growth per unit time per unit initial growth.

$$\text{Relative Growth Rate} = \frac{\text{Growth in given time period}}{\text{Measurement at start of time period}}$$

Suppose two leaves have grown by $5\text{ cm}^2$ in one day. Initial size of leaf A was $5\text{ cm}^2$ while that of leaf B was $50\text{ cm}^2$. Though their absolute growth is the same ($5\text{ cm}^2/\text{day}$), relative rate of growth is faster in leaf A ($\frac{5}{5}$) because of initial small size than in leaf B ($\frac{5}{50}$).

Question 13.4. List five main groups of natural plant growth regulators. Write a note on discovery, physiological functions and agricultural/ horticultural applications of any one of them.

Solution:

There are five main groups of natural plant growth regulators which are very much recognised as natural hormones in plants. These are:

1. Auxins
2. Gibberellins
3. Cytokinins
4. Abscisic acid
5. Ethylene

Discovery of auxin: In 1880, Charles Darwin and Francis Darwin worked with the coleoptile of canary grass (Phalaris sp.) and found the existence of a substance in coleoptile tip, which was able to recognise the light stimulus and leads to the bending of tip towards light. Boysen and Jensen ($1910\text{–}1913$) worked on Avena seedling and explained that the substances secreted in the tip are soluble in water (gelatin). Paal (1919) reported that the substances secreted in the tip are translocated downwards and caused cell elongation in half portion which was on the dark side and hence bending was observed in opposite direction. F.W. Went (1928) further refined this experiment and supported the observations of Paal. He was the first person to isolate and name these substances of tip as auxins (Greek Auxein – means ‘to grow’). In 1931, Kogl and Haagen-Smith isolated crystalline compounds from human urine. These were named as auxin-a, auxin-b and heteroauxin.

Physiological functions of auxins:
– Auxins induce cambial cell divisions, shoot cell elongation and early differentiation of xylem and phloem in tissue culture experiments.
– In general, auxins initiate rooting but inhibit the growth of roots. $\text{IBA}$ is the most potent root initiator.
– Auxins inhibit the growth of axillary buds (apical dominance) but enhance the size of carpel and hence earlier fruit formation.
– Application of auxins retards the process of senescence (last degradative phase), the abscission of leaves, fruits, branches, etc.
– Auxins induce feminisation, i.e., on male plant, female flowers are produced.

Agricultural/horticultural application of auxins:
– Application of auxins like $\text{IAA}$, $\text{IBA}$, $\text{NAA}$ induce rooting in stem cuttings of many plants. This method is widely used to multiply several economically useful plants.
– Normally, auxins inhibit flowering however in litchi and pineapple, application of auxin promotes flowering thus used in orchards.
– Auxin induces parthenocarpy in some plants including tomato, pepper, cucumber and Citrus, thus, produces seedless fruits of more economic value.
– Auxins like $2,4\text{-D}$ and $2,4,5\text{-T}$ are commercially used as weedicides, due to their low cost and greater chemical stability. They are selective herbicides (killing broad-leaved plants, but not grasses).
– For checking premature fruit drop, auxins are applied which prevent the formation of abscission zone in the petiole or just below the fruit.
– Auxin, produced in the apical bud, suppresses the development of lateral buds, i.e., apical dominance. Thus practically used in prolonging the dormancy period of potato tubers.
– Naphthalene acetamide is used to prevent the lodging or falling of crops.
– Auxin ($2,4\text{-D}$) promotes callus formation in tissue culture.

Question 13.5. What do you understand by photoperiodism and vernalisation? Describe their significance.

Solution:

The physiological mechanism for flowering is controlled by two factors: photoperiod or light period, i.e., photoperiodism and low temperature, i.e., vernalisation.

Photoperiodism is defined as the flowering response of a plant to relative lengths of light/dark period. Significance of photoperiodism:
(i) Photoperiodism determines the season in which a particular plant shall flower (e.g., short day plants, long day plants).
(ii) Knowledge of photoperiodic effect is useful in keeping some plants in vegetative growth to obtain higher yield of tubers, rhizomes etc.
(iii) A plant can be made to flower throughout the year by providing favourable photoperiod.
(iv) Helps plant breeders in effective cross-breeding.

Vernalisation is promotion or induction of flowering by exposing a plant to low temperature for some time. Significance of vernalisation:
(i) Crops can be grown earlier.
(ii) Plants can be grown in regions where normally they do not grow.
(iii) Yield of the plant is increased.
(iv) Resistance to cold, frost, and fungal diseases is increased.

Question 13.6. Why is abscisic acid also known as stress hormone?

Solution:

A fairly high concentration of abscisic acid ($\text{ABA}$) is found in leaves of plants growing under stress conditions, such as drought, flooding, injury, mineral deficiency etc. It is accompanied by loss of turgor and closure of stomata. When such plants are transferred to normal conditions, they regain normal turgor and $\text{ABA}$ concentration decreases. Since the synthesis of $\text{ABA}$ is accelerated under stress condition and the same is destroyed or inactivated when stress is relieved, it is also known as stress hormone.

Question 13.7. ‘Both growth and differentiation in higher plants are open’. Comment.

Solution:

Plant growth is generally indeterminate. Higher plants possess specific areas called meristems which take part in the formation of new cells. The body of plants is built on a modular fashion where structure is never complete because the tips (with apical meristem) are open-ended – always growing and forming new organs to replace the older or senescent ones. Growth is invariably associated with differentiation. Not only the growth of plants is open-ended, their differentiation is also open. The same apical meristem cells give rise to different types of cells at maturity, e.g., xylem, phloem, parenchyma, sclerenchyma fibres, collenchyma, etc. Thus, both the processes are indeterminate, unlimited and develop into different structures at maturity i.e., both are open.

Question 13.8. ‘Both a short day plant and a long day plant can produce flower simultaneously in a given place’. Explain.

Solution:

A short day plant ($\text{SDP}$) flowers only when it receives a long dark period and short photoperiod, e.g., Xanthium, Dahlia etc. On the other hand, a long day plant ($\text{LDP}$) will flower only when it receives a long photoperiod and short dark period, e.g., wheat, oat etc. Thus critical photoperiod is that continuous duration of light which must not be exceeded in $\text{SDP}$ and should always be exceeded in $\text{LDP}$ in order to bring them to flower. Xanthium requires light for less than $15.6\text{ hrs}$ and Henbane requires light for more than $11\text{ hrs}$. Xanthium (an $\text{SDP}$) and Henbane ($\text{LDP}$) will flower simultaneously in light period between $11$ to $15.6\text{ hrs}$.

Question 13.9. Which one of the plant growth regulators would you use if you are asked to
(a) induce rooting in a twig
(b) quickly ripen a fruit
(c) delay leaf senescence
(d) induce growth in axillary buds
(e) ‘bolt’ a rosette plant
(f) induce immediate stomatal closure in leaves.

Solution:

(a) Auxins like $\text{IBA}$, $\text{NAA}$.

(b) Ethylene

(c) Cytokinins

(d) Cytokinins

(e) Gibberellins

(f) Abscisic acid ($\text{ABA}$)

Question 13.10. Would a defoliated plant respond to photoperiodic cycle? Why?

Solution:

No, a defoliated plant would not respond to photoperiodic cycle because photoperiodic stimulus is picked up by the leaves only. Even one leaf or a part of it is sufficient for this purpose. For perception of photoperiodic cycle, there must be the presence of leaves under inductive photoperiod, so that, the hormone responsible for flowering can be produced.

Question 13.11. What would be expected to happen if:
(a) $\text{GA}_3$ is applied to rice seedlings
(b) dividing cells stop differentiating
(c) a rotten fruit gets mixed with unripe fruits
(d) you forget to add cytokinin to the culture medium.

Solution:

(a) The coleoptile will elongate rapidly, as $\text{GA}_3$ helps in cell growth.

(b) The development of callus (mass of undifferentiated cells) will take place.

(c) The unripe fruits will ripen quickly because of the increased rate of respiration due to emission of ethylene from rotten fruit.

(d) Cell division will retard and shoot will not initiate from the callus.

Question 12.1. Give the schematic representation of an overall view of Krebs’ cycle.

Solution:

Question 12.2. Differentiate between
(a) Respiration and Combustion
(b) Glycolysis and Krebs’ cycle
(c) Aerobic respiration and Fermentation

Solution:

(a) Differences between respiration and combustion:

(b) Differences between glycolysis and Krebs’ cycle:

(c) Differences between aerobic respiration and fermentation:

Question 12.3. What are respiratory substrates? Name the most common respiratory substrate.

Solution:

Respiratory substrates are those organic substances which are oxidised during respiration to liberate energy inside the living cells. The common respiratory substrates are carbohydrates, proteins, fats and organic acids. The most common respiratory substrate is glucose. It is a hexose monosaccharide.

Question 12.4. Give the schematic representation of glycolysis.

Solution:

Question 12.5. Explain ETS.

Solution:

An electron transport chain or system ($\text{ETS}$) is a series of coenzymes and cytochromes that take part in the passage of electrons from a chemical to its ultimate acceptor. Reduced coenzymes participate in electron transport chain. Electron transport takes place on cristae of mitochondria [oxysomes ($F_0\text{-}F_1$ particles) found on the inner surface of the membrane of mitochondria]. $\text{NADH}$ formed in glycolysis and citric acid cycle are oxidised by $\text{NADH}$ dehydrogenase (complex I) and the electrons are transferred to ubiquinone. Ubiquinone also receives reducing equivalents via $\text{FADH}_2$ through the activity of succinate dehydrogenase (complex II). The reduced ubiquinone is then oxidised by transfer of electrons of cytochrome $c$ via cytochrome $bc_1$ complex (complex III). Cytochrome $c$ acts as a mobile carrier between complex III and complex IV. Complex IV refers to cytochrome $c$ oxidase complex containing cytochromes $a$ and $a_3$ and two copper centres. When the electrons are shunted over the carriers via complex I to IV in the electron transport chain, they are coupled to $\text{ATP}$ synthetase (complex V) for the formation of $\text{ATP}$ from $\text{ADP}$ and $P_i$. Oxygen functions as the terminal acceptor of electrons and is reduced to water along with the hydrogen atoms. Reduced coenzymes do not combine directly with molecular $\text{O}_2$. Only their hydrogen or electrons are transferred through various substances and finally reach $\text{O}_2$. Both cytochrome $a$ and $a_3$ form a system called cytochrome oxidase. Copper is also present in cyt $a_3$ in addition to iron. The molecular oxygen that has accepted electrons now receives the protons that were liberated into the surrounding medium to give rise to a molecule of water. The liberated energy is utilised for the synthesis of $\text{ATP}$ from $\text{ADP}$ and $P_i$.

Question 12.6. What are the main steps in aerobic respiration? Where does it take place?

Solution:

Aerobic respiration is an enzymatically controlled release of energy in a stepwise catabolic process of complete oxidation of organic food into carbon dioxide and water with oxygen acting as terminal oxidant. The common pathway of aerobic respiration consists of three steps – glycolysis, Krebs’ cycle and terminal oxidation ($\text{ETS}$). Aerobic respiration takes place within mitochondria (though glycolysis occurs in the cytoplasm). The final product of glycolysis, pyruvate is transported from the cytoplasm into the mitochondria.

Question 12.7. What are the assumptions made during the calculation of net gain of $\text{ATP}$?

Solution:

It is possible to make calculations of the net gain of $\text{ATP}$ for every glucose molecule oxidised; but in reality this can remain only a theoretical exercise. These calculations can be made only on certain assumptions that:
– There is a sequential, orderly pathway functioning, with one substrate forming the next and with glycolysis, TCA cycle and ETS pathway following one after another.
– $\text{NADH}$ synthesised in glycolysis is transferred into the mitochondria and undergoes oxidative phosphorylation.
– None of the intermediates in the pathway are utilised to synthesise any other compound.
– Only glucose is being respired – no other alternative substrates are entering in the pathway at any of the intermediary stages.

These kinds of assumptions are not really valid in a living system; all pathways work simultaneously and do not take place one after another; substrates enter the pathways and are withdrawn from it as and when necessary; $\text{ATP}$ is utilised as and when needed; enzymatic rates are controlled by multiple means. Hence, there can be a net gain of 36 or $38\text{ ATP}$ molecules during aerobic respiration of one molecule of glucose.

Question 12.8. Distinguish between the following:
(a) Aerobic respiration and Anaerobic respiration.
(b) Glycolysis and Fermentation.
(c) Glycolysis and Citric acid cycle.

Solution:

(a) Differences between aerobic and anaerobic respiration:

(b) Differences between glycolysis and fermentation:

(c) Differences between glycolysis and citric acid cycle:

Question 12.9. Discuss “The respiratory pathway is an amphibolic pathway”.

Solution:

Amphibolic pathway is the one which is used for both breakdown (catabolism) and build-up (anabolism) reactions. Respiratory pathway is mainly a catabolic process which serves to run the living system by providing energy. The pathway produces a number of intermediates. Many of them are raw materials for building up both primary and secondary metabolites. Acetyl $\text{CoA}$ is helpful not only in Krebs’ cycle but is also a raw material for synthesis of fatty acids, steroids, terpenes, aromatic compounds and carotenoids. $\alpha$-ketoglutarate is an organic acid which forms glutamate (an important amino acid) on amination. $\text{OAA}$ (Oxaloacetic acid) on amination produces aspartate. Both aspartate and glutamate are components of proteins. Succinyl $\text{CoA}$ forms cytochromes and chlorophyll.

Hence, fatty acids would be broken down to acetyl $\text{CoA}$ before entering the respiratory pathway when it is used as a substrate. But when the organism needs to synthesise fatty acids, acetyl $\text{CoA}$ would be withdrawn from the respiratory pathway for it. Hence, the respiratory pathway comes into the picture both during breakdown and synthesis of fatty acids. Similarly, during breakdown and synthesis of proteins too, respiratory intermediates form the link. Breaking down processes within the living organism are catabolism, and synthesis is anabolism. Because the respiratory pathway is involved in both anabolism and catabolism, it is better to consider the respiratory pathway as an amphibolic pathway rather than strictly as a catabolic one.

Question 12.10. Define $\text{RQ}$. What is its value for fats?

Solution:

Respiratory quotient ($\text{RQ}$) is the ratio of the volume of carbon dioxide produced to the volume of oxygen consumed in respiration over a period of time.

$$\text{RQ} = \frac{\text{Volume of }\text{CO}_2\text{ evolved}}{\text{Volume of }\text{O}_2\text{ consumed}}$$

$\text{RQ}$ is less than one when the respiratory substrate is either fat or protein. For example, for tripalmitin (a fat):

$$\text{C}_{51}\text{H}_{98}\text{O}_6 + 145\text{O}_2 \rightarrow 102\text{CO}_2 + 98\text{H}_2\text{O}$$
$$\text{RQ} = \frac{102\text{CO}_2}{145\text{O}_2} \approx 0.7$$

$\text{RQ}$ is about $0.7$ for most of the common fats.

Question 12.11. What is oxidative phosphorylation?

Solution:

Oxidative phosphorylation is the synthesis of energy-rich $\text{ATP}$ molecules with the help of energy liberated during oxidation of reduced co-enzymes ($\text{NADH}, \text{FADH}_2$) produced in respiration. The enzyme required for this synthesis is called $\text{ATP}$ synthase (complex V). $\text{ATP}$ synthase is located in $F_1$ head piece of $F_0\text{-}F_1$ particles present in the inner mitochondrial membrane. $\text{ATP}$ synthase becomes active in $\text{ATP}$ formation only when there is a proton gradient having higher concentration of $\text{H}^+$ protons on the $F_0$ side as compared to the matrix side (chemiosmotic hypothesis).

Increased proton concentration is produced in the intermembrane space by the pumping of protons with the help of energy liberated by passage of electrons through the $\text{ETS}$. The flow of protons through the $F_0$ channel induces the $F_1$ particle to function as $\text{ATP}$ synthase, attaching a phosphate radical to $\text{ADP}$ to produce $\text{ATP}$. Oxidation of one molecule of $\text{NADH}$ produces $3\text{ ATP}$ molecules (or $2.5\text{ ATP}$ depending on shuttle), while oxidation of $\text{FADH}_2$ forms $2\text{ ATP}$ molecules ($1.5\text{ ATP}$).

Question 12.12. What is the significance of step-wise release of energy in respiration?

Solution:

The utility of step-wise release of energy in respiration are given as follows:

(i) There is a step-wise release of chemical bond energy which is very easily trapped in forming $\text{ATP}$ molecules.

(ii) Cellular temperature is not allowed to rise destructively.

(iii) Wastage of energy is reduced.

(iv) There are several intermediates which can be used in production of a number of biochemicals.

(v) Through their metabolic intermediates different substances can undergo respiratory catabolism.

(vi) Each step of respiration is controlled by its own enzyme, helping in controlling the rate of respiration and the amount of energy liberated.

Question 11.1. By looking at a plant externally can you tell whether a plant is $C_3$ or $C_4$? Why and how?

Solution:

It is not possible to distinguish externally between a $C_3$ and $C_4$ plant, but generally tropical plants are adapted for the $C_4$ cycle.

Question 11.2. By looking at which internal structure of a plant can you tell whether a plant is $C_3$ or $C_4$? Explain.

Solution:

$C_4$ plants live in hot moist or arid and nonsaline or saline habitats. Internally the leaves show Kranz anatomy. In Kranz anatomy, the mesophyll is undifferentiated and its cells occur in concentric layers around vascular bundles. Vascular bundles are surrounded by large-sized bundle sheath cells which are arranged in a wreath-like manner (Kranz – wreath). The mesophyll and bundle sheath cells are connected by plasmodesmata or cytoplasmic bridges. The chloroplasts of the mesophyll cells are smaller. They have well-developed grana and a peripheral reticulum but no starch. Mesophyll cells are specialised to perform light reaction, evolve $\text{O}_2$ and produce assimilatory power ($\text{ATP}$ and $\text{NADPH}$). They also possess enzyme PEPcase for initial fixation of $\text{CO}_2$. The chloroplasts of the bundle sheath cells are agranal.

Question 11.3. Even though very few cells in a $C_4$ plant carry out the biosynthetic – Calvin pathway, yet they are highly productive. Can you discuss why?

Solution:

Since, through the $C_4$ cycle, a plant can photosynthesise even in the presence of very low concentration of $\text{CO}_2$ (up to $10\text{ parts per million}$), the partial closure of stomata due to xeric conditions would not bring much effect. Therefore, the plants can adapt to grow at low water content, high temperature and bright light intensities. This cycle is specially suited to such plants which grow in dry climates of tropics and subtropics. Besides, the photosynthetic rate remains higher due to the absence of photorespiration in these plants. It can be visualised that both the $C_4$ cycle and photorespiration are the result of evolution or might have been one of the reasons of evolution for the adaptation of plants to different environments. $C_4$ plants are about twice as efficient as $C_3$ plants in converting solar energy into the production of dry matter.

Question 11.4. RuBisCO is an enzyme that acts both as a carboxylase and oxygenase. Why do you think RuBisCO carries out more carboxylation in $C_4$ plants?

Solution:

RuBisCO is an enzyme which acts both as carboxylase (carboxylation during photosynthesis) and oxygenase (during photorespiration). But RuBisCO carries out more carboxylation in $C_4$ plants. In $C_4$ plants, initial fixation of carbon dioxide occurs in mesophyll cells. The primary acceptor of $\text{CO}_2$ is phosphoenol pyruvate or PEP. It combines with carbon dioxide in the presence of PEP carboxylase or PEPcase to form oxaloacetic acid or oxaloacetate. Malic acid or aspartic acid is translocated to bundle sheath cells through plasmodesmata. Inside the bundle sheath cells they are decarboxylated (and deaminated in case of aspartic acid) to form pyruvate and $\text{CO}_2$. $\text{CO}_2$ is again fixed inside the bundle sheath cells through the Calvin cycle. RuBP of the Calvin cycle is called secondary or final acceptor of $\text{CO}_2$ in $C_4$ plants. Pyruvate is sent back to mesophyll cells.

Question 11.5. Suppose there were plants that had a high concentration of chlorophyll b, but lacked chlorophyll a, would it carry out photosynthesis? Then why do plants have chlorophyll b and other accessory pigments?

Solution:

Plants that do not possess chlorophyll a will not carry out photosynthesis because it is the primary pigment and acts as the reaction centre. It performs the primary reactions of photosynthesis or conversion of light into chemical or electrical energy. Other photosynthetic pigments are called accessory pigments. They absorb light energy of different wavelengths and hence broaden the spectrum of light absorbed by photosynthetic pigments. These pigments hand over the absorbed energy to chlorophyll a.

Question 11.6. Give comparison between the following:
(a) $C_3$ and $C_4$ pathways
(b) Cyclic and non-cyclic photophosphorylation
(c) Anatomy of leaf in $C_3$ and $C_4$ plants.

Solution:

(a) The differences between $C_3$ and $C_4$ pathways:
– In $C_3$ pathways, the first stable product is a 3-carbon compound ($3\text{-PGA}$), whereas in $C_4$ pathways, it is a 4-carbon compound ($\text{OAA}$).
– $C_3$ plants lack Kranz anatomy, whereas $C_4$ plants exhibit Kranz anatomy.

(b) The differences between cyclic and non-cyclic photophosphorylation:
– Cyclic photophosphorylation involves only Photosystem I, produces only $\text{ATP}$, and does not involve photolysis of water or $\text{NADPH}$ formation.
– Non-cyclic photophosphorylation involves both Photosystem I and II, produces $\text{ATP}$ and $\text{NADPH}$, and involves photolysis of water with evolution of $\text{O}_2$.

(c) Differences between the leaf anatomy of $C_3$ and $C_4$ plants:
– $C_3$ leaves do not have bundle sheath chloroplast dimorphism.
– $C_4$ leaves possess Kranz anatomy with distinct bundle sheath cells surrounding vascular bundles containing large chloroplasts.

Question 11.7. Look at leaves of the same plant on the shady side and compare it with the leaves on the sunny side. Or compare the potted plants kept in the sunlight with those in the shade. Which of them has leaves that are darker green? Why?

Solution:

The leaves of the shaded side are darker green than those kept in sunlight due to two reasons:

(i) The chloroplasts occur mostly in the mesophyll cells along their walls for receiving optimum quantity of incident light.

(ii) The chloroplasts align themselves in vertical position along the lateral walls in high light intensity and along tangential walls in moderate light.

Question 11.8. The given figure shows the effect of light on the rate of photosynthesis. Based on the graph, answer the following questions.

(a) At which point/s (A, B or C) in the curve is light a limiting factor?
(b) What could be the limiting factor/s in region A?
(c) What do C and D represent on the curve?

Solution:

(a) At regions A and B light is the limiting factor.

(b) In the region A, light can be a limiting factor.

(c) C is the region where the rate of photosynthesis is not increased when light intensity is increased. D is the point where some other factors become limiting.

Question 11.9. Why does the colour of a leaf kept in the dark frequently become yellow, or pale green? Which pigment do you think is more stable?

Solution:

Carotenoid pigments are found in all photosynthetic cells. They are accessory pigments also found in roots, petals etc. These pigments do not breakdown easily thus temporarily reveal their colour due to unmasking, following breakdown of chlorophylls. Thus the colour of leaf kept in dark is yellow or pale green.

Question 10.1. What is the average cell cycle span for a mammalian cell?

Answer:

The average cell cycle span for a mammalian cell is approximately $24\text{ hours}$. This cycle includes various stages such as interphase (which consists of $G_1$, $S$, and $G_2$ phases) and the mitotic phase ($M$ phase), during which cell division takes place.

Question 10.2. Distinguish cytokinesis from karyokinesis.

Answer:

Cytokinesis is the division of the cytoplasm that occurs at the end of the cell cycle, resulting in two separate daughter cells. It begins during the late stages of the mitotic phase.

Karyokinesis refers to the division of the cell’s nucleus during mitosis or meiosis, resulting in the separation of the genetic material into two sets, one for each daughter cell.

Question 10.3. Describe the events taking place during interphase.

Answer:

Interphase is the phase where the cell prepares for division and is divided into three stages:

– $G_1$ phase (Gap 1): The cell is metabolically active, growing in size and synthesizing proteins for DNA replication.

– $S$ phase (Synthesis): DNA replication occurs, doubling the DNA content of the cell. The chromosome number remains the same, but each chromosome consists of two chromatids.

– $G_2$ phase (Gap 2): The cell continues to grow and synthesizes proteins needed for mitosis. It also ensures that the DNA replication is accurate before cell division begins.

Question 10.4. What is the $G_0$ (quiescent phase) of the cell cycle?

Answer:

The $G_0$ phase is a resting phase where cells are no longer dividing. It occurs when cells exit the $G_1$ phase and enter a dormant state. Cells in the $G_0$ phase can either remain inactive indefinitely or re-enter the cell cycle when needed, such as during tissue repair or cell regeneration.

Question 10.5. Why is mitosis called equational division?

Answer:

Mitosis is called equational division because it results in two daughter cells that are genetically identical to the parent cell, with the same number of chromosomes. The chromosome number does not change, and each daughter cell receives an equal and identical set of chromosomes.

Question 10.6. Name the stage of cell cycle at which one of the following events occur:
(i) Chromosomes are moved to spindle equator.
(ii) Centromere splits and chromatids separate.
(iii) Pairing between homologous chromosomes takes place.
(iv) Crossing over between homologous chromosomes takes place.

Answer:

(i) Chromosomes are moved to the spindle equator in Metaphase. During this stage, chromosomes align along the metaphase plate, positioned equidistantly from the two poles of the spindle.

(ii) Centromere splits and chromatids separate in Anaphase. Here, the centromeres divide, and the sister chromatids are pulled toward opposite poles of the cell.

(iii) Pairing between homologous chromosomes takes place in the Zygotene stage of Prophase I in meiosis. During zygotene, homologous chromosomes pair up closely in a process known as synapsis.

(iv) Crossing over between homologous chromosomes takes place during the Pachytene stage of Prophase I in meiosis. In this stage, homologous chromosomes exchange segments of genetic material, contributing to genetic diversity.

Question 10.7. Describe the following:
(a) synapsis (b) bivalent (c) chiasmata
Draw a diagram to illustrate your answer.

Answer:

(a) Synapsis: Synapsis is the pairing of homologous chromosomes during Zygotene in Prophase I of meiosis, where they align closely and exchange genetic material.

(b) Bivalent: A bivalent or tetrad refers to a pair of homologous chromosomes, each made up of two sister chromatids, that are synapsed during Zygotene.

(c) Chiasmata: Chiasmata are X-shaped structures formed during Diplotene of meiosis, where crossing over occurs between homologous chromosomes, exchanging genetic material.

Question 10.8. How does cytokinesis in plant cells differ from that in animal cells?

Answer:

In plant cells, cytokinesis occurs by the formation of a cell plate, which eventually forms the new cell wall.

In animal cells, cytokinesis occurs by cleavage, where the cell membrane pinches inwards to divide the cell into two. The process begins at the periphery and moves inward, forming a midbody at the center of the cell.

Question 10.9. Find examples where the four daughter cells from meiosis are equal in size and where they are unequal in size.

Answer:

In the formation of male gametes (sperms) in humans, the four daughter cells formed during meiosis are equal in size. In the formation of female gametes (ova), three smaller polar bodies are produced, leaving one large ovum, resulting in unequal daughter cells.

Question 10.10. Distinguish anaphase of mitosis from anaphase I of meiosis.

Answer:

In Anaphase of mitosis, the centromere splits, and sister chromatids separate, moving toward opposite poles. In Anaphase I of meiosis, the centromere does not split. Instead, homologous chromosomes are pulled apart, and sister chromatids remain connected at their centromere.

Question 10.11. List the main differences between mitosis and meiosis.

Answer:

Mitosis: Occurs in somatic cells, results in two identical daughter cells, and maintains the same chromosome number.

Meiosis: Occurs in germ cells, results in four non-identical daughter cells, and halves the chromosome number to create gametes for sexual reproduction.

Question 10.12. What is the significance of meiosis?

Answer:

Meiosis is essential for sexual reproduction as it reduces the chromosome number by half, ensuring that offspring inherit the correct number of chromosomes. It promotes genetic diversity through crossing over and independent assortment, which is important for evolution.

Question 10.13. Discuss haploid insects and lower plants where cell division occurs.

Answer:

Haploid insects like drones of honeybees undergo cell division by mitosis.

Lower plants like Spirogyra, Chlamydomonas, and certain pteridophytes also undergo mitosis for reproduction and growth.

Question 10.14. Can there be mitosis without DNA replication in the S phase?

Answer:

No, DNA replication is essential for mitosis. Without replication during the $S$ phase, the cell would not have enough genetic material to divide properly.

Question 10.15. Can there be DNA replication without cell division?

Answer:

Yes, DNA replication can occur without cell division. For example, during the preparation for meiosis, DNA replication takes place without immediate cell division, allowing the chromosomes to duplicate.

Question 10.16. Analyse the events during every stage of the cell cycle and notice how the following two parameters change:
(i) the number of chromosomes ($N$) per cell
(ii) the amount of DNA content ($C$) per cell

Answer:

(i) The number of chromosomes ($N$) per cell:
– $G_1$ phase: The number of chromosomes remains the same as the parent cell (diploid in somatic cells). Each chromosome consists of a single chromatid.
– $S$ phase: DNA replication occurs, but the number of chromosomes does not change. Each chromosome now consists of two sister chromatids, but the chromosome count ($N$) stays the same.
– $G_2$ phase: The number of chromosomes remains the same, as the cell continues to grow and prepare for mitosis.
– $M$ phase (mitosis): The chromosome number stays the same as the chromatids separate during anaphase, ensuring that the two daughter cells receive an identical set of chromosomes ($N$).
Thus, throughout the cell cycle, the number of chromosomes ($N$) does not change, but the structure of the chromosomes changes from a single chromatid in $G_1$ to two sister chromatids in $S$ and $G_2$, before separating during mitosis.

(ii) The amount of DNA content ($C$) per cell:
– $G_1$ phase: The DNA content is at its baseline ($C$), representing one copy of the DNA per chromosome.
– $S$ phase: DNA replication takes place, and the DNA content doubles. Each chromosome now consists of two sister chromatids, effectively doubling the DNA content ($2C$).
– $G_2$ phase: The DNA content remains doubled ($2C$) as the cell prepares for division, and no further DNA replication occurs.
– $M$ phase: As the chromatids are separated into two daughter cells, the DNA content is halved back to the baseline ($C$) in each daughter cell.
Thus, the DNA content doubles during $S$ phase ($2C$) and then is halved in the $M$ phase ($C$) as the cell divides into two.