NCERT Solutions For Class 11 Biology Chapter 22
NCERT Solutions for Class 11 Biology chapter 22 Chemical Coordination and Integration PDF – eSaral

NCERT Solutions for Class 11 Biology chapter 22 Chemical Coordination and Integration PDF

Hey, are you a class 11 student and looking for ways to download NCERT Solutions for Class 11 Biology chapter 22 Chemical Coordination and Integration PDF? If yes. Then read this post till the end.

In this article, we have listed NCERT Solutions for Class 11 Biology chapter 22 Chemical Coordination and Integration in PDF that are prepared by Kota’s top Doctor’s Faculties by keeping Simplicity in mind.

If you want to learn and understand class 11 Biology chapter 22 “Chemical Coordination and Integration” in an easy way then you can use these solutions PDF.

NCERT Solutions helps students to Practice important concepts of subjects easily. Class 11 Biology solutions provide detailed explanations of all the NCERT questions that students can use to clear their doubts instantly.

If you want to score high in your class 11 Biology Exam then it is very important for you to have a good knowledge of all the important topics, so to learn and practice those topics you can use eSaral NCERT Solutions.

In this article, we have listed NCERT Solutions for Class 11 Biology chapter 22 Chemical Coordination and Integration PDF that you can download to start your preparations anytime.

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Download NCERT Solutions for Class 11 Biology chapter 22 Chemical Coordination and Integration PDF



Question1. Define the following:

(a) Exocrine gland

(b) Endocrine gland

(c) Hormone

Solution. (a) Exocrine gland: Exocrine glands secrete their products into ducts that lead to the target tissue. For example, enzymes. The salivary gland secretes saliva into the collecting duct, which leads to the mouth.

(b) Endocrine gland: Endocrine glands secrete their products directly into the blood. For example, hormones. Insulin is a peptide hormone produced in beta cells of the pancreas that allows your body to use glucose from the food that you eat for energy or to store glucose for future use.

(c) Hormone: Hormones are chemical organic substances secreted in plants and animals that act like messenger molecules in the body. After being made in one part of the body, they travel to other parts of the body where they control how cells and organs do their work. For example, Insulin is a peptide hormone produced in beta cells of the pancreas that allows your body to use glucose from the food that you eat for energy or to store glucose for future use.

Question 2. Diagrammatically indicate the location of the various endocrine glands in our body.

Solution.

NCERT Solutions for Class 11 Biology chapter 22 Chemical Coordination and Integration PDF Image 1

Figure: Location of the various endocrine glands in our body

Question 3. List the hormones secreted by the following: Solution. (a) Hypothalamus (b) Pituitary (c) Thyroid (d) Parathyroid (e) Adrenal (f) Pancreas

(g) Testis (h) Ovary (i) Thymus (j) Atrium (k) Kidney (I) G-I Tract

Solution.

NCERT Solutions for Class 11 Biology chapter 22 Chemical Coordination and Integration PDF Image 2 NCERT Solutions for Class 11 Biology chapter 22 Chemical Coordination and Integration PDF Image 3 NCERT Solutions for Class 11 Biology chapter 22 Chemical Coordination and Integration PDF Image 4 NCERT Solutions for Class 11 Biology chapter 22 Chemical Coordination and Integration PDF Image 5 NCERT Solutions for Class 11 Biology chapter 22 Chemical Coordination and Integration PDF Image 6

Question 4. Fill in the blanks: NCERT Solutions for Class 11 Biology chapter 22 Chemical Coordination and Integration PDF Image 7

Solution.

NCERT Solutions for Class 11 Biology chapter 22 Chemical Coordination and Integration PDF Image 8

Question 5. Write short notes on the functions of the following hormones:

(a) Parathyroid hormone (PTH) (b) Thyroid hormones (c) Thymosins (d) Androgens (e) Estrogens

(f) Insulin and Glucagon


Solution. a. Parathyroid hormone (PTH): Parathyroid glands secrete a peptide hormone named parathyroid hormone (PTH). It increases the level of calcium in the blood. It also promotes the reabsorption of calcium from nephrons and promotes the absorption of calcium from digested food. Therefore, it plays an important role in maintaining a balance of calcium in the body.

b. Thyroid hormones: The thyroid gland secretes thyroxine, triiodothyronine, and thyrocalcitonin.

Thyroxine $\left(\mathrm{T}_{4}\right)$ maintains the basal metabolic rate of the body and regulates the metabolism of carbohydrate, fat, and protein. Thyroid hormones also support the process of red blood cell formation and to maintain water and electrolyte balance.

Thyrocalcitonin or calcitonin lowers calcium level in blood plasma. It plays a significant role in calcium levels along with the parathyroid hormone.

Triiodothyronine $\left(\mathrm{T}_{3}\right)$ plays a significant role in almost all the vital functions of the body like metabolic rate, heart and digestive functions, muscle control, brain development and its function, and the maintenance of bones.

c. Thymosins: The thymus gland secretes thymosin. It plays a significant role in protecting the body against infectious agents. It helps in the differentiation of T-lymphocytes and also promotes the production of antibodies. Hence, it offers both humoral and cell-mediated immunity. Thymosin also helps in the development of sex glands.

d. Androgens: Leydig celis of iestis produre anürogens such as testosierone. Tesiosierone is a male sex hormone that controls the development of secondary sex characters like the growth of the reproductive organs, facial hairs, hoarse voice, etc. Androgens also regulate the functions, development, and maturation of male accessory organs such as epididymis and prostate glands. It stimulates spermatogenesis for the formation of mature sperms and also influences male sexual behaviour.

e. Estrogens: Estrogen is synthesised and secreted by the growing ovarian follicles. It is the female sex hormone that controls the development of secondary sex characters like enlargement of mammary glands, high pitch voice, etc. It plays an essential role in the growth and development of female secondary sex organs. It also helps in the development of growing ovarian follicles. It influences female sexual behaviour.

f. Insulin and Glucagon: The pancreas secretes insulin and glucagon to maintain the homeostasis of glucose levels in the blood.

\alpha-cells of the pancreas secrete glucagon, which increases the blood glucose during the hypoglycemic condition in the body. Additionally, glucagon is responsible for the process of gluconeogenesis/glycogenolysis, which contributes to hyperglycemia. Glucagon reduces cellular uptake and utilisation of glucose.

$\beta$-cells secrete insulin that lowers blood glucose during the hyperglycemic state in the body. Additionally, insulin is responsible for the conversion of glucose to glycogen (glycogenesis) in the target cells, which contributes to hyperglycemia. Insulin increases cellular uptake and utilisation of glucose.

Question 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) Hyperglycemic hormone and hypoglycemic hormone: Hyperglycemic hormone is glucagon, and the hypoglycemic hormone is insulin.

(b) Hypercalcemic hormone: Parathyroid hormone (PTH)

(c) Gonadotrophic hormones: Luteinizing hormone (LH) and follicle-stimulating hormone (FSH)

(d) Progestational hormone: Progesterone

(e) Blood pressure-lowering hormone: Nor-adrenalin

(f) Androgens and estrogens: An example of androgen is testosterone, and an example of estrogen is estradiol.

Question 7. Which hormonal deficiency is responsible for the following:

(a) Diabetes mellitus

(b) Goitre

(c) Cretinism

Solution. (a) Diabetes mellitus: It is characterised by abnormally high glucose levels in the blood due to the deficiency of insulin hormone secreted by the pancreas.

(b) Goitre: It is characterised by an abnormal enlargement of the thyroid gland due to the iodine deficiency that causes hypothyroidism due to reduced levels of $\mathrm{T}_{3}$ and $\mathrm{T}_{4}$ hormones secreted by the thyroid gland.

(c) Cretinism: It is characterised by stunted growth in the baby due to the deficiency of $\mathbf{T}_{3}$ and $\mathrm{T}_{4}$ hormones secreted by the thyroid gland (hypothyroidism) in pregnant women.

Question 8. Briefly mention the mechanism of action of FSH.

Solution. – Follicle-stimulating hormone (FSH) is secreted by the pars distalis region of the anterior pituitary. FSH is essential for pubertal development and the function of women’s ovaries and men’s testes.

– In females, FSH stimulates the growth and maturation of ovarian follicle. As the follicle grows and matures, the granulosa cells of the ovarian follicles in the ovaries release an inhibitory hormone known as inhibin that suppresses the process of FSH production.

– In males, FSH is required for spermatogenesis within the specialised cells of testes called Sertoli cells. Once mature sperms are produced, Sertoli cells release an inhibitory hormone known as inhibin that suppresses the process of FSH production.



Figure: Mode of action of FSH in females



Figure: Mode action of FSH in males

Question 9. Match the following:



Solution.



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NCERT Solutions For Class 11 Biology Chapter 21
NCERT Solutions for Class 11 Biology chapter 21 Neural Control and Coordination PDF – eSaral

NCERT Solutions for Class 11 Biology chapter 21 Neural Control and Coordination PDF

Hey, are you a class 11 student and looking for ways to download NCERT Solutions for Class 11 Biology chapter 21 Neural Control and Coordination PDF? If yes. Then read this post till the end.

In this article, we have listed NCERT Solutions for Class 11 Biology chapter 21 Neural Control and Coordination in PDF that are prepared by Kota’s top Doctor’s Faculties by keeping Simplicity in mind.

If you want to learn and understand class 11 Biology chapter 21 “Neural Control and Coordination” in an easy way then you can use these solutions PDF.

NCERT Solutions helps students to Practice important concepts of subjects easily. Class 11 Biology solutions provide detailed explanations of all the NCERT questions that students can use to clear their doubts instantly.

If you want to score high in your class 11 Biology Exam then it is very important for you to have a good knowledge of all the important topics, so to learn and practice those topics you can use eSaral NCERT Solutions.

In this article, we have listed NCERT Solutions for Class 11 Biology chapter 21 Neural Control and Coordination PDF that you can download to start your preparations anytime.

So, without wasting more time Let’s start.

Download NCERT Solutions for Class 11 Biology chapter 21 Neural Control and Coordination PDF



Question 1: Briefly describe the structure of the following:

1. Brain (b) Eye (c) Ear

Solution. (A) Brain: Brain is the main coordinating centre of the body. It is a part of nervous system that controls and monitors every organ of the body. It is well protected by cranial meninges that are made up of an outer layer called dura mater, a thin middle layer called arachnoid, and an inner layer called pia mater.

It is divided into three regions − forebrain, midbrain, and hindbrain.

Forebrain: It is the main thinking part of the brain. It consists of cerebrum, thalamus, and hypothalamus.

(a) Cerebrum:

Cerebrum is the largest part of the brain and constitutes about four-fifth of its weight. Cerebrum is divided into two cerebral hemispheres by a deep longitudinal cerebral fissure. These hemispheres are joined by a tract of nerve fibre known as corpus callosum. The cerebral hemispheres are covered by a layer of cells known as cerebral cortex or grey matter. Cerebrum has sensory regions known as association areas that receive sensory impulses from various receptors as well as from motor regions that control the movement of various muscles. The innermost part of cerebrum gives an opaque white appearance to the layer and is known as the white matter.

(b) Thalamus:

Thalamus is the main centre of coordination for sensory and motor signalling. It is wrapped by cerebrum.

(c) Hypothalamus:

It lies at the base of thalamus and contains a number of centres that regulate body temperature and the urge for eating and drinking. Some regions of cerebrum, along with hypothalamus, are involved in the regulation of sexual behaviour and expression of emotional reactions such as excitement, pleasure, fear, etc.

Midbrain:

It is located between the thalamus region of the forebrain and pons region of hindbrain. The dorsal surface of midbrain consists of superior and inferior corpora bigemina and four round lobes called corpora quadrigemina. A canal known as cerebral aqueduct passes through the midbrain. Midbrain is concerned with the sense of sight and hearing.

Hindbrain:

It consists of three regions − pons, cerebellum, and medulla oblongata.

(a) Pons is a band of nerve fibre that lies between medulla oblongata and midbrain. It connects the lateral parts of cerebellar hemisphere together.

(b) Cerebellum is a large and well developed part of hindbrain. It is located below the posterior sides of cerebral hemispheres and above medulla oblongata. It is responsible for maintaining posture and equilibrium of the body.

(c) Medulla oblongata is the posterior and simplest part of the brain. It is located beneath the cerebellum. Its lower end extends in the form of spinal cord and leaves the skull through foramen magnum.

(B) Eye: Eyes are spherical structures that consist of three layers.

(a) The outer layer is composed of sclera and cornea.

(i) Sclera is an opaque tissue that is usually known as white of the eye. It is composed of a dense connective tissue.

(ii) Cornea is a transparent anterior portion of eye that lacks blood vessels and is nourished by lymph from the nearby area. It is slightly bulged forward and helps in focusing light rays with the help of lens.

(b) The middle layer of eye is vascular in nature and contains choroid, ciliary body, and iris.

(i) Choroid lies next to the sclera and contains numerous blood vessels that provide nutrients and oxygen to the retina and other tissues.

(ii) Ciliary body: The choroid layer is thin over posterior region and gets thickened in the anterior portion to form ciliary body. It contains blood vessels, ciliary muscles, and ciliary processes.

(iii) Iris: At the junction of sclera and cornea, the ciliary body continues forward to form thin coloured partition called iris. It is the visible coloured portion of eye.

The eye contains a transparent, biconvex, and elastic structure just behind the iris. It is known as lens. The lens is held in position by suspensory ligaments attached to the ciliary body. The lens divides the eye ball into two chambers – an anterior aqueous and posterior vitreous chamber.

(c) The innermost nervous coat of eye contains retina. Retina is the innermost layer. It contains three layers of cells – inner ganglion cells, middle bipolar cells, and outermost photoreceptor cells. The receptor cells present in the retina are of two types – rod cells and cone cells.

(a) Rod cells –The rods contain the rhodopsin pigment (visual purple) that is highly sensitive to dim light. It is responsible for twilight vision.

(b) Cone cells –The cones contain the iodopsin pigment (visual violet) and are highly sensitive to high intensity light. They are responsible for daylight and colour visions.

The innermost ganglionic cells give rise to optic nerve fibre that forms optic nerve in each eye and is connected with the brain.

(C) Ear: Ear is the sense organ for hearing and equilibrium. It consists of three portions – external ear, middle ear, and internal ear.

1. External ear:

It consists of pinna, external auditory meatus, and a tympanic membrane.

(a) Pinna is a sensitive structure that collects and directs the vibrations into the ear to produce sound.

(b) External auditory meatus is a tubular passage supported by cartilage in external ear.

(c) Tympanic membrane is a thin membrane that lies close to the auditory canal. It separates the middle ear from external ear.

2. Middle ear:

It is an air-filled tympanic cavity that is connected with pharynx through eustachian tube. Eustachian tube helps to equalize air pressure in both sides of tympanic membrane. The middle ear contains a flexible chain of three middle bones called ear ossicles. The three ear ossicles are malleus, incus, and stapes that are attached to each other.

3. Internal ear:

It is also known as labyrinth. Labyrinth is divided into bony labyrinth and a membranous labyrinth. Bony labyrinth is filled with perilymph while membranous labyrinth is filled with endolymph. Membranous labyrinth is divided into 2 parts.

(a) Vestibular apparatus Vestibular apparatus is a central sac-like part that is divided into utriculus and sacculus. A special group of sensory cells called macula are present in sacculus and utriculus.

Vestibular apparatus also contains three semi-circular canals. The lower end of each semi-circular canal contains a projecting ridge called crista ampularis. Each ampulla has a group of sensory cells called crista. Crista and macula are responsible for maintaining the balance of body and posture.

(b) Cochlea:

Cochlea is a long and coiled outgrowth of sacculus. It is the main hearing organ. Cochlea consists of three membranes. The organ of corti, a hearing organ, is located on the basilar membrane that has hair cells.

Question 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) Central neural system (CNS) and Peripheral neural system (PNS)

NCERT Solutions for Class 11 Biology chapter 21 Neural Control and Coordination PDF Image 1

(b) Resting potential and action potential

NCERT Solutions for Class 11 Biology chapter 21 Neural Control and Coordination PDF Image 2

(c) Choroid and retina

NCERT Solutions for Class 11 Biology chapter 21 Neural Control and Coordination PDF Image 3

Question 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 of the membrane of a nerve fibre

move faster from inside to outside as compared to sodium ions. Therefore, the membrane becomes positively charged outside and negatively charged inside. This is known as polarization of membrane or polarized nerve.

NCERT Solutions for Class 11 Biology chapter 21 Neural Control and Coordination PDF Image 4

(b) Depolarisation of the membrane of a nerve fibre

When an electrical stimulus is given to a nerve fibre, an action potential is generated. The membrane becomes permeable to sodium ions than to potassium ions. This results into positive charge inside and negative charge outside the nerve fibre. Hence, the membrane is said to be depolarized.

NCERT Solutions for Class 11 Biology chapter 21 Neural Control and Coordination PDF Image 5

(c) Conduction of a nerve impulse along a nerve fibre

There are two types of nerve fibres – myelinated and non-myelinated. In myelinated nerve fibre, the action potential is conducted from node to node in jumping manner. This is because the myelinated nerve fibre is coated with myelin sheath. The myelin sheath is impermeable to ions. As a result, the ionic exchange and depolarisation of nerve fibre is not possible along the whole length of nerve fibre. It takes place only at some point, known as nodes of Ranvier, whereas in non-myelinated nerve fibre, the ionic exchange and depolarization of nerve fibre takes place along the whole length of the nerve fibre. Because of this ionic exchange, the depolarized area becomes repolarised and the next polarized area becomes depolarized.

(d) Transmission of a nerve impulse across a chemical synapse

Synapse is a small gap that occurs between the last portion of the axon of one neuron and the dendrite of next neuron. When an impulse reaches at the end plate of axon, vesicles consisting of chemical substance or neurotransmitter, such as acetylcholine, fuse with the plasma membrane. This chemical moves across the cleft and attaches to chemo-receptors present on the membrane of the dendrite of next neuron. This binding of chemical with chemo-receptors leads to the depolarization of membrane and generates a nerve impulse across nerve fibre.

The chemical, acetylcholine, is inactivated by enzyme acetylcholinestrase. The enzyme is present in the post synaptic membrane of the dendrite.

It hydrolyses acetylcholine and this allows the membrane to repolarise.

NCERT Solutions for Class 11 Biology chapter 21 Neural Control and Coordination PDF Image 6

Question 4: Draw labelled diagrams of the following:

(a) Neuron

(b) Brain

(c) Eye

(d) Ear

Solution. (a) Neuron

NCERT Solutions for Class 11 Biology chapter 21 Neural Control and Coordination PDF Image 7

(b) Brain

NCERT Solutions for Class 11 Biology chapter 21 Neural Control and Coordination PDF Image 8

(c) Eye

NCERT Solutions for Class 11 Biology chapter 21 Neural Control and Coordination PDF Image 9

(d) Ear

NCERT Solutions for Class 11 Biology chapter 21 Neural Control and Coordination PDF Image 10

Question 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 neural system provides rapid coordination among the organs of the body. This coordination is in the form of electric impulses and is quick and short lived. All the physiological processes in the body are closed linked and dependent upon each other. For example, during exercise, our body requires more oxygen and food. Hence, the breathing rate increases automatically and the heart beats faster. This leads to a faster supply of oxygenated blood to the muscles. Moreover, the cellular functions require regulation continuously. These functions are carried out by the hormones. Hence, the neural system along with the endocrine system control and coordinate the physiological processes.

(b) Forebrain

It is the main thinking part of the brain. It consists of cerebrum, thalamus, and hypothalamus.

(i) Cerebrum:

Cerebrum is the largest part of the brain and constitutes about four-fifth of its weight. Cerebrum is divided into two cerebral hemispheres by a deep longitudinal cerebral fissure. These hemispheres are joined by a tract of nerve fibres known as corpus callosum. The cerebral hemispheres are covered by a layer of cells known as cerebral cortex or grey matter. Cerebrum has sensory regions known as association areas that receive sensory impulses from various receptors as well as from motor regions that control the movement of various muscles. The innermost part of cerebrum gives an opaque white appearance to the layer and is known as the white matter.

(ii) Thalamus:

Thalamus is the main centre of coordination for sensory and motor signalling. It is wrapped by cerebrum.

(iii) Hypothalamus:

It lies at the base of thalamus and contains a number of centres that regulate body temperature and the urge for eating and drinking. Some regions of cerebrum, along with hypothalamus, are involved in the regulation of sexual behaviour and expression of emotional reactions such as excitement, pleasure, fear, etc.

(c) Midbrain

It is located between the thalamus region of the forebrain and pons region of hindbrain. The dorsal surface of midbrain consists of superior and inferior corpora bigemina and four round lobes called corpora quadrigemina. A canal known as cerebral aqueduct passes through the midbrain. Midbrain is concerned with the sense of sight and hearing.

(d) Hindbrain

It consists of three regions – pons, cerebellum, and medulla oblongata.

(i) Pons is a band of nerve fibres that lies between medulla oblongata and midbrain. It connects the lateral parts of cerebellar hemisphere together.

(ii) Cerebellum is a large and well developed part of hindbrain. It is located below the posterior sides of cerebral hemispheres and above the medulla oblongata. It is responsible for maintaining posture and equilibrium of the body.

(iii) Medulla oblongata is the posterior and simplest part of the brain. It is located beneath the cerebellum. Its lower end extends in the form of spinal cord and leaves the skull through foramen magnum.

(e) Retina

Retina is the innermost layer. It contains three layers of cells – inner ganglion cells, middle bipolar cells, and outermost photoreceptor cells. The receptor cells present in the retina are of two types – rod cells and cone cells.

(i) Rod cells –The rods contain rhodopsin pigment (visual purple), which is highly sensitive to dim light. It is responsible for twilight vision.

(ii) Cone cells –The cones contain iodopsin pigment (visual violet) and are highly sensitive to high intensity light. They are responsible for daylight and colour visions.

The innermost ganglionic cells give rise to optic nerve fibre that forms optic nerve in each eye and is connected with the brain. In this region, the photoreceptor cells are absent. Hence, it is known as the blind spot. At the posterior part, lateral to blind spot, there is a pigmented spot called macula lutea. This spot has a shallow depression at its middle known as fovea. Fovea has only cone cells. They are devoid of rod cells. Hence, it is the place of most distinct vision.

(f) Ear ossicles

The middle ear contains a flexible chain of three middle bones called ear ossicles. The three ear ossicles are as follows.

(i) Malleus

(ii) Incus

(iii) Stapes

The malleus is attached to tympanic membrane on one side and to incus on the other side. The incus is connected with stapes. Stapes, in turn, are attached with an oval membrane, fenestra ovalis, of internal ear. The ear ossicles act as a lever that transmits sound waves from external ear to internal ear.

(g) Cochlea

Cochlea is a long, coiled outgrowth of sacculus. It is the main hearing organ. The cochlea forms three chambers.

(i) Upper − scala vestibule

(ii) Middle − scala media

(iii) Lower − scale tympani

The floor of the scala media is basilar membrane while its roof is Reissner’s membrane. Reissner’s membrane gives out a projection called tectorial membrane. The organ of corti, a hearing organ, is located on the basilar membrane. Organ of corti contains receptor hair cells. The upper scala vestibule and lower scala tympani contain perilymph.

(h) Organ of corti

Organ of corti is the hearing organ. It is located on the basilar membrane that contains hair cells. Hair cells act as auditory receptors. They are present on the internal side of organ of corti.

(i) Synapse

Synapse is a junction between the axon terminal of one neuron and the dendrite of next neuron. It is separated by a small gap known as synaptic cleft.

There are two types of synapses.

(a) Electrical synapse

(b) Chemical synapse

In electrical synapses, the pre and post synaptic neurons lie in close proximity to each other. Hence, the impulse can move directly from one neuron to another across the synapse. This represents a faster method of impulse transmission.

In chemical synapses, the pre and post synaptic neurons are not in close proximity. They are separated by a synaptic cleft. The transmission of nerve impulses is carried out by chemicals such as neurotransmitters.

Question 6:

Give a brief account of:

(a) Mechanism of synaptic transmission

(b) Mechanism of vision

(c) Mechanism of hearing

Solution. (a) Mechanism of synaptic transmission

Synapse is a junction between two neurons. It is present between the axon terminal of one neuron and the dendrite of next neuron separated by a cleft.

There are two ways of synaptic transmission.

(1) Chemical transmission

(2) Electrical transmission

1. Chemical transmission – When a nerve impulse reaches the end plate of axon, it releases a neurotransmitter (acetylcholine) across the synaptic cleft. This chemical is synthesized in cell body of the neuron and is transported to the axon terminal. The acetylcholine diffuses across the cleft and binds to the receptors present on the membrane of next neuron. This causes depolarization of membrane and initiates an action potential.

2. Electrical transmission – In this type of transmission, an electric current is formed in the neuron. This electric current generates an action potential and leads to transmission of nerve impulse across the nerve fibre. This represents a faster method of nerve conduction than the chemical method of transmission.

(b) Mechanism of vision

Retina is the innermost layer of eye. It contains three layers of cells – inner ganglion cells, middle bipolar cells, and outermost photoreceptor cells. A photoreceptor cell is composed of a protein called opsin and an aldehyde of vitamin A called retinal. When light rays are focused on the retina through cornea, it leads to the dissociation of retinal from opsin protein. This changes the structure of opsin. As the structure of opsin changes, the permeability of membrane changes, generating a potential difference in the cells. This generates an action potential in the ganglionic cells and is transmitted to the visual cortex of the brain via optic nerves. In the cortex region of brain, the impulses are analysed and image is formed on the retina.

(c) Mechanism of hearing

The pinna of the external region collects the sound waves and directs it towards ear drum or external auditory canal. These waves strike the tympanic membrane and vibrations are created. Then, these vibrations are transmitted to the oval window, fenestra ovalis, through three ear ossicles, named as malleus, incus, and stapes. These ear ossicles act as lever and transmit the sound waves to internal ear. These vibrations from fenestra ovalis are transmitted into cochlear fluid. This generates sound waves in the lymph. The formation of waves generates a ripple in the basilar membrane. This movement bends the sensory hair cells present on the organ of corti against tectorial membrane. As a result of this, sound waves are converted into nerve impulses. These impulses are then carried to auditory cortex of brain via auditory nerves. In cerebral cortex of brain, the impulses are analysed and sound is recognized.



Question 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) Photoreceptors are cells that are sensitive to light. They are of two types – rods and cones. These are present in the retina. Cones help in distinguishing colours. There are three types of cone cells – those responding to green light, those responding to blue light, and those responding to red light. These cells are stimulated by different lights, from different sources. The combinations of the signals generated help us see the different colours.

(b) Vestibular apparatus is located in the internal ear, above the cochlea and helps in maintaining body balance. Crista and macula are the sensory spots of the vestibular apparatus controlling dynamic equilibrium.

(c) Pupil is the small aperture in the iris that regulates the amount of light entering the eye. Cornea, aqueous humour, lens, and vitreous humour act together and refract light rays, focussing them onto the photoreceptor cells of the retina.

Question 8:

Explain the following:

(a) Role of 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) Sodium ions play an important role in the generation of action potential. When a nerve fibre is stimulated, the membrane potential decreases. The membrane becomes more permeable to $\mathrm{Na}^{+}$ions than to $\mathrm{K}^{+}$ions. As a result, $\mathrm{Na}^{+}$diffuses from the outside to the inside of the membrane. This causes the inside of the membrane to become positivelycharged, while the outer membrane gains a negatively charge. This reversal of polarity across the membrane is known as depolarisation. The rapid inflow of $\mathrm{Na}^{+}$ions causes the membrane potential to increase, thereby generating an action potential.



(b) Retina is the innermost layer of the eye. It contains three layers of cells – inner ganglion cells, middle bipolar cells, and outermost photoreceptor cells. Photoreceptor cells are composed of a protein called opsin and an aldehyde of vitamin A called retinal. When light rays are focused on the retina through the cornea, retinal gets dissociated from opsin. As a result, the structure of opsin gets changed. This in turn causes the permeability of the membrane to change, thereby generating a potential difference in the cells. Consequently, an action potential is generated in the ganglion cells and is transmitted to the visual cortex of the brain via the optic nerves. In the cortex region of the brain, the impulses are analysed and the image is formed on the retina

(c) The pinna of the external ear collects the sound waves and directs them to the tympanic membrane (ear drum) via the external auditory canal. The ear drum then vibrates the sound waves and conducts them to the internal ear through the ear ossicles. The ear ossicles increase the intensity of the sound waves. These vibrating sound waves are conducted through the oval window to the fluid in the cochlea. Consequently, a movement is created in the lymph. This movement produces vibrations in the basilar membrane, which in turn stimulate the auditory hair cells. These cells generate a nerve impulse, conducting it to the auditory cortex of the brain via afferent fibres. The auditory cortex region interprets the nerve impulse and sound is recognised.

Question 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 and non-myelinated axons



(b) Dendrites and axons



(c) Rods and cones



(d) Thalamus and Hypothalamus



(e) Cerebrum and Cerebellum



Question 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) Cochlea determines the pitch of a sound.

(b) Forebrain is largest and the most developed part of the human brain.

(c) Hypothalamus acts as a master clock in the human body.

Question 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 chaisma

Solution. Answer: (c) Blind spot

Blind spot is the part where the optic nerve passes out of the retina. Photoreceptors are absent from this region.

Question 12: Distinguish between:

(a) afferent neurons and efferent neurons

(b) impulse conduction in a myelinated nerve fibre and unmyelinated nerve fibre

(c) aqueous humor and vitreous humorv

(d) blind spot and yellow spot

(f) cranial nerves and spinal nerves.

Solution. (a) Afferent neurons and efferent neurons



(b) Impulse conduction in a myelinated nerve fibre and an unmyelinated nerve fibre



c) Aqueous humour and vitreous humour



(d) Blind spot and yellow spot



(e) Cranial nerves and spinal nerves



Also Read,

Class 11 Chemistry Notes.

Class 11 Biology Book Chapterwise.

Class 11 Biology Exemplar Chapterwise.

If you have any Confusion related to NCERT Solutions for Class 11 Biology chapter 21 Neural Control and Coordination PDF then feel free to ask in the comments section down below.

To watch Free Learning Videos on Class 11 Biology by Kota’s top Doctor’s Faculties Install the eSaral App
NCERT Solutions For Class 11 Biology Chapter 20
NCERT Solutions for Class 11 Biology chapter 20 Locomotion and Movement PDF – eSaral

NCERT Solutions for Class 11 Biology chapter 20 Locomotion and Movement PDF

Hey, are you a class 11 student and looking for ways to download NCERT Solutions for Class 11 Biology chapter 20 Locomotion and Movement PDF? If yes. Then read this post till the end.

In this article, we have listed NCERT Solutions for Class 11 Biology chapter 20 Locomotion and Movement PDF that are prepared by Kota’s top Doctor’s Faculties by keeping Simplicity in mind.

If you want to learn and understand class 11 Biology chapter 20 “Locomotion and Movement” in an easy way then you can use these solutions PDF.

NCERT Solutions helps students to Practice important concepts of subjects easily. Class 11 Biology solutions provide detailed explanations of all the NCERT questions that students can use to clear their doubts instantly.

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In this article, we have listed NCERT Solutions for Class 11 Biology chapter 20 Locomotion and Movement PDF that you can download to start your preparations anytime.

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Question 1. Draw the diagram of a sarcomere of skeletal muscle showing different regions.

Solution.

NCERT Solutions for Class 11 Biology chapter 20 Locomotion and Movement PDF Image 1

Figure: Sarcomere of skeletal muscle

Question 2. Define sliding filament theory of muscle contraction.

Solution. Sliding filament theory is a mechanism of muscle contraction in which the myosin and actin filaments of striated muscle slide over each other to shorten the length of the muscle fibres.

[Note: When calcium ions bind to troponin molecules on the actin filaments, myosinbinding sites on the actin filaments are exposed. This forms a bridge between actin and

myosin, which requires ATP as an energy source. Hydrolysis of ATP in the heads of the myosin molecules causes the heads to change shape and bind to the actin filaments. The release of ADP from the myosin heads causes a further change in shape and generates mechanical energy that causes the actin and myosin filaments to slide over one another.]

Question 3. Describe the important steps in muscle contraction.

Solution. The process of muscular contraction occurs in four important steps:

i. Depolarisation and calcium ion release

ii. Actin and myosin cross-bridge formation

iii. Sliding mechanism of actin and myosin filaments

iv. Sarcomere shortening (muscle contraction)

i. Depolarisation and calcium ion release:

– Central nervous system (CNS) via a motor neuron sends a signal for muscle contraction.

– A motor neuron, along with its connected muscle fibres constitutes a motor unit. The junction between a motor neuron and the sarcolemma of the muscle fibre is called the motor-end plate or neuromuscular junction.

– A neurotransmitter (Acetylcholine) is released when a neural signal reaches this neuromuscular junction, which generates an action potential in the sarcolemma.

– Acetylcholine spreads through the muscle fibre and causes the release of calcium ions into the sarcoplasm.

ii. Actin and myosin cross-bridge formation

– When these calcium ions bind to troponin molecules on the actin filaments, myosin-binding sites on the actin filaments are exposed.

– This forms a cross-bridge between actin and myosin, which requires ATP as an energy source. Hydrolysis of ATP in the heads of the myosin molecules causes the heads to change shape and bind to the actin filaments.

iii. Sliding mechanism of actin and myosin filaments

– The release of ADP from the myosin heads causes a further change in shape and generates mechanical energy that causes the actin and myosin filaments to slide over one another.

iv. Sarcomere shortening (muscle contraction)

– The myosin heads pull the attached actin filaments towards the centre of ‘ $A$ ‘ band. The ‘Z’ line attached to these actins are also dragged inwards thereby causing a shortening of the sarcomere, i.e., contraction.

– So, during shortening of the muscle or muscle contraction, the ‘I’ bands get reduced, whereas the ‘ $A$ ‘ bands retain the length.

Question 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) Actin is present in thin filament: True

(Actin filaments are thinner as compared to the myosin filaments, hence are commonly called thin and thick filaments respectively. The actin filament is made of actin, troponin and tropomyosin.)

(b) H-zone of striated muscle fiber represents both thick and thin filaments: False

(This central part of thick filament, not overlapped by thin filaments is called the ‘ $\mathrm{H}$ ‘ zone.)

(c) Human skeleton has 206 bones: True

(In human beings, this system is made up of 206 bones and a few cartilages.)

(d) There are 11 pairs of ribs in man: False

(There are 12 pairs of ribs in men.)

(e) Sternum is present on the ventral side of the body: True

(Sternum is a flat bone on the ventral midline of the thorax.)

Question 5. Write the difference between

(a) Actin and Myosin

(b) Red and White muscles

(c) Pectoral and Pelvic girdle

Solution. (a) Actin and Myosin

NCERT Solutions for Class 11 Biology chapter 20 Locomotion and Movement PDF Image 2

(b) Red and White muscles

NCERT Solutions for Class 11 Biology chapter 20 Locomotion and Movement PDF Image 3

NCERT Solutions for Class 11 Biology chapter 20 Locomotion and Movement PDF Image 4

(c) Pectoral and Pelvic girdle



Question 6. Match Column I with Column II :

Solution.

NCERT Solutions for Class 11 Biology chapter 20 Locomotion and Movement PDF Image 6

Question 7. What are the different types of movements exhibited by the cells of human body?

Solution. Ciliary: Ciliary movement occurs in ciliated epithelium lined inside our tubular organs. The coordinated ciliary movements in the trachea help us in removing dust particles and foreign substances inhaled. The ciliary movement also facilitates passage of ova through the female reproductive tract.

Amoeboid: Amoeboid movement is exhibited by some specialised cells in our body like macrophages and leucocytes in blood. The pseudopodia formed by the streaming of protoplasm in these cells help in such movement. Microfilaments (cytoskeletal elements) also exhibit amoeboid movement.

Muscular: The movements of jaws, tongue, limbs, etc., show muscular movement. Majority of multicellular organisms can do locomotion and other movements due to the contractile property of muscles. Locomotion requires coordinated activity of skeletal, neural systems and muscular movements.

Question 8. How do you distinguish between a skeletal muscle and a cardiac muscle?

Solution.





Question 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) atlas/axis: Pivotal joint

(b) carpal/metacarpal of thumb: Saddle joint

(c) between phalanges: Hinge joint

(d) femur/acetabulum: Ball and socket joint

(e) between cranial bones: Fibrous joint

(f) between pubic bones in the pelvic girdle: Ball and socket joint

(f) between pubic bones in the pelvic girdle: Ball and socket joint

Question 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 2 ‘F’ actins and two other proteins namely ______________And________________.

(f) The human cranium is made of________________bones.

Solution (a) seven

(b) 14

(c) troponin and tropomyosin

(d) sarcoplasmic reticulum

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

(f) eight

Also Read,

Download Class 11 Chemistry Notes.

Download Class 11 Biology Book Chapterwise.

Download Class 11 Biology Exemplar Chapterwise.

If you have any Confusion related to NCERT Solutions for Class 11 Biology chapter 20 Locomotion and Movement PDF then feel free to ask in the comments section down below.

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NCERT Solutions For Class 11 Biology Chapter 19
NCERT Solutions for Class 11 Biology chapter 19 Excretory Products and their Elimination PDF – eSaral

NCERT Solutions for Class 11 Biology chapter 19 Excretory Products and their Elimination PDF

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In this article, we have listed NCERT Solutions for Class 11 Biology chapter 19 Excretory Products and their Elimination in PDF that are prepared by Kota’s top Doctor’s Faculties by keeping Simplicity in mind.

If you want to learn and understand class 11 Biology chapter 19 “Excretory Products and their Elimination” in an easy way then you can use these solutions PDF.

NCERT Solutions helps students to Practice important concepts of subjects easily. Class 11 Biology solutions provide detailed explanations of all the NCERT questions that students can use to clear their doubts instantly.

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In this article, we have listed NCERT Solutions for Class 11 Biology chapter 19 Excretory Products and their Elimination PDF that you can download to start your preparations anytime.

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Download NCERT Solutions for Class 11 Biology chapter 19 Excretory Products and their Elimination PDF



Question 1: Define Glomerular Filtration Rate (GFR)

Solution. Glomerular filtration rate is the amount of glomerular filtrate formed in all the nephrons of both the kidneys per minute. In a healthy individual, it is about $125 \mathrm{~mL} /$ minute. Glomerular filtrate contains glucose, amino acids, sodium, potassium, urea, uric acid, ketone bodies, and large amounts of water.

Question 2: Explain the autoregulatory mechanism of GFR.

Solution. The mechanism by which the kidney regulates the glomerular filtration rate is autoregulative. It is carried out by the juxtaglomerular apparatus. Juxtaglomerular apparatus is a microscopic structure located between the vascular pole of the renal corpuscle and the returning distal convoluted tubule of the same nephron.

It plays a role in regulating the renal blood flow and glomerular filtration rate. When there is a fall in the glomerular filtration rate, it activates the juxtaglomerular cells to release renin. This stimulates the glomerular blood flow, thereby bringing the GFR back to normal. Renin brings the GFR back to normal by the activation of the renin-angiotensin mechanism.

Question 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 4: Give a brief account of the counter current mechanism.

Solution. The counter current mechanism operating inside the kidney is the main adaptation for the conservation of water. There are two counter current mechanisms inside the kidneys They are Henle’s loop and vasa rectae.

Henle’s loop is a U-shaped part of the nephron. Blood flows in the two limbs of the tube in opposite directions and this gives rise to counter currents. The Vasa recta is an efferent arteriole, which forms a capillary network around the tubules inside the renal medulla. It runs parallel to Henley’s loop and is U-shaped. Blood flows in opposite directions in the two limbs of vasa recta. As a result, blood entering the renal medulla in the descending limb comes in close contact with the outgoing blood in the ascending limb.



The osmolarity increases from $300 \mathrm{mOsmoll}^{-1}$ in the cortex to $1200 \mathrm{mOsmoll}^{-1}$ in the inner medulla by counter current mechanism. It helps in maintaining the concentration gradient, which in turn helps in easy movement of water from collecting tubules. The gradient is a result of the movement of $\mathrm{NaCl}$ and urea.

Question $5:$ Describe the role of liver, lungs and skin in excretion.

Solution. Liver, lungs, and skin also play an important role in the process of excretion.

Role of the liver:

Liver is the largest gland in vertebrates. It helps in the excretion of cholesterol, steroid hormones, vitamins, drugs, and other waste materials through bile. Urea is formed in the liver by the ornithine cycle. Ammonia – a toxic substance $-$ is quickly changed into urea in the liver and thence eliminated from the body. Liver also changes the decomposed haemoglobin pigment into bile pigments called bilirubin and biliverdin

Role of the lungs:

Lungs help in the removing waste materials such as carbon dioxide from the body.

Role of the skin:

Skin has many glands which help in excreting waste products through pores. It has two types of glands – sweat and sebaceous glands.

Sweat glands are highly vascular and tubular glands that separate the waste products from the blood and excrete them in the form of sweat. Sweat excretes excess salt and water from the body.

Sebaceous glands are branched glands that secrete an oily secretion called sebum.

Question 6: Explain micturition.

Solution. Micturition is the process by which the urine from the urinary bladder is excreted. As the urine accumulates, the muscular walls of the bladder expand. The walls stimulate the sensory nerves in the bladder, setting up a reflex action. This reflex stimulates the urge to pass out urine. To discharge urine, the urethral sphincter relaxes and the smooth muscles of the bladder contract. This forces the urine out from the bladder. An adult human excretes about 1 – 1.5 litres of urine per day.

Question 7: Match the items of column I with those of column II:



Solution



Question 8: What is meant by the term osmoregulation?

Solution. Osmoregulation is a homeostatic mechanism that regulates the optimum temperature of water and salts in the tissues and body fluids. It maintains the internal environment of the body by water and ionic concentration.

Question 9: Terrestrial animals are generally either ureotelic or uricotelic, not ammonotelic, why?

Solution. Terrestrial animals are either ureotelic or uricotelic, and not ammonotelic. This is because of the following two main reasons:

(a) Ammonia is highly toxic in nature. Therefore, it needs to be converted into a less toxic form such as urea or uric acid.

(b) Terrestrial animals need to conserve water. Since ammonia is soluble in water, it cannot be eliminated continuously. Hence, it is converted into urea or uric acid. These forms are less toxic and also insoluble in water. This helps terrestrial animals conserve water.

Question 10: What is the significance of juxtaglomerular apparatus (JGA) in kidney function?

Solution. Juxtaglomerular apparatus (JGA) is a complex structure made up of a few cells of glomerulus, distal tubule, and afferent and efferent arterioles. It is located in a specialised region of a nephron, wherein the afferent arteriole and the distal convoluted tubule (DLT) come in direct contact with each other.

The juxtaglomerular apparatus contains specialised cells of the afferent arteriole known as juxtaglomerular cells. These cells contain the enzyme renin that can sense blood pressure. When glomerular blood flow (or glomerular blood pressure or glomerular filtration rate) decreases, it activates juxtaglomerular cells to release renin.

Renin converts the angiotensinogen in blood into angiotensin I and further into angiotensin II. Angiotensin II is a powerful vasoconstrictor that increases the glomerular blood pressure and filtration rate. Angiotensin II also stimulates the adrenal cortex of the adrenal gland to produce aldosterone. Aldosterone increases the rate of absorption of sodium ions and water from the distal convoluted tubule and the collecting duct. This also leads to an increase in blood pressure and glomerular filtration rate. This mechanism, known as renin-angiotensin mechanism, ultimately leads to an increased blood pressure.



Question 11: Name the following:

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) Amphioxus is an example of a chordate that has flame cells as excretory structures. Flame cell is a type of excretory and osmoregulatory system.

(b) The cortical portions projecting between the medullary pyramids in the human kidney are the columns of Bertini. They represent the cortical tissues present within the medulla.

(c) A loop of capillary that runs parallel to Henle’s loop is known as vasa rectae. Vasa rectae, along with Henle’s loop, helps in maintaining a concentration gradient in the medullary interstitium.

Question 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 the hormone vasopressin

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

(d) A healthy adult human excretes (on an average) $25-30 \mathrm{gm}$ of urea/day.

Also Read,

Download Class 11 Chemistry Notes PDF.

Download Class 11 Biology Book Chapterwise PDF.

Download Class 11 Biology Exemplar Chapterwise PDF.

If you have any Confusion related to NCERT Solutions for Class 11 Biology chapter 19 Excretory Products and their Elimination PDF then feel free to ask in the comments section down below.

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NCERT Solutions For Class 11 Biology Chapter 18
NCERT Solutions for Class 11 Biology chapter 18 Body Fluids and Circulation PDF – eSaral

NCERT Solutions for Class 11 Biology chapter 18 Body Fluids and Circulation PDF

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In this article, we have listed NCERT Solutions for Class 11 Biology chapter 18 Body Fluids and Circulation in PDF that are prepared by Kota’s top Doctor’s Faculties by keeping Simplicity in mind.

If you want to learn and understand class 11 Biology chapter 18 “Body Fluids and Circulation” in an easy way then you can use these solutions PDF.

NCERT Solutions helps students to Practice important concepts of subjects easily. Class 11 Biology solutions provide detailed explanations of all the NCERT questions that students can use to clear their doubts instantly.

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In this article, we have listed NCERT Solutions for Class 11 Biology chapter 18 Body Fluids and Circulation PDF that you can download to start your preparations anytime.

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Download NCERT Solutions for Class 11 Biology chapter 18 Body Fluids and Circulation PDF



Question 1. Name the components of the formed elements in the blood and mention one major function of each of them.

Solution. The components of the formed elements in the blood include:

a) Erythrocytes – It is also known as red blood cells (RBC). It functions in the transportation of respiratory gases in the body.

b) Leucocytes – It is also known as White blood cells (WBC). It is classified into granulocytes and agranulocytes. Granulocytes are Neutrophils, Eosinophils and Basophils. Agranulocytes are monocytes and lymphocytes. Neutrophils and monocytes are phagocytic in function. Basophils are involved in inflammatory functions. Eosinophils counter allergic reactions and lymphocytes are involved in immune responses.

c) Platelets – They are also known as thrombocytes. They are involved in coagulation or blood clotting mechanism.

Question 2. What is the importance of plasma proteins?

Solution. Plasma is a viscous, straw-coloured fluid which comprises of almost $55 \%$ of the blood. The major proteins in the blood are Fibrinogen, globulins and albumins. Fibrinogen is involved in coagulation or clotting of blood. Albumin is required for the osmotic balance Globulin is involved in the defence mechanisms of the body.

Question 3. Match Column I with Column II :

Column I Column II

(a) Eosinophils (i) Coagulation

(b) RBC (ii) Universal Recipient

(c) AB Group

(iii) Resist Infections

(d) Platelets (iv) Contraction of Heart

(e) Systole (v) Gas transport

Solution:

NCERT Solutions for Class 11 Biology chapter 18 Body Fluids and Circulation PDF Image 1

Question 4. Why do we consider blood as a connective tissue?

Solution. Blood is a fluid connective tissue formed from mesoderm of embryo. It is a connective tissue in the form of fluid and comprises of the fluid matrix, plasma and the formed elements. It has many roles which include transport of respiratory gases, defence mechanism, blood clotting, and maintaining osmotic balance.

Question 5. What is the difference between lymph and blood?

Solution.

NCERT Solutions for Class 11 Biology chapter 18 Body Fluids and Circulation PDF Image 2

Question 6. What is meant by double circulation? What is its significance?

Solution. In human beings, the blood enters the heart twice within a single cycle, and this type of circulation is called double circulation.

In double circulation, the pulmonary artery carries the deoxygenated blood from the right ventricle, and the exchange of gases takes place in the lungs after which the oxygenated blood enters the heart through the pulmonary veins into the left auricle. This circulation is called as pulmonary circulation.

The oxygenated blood is carried by the aorta to various parts of the body through the network of arteries, arterioles and capillaries into various tissues. The venules, veins and superior \& inferior vena cava carry deoxygenated blood and transfer it into the right auricle of the body. While oxygen and various nutrients are provided to the tissues, the carbon dioxide is taken away from the tissues through the blood. This is known as systemic circulation.

This type of blood circulation helps in maintaining body temperature, prevents mixing of oxygenated and deoxygenated blood and increases the efficiency of the circulatory system

NCERT Solutions for Class 11 Biology chapter 18 Body Fluids and Circulation PDF Image 3

Question 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) blood and lymph

NCERT Solutions for Class 11 Biology chapter 18 Body Fluids and Circulation PDF Image 4

b) Open and closed system of circulation

NCERT Solutions for Class 11 Biology chapter 18 Body Fluids and Circulation PDF Image 5

c) Systole and diastole

NCERT Solutions for Class 11 Biology chapter 18 Body Fluids and Circulation PDF Image 6

d) P-wave and T-wave





Question 8. Describe the evolutionary change in the pattern of heart among the vertebrates.

Solution. The muscular chambered heart is present in all the vertebrates. A fish has two-chambered heart, i.e. it has an auricle and a ventricle. It has a single circulation, where the gills oxygenates the deoxygenated blood pumped by the heart. It is then circulated to all body parts, from where the deoxygenated blood returns to the heart.

In amphibians and reptiles (excluding crocodiles), the three-chambered heart is present, which consist of two auricles and one ventricle. Incomplete double circulation is observed in these animals. The right auricle receives deoxygenated blood from all parts of the body, and left auricle receives oxygenated blood from lungs or skin or gills and blood form both auricles enter the single ventricle which pumps out mixed blood.

Birds, crocodiles and mammals possess four-chambered heart, which consists of two auricles and two ventricles. In birds and mammals, the blood from the right auricle enters right ventricle, and from the left auricle to the left ventricle, hence there is no mixing of blood. Hence, they have double circulation.

Question. 9. Why do we call our heart myogenic?

Solution. The heart is myogenic as the normal activities of the heart, such as contractions is autocontrolled. It is carried out by the specialised cardiac musculature known as nodal tissues.

Question 10. Sino-atrial node is called the pacemaker of our heart. Why?

Solution. Sino-atrial node (SAN) is called the pacemaker of our heart as it sets the pace for activities of the heart, such as the contraction and relaxation. It generates the maximum number of action potential per minute, which is $70-75$ per minute.

Question 11. What is the significance of atrio-ventricular node and atrio-ventricular bundle in the functioning of heart?

Solution. The atrio-ventricular node and atrio-ventricular bundle function in conduction of electrical impulses in the heart.

Atrio-ventricular node (AVN) is present in the lower-left corner of the right atrium and close to the atrio-ventricular septum and made of nodal tissue. The atrio-ventricular node conducts the electrical impulse received from the SAN, and it is passed on to the atrioventricular bundle (AV bundle) or Bundle of His, which branches out into tiny fibres called Purkinje fibres that are present throughout the ventricular musculature.

Question 12. Define a cardiac cycle and the cardiac output.

Solution. Cardiac cycle:

The sequential activity of heart such as contraction and relaxation that cyclically take place is called as cardiac cycle. Cardiac cycle denotes the atrial and ventricular systole and diastole. Each cardiac cycle is represented by a heartbeat, i.e. if the no. of heart beat per minute in an individual is 72 beats per minute, then the no. of cardiac cycles per minute is also 72 cycles.

Cardiac output:

Cardiac output is the stroke volume multiplied by heart rate. Cardiac output $=$ Stroke volume $^{*}$ heart rate,

Where, Stroke volume is the volume of blood pumped by the ventricles during every cardiac cycle.

Ventricles pump approximately $70 \mathrm{ml}$ of blood during every cardiac cycle. Heart rate is the no. of heartbeats per minute. The heart rate is 72 beats per minutes.

Hence, in a healthy individual, the cardiac output is on an average, $5000 \mathrm{ml}$ or $5 \mathrm{~L}$.

Question 13. Explain heart sounds.

Solution. There are two prominent heart sounds produced during every cardiac cycle. The sounds are lub ( first heart sound) and dub ( second heart sound).

The sound lub is produced when the bicuspid and tricuspid valves close, and the sound dub is produced when the semilunar valves close. They are prominent in the clinical diagnosis.

Question 14. Draw a standard ECG and explain the different segments in it.

Solution. ECG is used to represent the electrical activity of the heart graphically during a cardiac cycle.

A standard ECG wave can be studied as follows:



The P- wave of the ECG represents the electric de-polarisation of atria, i.e. it refers to the contraction of atria.

The QRS complex represents the electric de-polarisation of the ventricles, i.e. it refers to the contraction of ventricles, which indicates the beginning of systole.

The T-wave represents the electric re-polarisation of the ventricles, i.e. it refers to the relaxation of the ventricles. The end of T-wave indicates the end of systole.

The count of the number of QRS complex peaks for a minute helps in determining the heartbeat. The peaks of the ECG is usually of the same shape in the individuals, and any deviation in it could be an indication of abnormalities in the heart. Hence ECG has a clinical significance.

Also Read,

Download Class 11 Chemistry Notes Free PDF.

Download Class 11 Biology Book Chapterwise Free PDF.

Download Class 11 Biology Exemplar Chapterwise Free PDF.

If you have any Confusion related to NCERT Solutions for Class 11 Biology chapter 18 Body Fluids and Circulation PDF then feel free to ask in the comments section down below.

To watch Free Learning Videos on Class 11 Biology by Kota’s top Doctor’s Faculties Install the eSaral App
NCERT Solutions For Class 11 Biology Chapter 17
NCERT Solutions for Class 11 Biology chapter 17 Breathing and Exchange of Gases PDF – eSaral

NCERT Solutions for Class 11 Biology chapter 17 Breathing and Exchange of Gases PDF

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NCERT Solutions helps students to Practice important concepts of subjects easily. Class 11 Biology solutions provide detailed explanations of all the NCERT questions that students can use to clear their doubts instantly.

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Download NCERT Solutions for Class 11 Biology chapter 17 Breathing and Exchange of Gases PDF



Question 1: Define vital capacity. What is its significance?

Solution. Vital capacity is the maximum volume of air that can be exhaled after a maximum inspiration. It is about $3.5-4.5$ litres in the human body. It promotes the act of supplying fresh air and getting rid of foul air, thereby increasing the gaseous exchange between the tissues and the environment.

Question 2: State the volume of air remaining in the lungs after a normal breathing.

Solution. jolume of air remaining in the lungs after a normal expiration is known as functional residual capacity (FRC). It includes expiratory reserve volume (ERV) and residual volur (RV). ERV is the maximum volume of air that can be exhaled after a normal expiration. It is about $1000 \mathrm{~mL}$ to $1500 \mathrm{~mL} . \mathrm{RV}$ is the volume of air remaining in the lungs after maximum expiration. It is about $1100 \mathrm{~mL}$ to $1500 \mathrm{~mL}$.

$\therefore \mathrm{FRC}=\mathrm{ERV}+\mathrm{RV}$

$\cong 1500+1500$

$\cong 3000 \mathrm{~mL}$

Functional residual capacity of the human lungs is about $2500-3000 \mathrm{~mL}$.

Question 3: Diffusion of gases occurs in the alveolar region only and not in the other parts of respiratory system. Why?

Solution. Each alveolus is made up of highly-permeable and thin layers of squamous epithelial cells. Similarly, the blood capillaries have layers of squamous epithelial cells. Oxygen-rich ail enters the body through the nose and reaches the alveoli. The deoxygenated (carbon dioxide-rich) blood from the body is brought to the heart by the veins. The heart pumps it to the lungs for oxygenation. The exchange of $\mathrm{O}_{2}$ and $\mathrm{CO}_{2}$ takes place between the blood capillaries surrounding the alveoli and the gases present in the alveoli.

Thus, the alveoli are the sites for gaseous exchange. The exchange of gases takes place by simple diffusion because of pressure or concentration differences. The barrier between the alveoli and the capillaries is thin and the diffusion of gases takes place from higher partial pressure to lower partial pressure. The venous blood that reaches the alveoli has lower partial pressure of $\mathrm{O}_{2}$ and higher partial pressure of $\mathrm{CO}_{2}$ as compared to alveolar air. Hence, oxygen diffuses into blood. Simultaneously, carbon dioxide diffuses out of blood and into the alveoli.

Question 4 : What are the major transport mechanisms for $\mathrm{CO}_{2}$ ? Explain.

Solution. Plasma and red blood cells transport carbon dioxide. This is because they are readily soluble in water.

(1) Through plasma:

About $7 \%$ of $\mathrm{CO}_{2}$ is carried in a dissolved state through plasma. Carbon dioxide combines with water and forms carbonic acid.

NCERT Solutions for Class 11 Biology chapter 17 Breathing and Exchange of Gases PDF Image 1

Since the process of forming carbonic acid is slow, only a small amount of carbon dioxide is carried this way.

(2) Through RBCs:

About $20-25 \%$ of $\mathrm{CO}_{2}$ is transported by the red blood cells as carbaminohaemoglobin. Carbon dioxide binds to the amino groups on the polypeptide chains of haemoglobin and forms a compound known as carbaminohaemoglobin

(3) Through sodium bicarbonate:

About $70 \%$ of carbon dioxide is transported as sodium bicarbonate. As $\mathrm{CO}_{2}$ diffuses into the blood plasma, a large part of it combines with water to form carbonic acid in the presence of the enzyme carbonic anhydrase. Carbonic anhydrase is a zinc enzyme that speeds up the formation of carbonic acid. This carbonic acid dissociates into bicarbonate $\left(\mathrm{HCO}_{3}^{-}\right)$and hydrogen ions $\left(\mathrm{H}^{+}\right)$.

NCERT Solutions for Class 11 Biology chapter 17 Breathing and Exchange of Gases PDF Image 2

NCERT Solutions for Class 11 Biology chapter 17 Breathing and Exchange of Gases PDF Image 3

Question 5: What will be the $\mathrm{pO}_{2}$ and $\mathrm{pCO}_{2}$ in the atmospheric air compared to those in the alveolar air?

(i) $\mathrm{pO}_{2}$ lesser, $\mathrm{pCO}_{2}$ higher

(ii) $\mathrm{pO}_{2}$ higher, $\mathrm{pCO}_{2}$ lesser

(iii) $\mathrm{pO}_{2}$ higher, $\mathrm{pCO}_{2}$ higher

(iv) $\mathrm{pO}_{2}$ lesser, $\mathrm{pCO}_{2}$ lesser

Solution. Answer: (ii) $\mathrm{pO}_{2}$ higher, $\mathrm{pCO}_{2}$ lesser

The partial pressure of oxygen in atmospheric air is higher than that of oxygen in alveolar air. In atmospheric air, $\mathrm{pO}_{2}$ is about $159 \mathrm{~mm} \mathrm{Hg}$. In alveolar air, it is about $104 \mathrm{~mm} \mathrm{Hg}$.

The partial pressure of carbon dioxide in atmospheric air is lesser than that of carbon dioxide in alveolar air. In atmospheric air, $\mathrm{pCO}_{2}$ is about $0.3 \mathrm{mmHg} .$ In alveolar air, it is about $40 \mathrm{~mm} \mathrm{Hg}$

Question 6 : Explain the process of inspiration under normal conditions.

Solution.

NCERT Solutions for Class 11 Biology chapter 17 Breathing and Exchange of Gases PDF Image 4

Inspiration or inhalation is the process of bringing air from outside the body into the lungs. It is carried out by creating a pressure gradient between the lungs and the atmosphere.

When air enters the lungs, the diaphragm contracts toward the abdominal cavity, thereby increasing the space in the thoracic cavity for accommodating the inhaled air.

The volume of the thoracic chamber in the anteroposterior axis increases with the simultaneous contraction of the external intercostal muscles. This causes the ribs and the sternum to move out, thereby increasing the volume of the thoracic chamber in the dorsoventral axis.

The overall increase in the thoracic volume leads to a similar increase in the pulmonary volume. Now, as a result of this increase, the intra-pulmonary pressure becomes lesser than the atmospheric pressure. This causes the air from outside the body to move into the lungs.

Question 7: How is respiration regulated?

Solution. The respiratory rhythm centre present in the medulla region of the brain is primarily responsible for the regulation of respiration. The pneumotaxic centre can alter the function performed by the respiratory rhythm centre by signalling to reduce the inspiration rate.

The chemosensitive region present near the respiratory centre is sensitive to carbon dioxide and hydrogen ions. This region then signals to change the rate of expiration for eliminating the compounds.

The receptors present in the carotid artery and aorta detect the levels of carbon dioxide and hydrogen ions in blood. As the level of carbon dioxide increases, the respiratory centre sends nerve impulses for the necessary changes.

Question 8: What is the effect of $\mathrm{pCO}_{2}$ on oxygen transport?

Solution. $\mathrm{pCO}_{2}$ plays an important role in the transportation of oxygen. At the alveolus, the low $\mathrm{pCO}_{2}$ and high $\mathrm{pO}_{2}$ favours the formation of haemoglobin. At the tissues, the high $\mathrm{pCO}_{2}$ and low $\mathrm{pO}_{2}$ favours the dissociation of oxygen from oxyhaemoglobin. Hence, the affinity of haemoglobin for oxygen is enhanced by the decrease of $\mathrm{pCO}_{2}$ in blood. Therefore, oxygen is transported in blood as oxyhaemoglobin and oxygen dissociates from it at the tissues.

Question 9: What happens to the respiratory process in a man going up a hill?

Solution. As altitude increases, the oxygen level in the atmosphere decreases. Therefore, as a man goes uphill, he gets less oxygen with each breath. This causes the amount of oxygen in the blood to decline. The respiratory rate increases in response to the decrease in the oxygen content of blood. Simultaneously, the rate of heart beat increases to increase the supply of oxygen to blood

Question 10: What is the site of gaseous exchange in an insect?

Solution. In insects, gaseous exchange occurs through a network of tubes collectively known as the tracheal system. The small openings on the sides of an insect’s body are known as spiracles. Oxygen-rich air enters through the spiracles. The spiracles are connected to the network of tubes. From the spiracles, oxygen enters the tracheae. From here, oxygen diffuses into the cells of the body.

The movement of carbon dioxide follows the reverse path. The $\mathrm{CO}_{2}$ from the cells of the body first enters the tracheae and then leaves the body through the spiracles.

Question 11: Define oxygen dissociation curve. Can you suggest any reason for its sigmoidal pattern?

Solution. The oxygen dissociation curve is a graph showing the percentage saturation of oxyhaemoglobin at various partial pressures of oxygen.

The curve shows the equilibrium of oxyhaemoglobin and haemoglobin at various partial pressures.

In the lungs, the partial pressure of oxygen is high. Hence, haemoglobin binds to oxygen and forms oxyhaemoglobin.

Tissues have a low oxygen concentration. Therefore, at the tissues, oxyhaemoglobin releases oxygen to form haemoglobin.

The sigmoid shape of the dissociation curve is because of the binding of oxygen to haemoglobin. As the first oxygen molecule binds to haemoglobin, it increases the affinity for the second molecule of oxygen to bind. Subsequently, haemoglobin attracts more oxygen.

Question 12: Have you heard about hypoxia? Try to gather information about it, and discuss with your friends.

Solution. Hypoxia is a condition characterised by an inadequate or decreased supply of oxygen to the lungs. It is caused by several extrinsic factors such as reduction in $\mathrm{pO}_{2}$, inadequate oxygen, etc. The different types of hypoxia are discussed below

Hypoxemic hypoxia

In this condition, there is a reduction in the oxygen content of blood as a result of the low partial pressure of oxygen in the arterial blood

Anaemic hypoxia

In this condition, there is a reduction in the concentration of haemoglobin.

Stagnant or ischemic hypoxia

In this condition, there is a deficiency in the oxygen content of blood because of poor blood circulation. It occurs when a person is exposed to cold temperature for a prolonged period of time.

Histotoxic hypoxia

In this condition, tissues are unable to use oxygen. This occurs during carbon monoxide or cyanide poisoning.

Question 13: Distinguish between

(a) IRV and ERV

(b) Inspiratory capacity and Expiratory capacity

(c) Vital capacity and Total lung capacity

Solution. (a)

NCERT Solutions for Class 11 Biology chapter 17 Breathing and Exchange of Gases PDF Image 5

(b)



(c)



Question 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 during normal respiration.

It is about 6000 to $8000 \mathrm{~mL}$ of air per minute.

The hourly tidal volume for a healthy human can be calculated as:

Tidal volume $=6000$ to $8000 \mathrm{~mL} /$ minute

Tidal volume in an hour $=6000$ to $8000 \mathrm{~mL} \times(60 \mathrm{~min})$

$=3.6 \times 10^{5} \mathrm{~mL}$ to $4.8 \times 10^{5} \mathrm{~mL}$

Therefore, the hourly tidal volume for a healthy human is approximately $3.6 \times 10^{5} \mathrm{~mL}$ to $4.8 \times 10^{5} \mathrm{~mL}$.

Also Read,

Download Class 11 Chemistry Notes PDF.

Download Class 11 Biology Book Chapterwise PDF.

Download Class 11 Biology Exemplar Chapterwise PDF.

If you have any Confusion related to NCERT Solutions for Class 11 Biology chapter 17 Breathing and Exchange of Gases PDF then feel free to ask in the comments section down below.

To watch Free Learning Videos on Class 11 Biology by Kota’s top Doctor’s Faculties Install the eSaral App
NCERT Solutions For Class 11 Biology Chapter 16
NCERT Solutions for Class 11 Biology chapter 16 Digestion and Absorption PDF – eSaral

NCERT Solutions for Class 11 Biology chapter 16 Digestion and Absorption PDF

Hey, are you a class 11 student and looking for ways to download NCERT Solutions for Class 11 Biology chapter 16 Digestion and Absorption PDF? If yes. Then read this post till the end.

In this article, we have listed NCERT Solutions for Class 11 Biology chapter 16 Digestion and Absorption in PDF that are prepared by Kota’s top Doctor’s Faculties by keeping Simplicity in mind.

If you want to learn and understand class 11 Biology chapter 16 “Digestion and Absorption” in an easy way then you can use these solutions PDF.

NCERT Solutions helps students to Practice important concepts of subjects easily. Class 11 Biology solutions provide detailed explanations of all the NCERT questions that students can use to clear their doubts instantly.

If you want to score high in your class 11 Biology Exam then it is very important for you to have a good knowledge of all the important topics, so to learn and practice those topics you can use eSaral NCERT Solutions.

In this article, we have listed NCERT Solutions for Class 11 Biology chapter 16 Digestion and Absorption PDF that you can download to start your preparations anytime.

So, without wasting more time Let’s start.

Download NCERT Solutions for Class 11 Biology chapter 16 Digestion and Absorption PDF



Question 1: Choose the correct answer among the following:

(a) Gastric juice contains

(i) pepsin, lipase and rennin

(ii) trypsin lipase and rennin

(iii) trypsin, pepsin and lipase

(iv) trypsin, pepsin and renin

(b) Succus entericus is the name given to

(i) a junction between ileum and large intestine

(ii) intestinal juice

(iii) swelling in the gut

(iv) appendix

Solution. Answer (a): (i) Pepsin, lipase, and rennin

Gastric juice contains pepsin, lipase, and rennin. Pepsin is secreted in an inactive form as pepsinogen, which is activated by HCl. Pepsin digests proteins into peptones. Lipase breaks down fats into fatty acids. Rennin is a photolytic enzyme present in the gastric juice. It helps in the coagulation of milk

Answer (b): (ii) Intestinal juice Succus entericus is another name for intestinal juice. It is secreted by the intestinal gland. Intestinal juice contains a variety of enzymes such as maltase, lipases, nucleosidases, dipeptidases, etc.

Question 2: Match column I with column II

NCERT Solutions for Class 11 Biology chapter 16 Digestion and Absorption PDF Image 1

Solution.

NCERT Solutions for Class 11 Biology chapter 16 Digestion and Absorption PDF Image 2

Question 3: Answer briefly:

(a) Why are villi present in the intestine and not in the stomach?

(b) How does pepsinogen change into its active form?

(c) What are the basic layers of the wall of alimentary canal?

(d) How does bile help in the digestion of fats?

Solution.

NCERT Solutions for Class 11 Biology chapter 16 Digestion and Absorption PDF Image 3

(a) The mucosal wall of the small intestine forms millions of tiny finger-like projections known as villi. These villi increase the surface area for more efficient food absorption.

Within these villi, there are numerous blood vessels that absorb the digested products of proteins and carbohydrates, carrying them to the blood stream. The villi also contain lymph vessels for absorbing the products of fat-digestion. From the blood stream, the absorbed food is finally delivered to each and every cell of the body.

The mucosal walls of the stomach form irregular folds known as rugae. These help increase the surface area to volume ratio of the expanding stomach.

(b) Pepsinogen is a precursor of pepsin stored in the stomach walls. It is converted into pepsin by hydrochloric acid. Pepsin is the activated in the form of pepsinogen.

Pepsinogen $\stackrel{\mathrm{HCl}}{\longrightarrow}$ Pepsin + Inactive peptide

(Inactive) (Active)

(c) The walls of the alimentary canal are made up of four layers. These are as follows:

NCERT Solutions for Class 11 Biology chapter 16 Digestion and Absorption PDF Image 4

(i) Serosa is the outermost layer of the human alimentary canal. It is made up of a thin layer of secretory epithelial cells, with some connective tissues underneath

(ii) Muscularis is a thin layer of smooth muscles arranged into an outer longitudinal layer and an inner circular layer.

(iii) Sub-mucosa is a layer of loose connective tissues, containing nerves, blood, and lymph vessels. It supports the mucosa.

iv. Mucosa is the innermost lining of the lumen of the alimentary canal. It is mainly involved in absorption and secretion.

(d) Bile is a digestive juice secreted by the liver and stored in the gall bladder. Bile juice has bile salts such as bilirubin and biliverdin. These break down large fat globules into smaller globules so that the pancreatic enzymes can easily act on them. This process is known as emulsification of fats. Bile juice also makes the medium alkaline and activates lipase

Question 4: State the role of pancreatic juice in digestion of proteins.

Solution. Pancreatic juice contains a variety of inactive enzymes such as trypsinogen, chymotrypsinogen, and carboxypeptidases. These enzymes play an important role in the digestion of proteins.

Physiology of protein-digestion

The enzyme enterokinase is secreted by the intestinal mucosa. It activates trypsinogen into trypsin.

NCERT Solutions for Class 11 Biology chapter 16 Digestion and Absorption PDF Image 5

Trypsin then activates the other enzymes of pancreatic juice such as chymotrypsinogen and carboxypeptidase.

Chymotrypsinogen is a milk-coagulating enzyme that converts proteins into peptides.

NCERT Solutions for Class 11 Biology chapter 16 Digestion and Absorption PDF Image

NCERT Solutions for Class 11 Biology chapter 16 Digestion and Absorption PDF Image

Carboxypeptidase acts on the carboxyl end of the peptide chain and helps release the last amino acids.



Hence, it helps in the digestion of proteins

NCERT Solutions for Class 11 Biology chapter 16 Digestion and Absorption PDF Image 7

Thus, in short, we can say that the partially-hydrolysed proteins present in the chyme are acted upon by various proteolytic enzymes of the pancreatic juice for their complete digestion.

NCERT Solutions for Class 11 Biology chapter 16 Digestion and Absorption PDF Image

Question 5 : Describe the process of digestion of protein in stomach.

Solution. The digestion of proteins begins in the stomach and is completed in the small intestine. The digestive juice secreted in the gastric glands present on the stomach walls is called gastric juice. The food that enters the stomach becomes acidic on mixing with this gastric juice.

The main components of gastric juice are hydrochloric acid, pepsinogen, mucus, and rennin. Hydrochloric acid dissolves the bits of food and creates an acidic medium so that pepsinogen is converted into pepsin. Pepsin is a protein- digesting enzyme. It is secreted in its inactive form called pepsinogen, which then gets activated by hydrochloric acid. The activated pepsin then converts proteins into proteases and peptides.

NCERT Solutions for Class 11 Biology chapter 16 Digestion and Absorption PDF Image 8

Rennin is a proteolytic enzyme, released in an inactive form called prorennin. Rennin plays an important role in the coagulation of milk.



Question 6: Given the dental formula of human beings

Solution. The dental formula expresses the arrangement of teeth in each half of the upper jaw and the lower jaw. The entire formula is multiplied by two to express the total number of teeth.

The dental formula for milk teeth in humans is: $\frac{2102}{2102} \times 2=20$

Each half of the upper jaw and the lower jaw has 2 incisors, 1 canine, and 2 molars. Premolars are absent in milk teeth.

The dental formula for permanent teeth in humans is: $\frac{2123}{2123} \times 2=32$

Each half of the upper jaw and the lower jaw has 2 incisors, 1 canine, 2 premolars, and 3 molars. An adult human has 32 permanent teeth

Question 7: Bile juice contains no digestive enzymes, yet it is important for digestion. Why?

Solution. Bile is a digestive juice secreted by the liver. Although it does not contain any digestive enzymes, it plays an important role in the digestion of fats. Bile juice contains bile salts, bile pigments like bilirubin, biliverdin and phospholipids. Bile salts break down large fat globules into smaller globules so that the pancreatic enzymes can easily act on them. This process is known as emulsification of fats. Bile juice also makes the medium alkaline and activates lipase

Question 8 Describe the digestive role of chymotrypsin. What two other digestive enzymes of the same category are secreted by its source gland?

Solution. The enzyme trypsin (present in the pancreatic juice) activates the inactive enzyme chymotrypsinogen into chymotrypsin.



The activated chymotrypsin plays an important role in the further breakdown of the partially-hydrolysed proteins.

The other digestive enzymes of the same category are trypsinogen and carboxypeptidase. These are secreted by the same source-gland, pancreas.

Trypsinogen is present in an inactive form in the pancreatic juice. The enzyme enterokinase – secreted by the intestinal mucosa – activates trypsinogen into trypsin.



The activated trypsin then further hydrolyses the remaining trypsinogen and activates other pancreatic enzymes such as chymotrypsinogen and carboxypeptidase. Trypsin also helps in breaking down proteins into peptides.



Carboxypeptidases act on the carboxyl end of the peptide chain and help in releasing the last amino acids.



Question 9: How are polysaccharides and disaccharides digested?

Solution. The digestion of carbohydrates takes place in the mouth and the small intestine region of the alimentary canal. The enzymes that act on carbohydrates are collectively known as carbohydrases.

Digestion in the mouth:

As food enters the mouth, it gets mixed with saliva. Saliva – secreted by the salivary glands – contains a digestive enzyme called salivary amylase. This enzyme breaks down starch into sugar at $\mathrm{pH} 6.8$



Salivary amylase continues to act in the oesophagus, but its action stops in the stomach as the contents become acidic. Hence, carbohydrate-digestion stops in the stomach.

Digestion in the small intestine:

Carbohydrate-digestion is resumed in the small intestine. Here, the food gets mixed with the pancreatic juice and the intestinal juice. Pancreatic juice contains the pancreatic amylase that hydrolyses the polysaccharides into disaccharides.



(Polysaccharides)

Similarly, the intestinal juice contains a variety of enzymes (disaccharidases such as maltase, lactase, sucrase, etc.). These disaccharidases help in the digestion of disaccharides. The digestion of carbohydrates is completed in the small intestine.







Question 10: What would happen if HCl were not secreted in the stomach?

Solution. Hydrochloric acid is secreted by the glands present on the stomach walls. It dissolves bits of food and creates an acidic medium. The acidic medium allows pepsinogen to be converted into pepsin. Pepsin plays an important role in the digestion of proteins. Therefore, if HCl were not secreted in the stomach, then pepsin would not be activated. This would affect protein digestion. A $\mathrm{pH}$ of about $1.8$ is necessary for proteins to be digested. This $\mathrm{pH}$ is achieved by $\mathrm{HCl}$.

Question 11: How does butter in your food gets digested and absorbed in the body?

Solution. $\underline{\text { Digestion of fats: }}$

Butter is a fat product and gets digested in the small intestine. The bile juice secreted by the liver contains bile salts that break down large fat globules into smaller globules, so as to increase their surface area for the action of lipase. This process is referred to as emulsification of fats.

After this, the pancreatic lipase present in the pancreatic juice and the intestinal lipase present in the intestinal juice hydrolyse the fat molecules into triglycerides, diglycerides, monoglycerides, and ultimately into glycerol.



Diglycerides and monoglycerides $\stackrel{\text { Lipases }}{\longrightarrow}$ Fatty acids + Glycerol

Absorption of fats:

Fat absorption is an active process. During fat digestion, fats are hydrolysed into fatty acids and glycerol. However, since these are water insoluble, they cannot be directly absorbed by the blood. Hence, they are first incorporated into small droplets called micelles and then transported into the villi of the intestinal mucosa.

They are then reformed into small microscopic particles called chylomicrons, which are small, protein- coated fat globules. These chylomicrons are transported to the lymph vessels in the villi. From the lymph vessels, the absorbed food is finally released into the blood stream and from the blood stream, to each and every cell of the body.

Question 12: Discuss the main steps in the digestion of proteins as the food passes through different parts of the alimentary canal.

Solution. The digestion of proteins begins in the stomach and is completed in the small intestine. The enzymes that act on proteins are known as proteases.

Digestion in the stomach

The digestive juice secreted in the gastric glands present on the stomach walls is called gastric juice. The main components of gastric juice are HCl, pepsinogen, and rennin. The food that enters the stomach becomes acidic on mixing with this gastric juice.

The acidic medium converts inactive pepsinogen into active pepsin. The active pepsin then converts proteins into proteases and peptides.



The enzyme rennin plays an important role in the coagulation of milk.



The enzyme rennin plays an important role in the coagulation of milk.

Digestion in the small intestine:

The food from the stomach is acted upon by three enzymes present in the small intestine – pancreatic juice, intestinal juice (known as succus entericus), and bile juice.

Action of pancreatic juice

Pancreatic juice contains a variety of inactive enzymes such as trypsinogen, chymotrypsinogen, and carboxypeptidases. The enzymes are present in an inactivated state. The enzyme enterokinase secreted by the intestinal mucosa activates trypsinogen into trypsin.



The activated trypsin then activates the other enzymes of pancreatic juice.

Chymotrypsinogen is a proteolytic enzyme that breaks down proteins into peptides.





Carboxypeptidases act on the carboxyl end of the peptide chain and help in releasing the last amino acids



Action of bile juice

Bile juice has bile salts such as bilirubin and biliverdin which break down large, fat globules into smaller globules so that pancreatic enzymes can easily act on them. This process is known as emulsification of fats. Bile juice also makes the medium alkaline and activates lipase. Lipase then breaks down fats into diglycerides and monoglycerides.

Action of intestinal juice

Intestinal juice contains a variety of enzymes. Pancreatic amylase digests polysaccharides into disaccharides. Disaccharidases such as maltase, lactase, sucrase, etc., further digest the disaccharides.

The proteases hydrolyse peptides into dipeptides and finally into amino acids.



Pancreatic lipase breaks down fats into diglycerides and monoglycerides.

The nucleases break down nucleic acids into nucleotides and nucleosides.

Question 13: Explain the term thecodont and diphyodont.

Solution.



Thecodont is a type of dentition in which the teeth are embedded in the deep sockets of the jaw bone. Ankylosis is absent and the roots are cylindrical.

Examples include living crocodilians and mammals.

Diphyodont is a type of dentition in which two successive sets of teeth are developed during the lifetime of the organism. The first set of teeth is deciduous and the other set is permanent.

The deciduous set of teeth is replaced by the permanent adult teeth.

This type of dentition can be seen in humans.

Question 14: Name different types of teeth and their number in an adult human.

Solution. There are four different types of teeth in an adult human. They are as follows:

(i) Incisors

The eight teeth in the front are incisors. There are four incisors each in the upper jaw and the lower jaw. They are meant for cutting.

(ii) Canines

The pointy teeth on either side of the incisors are canines. They are four in number, two each placed in the upper jaw and the lower jaw. They are meant for tearing.

(iii) Premolars

They are present next to the canines. They are eight in number, four each placed in the upper jaw and the lower jaw. They are meant for grinding.

(iv) Molars

They are present at the end of the jaw, next to the premolars. There are twelve molars, six each placed in the upper jaw and the lower jaw.

Hence, the dental formula in humans is $\frac{2123}{2123} \times 2=32$

This means each half of the upper jaw and the lower jaw has 2 incisors, 1 canine, 2 premolars, and 3 molars. Hence, an adult human has 32 permanent teeth.

Question 15 : What are the functions of liver?

Solution. Liver is the largest and heaviest internal organ of the body. It is not directly involved in digestion, but secretes digestive juices. It secretes bile which plays a major role in the emulsification of fats.

Also Read,

Download Class 11 Chemistry Notes.

Download Class 11 Biology Book Chapterwise.

Download Class 11 Biology Exemplar Chapterwise.

If you have any Confusion related to NCERT Solutions for Class 11 Biology chapter 16 Digestion and Absorption PDF then feel free to ask in the comments section down below.

To watch Free Learning Videos on Class 11 Biology by Kota’s top Doctor’s Faculties Install the eSaral App
NCERT Solutions For Class 11 Biology Chapter 15
NCERT Solutions for Class 11 Biology chapter 15 Plant Growth and Development PDF – eSaral

NCERT Solutions for Class 11 Biology chapter 15 Plant Growth and Development PDF

Hey, are you a class 11 student and looking for ways to download NCERT Solutions for Class 11 Biology chapter 15 Plant Growth and Development PDF ? If yes. Then read this post till the end.

In this article, we have listed NCERT Solutions for Class 11 Biology chapter 15 Plant Growth and Development in PDF that are prepared by Kota’s top Doctor’s Faculties by keeping Simplicity in mind.

If you want to learn and understand class 11 Biology chapter 15 “Plant Growth and Development” in an easy way then you can use these solutions PDF.

NCERT Solutions helps students to Practice important concepts of subjects easily. Class 11 Biology solutions provide detailed explanations of all the NCERT questions that students can use to clear their doubts instantly.

If you want to score high in your class 11 Biology Exam then it is very important for you to have a good knowledge of all the important topics, so to learn and practice those topics you can use eSaral NCERT Solutions.

In this article, we have listed NCERT Solutions for Class 11 Biology chapter 15 Plant Growth and Development PDF that you can download to start your preparations anytime.

So, without wasting more time Let’s start.

Download NCERT Solutions for Class 11 Biology chapter 15 Plant Growth and Development PDF



Question 1. Define growth, differentiation, development, dedifferentiation, redifferentiation, determinate growth, meristem, and growth rate.

Solution: 1. Growth: Growth is referred as an irreversible, permanent increase in size of an organ or its parts or even of an individual cell.

2. Differentiation: Differentiation is a process in which cells undergo maturation to perform specific functions is referred to as differentiation.

Example: Meristematic cells matures to permanent cells such as xylem and phloem.

3.Development: Development is a process in which the changes occur in the organism in his entire life.

Example: Formation of entire plant through seeds.

4. Dedifferentiation: Dedifferentiation is a process in which the cells have lost the capacity to divide can regain the capacity of division under certain conditions.

For Example: Formation of meristems – interfascicular cambium and cork cambium from fully differentiated parenchyma cells.

5. Redifferentiation: Redifferentiation is a process in which certain tissues that were once able to divide and produce cells again lose the capacity to divide but are mature to perform specific functions.

6. Determinate growth: When the growth of the plant is terminated at a certain point, and maturation starts that condition is refer to as the determinate growth.

7. Meristem: Meristems are the regions in the plant cells through which it divides continuously throughout their life. These cells become permanent and looses the capacity to divide.

8. Growth Rate: Increase in growth of the plant per unit time is referred to as growth rate.

Question 2. Why is not any one parameter good enough to demonstrate growth throughout the life of a flowering plant?

Solution. Growth occurs when there is increase in the amount of the protoplasm. During the growth periods there are several parameters which are involved, such as increase in weight, area, length, volume, and number of the plant part. Hence if only one parameter is considered, it will not be enough for determining the growth of the plant.

Question 3. Describe briefly:

(a) Arithmetic growth

(b) Geometric growth

(c) Sigmoid growth curve

(d) Absolute and relative growth rate

Solution: (a) Arithmetic growth:

-In mitotic division, when one of the daughter cells continues to divide while the other daughter cell undergoes maturation is reffered to as the arithmetic growth.(1 marks) Example: Elongation of roots at the constant rate.

NCERT Solutions for Class 11 Biology chapter 15 Plant Growth and Development PDF Image 1

Diagram showing geometric growth (1 marks) Mathematical expression of Arithmetic growth is: (1 marks)

$\mathrm{Lt}=\mathrm{L}_{0}+\mathrm{rt}$ where,

Lt = Length at time ” $\mathrm{t}$ ”

$\mathrm{L}_{0}=$ Length at time “zero”$\mathrm{L}_{0}=$ Length at time “zero”

$\mathrm{r}=$ growth rate/elongation per unit time

NCERT Solutions for Class 11 Biology chapter 15 Plant Growth and Development PDF Image 2

(b) Geometric growth:

– Geometric growth refers to the growth in which initial growth is slow (lag phase), and it increases rapidly thereafter – at an exponential rate (log or exponential phase).

-In this type of growth, both the daughter cells undergo mitotic division.

Mathematical expression for the exponential growth is expressed as:

W1 = Wo ert

W1 = final size (weight, height, number etc.)

WO = initial size at the beginning of the period

$\mathrm{r}=$ growth rate

$\mathrm{t}=$ time of growth

$e$ = base of natural logarithms Here, $r$ is the relative growth rate and is also the measure of the ability of the plant to produce new plant material, referred to as efficiency index.

Hence, the final size of W1 depends on the initial size, Wo.



Diagram showing geometric growth

(c) Sigmoid growth curve:

-When growth is plotted against the time, it gives an S shaped curve which is referred to as the sigmoid growth curve. The shape curve due to the slow growth in the log phase, rapid increase in the exponential phase and decline in the senescence phase.

NCERT Solutions for Class 11 Biology chapter 15 Plant Growth and Development PDF Image 3

Graph showing sigmoid curve

(d) Absolute and relative growth rates: -Absolute growth rate refers to the growth per unit time.

-Relative growth rates refer to the growth that occurred as compared to the initial period of the growth.

Question 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. List of five main natural plant growth regulators are:

1-Auxin

2-Cytokinin

3-Gibberellin

4- Ethylene

5-Absisic acid

Short note on discovery, physiological functions and agricultural/horticultural applications:

Discovery:

Gibberellin was discovered by the scientist E. Kurosawa.

He observed the disease of rice seedling “balcane” which was caused by the fungal pathoger Gibberella fujikuroi. The symptoms appeared when the rice seeds were treated with the fungus. The active substance was isolated and were named as gibberellic acid.

Physiological function:

-They cause the increase in length from the internodes.

-It elongates and improves the shape of the fruit.

-It induces the maturity in the juvenile plants.

-It promotes bolting in rosette plant.

– They help in the delay of senescence. Agricultural application:

-It is applied to the grapes in order to increase the length of grape stalks.

-Application of gibberellins to increase the size of sugar cane stem hence increasing the sugar content.

-It is applied to fruits such as apple to elongate its shape.

-It is sprayed to the crops which increases their length hence increasing its yield.

-It is applied to juvenile conifers to speed up the maturity hence leading to early seed production.

-It is applied to plants such as beet and cabbages to convert the bolting into rosette habit.

Question 5. What do you understand by photoperiodism and vernalization?

Solution. Photoperiodism: Photoperiodism refers to the process of periodic exposure of light to the plants in to induce flowering.

There are three categories of the plant:

Long day Plants: When plants require the exposure to light for a period exceeding a well-define critical duration are referred to as the long day plants.

For example: Spinach, Beet root, radish, etc.

Short day Plants:

When the plants are exposed to light for a less than critical duration before the flowering than the plants are reffered to as short day plant.

For example: Rice, Cotton, tobacco, etc.

Day neutral Plants: When the flowering is independent of the exposure of light, such plants are day neutral plant.

For example: Tomato, Sunflower, pea, etc.

Vernalization: When the flowering is induced qualitatively and quantitatively on the basis of low temperature, that process is referred to as the vernalization.

Question 6. Why is abscisic acid also known as stress hormone?

Solution. Abscisic acid is works during the stress condition for example it regulates abscission during the autumn and dormancy during the spring.

It has of mainly has inhibitory effects on plant growth and development.

The inhibitory role performed by Abscisic acid are:

Inhibiting the seed germination

-Closing of stomata in the epidermis

-Seed development

-Maturation of the plant

-Breaking of seed dormancy

-Helps the seeds to withstand desiccation

Question 7. ‘Both growth and differentiation in higher plants are open’. Comment.

Plants consists three types of meristems which includes apical meristem, lateral meristem, and intercalary meristem which divides and contribute to the growth of the plant later the divided cells becomes undergo differentiate to form permanent tissue hence the growth and differentiation in higher plant is open.

Question 8. ‘Both a short day plant and a long day plant can produce flower simultaneously in a given place’. Explain.

Solution. In order to induce flowering in short day plants and long day plants at a given place, the plants should be provided the photoperiod according to their requirements.

Short day plants require less amount of light as per their photoperiod for flowering and long day plants require more amount of light for flowering hence they can produce flowers at the same place but the time duration, when they are flowering, will vary.

Question 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) induce rooting in a twig -Auxin

(b) quickly ripen a fruit -Ethylene

(c) delay leaf senescence-Cytokinin’s

(d) induce growth in axillary buds – Cytokinin’s

(e) ‘bolt’ a rosette plant -Gibberellic acid

(f) induce immediate stomatal closure in leaves. -Abscisic acid.

Question 10. Would a defoliated plant respond to photoperiodic cycle? Why?

Solution: – A defoliated plant will not respond to the photoperiodic cycle because leaves are the region which responds to the light and dark durations. It is hypothesized that there is hormonal substance which is responsible for flowering, which is synthesized in the leaves. It migrates from leaves to shoot apices and generates the flowering bud. When the plants with leaves is exposed to the necessary photoperiod, the flowering is induced.

Question 11. What would be expected to happen if:

(a) GA3 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) If there is GA3 application on the rice seedling, then the elongation between the node and internode of the rice plant will take place.

(b) When the dividing cells stops differentiating, then an unorganized mass is formed, which is known as callus.Since the differentiation is absent, then the formation of root and shoot will not take place.

(c) When the rotten fruits are mixed with ripen fruits, then the ethylene gas will be produced, which will hasten the ripening process in unripe fruits.

(d) If cytokinin is not there in the medium then there will not be any growth, differentiation and development of the cell in the culture medium.

Also Read,

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Class 11 Biology Book Chapterwise.

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NCERT Solutions For Class 11 Biology Chapter 14
NCERT Solutions for Class 11 Biology chapter 14 Respiration in Plants PDF – eSaral

NCERT Solutions for Class 11 Biology chapter 14 Respiration in Plants PDF

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Question 1: Differentiate between

(a) Respiration and Combustion

(b) Glycolysis and Krebs’ cycle

(c) Aerobic respiration and Fermentation

Solution. (a) Respiration and combustion



(b) Glycolysis and Krebs cycle



(c) Aerobic respiration and fermentation



Question 2: What are respiratory substrates? Name the most common respiratory substrate.

Solution. The compounds oxidised during the process of respiration are called respiratory substrates. Carbohydrates, especially glucose, act as respiratory substrates. Fats, proteins, and organic acids also act as respiratory substrates.

Question 3: Give the schematic representation of glycolysis?

Solution.



Question 4: What are the main steps in aerobic respiration? Where does it take place?

Solution. The major steps in aerobic respiration and the sites where they occur are listed in the given table.



Question 5: Give the schematic representation of an overall view of Krebs cycle.

Solution.



Question 6: Explain ETS.

Solution. ETS or electron transport system is located in the inner mitochondrial membrane. It helps in releasing and utilizing the energy stored in $\mathrm{NADH}+\mathrm{H}^{+}$and $\mathrm{FADH}_{2}$. NADH which is formed during gly and citric acid cycle, gets oxidized by NADH dehydrogenase (complex 1). The electrons so generated get transferred to ubiquinone through FMN. In a similar manner, FADH (complex II) qenerater citric acid ate fransferrer to ubicuinone. The electrons from ubiquinone are received by cytochrome $\mathrm{bc}_{1}$ (complex III) and further get transferred to cytochrome c. The cytochrome $\mathrm{c}$ acte bile and cytochrome co oxidase comnlex containino cutochrome a and $a_{3}$ alona with conner $c$

During the transfer of electrons from each complex, the process is accompanied by the production of ATP from ADP and inorganic phosphate by the action ATP synthase (complex $\mathbf{V}$ ). The amount of ATP produced depends on the molecule, which has been oxidized. 2 ATP molecules are

produced by the oxidation of one molecule of NADH. One molecule of FADH $_{2}$, on oxidation, gives 3 ATP molecules



Question 7: Distinguish between the following:

(a) Aerobic respiration and Anaerobic respiration

(b) Glycolysis and Fermentation

(c) Glycolysis and Citric acid Cycle

Solution. (a) Aerobic respiration and Anaerobic respiration



(b) Glycolysis and Fermentation



(c) Glycolysis and citric acid cycle

Question 8: What are the assumptions made during the calculation of net gain of ATP?

Solution. For theoretical calculation of ATP molecules, various assumptions are made, which are as follows.

(a) It is assumed that various parts of aerobic respiration such as glycolysis, TCA cycle, and ETS occur in a sequential and orderly pathway.

(b) NADH produced during the process of glycolysis enters into mitochondria to undergo oxidative phosphorylation.

(c) Glucose molecule is assumed to be the only substrate while it is assumed that no other molecule enters the pathway at intermediate stages.

(d) The intermediates produced during respiration are not utilized in any other process.

Question 9: Discuss “The respiratory pathway is an amphibolic pathway.”

Solution. Respiration is generally assumed to be a catabolic process because during respiration, various substrates are broken down for deriving energy. Carbohydrates are broken down to glucose before entering respiratory pathways. Fats get converted into fatty acids and glycerol whereas fatty acids get converted into acetyl CoA before entering the respiration. In a similar manner, proteins are converted into amino acids. which enter respiration after deamination

Therefore, respiration can be termed as amphibolic pathway as it involves both anabolism and catabolism.

Question 10: Define RQ. What is its value for fats?

Solution. Respiratory quotient $(\mathrm{RQ})$ or respiratory ratio can be defined as the ratio of the volume of $\mathrm{CO}_{2}$ evolved to the volume of $\mathrm{O}_{2}$ consumed during respiration. The value of respiratory quotient depends on the type of respiratory substrate. Its value is one for carbohydrates. However, it is always less than one for fats as fats consume more oxygen for respiration than carbohydrates

It can be illustrated through the example of tripalmitin fatty acid, which consumes 145 molecules of $\mathrm{O}_{2}$ for respiration while 102 molecules of $\mathrm{CO}_{2}$ are evolved. The $\mathrm{RQ}$ value for tripalmitin is $0.7$.

Question 11: What is oxidative phosphorylation?

Solution. Oxidative phosphorylation is a process in which electrons are transferred from electron donors to oxygen, which acts as electron acceptor. The oxidation-reduction reactions are involved in the formation proton gradient. The main role in oxidative phosphorylation is played by the enzyme ATP synthase (complex $\mathrm{V}$ ). This enzyme complex consists of $\mathrm{F}_{0}$ and $\mathrm{F}_{1}$ components. The $\mathrm{F}_{1}$ headpiece is a periphera orotein complex and contains the site for ATP synthesis from ADP and inorganic phosphate. $\mathrm{F}_{0}$ component is a part of membrane protein complex, which acts as a channel for crossing of the protons from inner

mitochondrial membrane to the mitochondrial matrix. For every two protons passing through $\mathrm{F}_{0}-\mathrm{F}_{1}$ complex, synthesis of one ATP molecule take

Question 12: What is the significance of step-wise release of energy in respiration?

Solution. The process of aerobic respiration is divided into four phases – glycolysis, TCA cycle, ETS, and oxidative phosphorylation. It is generally assumed that the process of respiration and production of ATP in each ‘ase takes place in a step-wise manner. The product of one pathway forms the substrate of the other pathway. Various molecules produced during respiration are involved in other biochemical processes. iratory substrates enter and withdraw from pathway on necessity. ATP gets utilized wherever required and enzymatic rates are generally controlled. Thus, the step-wise release of energy makes the phase takcol The respiratol stem more efficient in extracting and storing energy

Also Read,

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NCERT Solutions for Class 11 Biology chapter 13 Photosynthesis in Higher Plants PDF – eSaral

NCERT Solutions for Class 11 Biology chapter 13 Photosynthesis in Higher Plants PDF

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Question 1. By looking at a plant externally can you tell whether a plant is $\mathrm{C} 3$ or C4? Why and how?

Solution. C3 plants include sunflower, wheat, spinach, cotton and sugarcane and C4 plants includes pineapple, amaranth, maize etc. Both categories of plants include crop plants, fruits and flowering plants, which makes it very difficult to distinguish between C3 and C4 plants by only looking at the external appearance.

Question 2. By looking at which internal structure of a plant can you tell whether a plant is C3 or C4? Explain.

Solution. -If we will do the microscopic study of the tissues of C3 and C4 plant, then we will be able to classify the plants into $\mathrm{C} 3$ and $\mathrm{C} 4$.

-C3 plants and C4 plants can be distinguished on the basis of Kranz anatomy. C4 plants have Kranz anatomy, which means they have a wreath-like arrangement around the bundle sheath cells. Also, vascular bundles are surrounded by the bundle sheath cells. In leaves, mesophyll cells are not differentiated in the spongy layer and palisade layer. There is a lesser intracellular space and mesophyll have chloroplast. When the leaf tissue of C3 plants is viewed under a microscope. The mesophyll cells of leaves are distinguished into palisade and spongy parenchyma. If we will do the microscopic study of the tissues of $\mathrm{C} 3$ and $\mathrm{C} 4$ plant, then we will be able to classify the plants into $\mathrm{C} 3$ and $\mathrm{C} 4$.

Question 3. Even though a very few cells in a C4 plant carry out the biosynthetic – Calvin pathway, yet they are highly productive. Can you discuss why?

Solution. – In C4 plants, the dark reaction that is Calvin cycle occurs in bundle sheath cells. Though it has a few numbers of cells being involved in the Calvin cycle, C 4 plants have a high rate of photosynthesis due to the following factors:

-Rapid supply of CO2 from mesophyll cells taking part in initial carbon dioxide fixation.

– Since there is no photolysis of water in the bundle sheath cells hence photorespiration is also absent in them.

-Photosynthates are rapidly withdrawn by the bundle sheath cells as they are present in the bundle sheath cells.

– Photosynthesis continues even when stomata are closed due to fixation of $\operatorname{CO} 2$ released during respiration.

Question 4. RuBisCO is an enzyme that acts both as a carboxylase and oxygenase. Why do you think RuBisCO carries out more carboxylation in C4 plants?

Solution. -RuBisCo is referred to as the Ribulose-1,5-bisphosphate carboxylase oxygenase it is one of the most abundant enzymes present in the biosphere. RuBisCO is comparatively more affinitive for CO2 than for O2. It is the relative concentration of $O 2$ and CO2, which determines whether oxygen will bind or carbon dioxide will bind to the enzyme.That is RuBisCO functions as oxygenase only when there is a higher concentration of oxygen and a lower concentration of carbon dioxide. Also C4 plants, the enzyme RuBisCO is not present in the chloroplast of mesophyll cells. It is present only in bundle sheath chloroplasts which are receiving the supply of CO2 continuously even when the stomata are closed. during the decarboxylation of malic acid into pyruvic acid. Hence, RuBisCO usually acts as carboxylase rather than oxygenase.

Question 5. Suppose there were plants that had a high concentration of Chlorophyll b but lacked chlorophyll a, would it carry out photosynthesis. Then why do plants have chlorophyll $\mathrm{b}$ and other accessory pigments?

Solution. If chlorophyll a is absent in the plant then photosynthesis will not occur if chlorophyll a is absent from the system as chlorophyll is the primary pigment which is involved in the absorption of different wavelength of the visible spectrum that is it functions in the trapping of light for photosynthesis further they convert light energy into chemical energy. Plants have accessory pigments such as chlorophyll, xanthophyll and carotenoids these pigments absorb the light energy and pass it on to the chlorophyll molecule. The accessory pigments also prevent the enzymes from photooxidation.

Question 6. Why is the colour of a leaf kept in the dark frequently yellow, or pale green? Which pigment do you think is more stable?

Solution. When leaves are kept in dark, then the photosynthesis stops the green-colored chlorophyll molecule is degraded in the absence of light. The accessory pigments such as xanthophyll and carotenoids become prominent, which are yellow in colour; hence, the colour of leaves changes into yellow from green.

Question 7. Look at leaves of the same plant on the shady side and compare it with the leaves on the sunny side. Or, compare the potted plants kept in the sunlight with those in the shade. Which of them has leaves that are darker green? Why?

Solution. The chlorophyll content will be more in the shady side of the leaves because the side of the leaf not exposed in sunlight will have more aggregation and concentration of chlorophyll molecule hence it is darker. whereas the leaves exposed in sunlight will lose chlorophyll due to photo-oxidation of the chlorophyll molecule; hence the leaf exposed in sunlight will have a lighter colour.

Question 8. Figure shows the effect of light on the rate of photosynthesis. Based on the graph, answer the following questions:

(a) At which point/s (A, B or C) in the curve is light a limiting factor?

(b) What could be the limiting factor/s in region A?

(c) What do $\mathrm{C}$ and $\mathrm{D}$ represent on the curve?

Solution. a) Light is the limiting factor because the rate of photosynthesis is increasing with the intensity of light at the point an and 50 per cent at $\mathrm{B}$.

b) Besides light, other limiting factors will be $\mathrm{CO} 2$ and $\mathrm{H} 2 \mathrm{O}$.

c) C represents a stage beyond which light is not a limiting factor and $D$ is the line beyond which the intensity of light does not affect the rate of photosynthesis.

Question 9. Give comparison between the following:

(a) $\mathbf{C}_{ }$and $\mathbf{C}_{4}$ pathways (b) Cyclic and non-cyclic photophosphorylation

(c) Anatomy of leaf in $\mathrm{C}_{3}$ and $\mathrm{C}$. Plants

Solution. (a) $\mathrm{C}$ and $\mathrm{C}$. pathways:



(b) Cyclic and non-cyclic photophosphorylation:



(c) Anatomy of leaf in $\mathrm{C}$ and $\mathrm{C}_{+}$plants:



Also Read,

Class 11 Chemistry Notes pdf Download.

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NCERT Solutions for Class 11 Biology chapter 12 Mineral Nutrition PDF – eSaral

NCERT Solutions for Class 11 Biology chapter 12 Mineral Nutrition PDF

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Question $1:$ ‘All elements that are present in a plant need not be essential to its survival’. Comment.

Solution. Plants tend to absorb different kinds of nutrients from soil. However, a nutrient is inessential for a plant if it is not involved in the plant’s physiology and metabolism. For example, plants growing near radioactive sites tend to accumulate radioactive metals. Similarly, gold and selenium get accumulated in plants growing near mining sites. However, this does not mean that radioactive metals, gold, or selenium are sential nutrients for the survival of these plants

Question 2: Why is purification of water and nutrient salts so important in studies involving mineral nutrition using hydroponics?

Solution. Hydroponics is the art of growing plants in a nutrient solution in the absence of soil. Since the plant roots are exposed to a limited amount of the solution, there are chances that the concentrations of oxygen and other minerals in the plant roots would reduce. Therefore, in studies involving mineral nutrition using hydroponics, purification of water and nutrient salts is essential so as to maintain an optimum growth of th plants.

Question $3:$ Explain with examples: macronutrients, micronutrients, beneficial nutrients, toxic elements and essential elements

Solution. Macronutrients: They are the nutrients required by plants in large amounts. They are present in plant tissues in amounts more than 10 mmole $\mathrm{kg}^{-1}$ of dry matter. Examples include hydrogen, oxygen, and nitrogen.

Micronutrients: They are also called trace elements and are present in plant bodies in very small amounts, i.e., amounts less than 10 mmole $\mathrm{kg}^{-1}$ of dry matter. Examples include cobalt, manganese, zinc, etc.

Beneficial nutrients: They are plant nutrients that may not be essential, but are beneficial to plants. Sodium, silicon, cobalt and selenium are beneficial to higher plants.

Toxic elements: Micronutrients are required by plants in small quantities. An excess of these nutrients may induce toxicity in plants. For example, when manganese is present in large amounts, it induces deficiencies of iron, magnesium, and calcium by interfering with their metabolism.

Essential elements: These elements are absolutely necessary for plant growth and reproduction. The requirement of these elements is specific and non-replaceable. They are further classified as macro and micro-nutrients.

Question 4: Name at least five different deficiency symptoms in plants. Describe them and correlate them with the concerned mineral deficiency.

Solution. The five main deficiency symptoms arising in plants are:

– Chlorosis

– Necrosis

– Inhibition of cell division

– Delayed flowering

– Stunted plant growth

Chlorosis or loss of chlorophyll leads to the yellowing of leaves. It is caused by the deficiencies of nitrogen, potassium, magnesium, sulphur, iron, manganese, zinc, and molybdenum.

Necrosis is the death of plant tissues as a result of the deficiencies of calcium, magnesium, copper, and potassium.

Inhibition of cell division is caused by the deficiencies of nitrogen, potassium, sulphur, and molybdenum.

Delayed flowering is caused by the deficiencies of nitrogen, sulphur, and molybdenum.

Stunted plant growth is a result of the deficiencies of copper and sulphur.

Question 5: If a plant shows a symptom which could develop due to deficiency of more than

one nutrient, how would you find out experimentally, the real deficient mineral element?


Solution. In plants, the deficiency of a nutrient can cause multiple symptoms. For example, the deficiency of nitrogen causes chlorosis and delayed flowering.

In a similar way, the deficiency of a nutrient can cause the same symptom as that caused by the deficiency of another nutrient. For example, necrosis is caused by the deficiency of calcium, magnesium, copper, and potassium.

Another point to be considered is that different plants respond in different ways to the deficiency of the same nutrient.

Hence, to identify the nutrient deficient in a plant, all the symptoms developed in its different parts must be studied and compared with the available standard tables.

Question 6: Why is that in certain plants deficiency symptoms appear first in younger parts of the plant while in others they do so in mature organs?

Solution. Deficiency symptoms are morphological changes in plants, indicating nutrient deficiency. Deficiency symptoms vary from one element to another. The plant part in which a deficiency symptom occurs depends on the mobility of the deficient element in the plant. Elements such as nitrogen, potassium, and magnesium are highly mobile. These elements move from the mature organs to the younger parts of a plant. Therefore, the symptoms for the deficiencies of these elements first appear in the older parts of the plant. Elements such as calcium and sulphur are relatively immobile. These elements are not transported ol of the older parts of a plant. Therefore, the symptoms for the deficiencies of these elements first appear in the younger parts of the plant.

Question 7: How are the minerals absorbed by the plants?

Solution. The absorption of soil nutrients by the roots of plants occurs in two main phases – apoplast and symplast.

During the initial phase or apoplast, there is a rapid uptake of nutrients from the soil into the free spaces of plant cells. This process is passive and it usually occurs through trans-membrane proteins and ion-channels.

In the second phase or symplast, the ions are taken slowly into the inner spaces of the cells. This pathway generally involves the expenditure of energy in the form of ATP.

Question 8: What are the conditions necessary for fixation of atmospheric nitrogen by Rhizobium. What is their role in $\mathrm{N}_{2}$-fixation?

Solution. Rhizobium is a symbiotic bacterium present in the root nodules of leguminous plants. The basic requirements for Rhizobium to carry out nitrogen fixation are as follows:

(a) Presence of the enzyme nitrogenase

(b) Presence of leg-haemoglobin

(c) Non-haem iron protein, ferrodoxin as the electron-carrier

(d) Constant supply of ATP

(e) $\mathrm{Mg}^{2+}$ ions as co-factors

Rhizobium contains the enzyme nitrogenase – a Mo-Fe protein – that helps in the conversion of atmospheric free nitrogen into ammonia.

The reaction is as follows:

$\mathrm{N}_{2}+8 \mathrm{e}^{-}+8 \mathrm{H}^{+}+16 \mathrm{ATP} \rightarrow 2 \mathrm{NH}_{3}+\mathrm{H}_{2}+16 \mathrm{ADP}+16 \mathrm{Pi}$

The Rhizobium bacteria live as aerobes under free-living conditions, but require anaerobic conditions during nitrogen fixation. This is because the enzyme nitrogenase is highly sensitive to molecular oxygen. The nodules contain leg-haemoglobin, which protects nitrogenase from oxygen.

Question 9: What are the steps involved in formation of a root nodule?

Solution. Multiple interactions are involved in the formation of root nodules. The Rhizobium bacteria divide and form colonies. These get attached to the root hairs and epidermal cells. The root hairs get curled and are invaded by the bacteria. This invasion is followed by the formation of an infection thread that carries the bacteria into the cortex of the root. The bacteria get modified into rod-shaped hacterniclec A. the nericurle underan division leadina to the formation of root nodules. The nodules finally get connected with the vascular tissues of the roots for nutrient exchange.

Question 10: Which of the following statements are true? If false, correct them:

(a) Boron deficiency leads to stout axis.

(b) Every mineral element that is present in a cell is needed by the cell.

(c) Nitronen as a nutrient element is hiahly immobile in the nlants

(d) It is very easy to establish the essentiality of micronutrients because they are required only in trace quantities.

Solution. (a) True

(b) All the mineral elements present in a cell are not needed by the cell. For example, plants growing near radioactive mining sites tend to accumulate large amounts of radioactive compounds. These compounds are not essential for the plants.

(c) Nitrogen as a nutrient element is highly mobile in plants. It can be mobilised from the old and mature parts of a plant to its younger parts.

(d) True

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NCERT Solutions For Class 11 Biology Chapter 11
NCERT Solutions for Class 11 Biology chapter 11 Transport in Plants PDF – eSaral

NCERT Solutions for Class 11 Biology chapter 11 Transport in Plants PDF

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Question 1. What are the factors affecting the rate of diffusion?

Solution. -Diffusion refers to the movement of a substance (solid, liquid, gases) from the region of higher concentration to the region of lower concentration.

Factors that affect the rate of diffusion are:

-Concentration gradient

-Membrane permeability

-Temperature

– Pressure

Question 2. What are porins? What role do they play in diffusion?

Solution. Porins are the proteins that form large pores in the outer membranes of certain cell organelles such as plastids, mitochondria and some bacteria they allow molecules up to the size of small proteins to pass through them.

The role played by porins in diffusion:

– Certain bigger molecules take longer time to cross the biological membrane based on the concentration gradient; hence, their exchange occurs with the help of the porin proteins.

– There are special porin proteins which carry out the diffusion of water they are known as aquaporins.

Question 3. Describe the role played by protein pumps during active transport in plants.

Solution. The role played by the protein pumps during active transport is as follows:

-Protein pump transfers the substances against the concentration gradient.

-They transfer the movement of the molecule from the lower concentration gradient to a higher concentration gradient (uphill transport). They consume energy in order to carry out the transport.

-In root hairs, they pump ions from soil to the cytoplasm of the epidermal cells.

Question 4. Explain why pure water has the maximum water potential.

Solution. Water potential is defined as the concentration of water in the system. It is directly proportional to the kinetic energy possessed by the water.

When the water is pure, and no solutes are dissolved in water, the motion of the molecules is random. When the solute is added it the random motion of the molecule reduces hence reducing the water potential.

Hence it is considered that at standard temperature and pressure, the water potential of the molecules is zero. It is represented by the symbol psi $\left(\Psi_{w}\right)$.

Question 5. Differentiate between the following:

Solution (a) Diffusion and Osmosis

NCERT Solutions for Class 11 Biology chapter 11 Transport in Plants PDF Image 1

(b) Transpiration and Evaporation (3marks)

NCERT Solutions for Class 11 Biology chapter 11 Transport in Plants PDF Image 2

(c) Osmotic Pressure and Osmotic Potential

NCERT Solutions for Class 11 Biology chapter 11 Transport in Plants PDF Image 3

(d) Imbibition and Diffusion

NCERT Solutions for Class 11 Biology chapter 11 Transport in Plants PDF Image 4

(e) Apoplast and Symplast pathways of movement of water in plants.

NCERT Solutions for Class 11 Biology chapter 11 Transport in Plants PDF Image 5

(f) Guttation and Transpiration





Question 6. Briefly describe water potential. What are the factors affecting it?

Solution. Water potential is defined as the concentration of water in the system. It is directly proportional to the kinetic energy possessed by the water.

When the water is pure, and no solutes are dissolved in water, the motion of the molecules is random. When the solute is added it the random motion of the molecule reduces hence Teducing the water potential. Hence it is considered that at standard temperature and pressure, the water potential of the molecules is zero. It is represented by the symbol psi $\left(\Psi_{w}\right)$

– Factors affecting the water potential are as follows:

1. Solute potential

$\left(\psi_{\mathrm{s}}\right)$

2. Pressure

potential $\left(\psi_{\mathrm{p}}\right)$

3. Matric potential

4. Temperature

5. Pressure

6. Loss or gain of water

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

Solution. If the pressure greater than the atmospheric pressure is applied, then the water potential of the solution increases.

This happens because when the water enters the plant, then the pressure is built up against the cell wall. When the water enters the cell wall then it becomes turgid. The value of the potential of water is positive and is measured in bars.

Question 8. (a) With the help of well-labelled diagrams, describe the process of plasmolysis in plants, giving appropriate examples.

(b) Explain what will happen to a plant cell if it is kept in a solution having higher water potential.

Solution. A solution having higher water potential is referred to as the hypotonic solution. If the plant cell is kept in the solution having higher water potential then the plant cell will absorb water which is referred to endosmosis. This will make the plant cell turgid. When the plant cell is saturated with the water the protoplasm will exert the pressure on the walls hence causing the movement to stop.

a) Plasmolysis is the shrinkage of protoplast from the cell wall due to the excessive loss of water from the plants



-When a plant cell is placed in a hypertonic solution, the plant cell loses water which reduces the

turgor pressure making the plant cell flaccid. The loss of water from the plant is known as

wilting.

-When the plant is watered and the cells absorb water they become turgid.

b) A solution having high water potential is referred to as the hypotonic solution. If the plant cell is kept in the solution having higher water potential then the plant cell will absorb water which is referred to endosmosis. This will make the plant cell turgid. When the plant cell is saturated with the water the protoplasm will exert the pressure on the walls hence causing the movement to stop.

Question 9. How is the mycorrhizal association helpful in absorption of water and minerals in plants?

Solution. Mycorrhizae is the symbiotic association between the fungus and the roots of higher plants. It is formed as a network of fungal filaments around the young roots or by penetrating root cells.

Role of the fungus in mycorrhizae:

– Fungal hyphae help in the absorption of water where the roots of higher plants cannot penetrate. Since the pore size of hyphae is larger it is able to extract more amount of water from the ground.

Role of roots in symbiotic association:

– The roots of higher plants provide sugar and nitrogen-containing compounds.

– Other significance mycorrhizae help in germination of seeds of certain plants such as “Pinus” without the association of mycorrhizae the plant is unable to germinate.

Question 10. What role does root pressure play in water movement in plants?

Solution. Root pressure is the positive pressure that develops in the roots of plants by the active absorption of nutrients from the soil.

The role played by the root pressure in the transport of water:

– Root pressure is responsible to push up water up to small heights in the stem of a plant.

– It can provide a modest push in the process of water transport and establishes the continuous chain of water molecules in the xylem.

– Root pressure does not account for the majority of water transport; most plants meet their need by the transpiratory pull.

Question 11. Describe the transpiration pull model of water transport in plants. What are the factors

influencing transpiration? How is it useful to plants?

Solution. Transpiration pull model/Cohesion tension model

In tall trees, the transportation of water occurs with the help of the transpirational pull generated by high temperature or loss of water from the stomatal pores. This is called the cohesion-tension model of water transport.

– During the daytime, the water is lost through transpiration (by the leaves to the surroundings) causes the guard cells and other epidermal cells to become flaccid. The water is transported from the xylem and negative pressure or tension in the xylem vessels is created from the surfaces of the leaves to the tips of the roots, through the stem.

Ls a result, the water present in the xylem is pulled as a single column from the stem. The cohesion and adhesion forces of the water molecules and the cell walls of the xylem vessels prevent the water column from splitting.

Factors affecting the rate of transpiration are as follows:

1. Temperature: The rate of transpiration increases with the increase in atmospheric temperature because the temperature increases the rate of water evaporation from the cell surface, opens the stomata and decreases the relative humidity of the atmosphere.

2. Relative humidity: Relative humidity is the percentage of water vapour present in the air at a given time and temperature relative to the amount required to be present to make the air saturated at that temperature. The rate of transpiration is inversely proportional to the relative humidity.

3. Light: In most plants, the stomata open in the presence of light and close in darkness. The rate of transpiration increases in the presence of light and decreases in darkness.

4. Wind: Transpiration is lower in still air because the water vapour accumulates around the transpiring organs. The movement of air increases the rate of transpiration by removing the saturated air around the leaves.

Importance of transpiration:

1. The ascent of xylem sap: Ascent of sap mostly occurs due to the transpiration pull exerted by

the transpiration of water.

2. Maintaining the cell shape: Transpiration maintains the shape and structure of plant parts

by keeping the cells turgid.

3. Removal of excess water: Plants absorb far more amount of water than actually required

by them. Thus, transpiration removes the excess of water.

Question 12. Discuss the factors responsible for ascent of xylem sap in plants.

Solution. Factors affecting the ascent of xylem sap in plants:

1. Cohesion: It is the mutual attraction between water molecules.

2. Adhesion: It is the attraction of water molecules to polar surfaces (such as the surface of treachery elements).

3. Surface tension: Water molecules are attracted to each other in the liquid phase more than to

water in the gas phase.

4. Root pressure: It is positive pressure which pushes sap from below due to the active absorption by roots.

5. Transpiration pull: Transpiration in aerial parts brings the xylem sap under negative pressure or tension due to their continuous withdrawal of water.

Question 13. What essential role does the root endodermis play during mineral absorption in plants?

Solution. – Endodermal cells play an essential role in the transport of a substance. It carries out active and passive transport of the substances across the membrane.

-The selected minerals pass through it due to the deposition of suberin and lignin.

-The endodermal cells also contain the transport proteins known as porins they help the plant to carry out the quick transport of the substances.

-The presence of these proteins helps in the quick transport of the materials across the membrane.

Question 14. Explain why xylem transport is unidirectional and phloem transport bi-directional.

Solution. During the growth of a plant, its leaves act as the source of food as they carry out photosynthesis. The phloem conducts the food from the source to the sink (the part of the plant requiring or storing food).

During spring, this process is reversed as the food stored in the sink is mobilised toward the growing buds of the plant, through the phloem. Thus, the movement of food in the phloem is bidirectional (i.e., upward and downward).

The transport of water in the xylem takes place only from the roots to the leaves. Therefore, the movement of water and nutrients in the xylem is unidirectional.

Question 15. Explain pressure flow hypothesis of translocation of sugars in plants.

Solution – Leaves are referred to as the food factory of the plants the food is synthesised in the mesophyll cells of the leaves. The glucose synthesised during the process of photosynthesis is converted into the circulating sugar sucrose. The sucrose is then moved into the source cells this happens when the water moves from xylem vessels to the adjacent phloem vessels hence increasing the hydrostatic pressure and transporting the sucrose into the sieve cells of the phloem.

– The sucrose present in the sink gets converted into storage sugar starch or cellulose. This reduces the hydrostatic pressure in the cells of the sink region. Due to the difference in the pressure, the sugar is transported from the source region to the sink region.

Question 16. What causes the opening and closing of guard cells of stomata during transpiration?

Solution. – Opening and closing of stomata are caused by the exosmosis and endosmosis of water in the guard cells. In normal condition, the stomata tend to open during the day time in response to light and to close at night.

– The opening or closing of the stomata is a change in the turgidity of the guard cells. When water flows inside the guard cells it becomes turgid and kidney-shaped as the thin walls get extended and thick walls become slightly concave causing the opening of the stomal aperture.

– When the guard cells lose water turgor pressure reduces due to water loss (or water stress), the elastic inner walls regain their original shape, the guard cells become flaccid and dumbbell-shaped causing the closure of stomata.

Also Read,

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Class 11 Biology Book Chapterwise Free pdf.

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NCERT Solutions for Class 11 Biology chapter 10 Cell Cycle and Cell Division PDF- eSaral

NCERT Solutions for Class 11 Biology chapter 10 Cell Cycle and Cell Division PDF

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Question 1: What is the average cell cycle span for a mammalian cell?

Solution. The average cell cycle span for a mammalian cell is approximately 24 hours.

Question 2: Distinguish cytokinesis from karyokinesis.

Solution.

NCERT Solutions for Class 11 Biology chapter 10 Cell Cycle and Cell Division PDF Image 1

Question 3: Describe the events taking place during interphase.

Solution. Interphase involves a series of changes that prepare a cell for division. It is the period during which the cell experiences growth and DNA replication in an orderly manner. Interphase is divided into three phases.

(i) $G_{1}$ phase

(ii) S phase

(iii) $G_{2}$ phase

$\mathbf{G}_{1}$ phase $-$ It is the stage during which the cell grows and prepares its DNA for replication. In this phase, the cell is metabolically active

S phase $-\mathrm{It}$ is the stage during which DNA synthesis occurs. In this phase, the amount of DNA (per cell) doubles, but the chromosome number remains the same

$\mathbf{G}_{2}$ phase $-$ In this phase, the cell continues to grow and prepares itself for division. The proteins and RNA required for mitosis are synthesised during this stage.

NCERT Solutions for Class 11 Biology chapter 10 Cell Cycle and Cell Division PDF Image 2

Question 4: What is $\mathrm{G}_{0}$ (quiescent phase) of cell cycle?

Solution. $\mathrm{G}_{0}$ or quiescent phase is the stage wherein cells remain metabolically active, but do not proliferate unless called to do so. Such cells are used for replacing the cells lost during injury.

Question 5: Why is mitosis called equational division?

Solution. Mitosis is the process of cell division wherein the chromosomes replicate and get equally distributed into two daughter cells. The chromosome number in each daughter cell is equal to that in the parent cell, i.e., diploid. Hence, mitosis is known as equational division.

Question 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

Solution. (i) Metaphase

(ii) Anaphase

(iii) Zygotene of meiosis I

(iv) Pachytene of meiosis I

Question 7: Describe the following:

(a) synapsis

(b) bivalent

(c) chiasmata

Draw a diagram to illustrate your answer.

Solution. (a) Synapsis

The pairing of homologous chromosomes is called synapsis. This occurs during the second stage of prophase I or zygotene.

NCERT Solutions for Class 11 Biology chapter 10 Cell Cycle and Cell Division PDF Image 3

(b) Bivalent

Bivalent or tetrad is a pair of synapsed homologous chromosomes. They are formed during the zygotene stage of prophase I of meiosis.

NCERT Solutions for Class 11 Biology chapter 10 Cell Cycle and Cell Division PDF Image 4

(c) Chiasmata

Chiasmata is the site where two non-sister chromatids of homologous chromosomes have crossed over. It represents the site of cross-over. It is formed during the diplotene stage of prophase I of meiosis.

NCERT Solutions for Class 11 Biology chapter 10 Cell Cycle and Cell Division PDF Image 5

Question 8: How does cytokinesis in plant cells differ from that in animal cells?

Solution.



Question $9:$ Find examples where the four daughter cells from meiosis are equal in size and where they are found unequal in size.

Solution. (a) Spermatogenesis or the formation of sperms in human beings occurs by the process of meiosis. It results in the formation of four equal-sized daughter cells

(b) Oogenesis or the formation of ovum in human beings occurs by the process of meiosis. It results in the formation of four daughter cells which are unequal in size.

Question 10: Distinguish anaphase of mitosis from anaphase I of meiosis.

Solution.



Question 11: List the main differences between mitosis and meiosis.

Solution.



Question 12: What is the significance of meiosis?

Solution. Meiosis is the process involving the reduction in the amount of genetic material. It comprises two successive nuclear and cell divisions, with a single cycle of DNA replication. As a result, at the end of meiosis II, four haploid cells are formed.

Significance of meiosis

1. Meiosis maintains the chromosome number from generation to generation. It reduces the chromosome number to half so that the process of fertilisation restores the original number in the zygote.

2. Variations are caused by the cross-over and the random distribution of homologous chromosomes between daughter cells. Variations play an important role in evolution.

3. Chromosomal mutations are brought about by the introduction of certain abnormalities. These chromosomal mutations may be advantageous for an individual.

Question 13: Discuss with your teacher about

(i) haploid insects and lower plants where cell-division occurs, and

(ii) some haploid cells in higher plants where cell-division does not occur.

Solution. Mitotic cell division cannot take place without DNA replication in S phase. Two important events take place during $S$ phase – one is the synthesis or duplication of DNA and the other is the duplication of the centriole. DNA duplication is important as it maintains the chromosome number in the daughter cells. Mitosis is an equational division. Therefore, the duplication of DNA is an important step.

Question 15: Can there be DNA replication without cell division?

Solution. There can be DNA replication without cell division. During cell division, the parent cell gets divided into two daughter cells. However, if there is a repeated replication of DNA without any cell division, then this DNA will keep accumulating inside the cell. This would increase the volume of the cell nucleus, thereby causing cell expansion. An example of DNA duplication without cell division is commonly observed in the alivary glands of Drosophila. The chromosome undergoing repeated DNA duplication is known as polytene chr

Question 16: Analyse the events during every stage of cell cycle and notice how the following two parameters change

(i) Number of chromosomes (N) per cell

(ii) Amount of DNA content (C) per cell

Solution. During meiosis, the number of chromosomes and the amount of DNA in a cell change.

(i) Number of chromosomes (N) per cell

During anaphase I of the meiotic cycle, the homologous chromosomes separate and start moving toward their respective poles. As a result, the bivalents get divided into two sister chromatids and receive half the chromosomes present in the parent cell. Therefore, the number of chromosomes reduces in anaphase $\mathrm{I}$.

(ii) Amount of DNA content (C) per cell

During anaphase II of the meiotic cycle, the chromatids separate as a result of the splitting of the centromere. It is the centromere that holds together the sister chromatids of each chromosome. As a

result, the chromatids move toward their respective poles. Therefore, at each pole, a haploid number of chromosomes and a haploid amount of DNA are present.

During mitosis, the number of chromosomes remains the same. The DNA duplicated in the S phase gets separated in the two daughter cells during anaphase. As a result, the DNA content (C) of the two newlyformed daughter cells remains the same.

Also Read,

Download Class 11 Chemistry Notes Free pdf.

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NCERT Solutions for Class 11 Biology chapter 9 Biomolecules PDF- eSaral

NCERT Solutions for Class 11 Biology chapter 9 Biomolecules PDF

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In this article, we have listed NCERT Solutions for Class 11 Biology chapter 9 Biomolecules PDF that you can download to start your preparations anytime.

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Question 1. What are macromolecules? Give examples.

Solution: -Macromolecules are the molecules which have the molecular weight in the range of 1000 daltons (Da) or above 1000 Dalton and are formed by the polymerization of several similar or different types of molecules.

– They are found in the acid-insoluble fraction of the tissue extract.

-Examples of the macromolecules are:

1. Proteins

2. Lipids

3. Nucleic acid

4. Polysaccharide

Question 2. Illustrate a glycosidic, peptide and phosphodiester bond.

Solution:

NCERT Solutions for Class 11 Biology chapter 9 Biomolecules PDF Image 1

NCERT Solutions for Class 11 Biology chapter 9 Biomolecules PDF Image 2

Question 3. What is meant by the tertiary structure of proteins?

Solution. The tertiary structure is a structure formed in proteins when the beta-pleated sheets or alpha helix (secondary structures of proteins) folds further to form a hollow woolen balllike structure. (1marks for correct explanation 1 marks for illustration)

Diagram showing the folding of proteins in the tertiary structure.

NCERT Solutions for Class 11 Biology chapter 9 Biomolecules PDF Image 3

Question 4. Find and write down structures of 10 interesting small molecular weight biomolecules. Find if there is any industry which manufactures the compounds by isolation. Find out who are the buyers.

Solution.(a)

NCERT Solutions for Class 11 Biology chapter 9 Biomolecules PDF Image 4

NCERT Solutions for Class 11 Biology chapter 9 Biomolecules PDF Image 5

NCERT Solutions for Class 11 Biology chapter 9 Biomolecules PDF Image 6

Question 5. Proteins have primary structure. If you are given a method to know which amino acid is at either of the two termini (ends) of a protein, can you connect this information to purity or homogeneity of a protein.

Solution. The primary structure of protein constitutes amino acids. The two terminal of proteins are amino terminal and carboxyl terminal. If we have the information about the amino acids present at the two ends of the protein the upon comparing the given information about the terminal amino acid. It can be said that the protein which is isolated is pure or homogeneous if it has the same amino acids at the amino terminal and carboxyl terminal.

Question 6. Find out and make a list of proteins used as therapeutic agents. Find other applications of proteins (e.g. cosmetics etc.).

Solution. List of proteins used as therapeutic agents:

1. Insulin

2.Oxytocin

3.Antidiuretic Hormone (ADH),

4. Thrombin

5.Fibrinogen

6.Renin

7.Immunoglobulin

8. Diastase

9.Streptokinase

Other applications:

i. Cosmetics:- Proteins such as collagen and elastin are used in anti-ageing creams. Keratin and cystine protein is present in shampoos, which prevents the hair from damage.

ii. Sweeteners:-Certain proteins are artificial sweeteners, and taste modifiers. There are seven known proteins which are artificial sweeteners. Examples: Razzein ,Thaumatin, Monelin, Curculin, Mabinlin, Miraculin and Pentadin.

iii. Dietary proteins:- Protein is essential for body-building. Protein isolates such as whey, pea, soy, casein etc. are consumed before or after extensive exercise as they help in the formation of muscle.

Question 7. Explain the composition of triglyceride.

Solution. When three fatty acids are esterified with glycerol, it is known as triglycerides.

Glycerol is the compound which has three carbon atoms having three alcohol attached to. them. Whereas fatty acids are the high carbon containing acids.

– When the alcohol groups of glycerol and carboxylic groups of fatty acids react to form an ester bond with the elimination of water molecule which is formed is triglycerides.

Diagram showing composition and bonding in triglycerides



Question 8. Can you describe what happens when milk is converted into curd or yoghurt, from your understanding of proteins?

Solution: -When the inoculum of Lactobacillus bacteria is added to lukewarm milk, the formation of curd takes place. The Lactobacillus multiplies and secretes lactic acid. The lactic acid decreases the $\mathrm{pH}$ of the milk and makes it acidic.

-Due to this decrease in $\mathrm{pH}$, the secondary and tertiary structure of milk protein “casein” gets denatured and gets precipitated.

Question 9. Can you attempt building models of biomolecules using commercially available atomic models (ball and stick models)?

Solution. -Structure of biomolecules can be represented using ball and stick model.

-The bonds which hold the atoms are represented by sticks, whereas the atoms are depicted as balls. In order to show different molecules, different colour of balls can be used.

Question 10. Attempt titrating an amino acid against a weak base and discover the number of dissociating (ionizable) functional groups in the amino acid.

Solution. -When the weak base is added to the amino acid solution, the carboxylic group of amino acid dissociates and releases the hydrogen ion in order to balance the $\mathrm{pH}$ of the medium.

-When the $\mathrm{pH}$ of the medium reduces and becomes acidic, the amino group takes up the hydrogen ion from the medium and balances the $\mathrm{pH}$.

– When all the amino groups are positively charged, and all the carboxylic groups are aegatively charged in an amino acid, then that state of amino acid is referred to as “Zwitter-ion.”

$\underline{\text { Diagram showing titration of amino acid with a weak base }}$



Question 11. Draw the structure of the amino acid alanine.

Solution. Diagram showing structure of Alanine amino acid



Question 12. What are gums made of? Is fevicol different?

Solution. Gums are made up of polymeric substances which are secreted from the plants. It is a secondary metabolite which is formed as a result of a metabolic pathway of the plants. Fevicol is an adhesive which has a similar function as that of gum the difference between the two is that gum is a plant-based product, and fevicol is a synthetic product.

Question 13. Find out a qualitative test for proteins, fats and oils, amino acids and test any fruit juice, saliva, sweat and urine for them.

Solution: Qualitative tests for detecting the proteins, amino acids and fats are as follows:

Biuret test for proteins: The biuret test is the chemical test performed in order to detect the presence of protein in the sample.

-Biuret reagent reacts with peptide bond of proteins and forms a purple coloured complex.

ii. Grease test for Fats: When the drop of oil or fat is applied on a brown paper, it makes the paper translucent. This confirms the presence of fats or oil in the sample.

iii. Ninhydrin test for amino acids: Ninhydrin reagent (2,2-dihydroxyindane-1,3-dione) reacts with the primary and secondary amino group of amino acids.

It forms a dark blue coloured dye which is known as Ruheman’s purple, confirming the presence of amino acid in the sample.

Exception: Proline amino acid forms a yellow complex with ninhydrin due to the presence of the guanidine ring.





Question 14. Find out how much cellulose is made by all the plants in the biosphere and compare it with how much of paper is manufactured by man, and hence, what is the consumption of plant material by man annually. What a loss of vegetation?

Solution. -Cellulose is the structural component of the plants. The main cover of the biosphere is made up of plants. There are approximately $3.04$ trillion trees present on the biosphere and the amount of cellulose that is produced by them is 100 billion tonnes.

– It requires $16-20$ trees to make a ton of paper. Total 180 million tonnes of cellulose is produced every year, which involves the sacrifice of approximate 5 crore trees. Besides, that there is a dependency on plants for the plant products such as timber, food, medicines etc., which is also contributing to the loss of vegetation.

Question 15. Describe the important properties of enzymes.

Solution. Properties of enzymes are described as follows:

1. Chemical nature:

– Enzymes are the nitrogenous compounds that are made up of protein. All enzymes are protein in nature, but not all proteins are enzymes.

-Enzymes work as biocatalyst which means they help in accelerating the biochemical reaction by forming the enzyme-substrate complex and decreasing the activation energy of the reaction.

2. Molecular welght:

-Enzymes are categorized as macromolecules as they are proteins.

-The molecular weight of the enzyme ranges from 1000 dalton to 1,00,000 Dalton or above.

3.Changeless form:

-Enzymes are not consumed in the chemical reaction.

-They combine with the substrate forms enzyme-substrate complex and ronverts into the product.

4. Substrate specificity:

– Different biochemical reactions are catalyzed by a different enzyme. Every enzyme has its own substrate like every lock has its own key. The theory related to the enzyme-substrate complex is referred to as a lock and key hypothesis. $\quad$

5. Heat sensitivity:

-Enzymes are sensitive to temperature. They work best at the optimum temperature that is from $30-40$ degree Celsius. Enzymes get inhibited at low temperature, and they get denatured at high temperature.

6. pH sensitivity:

Enzymes are sensitive to $\mathrm{pH}$ they activity of enzymes is best at the optimum temperature they undergo denaturation when they are subjected to high $\mathrm{pH}$ or low $\mathrm{pH}$.

Also Read,

Download Class 11 Chemistry Notes pdf.

Download Class 11 Biology Book Chapterwise pdf.

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NCERT Solutions For Class 11 Biology Chapter 8
NCERT Solutions for Class 11 Biology chapter 8 Cell The Unit of Life PDF – eSaral

NCERT Solutions for Class 11 Biology chapter 8 Cell The Unit of Life PDF

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In this article, we have listed NCERT Solutions for Class 11 Biology chapter 8 Cell The Unit of Life PDF that you can download to start your preparations anytime.

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Download NCERT Solutions for Class 11 Biology chapter 8 Cell The Unit of Life PDF



Question 1: Which of the following is not correct?

(a) Robert Brown discovered the cell.

(b) Schleiden and Schwann formulated the cell theory.

(c) Virchow explained that cells are formed from pre-existing cells.

(d) A unicellular organism carries out its life activities within a single cell.

Solution. (a) Robert Brown did not discover the cell. The cell was discovered by Robert Hook.

Question 2: New cells generate from

(a) bacterial fermentation

(b) regeneration of old cells

(c) pre-existing cells

(d) abiotic materials

Solution. Answer: (c)

According to the biogenic theory, new cells can only arise from pre-existing cells. Only complete cells, in favourable conditions, can give rise to new cells.

Question 3: Match the following

(a) Cristae

(i) Flat membranous sacs in stroma

(b) Cisternae

(ii) Infoldings in mitochondria

(c) Thylakoids

(iii) Disc-shaped sacs in Golgi apparatus

Solution.

NCERT Solutions for Class 11 Biology chapter 8 Cell The Unit of Life PDF Image 1

Question 4: Which of the following is correct:

(a) Cells of all living organisms have a nucleus.

(b) Both animal and plant cells have a well defined cell wall.

(c) In prokaryotes, there are no membrane bound organelles.

(d) Cells are formed de novo from abiotic materials.

Solution Answer: (c)

Membrane-bound organelles are organelles surrounded by a double membrane. Nucleus, mitochondria, chloroplasts, etc., are examples of such organelles. These cell organelles are absent from prokaryotes.

(a) Only eukaryotic cells have nuclei. They are absent from prokaryotes.

(b) Cell walls are only present in plant cells. They are absent from all animal cells.

(d) All cells arise from pre-existing cells.

Question 5: What is a mesosome in a prokaryotic cell? Mention the functions that it performs.

Solution. Mesosome is a convoluted membranous structure formed in a prokaryotic cell by the invagination of the plasma membrane. Its functions are as follows:

(1) These extensions help in the synthesis of the cell wall, replication of DNA. They also help in the equal distribution of chromosomes into the daughter cells.

(2) It also increases the surface area of the plasma membrane to carry out various enzymatic activities.

(3) It helps in secretion processes as well as in bacterial respiration.

Question $6:$ How do neutral solutes move across the plasma membrane? Can the polar molecules also move across it in the same way? If not, then how are these transported across the membrane?

Solution. Plasma membrane is the outermost covering of the cell that separates it from the environment. It regulates the movement of substances into the cell and out from it. It allows the entry of only some substances and prevents the movement of other materials. Hence, the membrane is selectively-permeable

Movement of neutral solutes across the cell membrane $-$ Neutral molecules move across the plasma membrane by simple passive diffusion. Diffusion is the movement of molecules from a region of higher concentration to a region of lower concentration.

Movement of polar molecules across the cell membrane – The cell membrane is made up of a phospholipid bilayer and proteins. The movement of polar molecules across the nonpolar lipid bilayer requires carrier-proteins. Carrier-proteins are integral protein particles having certain affinity for specific solutes. As a result, they facilitate the transport of molecules across the membrane.

Question 7: Name two cell-organelles that are double membrane bound. What are the characteristics of these two organelles? State their functions and draw labelled diagrams of both.

Solution. Mitochondria and chloroplasts are the two organelles that are double-membrane-bound.

NCERT Solutions for Class 11 Biology chapter 8 Cell The Unit of Life PDF Image 1

Characteristics of the mitochondria

Mitochondria are double- membrane-bound structures. The membrane of a mitochondrion is divided into the inner and outer membranes, distinctly divided into two aqueous compartments – outer and inner compartments. The outer membrane is very porous (containing the organelle), while the inner membrane is deeply-folded.

These folds are known as cristae. Cristae increase the surface area inside the cell. They are the sites for ATP-generating chemical reactions. The membrane of a mitochondrion contains specific enzymes meant for specific mitochondrial functions. Hence, the mitochondria are the sites for aerobic respiration. They have their own DNA and ribosomes. Thus, they are able to make their own proteins. This is why they are considered as semi-autonomous organelles

Characteristics of chloroplasts

Chloroplasts are double-membrane-bound structures.

They are divided into outer and inner membranes, further divided into two distinct regions:

(i) Grana are stacks of flattened discs containing chlorophyll molecules. The flattened membranous sacs are called thylakoids. The thylakoids of adjacent grana are connected by membranous tubules called stroma lamellae.

(ii) Stroma is a homogenous mixture in which grana are embedded. It contains several enzymes that are used for the synthesis of carbohydrates and proteins. It also contains its own DNA and ribosomes.

NCERT Solutions for Class 11 Biology chapter 8 Cell The Unit of Life PDF Image 3

Functions of the mitochondria:

(i) They are the sites for cellular respiration.

(ii) They provide energy in the form of ATP for all vital activities of living cells.

(iii) They have their own DNA and ribosomes. Hence, they are regarded as semi-autonomous organelles.

(iv) They have several enzymes, intermediately required for the synthesis of various chemicals such as fatty acids, steroids, and amino acids.

Functions of chloroplasts:

(i) They trap solar energy and utilise it for manufacturing food for plants. Hence, they are involved in the process of photosynthesis.

(ii) They contain the enzymes required for the synthesis of carbohydrates and proteins.

Question 8: What are the characteristics of prokaryotic cells?

Solution. Prokaryotic cell is a unicellular organism lacking membrane-bound organelles.

The characteristics of prokaryotic cells are as follows:

(i) Most of them are unicellular.

(ii) They are generally small in size. The size of a prokaryotic cell varies from $0.5$ – 5 \mum.

(iii) The nuclear region of a prokaryotic cell is poorly-defined because of the absence of a nuclear membrane. Hence, a prokaryotic cell lacks a true nucleus.

(iv) The genetic materials of prokaryotic cells are naked. They contain single, circular chromosomes. In addition to the genomic DNA, they have a small, circular plasmid DNA.

(v) They have specialised membranous structures called mesosomes. Mesosomes are formed by the invagination of the cell membrane. These extensions help in the synthesis of the cell wall, replication of DNA. They also help in the equal distribution of chromosomes into the daughter cells

(vi) Membrane-bound cell organelles such as mitochondria, plastids, and endoplasmic reticulum are absent from a prokaryotic cell.

(vii) Most prokaryotic cells contain a three-layered structure – outermost glycocalyx, middle cell wall, and the innermost plasma membrane. This structure acts as a protective unit.Examples of prokaryotic cells include blue green algae, bacteria, etc.

Question 9: Multicellular organisms have division of labour. Explain.

Solution. Multicellular organisms are made up of millions and trillions of cells. All these cells perform specific functions. All the cells specialised for performing similar functions are grouped together as tissues in the body. Hence, a particular function is carried out by a group of cells at a definite place in the body. Similarly, different functions are carried out by different groups of cells in an organism. This is known as division of labour in multicellular organisms

Question 10: Cell is the basic unit of life. Discuss in brief.

Solution. Cells are the basic units of life capable of doing all the required biochemical processes that a normal cell has to do in order to live. The basic needs for the survival of all living organisms are the same. All living organisms need to respire, digest food for obtaining energy, and get rid of metabolic wastes.

Cells are capable of performing all the metabolic functions of the body. Hence, cells are called the functional units of life.

Question 11: What are nuclear pores? State their function.

Solution. Nuclear pores are tiny holes present in the nuclear membrane of the nucleus. They are formed by the fusion of two nuclear membranes.

These holes allow specific substances to be transferred into a cell and out from it. They allow molecules such as RNA and proteins to move in both directions, between the nucleus and the cytoplasm.

Question 12 : Both lysosomes and vacuoles are endomembrane structures, yet they differ in terms of their functions. Comment.

Solution. Lysosomes are membrane-bound vesicular structures holding a variety of enzymes such as lipases, proteases, and carbohydrases. The purpose of lysosomes is to digest worn out cells. They are involved in the intracellular digestion of foreign food particles and microbes. Sometimes, they also act as suicidal bags. They are involved in the self digestion of cells. They are a kind of waste disposal systems of a cell. On the other hand, vacuoles are storage sacs found in cells. They might store the waste products of cells. In unicellular organisms, the food vacuole contains the consumed food particles. It also plays a role in expelling excess water and some wastes from the cell

Question 13: Describe the structure of the following with the help of labelled diagrams.

(i) Nucleus (ii) Centrosome

Solution. (i) Nucleus

Nucleus controls all the cellular activities of the cell. It is spherical in shape. It is composed of the following structures:

Nucleoplasm/Nuclear matrix: It is a homogenous granular fluid present inside the nucleus. It contains the nucleolus and chromatin. Nucleolus is a spherical structure that is not and some basic proteins called histones

NCERT Solutions for Class 11 Biology chapter 8 Cell The Unit of Life PDF Image 4

(ii) Centrosome

Centrosome consists of two cylindrical structures called centrioles. Centrioles lie perpendicular to each other. Each has a cartwheel-like organisation.

A centriole is made up of microtubule triplets that are evenly spaced in a ring. The adjacent triplets are linked together. There is a proteinaceous hub in the central part of a of cilia and flagella.

NCERT Solutions for Class 11 Biology chapter 8 Cell The Unit of Life PDF Image 5

Question 14: What is a centromere? How does the position of centromere form the basis of classification of chromosomes. Support your answer with a diagram showing the position of centromere on different types of chromosomes.

Solution. Centromere is a constriction present on the chromosomes where the chromatids are held together.

Chromosomes are divided into four types based on the position of the centromere.

(i) Metacentric chromosome

The chromosomes in which the centromere is present in the middle and divides the chromosome into two equal arms is known as a metacentric chromosome



(ii) Sub-metacentric chromosome

The chromosome in which the centromere is slightly away from the middle region is known as a sub-metacentric chromosome. In this, one arm is slightly longer than the other.



(iii) Acrocentric chromosome

The chromosome in which the centromere is located close to one of the terminal ends is known as an acrocentric chromosome. In this, one arm is extremely long and the other is extremely short.



(iv) Telocentric chromosome

The chromosome in which the centromere is located at one of the terminal ends is known as a telocentric chromosome.



Also Read,

Download Class 11 Chemistry Notes.

Download Class 11 Biology Book Chapterwise.

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NCERT Solutions For Class 11 Biology Chapter 7
NCERT Solutions for Class 11 Biology chapter 7 Structural Organization in Animals PDF – eSaral

NCERT Solutions for Class 11 Biology chapter 7 Structural Organization in Animals PDF

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Question 1. Answer in one word or one line.

(i) Give the common name of Periplanata americana.

(ii) How many spermathecae are found in earthworm?

(iii) What is the position of ovaries in cockroach?

(iv) How many segments are present in the abdomen of cockroach?

(iv) How many segments are present in the abdomen of cockroach?

Solution: i) American cockroach / Cockroach

ii) Four pairs of spermathecae are found in the earthworm.

iii) Two large ovaries are found laterally lying in the $2^{\text {nd }}$ to $6^{\text {th }}$ abdominal segments.

iv) Ten segments are present in the abdomen of cockroach.

v) Malpighian tubules are present in the junction of midgut and hindgut.\

Question 2. Answer the following:

(i) What is the function of nephridia?

(ii) How many types of nephridia are found in earthworm based on their location?

Solution: i) The function of nephridia in earthworm is similar to the kidney in vertebrates. It is involved in the regulation of composition and volume of the body fluids.

ii) There are three types of nephridia, found in the earthworm based on their location. They are:

a) Septal nephridia $-$ it is found on both sides of intersegmental septa of segment 15.

b) Integumentary nephridia $-$ it is found attached to the body wall of segment 3.

c) Pharyngeal nephridia $-$ found in the form of three paired tufts in the segments $-4,5$ and 6 .

Question 3. Draw a labelled diagram of the reproductive organs of an earthworm.

NCERT Solutions for Class 11 Biology chapter 7 Structural Organization in Animals PDF Image 1

Question 4. Draw a labelled diagram of alimentary canal of a cockroach.

NCERT Solutions for Class 11 Biology chapter 7 Structural Organization in Animals PDF Image 2

Question 5. Distinguish between the followings

(a) Prostomium and peristomium

(b) Septal nephridium and pharyngeal nephridium

Solution. a) Prostomium and peristomium

NCERT Solutions for Class 11 Biology chapter 7 Structural Organization in Animals PDF Image 3

NCERT Solutions for Class 11 Biology chapter 7 Structural Organization in Animals PDF Image 4

b) Septal nephridium and pharyngeal nephridium.

NCERT Solutions for Class 11 Biology chapter 7 Structural Organization in Animals PDF Image 5

Question 6. What are the cellular components of blood?

Solution: Blood is mainly composed of plasma and cells. The cellular components include Red blood cells or erythrocytes, White blood cells or leucocytes and platelets.

Question 7. What are the following and where do you find them in animal body.

(a) Chondriocytes

(b) Axons

(c) Ciliated epithelium

Solution: a) Chondrocytes – They are the cells of cartilage. They are found enclosed in the small cavities within the matrix of the cartilage. They are found in the outer ear joints, between the adjacent bones of the vertebral column, nose tip, limbs and hands of adults.

b) Axons $-$ Axons are the tail-like structure found in the neuron. It is long and cylindrical and is involved in the transmission of nerve impulses from the cell body to the other neuron. The axon is covered by a myelin sheath, which helps in the transmission of impulses.

c) Ciliated epithelium – Ciliated epithelium is the columnar or cuboidal epithelium which bears cilia towards their free surface. They are involved in the passing or moving particles or mucus in a specific path of conduction. They are found in the fallopian tubes in the female reproductive system and bronchioles of lungs.

Question 8. Describe various types of epithelial tissues with the help of labelled diagrams.

Solution: Epithelial tissue or epithelium are the tissues found either facing the body fluid or the external environment. It provides covering or lining to certain parts of the body. The cells of the epithelium are closely packed with little intercellular matrix. The two main types of epithelial tissues are:

i) simple epithelium

ii) compound epithelium

1) Simple epithelium is made of a single layer of cells and found in the lining of body cavities, tubes and ducts. There are three types of simple epithelial tissues based on the structure of cell present in them. The types include:

a) Squamous epithelium

b) Cuboidal epithelium

c) Columnar epithelium

a) Squamous epithelium

It is made of a single layer of thin flattened cells and resembles the pavement tiles. Hence it is also called as pavement tissue. The cells have irregular boundaries. It is found in the walls of blood vessels, air sacs of lungs and also functions in the formation of diffusion boundary.

NCERT Solutions for Class 11 Biology chapter 7 Structural Organization in Animals PDF Image 6

b) Cuboidal epithelium

It is made of a single layer of cube-shaped cells and is involved in functions such as secretion and absorption. It is found in the tubular parts of the nephron and the ducts of glands.

NCERT Solutions for Class 11 Biology chapter 7 Structural Organization in Animals PDF Image 7

c) Columnar epithelium

It is made of a single layer of long and slender cells. The nuclei are present at the base of these cells.

They are involved in secretion and absorption processes. They may have cilia.

NCERT Solutions for Class 11 Biology chapter 7 Structural Organization in Animals PDF Image 8

The columnar or cuboidal epithelium bearing cilia on free surface are called ciliated epithelium. They are involved in the passing or moving particles or mucus in a specific path of conduction. They are found in the fallopian tubes in the female reproductive system and bronchioles of lungs.

ii) Compound epithelium The compound epithelium has many layers of cells stacked one above the other. It is neither involved in secretion nor absorption. It functions in protecting against mechanical and chemical stresses. It is found on the skin surface, the moist surface of the mouth or buccal cavity, the inner lining of the salivary gland and pancreatic ducts, pharynx.

NCERT Solutions for Class 11 Biology chapter 7 Structural Organization in Animals PDF Image 9

NCERT Solutions for Class 11 Biology chapter 7 Structural Organization in Animals PDF Image 10

Question 9. Distinguish between

(a) Simple epithelium and compound epithelium

(b) Cardiac muscle and striated muscle [10 marks]

(c) Dense regular and dense irregular connective tissues

(d) Adipose and blood tissue

(e) Simple gland and compound gland

Solution:











Question 10. Mark the odd one in each series:

(a) Areolar tissue; blood; neuron; tendon

(b) RBC; WBC; platelets; cartilage

(c) Exocrine; endocrine; salivary gland; ligament

(d) Maxilla; mandible; labrum; antennae

(e) Protonema; mesothorax; metathorax; coxa

Solution: a) Neuron, since it is not a connective tissue.

b) Cartilage, since it is not a component of blood.

c) Ligament, since it is not a gland.

d) Antennae, since it is not one of part of the mouth of cockroach.

e)Protonema, since all others are parts of cockroach.

Question 11. Match the terms in column I with those in column II:





Solution.



Question 12. Mention breifly about the circulatory system of earthworm.

Solution. The vascular system of Earthworm is of closed type. Hence blood is confined to heart and blood vessels. It consists of blood vessels, capillaries and heart. The circulation is uni directional due to the contractions. The blood is conducted to the gut, body wall and nerve cord by the smaller blood vessels. The Blood glands present in $4^{\text {th }}, 5^{\text {th }}$ and $6^{\text {th }}$ segments are involved in the production of blood cells and haemoglobin. The haemoglobin is dissolved in blood plasma. There is an absence of specialized breathing components hence, the moist body surface of the earthworm is involved in the respiratory exchange, which is given into the bloodstream.

Question 13. Draw a neat diagram of digestive system of frog.

Solution.



Question 14. Mention the function of the following

(a) Ureters in frog

(b) Malpighian tubules

(c) Body wall in earthworm

Solution. a) Ureters in frog: It is a component of the excretory system in a frog. In male frogs, the urine and spermatozoa are both sent out through cloaca. Hence it acts as a urinogenital duct. In female frogs, the oviduct and ureters open out separately into the cloaca.

b) Malpighian tubules: It is found in the junction of midgut and hindgut and is involved in the excretory function. Each of the tubules is lined with ciliated and glandular cells. These tubules absorb nitrogenous wastes which are converted to uric acid and is excreted through the hindgut.

c) Body wall in Earthworm: The function of the body wall is to protect the delicate internal organs against injury. It also helps in burrowing and locomotion.

Also Read,

Class 11 Chemistry Notes Download.

Class 11 Biology Book Chapterwise Download.

Class 11 Biology Exemplar Chapterwise Download.

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NCERT Solutions For Class 11 Biology Chapter 6
NCERT Solutions for Class 11 Biology chapter 6 Anatomy of Flowering Plant PDF – eSaral

NCERT Solutions for Class 11 Biology chapter 6 Anatomy of Flowering Plant PDF

Hey, are you a class 11 student and looking for ways to download NCERT Solutions for Class 11 Biology chapter 6 Anatomy of Flowering Plant PDF? If yes. Then read this post till the end.

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Download NCERT Solutions for Class 11 Biology chapter 6 Anatomy of Flowering Plant PDF



Question 1: State the location and function of different types of meristem.

Solution Meristems are specialised regions of plant growth. The meristems mark the regions where active cell division and rapid division of cells take place. Meristems are of three types depending on their location.

Apical meristem

It is present at the root apex and the shoot apex. The shoot apical meristem is present at the tip of the shoots and its active division results in the elongation of the stem and formation of new leaves. The root apical meristem helps in root elongation.

Intercalary meristem

present between the masses of mature tissues present at the bases of the leaves of grasses. It helps in the regeneration of grasses after they have been grazed by herbivores Since the intercalary meristem and the apical meristem appear early in a plant’s life, they constitute the primary meristem.

Lateral meristem

It appears in the mature tissues of roots and shoots. It is called the secondary meristem as it appears later in a plant’s life. It helps in adding secondary tissues to the plant body and in increasing the girth of plants. Examples include fascicular cambium, interfascicular cambium, and cork cambium

Question 2: Cork cambium forms tissues that form the cork. Do you agree with this statement? Explain.

Solution. When secondary growth occurs in the dicot stem and root, the epidermal layer gets broken. There is a need to replace the outer epidermal cells for providing protection to the stem and root from infections. Therefore, the cork cambium develops from the cortical region. It is also known as phellogen and is composed of thin-walled rectangular cells. It cuts off cells toward both sides. The cells on the outer side get differentiated into the cork or phellem, while the cells on the inside give rise to the secondary cortex or phelloderm. The cork is impervious to water, but allows gaseous exchange through the lenticels. Phellogen, phellem, and phelloderm together constitute the periderm.

Question $3:$ Explain the process of secondary growth in stems of woody angiosperm with help of schematic diagrams. What is the significance?

Solution. In woody dicots, the strip of cambium present between the primary xylem and phloem is called the interfascicular cambium. The interfascicular cambium is formed from the cells of the medullary rays adjoining the interfascicular cambium. This results in the formation of a continuous cambium ring. The cambium cuts off new cells toward its either sides. The cells present toward the outside differentiate into the secondary phloem, while the cells cut off toward the pith give rise to the secondary xylem. The amount of the secondary xylem produced is more than that of the secondary phloem

NCERT Solutions for Class 11 Biology chapter 6 Anatomy of Flowering Plant PDF Image 1

NCERT Solutions for Class 11 Biology chapter 6 Anatomy of Flowering Plant PDF Image 2

Stages of secondary growth in dicot stem

The secondary growth in plants increases the girth of plants, increases the amount of water and nutrients to support the growing number of leaves, and also provides support to plants.

Question 4 : Draw illustrations to bring out anatomical difference between

(a) Monocot root and dicot root

(b) Monocot stem and dicot stem

Solution. (a)Monocot root and dicot root

NCERT Solutions for Class 11 Biology chapter 6 Anatomy of Flowering Plant PDF Image 3

T.S. of dicot root (Primary)

NCERT Solutions for Class 11 Biology chapter 6 Anatomy of Flowering Plant PDF Image 4

T.S. of monocot root

(b) Monocot stem and dicot stem

NCERT Solutions for Class 11 Biology chapter 6 Anatomy of Flowering Plant PDF Image 5

T.S. of dicot stem

NCERT Solutions for Class 11 Biology chapter 6 Anatomy of Flowering Plant PDF Image 6

T.S. of monocot stem

Question $5:$ Cut a transverse section of young stem of a plant from your school garden and observe it under the microscope. How would you ascertain whether it is a monocot stem or dicot stem? Give reasons.

Solution. The dicot stem is characterised by the presence of conjoint, collateral, and open vascular bundles, with a strip of cambium between the xylem and phloem. The vascular bundles are arranged in the form of a ring, around the centrally-located pith. The ground tissue is differentiated into the collenchyma, parenchyma, endodermis, pericycle, and pith. Medullary rays are present between the vascular bundles.

NCERT Solutions for Class 11 Biology chapter 6 Anatomy of Flowering Plant PDF Image 7

T.S. of dicot stem

The monocot stem is characterised by conjoint, collateral, and closed vascular bundles, scattered in the ground tissue containing the parenchyma. Each vascular bundle is surrounded by sclerenchymatous bundle-sheath cells. Phloem parenchyma is absent and water-containing cavities are present.



T.S of monocot stem

Question 6: The transverse section of a plant material shows the following anatomical features, (a) the vascular bundles are conjoint, scattered and surrounded by sclerenchymatous bundle sheaths (b) phloem parenchyma is absent. What will you identify it as?

Solution. The monocot stem is characterised by conjoint, collateral, and closed vascular bundles, scattered in the ground tissue containing the parenchyma. Each vascular bundle is surrounded by sclerenchymatous bundle-sheath cells. Phloem parenchyma and medullary rays are absent in monocot stems.

Question 7: Why are xylem and phloem called complex tissues?

Solution. Xylem and phloem are known as complex tissues as they are made up of more than one type of cells. These cells work in a coordinated manner, as a unit, to perform the various functions of the xylem and phloem.

Xylem helps in conducting water and minerals. It also provides mechanical support to plants. It is made up of the following components:

– Tracheids (xylem vessels and xylem tracheids)

– Xylem parenchyma

– Xylem fibres

Tracheids are elongated, thick-walled dead cells with tapering ends. Vessels are long, tubular, and cylindrical structures formed from the vessel members, with each having lignified walls and large central cavities. Both tracheids and vessels lack protoplasm. Xylem fibres consist of thick walls with an almost insignificant lumen. They help in providing mechanical support to the plant. Xylem parenchyma is made up of thin-walled parenchymatous cells that help in the storage of food materials and in the radial conduction of water

Phloem helps in conducting food materials. It is composed of:

– Sieve tube elements

– Companion cells

– Phloem parenchyma

– Phloem fibres

Sieve tube elements are tube-like elongated structures associated with companion cells. The end walls of sieve tube elements are perforated to form the sieve plate. Sieve tube elements are living cells containing cytoplasm and nucleus. Companion cells are parenchymatous in nature. They help in maintaining the pressure gradient in the sieve tube elements. Phloem parenchyma helps in the storage of food and is made up of long tapering cells, with a dense cytoplasm. Phloem fibres are made up of elongated sclerenchymatous cells with thick cell walls

Question 8: What is stomatal apparatus? Explain the structure of stomata with a labelled diagram.

Solution. Stomata are small pores present in the epidermis of leaves. They regulate the process of transpiration and gaseous exchange. The stomatal pore is enclosed between two bean-shaped guard cells. The inner walls of guard cells are thick, while the outer walls are thin. The guard cells are surrounded by subsidiary cells. These are the specialised epidermal cells present around the guard cells. The pores, the guard cells, and the subsidiary cells together constitute the stomatal apparatus.



Question 9: Name the three basic tissue systems in the flowering plants. Give the tissue names under each system.

Solution.



Question 10: How is the study of plant anatomy useful to us?

Solution. The study of plant anatomy helps us to understand the structural adaptations of plants with respect to diverse environmental conditions. It also helps us to distinguish between the strength of wood. This is useful in utilising it to its potential. The study of various plant fibres such as jute, flax, etc., helps in their commercial exploitation.

Question 11: What is periderm? How does periderm formation take place in dicot stem?

Solution. Periderm is composed of the phellogen, phellem, and phelloderm.

During secondary growth, the outer epidermal layer and the cortical layer are broken because of the cambium. To replace them, the cells of the cortex turn meristematic, giving rise to cork cambium or phellogen. It is composed of thin-walled, narrow and rectangular cells.

Phellogen cuts off cells on its either side. The cells cut off toward the outside give rise to the phellem or cork. The suberin deposits in its cell wall make it impervious to water. The inner cells give rise to the secondary cortex or phelloderm. The secondary cortex is parenchymatous.

Question 12: Describe the internal structure of a dorsiventral leaf with the help of labelled diagrams.

Solution. Dorsiventral leaves are found in dicots. The vertical section of a dorsiventral leaf contains three distinct parts.

[1] Epidermis:

Epidermis is present on both the upper surface (adaxial epidermis) and the lower surface (abaxial epidermis). The epidermis on the outside is covered with a thick cuticle. Abaxial epidermis bears more stomata than the adaxial epidermis.

[2] Mesophyll:

Mesophyll is a tissue of the leaf present between the adaxial and abaxial epidermises. It is differentiated into the palisade parenchyma (composed of tall, compactly-placed cells) and the spongy parenchyma (comprising oval or round, loosely-arranged cells with inter cellular spaces). Mesophyll contains the chloroplasts which perform the function of photosynthesis.

[3] Vascular system:

The vascular bundles present in leaves are conjoint and closed. They are surrounded by thick layers of bundle-sheath cells.



Also Read,

Class 11 Chemistry Notes PDF Download.

Class 11 Biology Book Chapterwise PDF Download.

Class 11 Biology Exemplar Chapterwise PDF Download.

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NCERT Solutions For Class 11 Biology Chapter 5
NCERT Solutions for Class 11 Biology chapter 5 Morphology of Flowering Plant PDF – eSaral

NCERT Solutions for Class 11 Biology chapter 5 Morphology of Flowering Plant PDF

Hey, are you a class 11 student and looking for ways to download NCERT Solutions for Class 11 Biology chapter 5 Morphology of Flowering Plant PDF? If yes. Then read this post till the end.

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In this article, we have listed NCERT Solutions for Class 11 Biology chapter 5 Morphology of Flowering Plant PDF that you can download to start your preparations anytime.

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Question 1. What is meant by modification of root? What type of modification of root is found in the:

(a) Banyan tree (b) Turnip (c) Mangrove trees

Solution: When roots of certain plants change its form, origin and role in order to perform functions such as storage of food, support and respiration it is refer to as modification of roots.

-They arise from the different parts of the plant, and such types of the root are referred to as the “adventitious roots.”

Root modification found in the following plants:

(a) Banyan tree: In Banyan tree roots are referred to as “prop roots.” They arise from the stem and reaches to the ground; the primary purpose of these roots is to provide support to the plant as it grows.

(b) Turnip: In turnip, the taproot of the plant gets modifies to store food. A similar type of modification is present other plants such as carrot, radish, and adventitious root of sweet potato.

(c) Mangroves trees:- Mangroves trees (Rhizophora) grows in the marshy and swampy area due to these roots are unable to perform respiration.

-Roots of these plants come out from the ground and grow vertically upwards to absorb oxygen for respiration, such types of modified roots are known as “pneumatophores.”

Question 2. Justify the following statements on the basis of external features:

(i) Underground parts of a plant are not always roots.

(ii) Flower is a modified shoot.

Solution: (i) Underground parts of a plant are not always roots.

-In certain plants such as potato, ginger turmeric, zamikand, colocasia, stems grow underground to perform the function of food storage and vegetative propagation.

-Underground stem serves as the organ of perennation and prevents the plant from the unfavorable condition.

(ii) Flower is a modified shoot.

The apical meristem of stem gets modified into floral meristem a and starts bearing flower then it is referred to as the modification of shoot into the flower.

The changes happen during modification are:

-The elongation of internodes stops.

– Condensation of the axis takes place.

– Position of the leaves on apex and nodes is replaced by floral meristem.

– Formation of floral appendages takes place on the apex and the on nodes in the lateral direction.

Question 3. How is a pinnately compound leaf different from a palmately compound leaf?

Solution: -Pinnately compound leaf and palmately compound leaf are the modification of compound leaf.

-Following are the differences between pinnately compound leaf and palmately compound leaf



Diagram showing pinnately compound leaves and palmately compound leaf.



Question 4. Explain with suitable examples of the different types of phyllotaxy.

Solution: -Phyllotaxy refers to the pattern in which leaves are arranged on the stem or branch.

-There are three types of arrangement are as follows





Question 5. Define the following terms: (2marks each definition)

(a) aestivation

(b) placentation

(c) actinomorphic

(d) zygomorphic

(e) superior ovary

(f) perigynous flower

(g) epipetalous stamen

Solution: (a) Aestivation :

Aestivation refers to the mode of arrangement of sepals and petal in the floral bud in comparison to the other members of the same whorl.

There are five types of aestivation found in Corolla:

– Valvate in Calotropis

– Twisted in China rose lady finger and cotton

– Imbricate in Casia and Gulmohur

– Vexillary in pea and bean flowers

(b) Placentation:

Placentation refers to the pattern in which ovules are arranged in the ovary.

– Types of placentation are as follows:

-Marginal in peas

-Axile in China rose tomato and lemon

-Parietal in mustard and argemone

-Free central in Dianthus and Primrose

-Basal in sunflower and mustard

(c) Actinomorphic

-When the symmetry of a flower is radial, that is when it can be divided into two radial halves in a radial plane passing through the center, then it referred to as the

“Actinomorphic flower.”

-Examples of Actinomorphic flower are:

-Mustad, Chilli, Dhatura

It is represented by the symbol $\oplus$ when explained in floral formula and diagram.

(d) Zygomorphic:

-When the symmetry of the flower is bilateral, which means when the flower is divided into two half, each half is the mirror image of another half then it is referred to as the

“Zygomorphic flower.”

– Examples of the zygomorphic flower are a pea, gulmohur, bean, and cassia

– It is represented by the symbol \% when explained in floral formula and diagram.

(e) Superior ovary:

-When the gynoecium occupies the highest position while the other parts are situated below as the ovary also occupies the highest position are referred to as the superior ovary.

– Flowers in which superior ovary is present are known as hypogynous flowers.

-Examples are mustard, China rose, and brinjal.



Diagram showing superior ovary

(f) Perigynous Flower:

-Flower in which gynoecium is present at the center and other parts of the flower such as calyx corolla and androecium are present at the edge of the thalamus then it is known as a perigynous flower.

-Examples of plant having perigynous flower are plum, rose, and peach.

Diagram showing perigynous flower



(g) Epipetalous stamen:

In certain plants where stamen (androecium) is attached to the petals, such type of conditions are referred to as epipetalous stamen.

– Example of flower having epipetalous stamen is brinjal.

Question 6. Differentiate between

(a) Racemose and cymose inflorescence

(b) Fibrous root and adventitious root

(c) Apocarpous and syncarpous ovary

Solution: (a) Racemose and cymose inflorescence



(b) Fibrous root and adventitious root





Apocarpous and syncarpous ovary



Question 7. Draw the labelled diagram of the following:

(i) gram seed

(ii) V.S. of maize seed

Solution: Diagram showing gram seeds



Diagram showing V.S. of maize seeds.



Question 8. Describe modifications of stem with suitable examples

Sôlution: When stem changes its form origin and shape to perform functions such as storage, protection, vegetative propagation, it is then referred to as the modified stem.

Modification in stem and function performed by them are as follows:

Food storage stems-

In certain plants such as the roots perform the function of storage of food. The storage roots also serve as the organ of perennation in order to prevent the plant from the unfavorable condition.

Examples of food storage stems are potato, ginger, turmeric, zaminkand, colocasia etc.

-Stem Tendrils for support-

Tendrils are the modification of axillary, but they are slender and spirally coiled.

-They help the plant to climb along with the support.

-Examples of the plants having stem tendrils are gourds such as cucumber, pumpkins, watermelon and grapevines.

-Thorns for protection:-

-When the axillary bud gets modified into a woody, a straight prickly and pointed structure such types of modification in the stem is referred to as thorns. The function of such modification is to prevent the plant from getting consumed by animals.

-Examples of plants having thorns are Citrus, Bougainvillea.

Water storage stems-(1 marks for correct description and example)

– Certain plants that grow in the arid region in such types of plants stem gets flattened or turns fleshy cylindrical in order to store water.

Examples are Euphorbia and opuntia

-Stems for vegetative propogation:-

– In certain plants stems get buried and spread underground, when the old part of the plant dies it leads to the formation of the new plant.

-Examples are grass and strawberry.

-In certain plants, the soft and slender lateral branches that arise from the main branch. After growing for a certain time, it gets buried underground and starts bearing roots for absorbing water after some time the lateral branch serves as the main plant. Examples are mint and jasmine

-In aquatic plants, the node and internode bear the tuft of roots. Examples are Pistia and Eichhornia

-Lateral stems of plants originate from the main axis grows beneath the soil, comes vertically upwards and further bears leaves.

– Example are pineapple, banana and chrysanthemum

Question 9. Take one flower each of the families Fabaceae and Solanaceae and write its semi-technical description. Also, draw their floral diagram after studying them.

Flower of the family Fabaceae



Technical description for Fabaceae:

Fabaceae family was earlier known as Papilionoideae

Fabaceae is the sub family of family Leguminosae

Vegetative characters of the sub-famliy are as follows:

Stem: The stem is erect and is mainly a climber.

Leaves: Phyllotaxy is alternate; they can be pinnately compound or simple compound.

-Leaf base is pulvinate and has stipule.

-Venetion is reticulate.

Floral

character

s of the

family

Fabacae

Inflorescence: Arrangement of the flower is racemose that is plant bear flowers on the lateral region.

Flower type -Flower is bisexual and bilaterally symmetrical known as “zygomorphic”.

Calyx:

-Sepals are five and are fused, which means gamosepalous.

-Aestivation of sepals is either valvate or imbricate.

Corolla:-

-Petals are five in number that is polypetalous, consisting of a posterior standard, two lateral wings, two anterior ones forming a keel (enclosing stamens and pistil),

Aestivation is vexillary

Androecium:

-Stamens are ten in number and are diadelphous ( 9 fused and 1 free)

-Anther has two layers and is known as dithecous

Gynoecium:

-Ovary is superior

-It has single carpel mono carpellary, unilocular with many ovules, and style is single

Fruit: It is mainly legume; seeds in fruit can be one to many. The endosperm is absent hence non-endospermic

Floral Formula: (1 marks for correct formulae)

$\% \mathrm{~K}_{(5)} \mathrm{C}_{1+2+(2)} \mathrm{A}_{(9)+1}$

$\mathrm{G}_{1}$

Floral Diagram



Examples and economic

Importance

-Pulses, which includes gram arhar moong soyabean, are consumed as the protein source. – Soya bean and groundnut provides edible oil.

-Indigofera provides dye.

-SunHemp provides fiber.

-Sesbania and trifolium are the sources of fodder.

-Mulathi has medicinal properties.

-Ornamental plants are lupin and sweet pea.

Flower of Family Solanacae



The technical description is as follows:

Family Solanacae is large and is known as ‘potato family’.

Its distribution is in tropics, subtropics and even temperate zones.

Vegetative characters of family Solanacae are as follows:

Plants include herbs, shrubs and small trees.

Stem:

-The stem is herbaceous, rarely woody, aerial; erect, cylindrical, branched, solid or hollow, hairy or glabrous,

-The stem is underground in potato (Solanum tuberosum)

Leaves:

-Phyllotaxy is alternate. Leaves are simple, pinnately compound and exstipulate.

-Venation in leaves is reticulate.

Floral Characters of family Solanacae are:

Inflorescence : Arrangement of the flower is cymose.

Flower:

The flower is bisexual and symmetry is radial type known as actinomorphic

Calyx:

-Sepals are five in number and are united.

– Aestivation of sepals is valvate

Corolla:

– Petals are five in number, and they are united.

-Aestivation of petals is valvate

Androecium:

Stamens are five and are epipetalous.

Gynoecium:

The gynoecium is bicarpellary, and carpels are fused known as syncarpous;

The ovary is superior and bilocular.

The placenta is swollen with many ovules.

Fruits: Fruit type is berry or capsule.

Seeds: Seeds are multiple and contain endosperm.

Floral Formula: (1 marks for correct formulae)

$\oplus_{+}^{*} \mathrm{~K}_{(5)} \widehat{\mathrm{C}_{(5)}} \mathrm{A}_{5} \underline{\mathrm{G}}_{(2)}$

Floral Diagram:



Examples with economic importance:

They serve as sources of food in tomato, brinjal, potato

They are spices such as chilly

Plants for medicinal importance are belladonna and ashwagandha

They serve the fumigatory purpose in tobacco

They are used for ornamental purposes in petunia.

Question 10. Describe the various types of placentations found in flowering plants.

Solution. Placentation refers to the pattern in which ovules are arranged in the ovary.

-Types of placentation are as follows:





Question 11. What is a flower? Describe the parts of a typical angiosperm flower.

Solution. When the apical meristem of the shoot takes responsibility for sexual reproduction, then it gets converted to floral meristem then it gives rise to flower in angiosperms. (1 mark for explaining flower)

-A flower is formed when elongation of internodes stops, and condensation of the main axis takes place finally the position of the leaves on apex and nodes is replaced by the floral meristem.

If the flower is inverted, then it has four whorls from outside to inside they are calyx (sepals), corolla (petals), androecium and gynoecium.

Calyx and corolla are the accessory organs, whereas androecium and gynoecium are the reproductive organs.

In certain plants, the calyx and corolla are fused; this condition is known as perianth.

Sarts of the flowers are as follows:

i. Calyx: The outermost part of the flower and the lowermost whorl is refer as the calyx.

-The calyx is also called as sepals which are green in color and shape is leaf like.

– The major function of sepal’s protection of flower in the bud stage and protection of reproductive structures of the flower.

ii. Corolla: Corolla is also referred to as petals and forms the second whorl of the flower that is present inside the sepals and outside the stamens. Corolla is the brightest and colored part of the flower, and its shape is flat and leaf like. The corolla can be fused or united and also possess scent. The main function of the corolla is to attract pollinators for pollination.

iii. Androecium: The androecium refers to the male reproductive part pf the plant. Parts of the stamens are anther and filament. The anther is di- thecous that is it has two layers and is a bilobed structure each lobe has two chambers known as pollen sacs. The pollen sacs carry out the production of pollen grains and stores it till dehiscence.

iv. Gynoecium: The innermost whorl of the flower is the female reproductive part known as gynoecium.

-It is referred to as carpel or pistil. Carpel consists of three main parts, which are stigma, style and ovary.

-Stigma is the region which receives the pollen grains, and style connects stamen to the ovary. The basal part of the the gynoecium is ovary each ovary contains one to many ovules attached to a flat cushion like structure placenta. Male gamete in the pollen grain fuse egg cell present in ovary later ovary develops into fruit and ovule into seeds

-Carpels are referred to apocarpous if there are multiple free carpels for examples lily and rose and it is referring to as syncarpous when it is fused and is single, for example, mustard and tomato.



Diagram showing different parts of the flower

Question 12. How do the various leaf modifications help plants?

Solution. The primary function of the leaves is to perform photosynthesis, but they undergo modification in order to support the plant for various other purposes such as:

1. Tendril for support: Leaves in certain plants get converted into the spirally coiled structure in order to support the plant. Example is pea plant.

2. Spines for protection: In certain xerophytic plants the leaves are modified into spines in order to prevent water loss through transpiration and preventing the plant from getting browsed by the animals. Examples are opuntia and Euphorbia.

3. Fleshy leaves for storage: Leaves becomes fleshy a swollen and stores food in the form of carbohydrates. Examples are onion and garlic

4. Nutrient deriving leaves: Leaves of certain insectivorous plants gets modified in order to obtain nitrogen from insects, examples are pitcher plant, Venus-fly trap .

Question 13. Define the term inflorescence. Explain the basis for the different types of inflorescence in flowering plants Inflorescence is reffered to as the arrangement of flowers on the stem.

Solution: Defination:- Inflorescence referred to as the arrangement or the pattern of the flowers on the floral axis.

The inflorescence is of two types:

i. Racemose inflorescence: In the racemose type of inflorescence, the later branch bears flowers in a pattern where old flowers are at the base and new flowers at the top. The main axis does not get converted into a floral axis and continues to grow. The succession in flowers grows from top to bottom is known as acropetal succession.

ii. Cymose inflorescence: In the cymose type of inflorescence, the main axis bears flower hence limiting the growth of the main axis. The flowers are arranged in such a manner in which old flower is present at the apex and young flower at the base, such type of succession is known as basipetal succession.



Question 14. Write the floral formula of the actinomorphic, bisexual, hypogynous flower with five united sepals, five free petals, five free stamens and two united carpleswith the superior ovary and axile placentation.

Solution. The description is given for the floral formulae for family Solanaceae



Question 15. Describe the arrangement of floral members in relation to their insertion on thalamus.

Solution: Based on the position of calyx, corolla and androecium in respect of the ovary on the thalamus, the flowers are of three types:

Hypogynous Flower:

In hypogynous flower, the carpel is present at the highest position, and other parts such as calyx corolla and androecium are present below it, such types of flowers are referred to as hypogynous.

The ovary in flowers is said to be a superior ovary.

Examples are mustard, china rose and brinjal.

Diagram showing hypogynous flower



Perigynous flower

In perigynous flower, the gynoecium is situated in the centre, and other parts of the flower are located on the rim of the thalamus almost at the same level, it is called perigynous.

The ovary in such type of flower is known as half inferior, examples are plum rose, peach.

Diagram showing perigynous flower



Epigynous flower: ( 1 mark for correct explanation and $0.5,0.5$ marks for the position of ovary and example)

In epigynous flowers, the margin of thalamus grows upward enclosing the ovary completely and getting fused with it, the other parts of flower arise above the ovary.

The position of ovary in such type of plant is said to be inferior.

Example of flowers is guava and cucumber, and the ray florets of sunflower.

Diagram showing Epigynous flower



Also Read,

Class 11 Chemistry Notes free PDF Download.

Class 11 Biology Book Chapterwise free PDF Download.

Class 11 Biology Exemplar Chapterwise free PDF Download.

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NCERT Solutions For Class 11 Biology Chapter 4
NCERT Solutions for Class 11 Biology chapter 4 Animal Kingdom PDF – eSaral

NCERT Solutions for Class 11 Biology chapter 4 Animal Kingdom PDF

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If you want to learn and understand class 11 Biology chapter 4 “Animal Kingdom” in an easy way then you can use these solutions PDF.

NCERT Solutions helps students to Practice important concepts of subjects easily. Class 11 Biology solutions provide detailed explanations of all the NCERT questions that students can use to clear their doubts instantly.

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In this article, we have listed NCERT Solutions for Class 11 Biology chapter 4 Animal Kingdom PDF that you can download to start your preparations anytime.

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Question 1: What are the difficulties that you would face in classification of animals, if common fundamental features are not taken into account?

Solution. For the classification of living organisms, common fundamental characteristics are considered.

If we consider specific characteristics, then each organism will be placed in a separate group and the entire objective of classification would not be achieved.

Classification of animals is also important in comparing different organisms and judging their individual evolutionary significance. If only a single characteristic is considered, then this objective would not be achieved.

Question 2: If you are given a specimen, what are the steps that you would follow to classify it?

Solution. There is a certain common fundamental feature that helps in classification of living organisms. The features that can be used in classification are as follows.

(1)



(2)



(3)



(4)



(5)



On the basis of above features, we can easily classify a specimen into its respective category.

Question $3:$ How useful is the study of the nature of body cavity and coelom in the classification of animals?

Solution. Coelom is a fluid filled space between the body wall and digestive tract. The presence or absence of body cavity or coelom plays a very important role in the classification of are examples of coelomates. On the other hand, the animals in which the body cavity is not lined by mesoderm are known as pseudocoelomates. In such animals, mesoderm is An example of acoelomates is platyhelminthes.



Question 4: Distinguish between intracellular and extracellular digestion?

Solution.



Question 5: What is the difference between direct and indirect development?

Solution.



Question 6: What are the peculiar features that you find in parasitic platyhelminthes?

Solution. Taenia (Tapeworm) and Fasciola (liver fluke) are examples of parasitic platyhelminthes.

Peculiar features in parasitic platyhelminthes are as follows.

1. They have dorsiventrally flattened body and bear hooks and suckers to get attached inside the body of the host.

2. Their body is covered with thick tegument, which protects them from the action of digestive juices of the host.

3. The tegument also helps in absorbing nutrients from the host’s body.

Question 7: What are the reasons that you can think of for the arthropods to constitute the largest group of the animal kingdom?

Solution. The phylum, Arthropoda, consists of more than two-thirds of the animal species on earth. The reasons for the success of arthropods are as follows.

i. Jointed legs that allow more mobility on land

ii. Hard exoskeleton made of chitin that protects the body

iii. The hard exoskeleton also reduces water loss from the body of arthropods making them more adapted to terrestrial conditions

Question 8: Water vascular system is the characteristic of which group of the following:

(a) Porifera

(b) Ctenophora

(c) Echinodermata

(d) Chordata

Solution. Water vascular system is a characteristic feature of the phylum, Echinodermata. It consists of an array of radiating channels, tube feet, and madreporite. The water vascular system helps in locomotion, food capturing, and respiration.

Question 9: “All vertebrates are chordates but all chordates are not vertebrates”. Justify the statement.

Solution. The characteristic features of the phylum, Chordata, include the presence of a notochord and paired pharyngeal gill slits. In sub-phylum Vertebrata, the notochord present in embryos gets replaced by a cartilaginous or bony vertebral column in adults. Thus, it can be said that all vertebrates are chordates but all chordates are not vertebrates.

Question 10: How important is the presence of air bladder in Pisces?

Solution. Gas bladder or air bladder is a gas filled sac present in fishes. It helps in maintaining buoyancy. Thus, it helps fishes to ascend or descend and stay in the water current.

Question 11: What are the modifications that are observed in birds that help them fly?

Solution. Birds have undergone many structural adaptations to suit their aerial life. Some of these adaptations are as follows.

(i) Streamlined body for rapid and smooth movement

(ii) Covering of feathers for insulation

(iii) Forelimbs modified into wings and hind limbs used for walking, perching, and swimming

(iv) Presence of pneumatic bones to reduce weight

(v) Presence of additional air sacs to supplement respiration

Question 12: Could the number of eggs or young ones produced by an oviparous and viviparous mother be equal? Why?

Solution. The numbers of eggs produced by an oviparous mother will be more than the young ones produced by a viviparous mother. This is because in oviparous animals, the development of young ones takes place outside the mother’s body. Their eggs are more prone to environmental conditions and predators. Therefore, to overcome the loss, more eggs are produced by mothers so that even under harsh environmental conditions, some eggs might be able to survive and produce young ones. On the other hand, in viviparous organisms, the development of young ones takes place in safe conditions inside the body of the mother. They are less exposed to environmental conditions and predators. Therefore, there are more chances of their survival and hence, less number of young ones is produced compared to the number of eggs

Question 13: Segmentation in the body is first observed in which of the following:

(a) Platyhelminthes

(b) Aschelminthes

(c) Annelida

(d) Arthropoda

Solution. The body segmentation first appeared in the phylum, Annelida (annulus meaning little ring).

Question 14: Match the following:



Solution.



Question 15: Prepare a list of some animals that are found parasitic on human beings.

Solution.



Also Read,

Class 11 Chemistry Notes free PDF.

Class 11 Biology Book Chapterwise free PDF.

Class 11 Biology Exemplar Chapterwise free PDF.

If you have any Confusion related to NCERT Solutions for Class 11 Biology chapter 4 Animal Kingdom PDF then feel free to ask in the comments section down below.

To watch Free Learning Videos on Class 11 Biology by Kota’s top Doctor’s Faculties Install the eSaral App
NCERT Solutions For Class 11 Biology Chapter 3
NCERT Solutions for Class 11 Biology chapter 3 Plant Kingdom PDF – eSaral

NCERT Solutions for Class 11 Biology chapter 3 Plant Kingdom PDF

Hey, are you a class 11 student and looking for ways to download NCERT Solutions for Class 11 Biology chapter 3 Plant Kingdom PDF? If yes. Then read this post till the end.

In this article, we have listed NCERT Solutions for Class 11 Biology chapter 3 Plant Kingdom in PDF that are prepared by Kota’s top Doctor’s Faculties by keeping Simplicity in mind.

If you want to learn and understand class 11 Biology chapter 3 “Plant Kingdom” in an easy way then you can use these solutions PDF.

NCERT Solutions helps students to Practice important concepts of subjects easily. Class 11 Biology solutions provide detailed explanations of all the NCERT questions that students can use to clear their doubts instantly.

If you want to score high in your class 11 Biology Exam then it is very important for you to have a good knowledge of all the important topics, so to learn and practice those topics you can use eSaral NCERT Solutions.

In this article, we have listed NCERT Solutions for Class 11 Biology chapter 3 Plant Kingdom PDF that you can download to start your preparations anytime.

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Download NCERT Solutions for Class 11 Biology chapter 3 Plant Kingdom PDF



Question 1. What is the basis of classification of algae?

Solution: Algae are classified into three main classes – Chlorophyceae, Phaeophyceae, and

Rhodophyceae based on the following factors:

(a) Major photosynthetic pigment present

(b) Cell wall composition

(c) Form of stored food

(d) Number of flagella and position of the insertion



Question 2. When and where does reduction division take place in the life cycle of a liverwort, a moss, a fern, a gymnosperm and an angiosperm?

Solution:











Question 3. Name three groups of plants that bear archegonia. Briefly describe the life cycle of any one of them.

Solution: Archegonium are the female sex organs which produce the female gametes or eggs. They are present in the life cycles of bryophytes, pteridophytes, and gymnosperms.

The life cycle of a fern (Dryopteris)

– Dryopteris are common ferns with pinnate-compound leaves.

– Their main plant-body is sporophytic.

-The sporangia are borne on the ventral surfaces of mature leaves.

– Each sporangium has spore mother cells, and they undergo meiosis to produce haploid spores.

– Upon maturation, these spores dehisce and germinate to give rise to a heart-shaped gametophyte called prothallus.

– The prothallus has both male and female sex organs called antheridia and archegonia, respectively.

– The antheridia produce sperms which swim in the water to reach to the archegonia. The archegonia produce the egg.

– A zygote is formed as a result of fertilisation. The zygote matures into an embryo which in turn develops into a new sporophyte. The young plant comes out of the archegonium of the parent gametophyte.



Question 4. Mention the ploidy of the following: protonemal cell of a moss; primary endosperm nucleus in dicot, leaf cell of a moss; prothallus cell of a fern; gemma cell in Marchantia; meristem cell of monocot, ovum of a liverwort, and zygote of a fern.

Solution. – protonema cell of a moss: Haploid

– primary endosperm nucleus in dicot: Triploid

– leaf cell of a moss: Haploid

– prothallus cell of a fern: Haploid

– gemma cell in Marchantia: Haploid

– meristem cell of monocot: Diploid

– ovum of a liverwort: Haploid

– zygote of a fern: Diploid

Question 5. Write a note on economic importance of algae and gymnosperms.

Solution: Economic importance of Algae:

a. Agar is obtained commercially from Gelidium and Gracilaria, which is widely used in the preparation of jellies, puddings, cream, etc. Also, agarose, which is obtained from agar, is widely used in biotechnology labs for making gels to run DNA samples.

b. Carrageenin is used as an emulsifier in chocolates, paints, and toothpaste. It is obtained from the red algae.

c. Some of the marine algae such as Porphyra, Sargassum, and Laminaria are edible. Chlorella and Spirulina are rich in proteins. Hence, they are used as food supplements.

d. Antibiotic chlorellin is extracted out from Chlorella. Red algae – Corallina is used in treating worm infections.

Economic importance of Gymnosperm:

a. Conifers provide an enormous amount of softwood for construction, plywood, paper industry etc.

b. Seeds of Pinus (Chilgoza) are edible.

c. Ephedrine drug is obtained from Ephredra used in asthma and bronchitis. An anticancer drug Taxol is obtained from Taxus.

d. Resins are used commercially for manufacturing sealing waxes and water-proof paints. A type of resin known as turpentine is obtained from various species of Pinus. Sawdust of conifers is used in making linoleum and plastics.

Question 6. Both gymnosperms and angiosperms bear seeds, then why are they classified separately?

Solution: – The seeds of the gymnosperm are naked, whereas a membrane covers the seeds of the angiosperm. Hence, both gymnosperms and angiosperms are classified separately.

A. The other difference between angiosperms and gymnosperms is that the diversity of angiosperm is greater than the gymnosperm. The higher diversity indicated the angiosperms are adaptive to terrestrial ecosystems.

Question 7. What is heterospory? Briefly comment on its significance. Give two examples.

Solution: – Heterospory is the presence of two kinds of spores on the same plant.

– These spores differ in size. The smaller one is known as microspore (germinates to form the male gametophyte-pollen grain), and the larger one is known as megaspore (germinates to form the female gametophyte-egg). The male gametophyte releases the pollens, and these reach the female gametophyte to fuse with the egg. Inside the female gametophyte, fertilisation results in the development of the zygote.

– This retention and germination of the megaspore within the megasporangium ensure proper development of the zygote into a young embryo. The evolution of the seed habit is related to the retention of the megaspore. Heterospory is thus considered an important step in evolution as it is a precursor to the seed habit. The heterospory is the first step of evolution of seed development in gymnosperms and angiosperms.

– Heterospory evolved first in pteridophytes such as Selaginella and Salvinia.

Question 8. Explain briefly the following terms with suitable examples:-

(i) protonema,

(ii) antheridium,

(iii) archegonium,

(iv) diplontic,

(v) sporophyll,

(vi) isogamy

Solution (i) Protonema: The predominant-first stage of the life cycle of a moss is the protonema stage, which develops directly from a spore. It is a creeping, green, branched and frequently filamentous stage. Example: Sphagnum.

(ii) Antheridium: It is the male sex organ of bryophytes and pteridophytes, which is surrounded by a jacket of sterile cells. They produce biflagellate antherozoids.

Example: Selaginella.

(iii) Archegonium: It is the female sex organ of bryophytes, pteridophytes, and gymnosperms. They produce biflagellate antherozoids.

Example: Cycas.

(iv) Diplontic: It represents the gametophytic phase during the lifecycle of angiosperm and gymnosperm (seed-bearing plants). In this, the diploid sporophyte is dominant. If the gametophytic phase is represented by the single to the few-celled haploid gametophyte, such kind of lifecycle is termed as diplontic.

Example: Pinus.

(v) Sporophyll: The main plant body of the pteridophytes is sporophytic. The sporophytes bear sporangia that are subtended by leaf-like appendages called sporophylls.

Example: Nephrolepis.

(vi) Isogamy: It is a mode of reproduction in which there is a union or fusion of gametes of the same size and structure.

Example: Spirogyra.

Question 9. Differentiate between the following:-

(i) red algae and brown algae

(ii) liverworts and moss

(iii) homosporous and heterosporous pteridophyte

(iv) syngamy and triple fusion

Solution: (i) red algae and brown algae



(ii) liverworts and moss



(iii) homosporous and heterosporous pteridophyte



(iv) syngamy and triple fusion



Question 10. How would you distinguish monocots from dicots?

Solution.





Question 11. Match the following (column I with column II)



Solution.



Question 12. Describe the important characteristics of gymnosperms.

Solution. – The term gymnosperm is a Greek word (gymnos – naked, sperma – seeds), i.e., the seeds of these plants are naked – not enclosed in fruits. Their ovules are not enclosed by any ovary wall and remain exposed, both before and after fertilisation.

– They consist of tap roots. The roots of Pinus show symbiotec association with mycorrhiza, roots of Cycas are specialized roots called coralloid roots and are associated with $\mathrm{N}_{2}$ fixing cyanobacteria.

-The stems are branched (Pinus, Cedrus) or unbranched (Cycas). The body of gymnosperm ranges from medium to tall trees and shrubs. The redwood tree Sequoia is one of the tallest trees in the world.

-The leaves can be compound (pinnate in Cycas) or simple (in Pinus). Their leaves are needle-like, have a thick cuticle and sunken stomata – these help in preventing water loss.

– Gymnosperms are heterosporous as they bear two kinds of spores – microspores and megaspores.

– The flowers are absent in gymnosperms. The microsporophylls and megasporophylls are arranged to form compact male and female cones.

– Mostly pollination occurs through wind and pollen grains reach the pollen chamber of the ovule through the micropyle.

– The female and male gametophytes are dependent on the sporophyte. The seeds contain haploid endosperms and remain uncovered.

Also Read,

Download Class 11 Chemistry Notes free PDF.

Download Class 11 Biology Book Chapterwise free PDF.

Download Class 11 Biology Exemplar Chapterwise free PDF.

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