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CLASS 11th CHAPTER -22 Chemical Coordination and Integration |Biology | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Biology Chapter -22 Chemical Coordination and Integration includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Biology exams must go through NCERT Solutions for Class 11 Chapter -22 Chemical Coordination and Integration . NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

Class 11th Chapter -22 Chemical Coordination and Integration | NCERT BIOLOGY SOLUTION |

Page No: 341

Exercises

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

Answer
(a) Exocrine gland: Glands that discharge their secretions into ducts are known as exocrine glands.

(b) Endocrine gland: Glands that do not discharge their secretions into ducts are known as endocrine glands. These glands discharge their secretions directly into the blood.

(c) Hormone: Hormones are non-nutrient chemicals which act as intercellular messengers and are produced in trace amounts.
 
2. Diagrammatically indicate the location of the various endocrine glands in our body.
 
Answer
Endocrine glands in Human
 
3. List the hormones secreted by the following:
(a) Hypothalamus (b) Pituitary (c) Thyroid
(d) Parathyroid (e) Adrenal (f) Pancreas
(g) Testis (h) Ovary (i) Thymus
(j) Atrium (k) Kidney (l) G-I Tract
 
Answer

(a) Hypothalamus
→ Thyrotropin Releasing Hormone (TRH)
→ Gonadotrophin releasing hormone (GnRH)
→ Somatotropin-releasing hormone
→ Adrenocorticotrophin-releasing hormone
→ Prolactin releasing Hormone
→ Growth-inhibiting hormone
→ Melanocyte-inhibiting hormone

(b) Pituitary
→ Growth Hormone (GH)
→ Prolactin (PRL)
→ Thyroid Stimulating Hormone (TSH)
→ Adrenocorticotrophic Hormone (ACTH)
→ Luteinizing Hormone (LH)
→ Follicle Stimulating Hormone (FSH)
→ Melanocyte Stimulating hormone (MSH)
• Neurohypophysis also known as posterior pituitary
→ Oxytocin
→ Vasopressin

(c) Thyroid
→ Thyroxin
→ Triiodothyronin
→ Calcitonin

(d) Parathyroid
→ Parathyroid Hormone

(e) Adrenal
→ Mineralocorticoids (mainly aldosterone)
→ Glucocorticoids (mainly cortisol)
→ Adrenaline
→ Nor-adrenaline

(f) Pancreas
→ Insulin
→ Glucagon

(g) Testis
→ Testosterone

(h) Ovary
→ Estrogen
→ Progesterone

(i) Thymus
→ Thymosins

(j) Atrium
→ Atrial Natriuretic factor

(k) Kidney
→ Erythropoietin

(l) G-I Tract
→ Gastrin
→ Secretin
→ Cholecystokinin (CCK)
→ Gastric inhibitory peptide (GIP)

4. Fill in the blanks:
Hormones Target gland

(a) Hypothalamic hormones __________________
► Pituitary

(b) Thyrotrophin (TSH) __________________
► Thyroid

(c) Corticotrophin (ACTH) __________________
► Adernal

(d) Gonadotrophins (LH, FSH) __________________
► Ovary, Testis

(e) Melanotrophin (MSH) __________________
► Melanoycte

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

Answer

(a) Parathyroid hormone (PTH): It is secreted by Parathyroid gland. It increases the Ca2+ levels in the blood. PTH promotes the reabsorption of calcium from nephrons and also, promotes the absorption of calcium from digested food. Thus, it plays an important role in calcium balance in the body.

(b) Thyroid hormones: Thyroid hormones play an important role in the regulation of the basal metabolic rate. These hormones also support the process of red blood cell formation. Thyroid hormones control the metabolism of carbohydrates, proteins and fats. Maintenance of water and electrolyte balance is also influenced by thyroid hormones. Thyroid gland also secretes a protein

hormone called thyrocalcitonin (TCT) which regulates the blood calcium levels.

(c) Thymosins: It play a major role in the differentiation of T-lymphocytes, which provide cell-mediated immunity. It also promote production of antibodies to provide humoral immunity.
(d) Androgens: It regulate the development, maturation and functions of the male accessory sex organs like epididymis, vas deferens, seminal vesicles, prostate gland, urethra etc. Androgens mainly testosterone stimulate muscular growth, growth of facial and axillary hair, aggressiveness, low pitch of voice etc. It stimulates spermatogenesis and formation of mature sperms and also influences male sexual behaviour. These hormones also produce synthetic effects on protein and carbohydrate metabolism.

(e) Estrogens: It produce wide ranging actions such as stimulation of growth and activities of female secondary sex organs, development of growing ovarian follicles, appearance of female secondary sex characters (e.g., high pitch of voice, etc.), mammary gland development. Estrogens also regulate female sexual behaviour.

(f) Insulin and Glucagon: These hormones are secreted by pancreas and helps in regulation of glucose level in body.

Insulin plays a major role in the regulation of glucose homeostasis. It acts mainly on hepatocytes and adipocytes (cells of adipose tissue), and enhances cellular glucose uptake and utilisation. It also stimulates conversion of glucose to glycogen (glycogenesis) in the target cells.
Glucagon plays an important role in maintaining the normal blood glucose levels. Glucagon acts mainly on the liver cells (hepatocytes) and stimulates glycogenolysis resulting in an increased blood sugar (hyperglycemia). It also stimulates the process of gluconeogenesis which also contributes to
hyperglycemia.

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

Answer

(a) Hyperglycemic hormone → Glucagon
     hypoglycemic hormone → Insulin

(b) Hypercalcemic hormone → Parathyroid hormone (PTH)

(c) Gonadotrophic hormones → Luteinizing hormone and Follicle stimulating hormone

(d) Progestational hormone → Progesterone

(e) Blood pressure lowering hormone → Nor-adrenaline

(f) Androgens and estrogens → Testosterone

Page No: 342

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

Answer

(a) Diabetes mellitus → Insulin
(b) Goitre → Thyroxin hormone
(c) Cretinism → Thyroxin hormone

8. Briefly mention the mechanism of action of FSH.

Answer

Follicle stimulating hormone (FSH) stimulates growth and development of the ovarian follicles in females.
FSH produces its effect by binding to its specific receptors present on the ovarian cell membrane. Binding of FSH hormone to its specific receptor present over the cell surface activates an enzyme called adenylate cyclase. This enzyme converts ATP into cyclicAMP. The cyclicAMP is called the second messenger and carries out various biochemical responses in the cell by activating the existing enzyme system in the cell. The biochemical responses then lead to physiological responses like ovarian growth.
Action of FSH
 
9. Match the following:


Column I
Column II
(a) T4(i) Hypothalamus
(b) PTH(ii) Thyroid
(c) GnRH(iii) Pituitary
(d) LH(iv) Parathyroid

Answer

Column I
Column II
(a) T4(ii) Thyroid
(b) PTH(iv) Parathyroid
(c) GnRH(i) Hypothalamus
(d) LH(iii) Pituitary
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CLASS 11th CHAPTER -21 Neural Control and Coordination |Biology | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Biology Chapter -21 Neural Control and Coordination includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Biology exams must go through NCERT Solutions for Class 11 Chapter -21 Neural Control and Coordination . NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

Class 11th Chapter -21 Neural Control and Coordination  | NCERT BIOLOGY SOLUTION |

Page No: 328

Exercises

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

Answer
(a) Brain: The human brain is well protected by the skull. Inside the skull, the brain is covered by cranial meninges consisting of an outer layer called dura mater, a very thin middle layer called arachnoid and an inner layer (which is in contact with the brain tissue) called pia mater. The brain can be divided into three major parts:
(i) Forebrain: The forebrain consists of cerebrum, thalamus and hypothalamus.
(ii) Midbrain: It is located between the thalamus/hypothalamus of the forebrain and pons of the hindbrain.
(iii) Hindbrain: The hindbrain comprises pons, cerebellum and medulla.

(b) Eye: The adult human eye ball is nearly a spherical structure. The wall of the eye ball is composed of three layers. The external layer is composed of a dense connective tissue and is called the sclera. The anterior portion of this layer is called the cornea. The middle layer, choroid, contains many blood vessels and looks bluish in colour. The choroid layer is thin over the posterior two-thirds of the eye ball, but it becomes thick in the anterior part to form the ciliary body. The ciliary body itself continues forward to form a pigmented and opaque structure called the iris which is the visible coloured portion of the eye. The eye ball contains a transparent crystalline lens which is held in place by ligaments attached to the ciliary body. In front of the lens, the aperture surrounded by the iris is called the pupil whose diameter is regulated by the muscle fibres of iris.
The inner layer is the retina and it contains three layers of neural cells from inside to outside – ganglion cells, bipolar cells and photoreceptor cells. There are two types of photoreceptor cells, namely, rods and cones.  The daylight (photopic) vision and colour vision are functions of cones and the twilight (scotopic) vision is the function of the rods. 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: It perform two sensory functions, hearing and maintenance of body balance. It can be divided into three major sections called the outer ear, the middle ear and the inner ear:

→ Outer ear: It consists of the pinna and external auditory meatus (canal). The pinna collects the vibrations in the air which produce sound. The external auditory meatus leads inwards and extends up to the tympanic membrane (the ear drum). There are very fine hairs and wax-secreting glands in the skin of the pinna and the meatus. The tympanic membrane is composed of connective tissues covered with skin outside and with mucus membrane inside.

→ Middle ear: It contains three ossicles called malleus, incus and stapes which are attached to one another in a chain-like fashion. The malleus is attached to the tympanic membrane and the stapes is attached to the oval window of the cochlea. The ear ossicles increase the efficiency of transmission of sound waves to the inner ear. An Eustachian tube connects the middle ear cavity with the pharynx. The Eustachian tube helps in equalising the pressures on either sides of the ear drum.

→ Inner 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 two parts – Vestibular apparatus and Cochlea.
The vestibular apparatus is composed of three semi-circular canals and the otolith (macula is the sensory part of saccule and utricle). Each semi-circular canal lies in a different plane at right angles to each other. The membranous canals are suspended in the perilymph of the bony canals. The base of canals is swollen and is called ampulla, which contains a projecting ridge called crista ampullaris which has hair cells. The saccule and utricle contain a projecting ridge called macula. The crista and macula are the specific receptors of the vestibular apparatus responsible for maintenance of balance of the body and posture.
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.

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

Answer
(a) Central neural system (CNS) and Peripheral neural system (PNS)

Central neural system (CNS)Peripheral neural system (PNS)
It is the main coordinating centre of the body.It is not the main coordinating centre of the body.
It lies inside the skull.It does not lie inside the skull.
This includes brain and spinal cord.This includes all the nerves of the body associated with the CNS (brain and spinal cord).

(b) Resting potential and action potential

Resting potential
Action potential
It is the potential difference across the nerve fibre when there is no conduction of nerve impulse.It is the potential difference across nerve fibre when there is conduction of nerve impulse.
The interior of the neuron is electronegative and the exterior is electropositive.The interior of the neuron is electropositive and the exterior is electronegative.
An active sodium pump operates.No sodium pump operates.

(c) Choroid and retina

ChoroidRetina
Choroid is the middle vascular layer of eye.Retina is the innermost nervous coat of eye.
It is rich in blood cellsIt is rich in neurons
Its function is to supply nutrients and oxygen to other parts of eye like retinaIts function is to form an image of an object over it.

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

Answer
(a) When the resting potential of the membrane changes it becomes polarized. During resting condition, the axoplasm inside the axon contains high concentration of K+ and negatively charged proteins and low concentration of Na+. As a result, the potassium ions 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.

(b) 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.

(c) There are two types of nerve fibres – myelinated and non-myelinated. In myelinated nerve fibre, the impulse is conducted from node to node in jumping manner as myelinated nerve fibre is are enveloped with Schwann cells, which form a myelin sheath around the axon. 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. 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) At a chemical synapse, the membranes of the pre- and post-synaptic neurons are separated by a fluid-filled space called synaptic cleft. When an impulse arrives at the axon terminal, it stimulates the movement of the synaptic vesicles towards the membrane where they fuse with the plasma membrane and release their neurotransmitters in the synaptic cleft. The released neurotransmitters bind to their specific receptors, present on the post-synaptic membrane. This binding opens ion channels allowing the entry of ions which can generate a new potential in the post-synaptic neuron. The new potential developed may be either excitatory or inhibitory.

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

Answer
(a) Neuron
Diagram of neuron
 
(b) Brain
Diagram of Brain
 
(c) Eye
Diagram of Eye
 
(d) Ear
Diagram of Ear
 
Page No: 329
 
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
 
Answer
(a) The process through which two or more organs interact and complement the functions of one another through the neural system, it is called neural coordination. All the physiological processes in the body are closed linked and dependent upon each other. The neural system and the endocrine system jointly coordinate and integrate all the activities of the organs so that they function in a synchronised fashion. The neural system provides an organised network of point-to-point connections for a quick coordination. The endocrine system provides chemical integration through hormones.
 
(b) The forebrain consists of cerebrum, thalamus and hypothalamus. 
 
→ Cerebrum forms the major part of the human brain. A deep cleft divides the cerebrum longitudinally into two halves, which are termed as the left and right cerebral hemispheres. The hemispheres are connected by a tract of nerve fibres called corpus callosum. The layer of cells which covers the cerebral hemisphere is called cerebral cortex and is thrown into prominent folds. The cerebral cortex is referred to as the grey matter due to its greyish appearance. The cerebral cortex contains motor areas, sensory areas and large regions that are neither clearly sensory nor motor in function. These regions called as the association areas are responsible for complex functions like intersensory associations, memory and communication.Fibres of the tracts are covered with the myelin sheath, which constitute the inner part of cerebral hemisphere. They give an opaque white appearance to the layer and, hence, is called the white matter.
 
→ Thalamus: It is a region present at the centre of the forebrain and wrapped by cerebrum. It is coordination center for sensory and motor signalling.
 
→ Hypothalamus: It lies at the base of the thalamus which contains a number of centres which control body temperature, urge for eating and drinking. It also contains the nerve centres for temperature regulation, hunger, thirst, heart beat and respiration regulation and emotions such as anger, love, cool, etc. It has connection with pituitary gland hence also controls growth and sexual behaviour.
(c) The midbrain is located between the thalamus/hypothalamus of the forebrain and pons of the hindbrain. A canal called the cerebral aqueduct passess through the midbrain. The dorsal portion of the midbrain consists mainly of four round swellings (lobes) called corpora quadrigemina. Midbrain and hindbrain form the brain stem.

(d) The hindbrain comprises pons, cerebellum and medulla.
→ Pons consists of fibre tracts that interconnect different regions of the brain.
→ Cerebellum has very convoluted surface in order to provide the additional space for many more neurons.
→ The medulla of the brain is connected to the spinal cord. The medulla contains centres which control respiration, cardiovascular reflexes and gastric secretions.

(e) Retina is the innermost layer which contains three layers of neural cells – from inside to outside – ganglion cells, bipolar cells and photoreceptor cells. There are two types of photoreceptor cells, namely, rods and cones. The daylight vision and colour vision are functions of cones and twilight vision is the function of the rods. The light enters through cornea, the lens and the images of objects are formed on the retina.
(f) The middle ear contains three ear ossicles called malleus, incus and stapes which are attached to one another in a chain-like fashion. The malleus is attached to the tympanic membrane, incus is connected with stapes. and the stapes is attached to the oval window of the cochlea.The ear ossicles increase the efficiency of transmission of sound waves to the inner ear.

(g) The coiled portion of the labyrinth is called cochlea. The membranes constituting cochlea, the reissner’s and basilar, divide the surounding perilymph filled bony labyrinth into an upper scala vestibuli and a lower scala tympani. The space within cochlea called scala media is filled with endolymph. At the base of the cochlea, the scala vestibuli ends at the oval window, while the scala tympani terminates at the round window which opens to the middle ear

(h) The organ of corti is a structure located on the basilar membrane which contains hair cells that act as auditory receptors. The hair cells are present in rows on the internal side of the organ of corti.

(i) A synapse is formed by the membranes of a pre-synaptic neuron and a post-synaptic neuron, which may or may not be separated by a gap called synaptic cleft. These are of two types electrical synapses and chemical synapses.

6. Give a brief account of:
(a) Mechanism of synaptic transmission
(b) Mechanism of vision
(c) Mechanism of hearing

Answer

(a) A nerve impulse is transmitted from one neuron to another through junctions called synapses which is formed by the membranes of a pre-synaptic neuron and a post-synaptic neuron may or may not be separated by synaptic cleft. There are two types of synapses, namely, electrical synapses and chemical synapses.
At electrical synapses, the membranes of pre- and post-synaptic neurons are in very close proximity so electrical current can flow directly from one neuron into the other across these synapses. Transmission of an impulse across electrical synapses is very similar to impulse conduction along a single axon and transmission is always faster than that across a chemical synapse however it is not common in human body.

At a chemical synapse, the membranes of the pre- and post-synaptic neurons are separated by a fluid-filled space called synaptic cleft. Chemicals called neurotransmitters are involved in the transmission of impulses at these synapses.

(b) The light rays in visible wavelength focussed on the retina through the cornea and lens generate impulses in rods and cones. The photosensitive compounds in the human eyes is composed of opsin and retinal. Light induces dissociation of the retinal from opsin resulting in changes in the structure of the opsin. This causes membrane permeability changes therefore potential differences are generated in the photoreceptor cells. This produces a signal that generates action potentials in the ganglion cells through the bipolar cells. These impulses are transmitted by the optic nerves to the visual cortex area of the brain, where the neural impulses are analysed and the image formed on the retina is recognised based on earlier memory and experience.


(c) The external ear receives sound waves and directs them to the ear drum.The ear drum vibrates in response to the sound waves and these vibrations are transmitted through the ear ossicles to the oval window. The vibrations are passed through the oval window on to the fluid of the cochlea, where they generate waves in the lymphs. The waves in the lymphs induce a ripple in the basilar membrane. These movements of the basilar membrane bend the hair cells, pressing them against the tectorial membrane therefore nerve impulses are generated in the associated afferent neurons. These impulses are transmitted by the afferent fibres via auditory nerves to the auditory cortex of the brain, where the impulses are analysed and the sound is recognised.

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?

Answer

(a) The cones are responsible for color vision. There are three types of cones cells that respond to green light, blue light and red light according to their characteristics. These cells are stimulated by different lights, from different sources. The combinations of the signals generated help us see the different colours.

(b) The Inner ear has three semi-circular canals forming cochlea. Cochlea is responsible for maintaining the body balance.

(c) Pupil is the small aperture surrounded by the the iris that regulates the amount of light entering the eye. This expand in case of low light and contract in case of intense light thereby regulating the amount of light falling on the retina.

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.

Answer

(a) The action potential is determined by Na+ ions. The Na+ channels which are closed in the resting state, open and cause the inflow of Na+ ions by diffusion into the inside of axoplasm. The electrical potential of the membrane changes from 70 mV towards zero and then the membrane is said to be depolarised.

(b) The photopigments of the retina are photosensitive compounds in the eye that are composed of retinal and opsin. Light induces dissociation of retinal from opsin which changes the structure of the opsin. It generates action potential in the bipolar neurons. These impulses/action potential are transmitted by the optic nerves to the visual cortex of the brain where the neural impulses are analysed and the erect image is recognised.

(c) When sound falls over the ear drum, it is then transmitted to the inner ear by ear ossicles. The vibrations are passed through the oval window onto the fluid of the cochlea, where they generate waves in the lymphs. The waves induce a ripple in the basilar membrane that bend the hair cells, pressing them against the techtonial membrane. As a result nerve impulses are generated in the associated afferent neurons and transmitted to auditory cortex of brain via auditory nerves, where the impulses are analysed and the sound is recognised.

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

Answer

(a) Myelinated and non-myelinated axons

Myelinated axons
Non myelinated axons
These appear whitish in colour.These appear greyish in colour.
Myelin sheath is presentMyelin sheath is absent
Nodes of Ranvier are present at intervals.Nodes of Ranvier are absent.
Transmission of nerve impulse is faster.Transmission of nerve impulse is slower
Ion exchange can occur only at the nodes.Ion exchange occurs all over the surface.

(b) Dendrites and axons

Dendrites
Axons
These are extension of cyton present at anterior position.These are extension of cyton present at posterior position.
These conduct impulses towards the cell body.These conduct impulses away from the cell body.
Dendrites are always non-myelinated.Axons can be myelinated or non-myelinated.
The terminals of dendrites become receptors.The terminals of axon ends in a group of branches called terminal arborisations.

(c) Rods and cones

Rods
Cones
Rods are sensitive to dim light.Cones are sensitive to bright light only.
They have visual purple pigment called rhodopsin.They have visual violet pigment called iodopsin.
Colours are not visible with rod cells.Cones help to see colours.

(d) Thalamus and Hypothalamus

Thalamus
Hypothalamus
It represent the side of diencephalon.It represents the lower part of diencephalon.
It is the major coordinating centre for sensory and motor signalling.It is the major centre for regulation of body temperature, thirst, hunger etc.
It does not secrete any hormone.It secretes several hormones.

(e) Cerebrum and Cerebellum

Cerebrum
Cerebellum
It is part of fore brainIt is part of hind brain
It consists of two cerebral hemispheres.It consists of two cerebellar hemispheres and a median vermis.
It initiates voluntary movements.It maintains posture and equilibrium.

10. Answer the following:

(a) Which part of the ear determines the pitch of a sound?
► Cochlea

(b) Which part of the human brain is the most developed?
► Cerebrum

(c) Which part of our central neural system acts as a master clock?
► Hypothalamus

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
► (c) blind spot

12. Distinguish between
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

Answer
a) Afferent neurons and Efferent neurons

Afferent neurons
Efferent neurons
They conduct impulses towards the central neural system.They conduct impulses away from the central neural system
It provides stimulus and evoke senses.It results in the response by the effectors.
They are sensory in nature.They are motor in nature.
These take information from the receptors.These take information to effectors.

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

Impulse conduction in a myelinated nerve fibre
Impulse conduction in a unmyelinated nerve fibre
The depolarisation occurs only in the nodes of Ranvier where the myelin sheath is absent.The depolarisation occurs all along the length of the nerve fibre.
Action potential jumps from one node of Ranvier to another.Action potential travels along the entire length of the fibre.
Conduction is faster.Conduction is slower.
Less amount of energy is required.More amount of energy is required.

c) Aqueous humour and Vitreous humour

Aqueous humour
Vitreous humour
It is a watery and transparent fluid.It is thick gelly-like transparent fluid 
It is present between lens and cornea.It is present between lens and retina.
It is continuously secreted by ciliary processes and drained out of the eyes.It is not replaced.
It provides nourishment to lens, cornea and other parts of anterior chamber.It does not have nutritive value.

d) Blind spot and Yellow spot

Blind spot
Yellow spot
Blind spot is a spot on the retina present at the point of origin of the optic nerve.Yellow spot is a small area on the retina present at the posterior pole of the eye, lateral to the blind spot.
It does not have a shallow depression.It has a shallow depression called fovea centralis.
Photoreceptor cells are absent from this region.Only cones are present in this region.
It is insensitive to light.It is sensitive to bright light.
Photoreceptor cells are absent from this region.Only cones are present in this region.
No image is formed at the blind spot.An image is formed at the yellow spot.

e) Cranial nerves and Spinal nerves

Cranial nerves
Spinal nerves
They originate from the brain.They originate from the spinal cord.
There are 12 pairs of cranial nerves.There are 31 pairs of spinal nerves.
These are sensory, motor or mixed in nature.They are mixed in nature.
They extend between the brain and body parts.They extend between the spinal cord and body parts.
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CLASS 11th CHAPTER -20 Locomotion and Movement |Biology | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Biology Chapter -20 Locomotion and Movement  includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Biology exams must go through NCERT Solutions for Class 11 Chapter -20 Locomotion and Movement  . NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

Class 11th Chapter -20 Locomotion and Movement   | NCERT BIOLOGY SOLUTION |

Exercises
 
Page No: 313
 
1. Draw the diagram of a sarcomere of skeletal muscle showing different regions.
 
Answer
 
Sarcomere
 
2. Define sliding filament theory of muscle contraction.
 
Answer
Sliding filament theory of muscle contraction states that contraction of a muscle fibre takes place by the sliding of the thin filaments over the thick filaments.
 
3. Describe the important steps in muscle contraction.
 
Answer
 
The important steps muscle contraction:

→ Muscle contraction is initiated by a signal sent by the central nervous system (CNS) via a motor neuron and reach the neuromuscular junction. As a result, neurotransmitter (Acetyl choline) which generates an action potential in the sarcolemma.

→ This spreads through the muscle fibre and causes the release of calcium ions into the sarcoplasm.

→ Increase in Calcium level leads to the binding of calcium with a subunit of troponin on actin filaments and thereby remove the masking of active sites for myosin. Utilising the energy from ATP hydrolysis, the myosin head now binds to the exposed active sites on actin to form a cross bridge.

→The actin filaments are pulled. As a result, the H-zone reduces. It is at this stage that the contraction of the muscle occurs.
 
→ After muscle contraction, the myosin head pulls the actin filament and releases ADP along with inorganic phosphate. ATP molecules bind and detach myosin and the cross bridges are broken and decreases the calcium ions contraction. As a result, masking the actin filaments and leading to muscle relaxation.

4. Write true or false. If false change the statement so that it is true.

(a) Actin is present in thin filament
► True

(b) H-zone of striated muscle fibre represents both thick and thin filaments.
► False, H -zone of striated muscle fibre represents thick filament.

(c) Human skeleton has 206 bones.
► True

(d) There are 11 pairs of ribs in man.
► False, There are 12 pairs of ribs in man.

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

5. Write the difference between:
(a) Actin and Myosin
(b) Red and White muscles
(c) Pectoral and Pelvic girdle

Answer

(a) Actin and Myosin

Actin
Myosin
Actin is a thin contractile protein.Myosin is a thick contractile protein.
The light bands called I-band or Isotropic band contain actin.The dark band called ‘A’ or Anisotropic band contains myosin.

(b) Red and White muscles

Red Muscles
White Muscles
Red muscle fibres are thin and smaller in size.White muscle fibres are thick and larger in size.
They are red in colour as they contain large amounts of myoglobin.They are white in colour as they contain small amounts of myoglobin.
They contain plenty of mitochondria.They contain less number of mitochondria.
They provide energy by aerobic respiration.They provide energy by anaerobic respiration.

(c) Pectoral and Pelvic girdle

Pectoral
Pelvic girdle
Pectoral girdle is situated in the pectoral region of the body.Pelvic girdle is situated in the pelvic region of the body.
It is composed of two bones namely, clavicle or collar bones and scapula or shoulder bone.It is composed of three bones, upper ileum, inner pubic, and ischium.
It has no articulation with the vertebral column.It has articulation with the vertebral column.
These perform functions like holding, lifting etc.These perform functions like running, standing, jumping etc.

6. Match Column I with Column II:
 Column I Column II
(a)Smooth muscle(i)Myoglobin
(b)Tropomyosin(ii)Thin filament
(c)Red muscle(iii)Sutures
(d)Skull(iv)Involuntary

Answer

 Column I Column II
(a)Smooth muscle(iv)Involuntary
(b)Tropomyosin(ii)Thin filament
(c)Red muscle(i)Myoglobin
(d)Skull(iii)Sutures

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

Answer

Movement is one of the significant features of living beings. The different types of movements exhibited by the cells of human body are:

→ Amoeboid movement: Some specialised cells in our body like macrophages and leucocytes in blood exhibit amoeboid movement. It is effected by pseudopodia formed by the streaming of protoplasm (as in Amoeba). Cytoskeletal elements like microfilaments are also involved in amoeboid movement.

→ Ciliary Movement: It occurs in most of our internal tubular organs which are lined by ciliated epithelium. The coordinated movements of cilia in the trachea help us in removing dust particles and some of the foreign substances inhaled alongwith the atmospheric air. Passage of ova through the female reproductive tract is also facilitated by the ciliary movement.

→ Muscular movement: Muscle cells show muscular movement such as limbs, jaws, tongue, etc,

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

Answer

Skeletal Muscle
Cardiac Muscle
The cells of skeletal muscles are unbranched.The cells of cardiac muscles are branched.
Intercalated disks are absent.The cells are joined with one another by intercalated disks that help in coordination or synchronization of the heart beat.
Alternate light and dark bands are present.Faint bands are present.
They are voluntary in nature.They are involuntary in nature.
They contract rapidly and get fatigued in a short span of time.They contract rapidly but do not get fatigued easily.
They are present in body parts such as the legs, tongue, hands, etc.These muscles are present in the heart and control the contraction and relaxation of the heart.

9. Name the type of joint between the following:-

(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

(f) between pubic bones in the pelvic girdle
► Cartilaginous joint

10. Fill in the blank spaces:

(a) All mammals (except a few) have __________ cervical vertebra.
► seven

(b) The number of phalanges in each limb of human is __________
► 14

(c) Thin filament of myofibril contains 2 ‘F’ actins and two other proteins namely __________ and __________.
► troponin, tropomyosin

(d) In a muscle fibre Ca++ is stored in __________
► sarcoplasmic reticulum

(e) __________ and __________ pairs of ribs are called floating ribs.
► 11th and 12th

(f) The human cranium is made of __________ bones.
► eight
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CLASS 11th CHAPTER -19 Excretory Products and their Elimination |Biology | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Biology Chapter -19 Excretory Products and their Elemination  includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Biology exams must go through NCERT Solutions for Class 11 Chapter -19 Excretory Products and their Elemination . NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

Class 11th Chapter -19 Excretory Products and their Elemination  | NCERT BIOLOGY SOLUTION |

Exercises
 
Page No: 300
 
1. Define Glomerular Filtration Rate (GFR)
 
Answer
The amount of the filtrate formed by the kidneys per minute is called glomerular filtration rate (GFR).
 
2. Explain the autoregulatory mechanism of GFR.
 
Answer
The kidneys have built-in mechanisms for the regulation of glomerular filtration rate. One such efficient mechanism is carried out by juxtabglomerular apparatus (JGA). JGA is a special sensitive region formed by cellular modifications in the distal convoluted tubule and the afferent arteriole at the location of their contact. A fall in GFR can activate the JG cells to release renin which can stimulate the glomerular blood flow and thereby the GFR back to normal.
 
3. Indicate whether the following statements are true or false:

(a) Micturition is carried out by a reflex.
► True

(b) ADH helps in water elimination, making the urine hypotonic.
► False

(c) Protein-free fluid is filtered from blood plasma into the Bowman’s capsule.
► True

(d) Henle’s loop plays an important role in concentrating the urine.
► True

(e) Glucose is actively reabsorbed in the proximal convoluted tubule.
► True
 
4. Give a brief account of the counter current mechanism.
 
Answer
The counter current mechanism operating inside the kidney is the main adaptation for the conservation of water. The Henle’s loop and vasa rectaare two counter current mechanisms inside the kidneys. The flow of filtrate in the two limbs of Henle’s loop is in opposite directions and thus forms a counter current. 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 mOsmolL-1 in the cortex to 1200 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 NaCl and urea.
Counter current mechanism
 
5. Describe the role of liver, lungs and skin in excretion.
 
Answer
• Role of Lungs: This remove large amounts of CO2 and also significant quantities of water every day.
 
• Role of Liver: This is the largest gland in our body, secretes bile-containing substances like bilirubin, biliverdin, cholesterol, degraded steroid hormones, vitamins and drugs. Most of these substances ultimately pass out alongwith digestive wastes.
 
• Role of Skin: Skin has sweat and sebaceous glands that can eliminate certain substances through their secretions. 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 eliminate certain substances like sterols, hydrocarbons and waxes through sebum.
 
6. Explain micturition.

Answer
The process of release of urine is called micturition. Urine formed by the nephrons is ultimately carried to the urinary bladder where it is stored till a voluntary signal is given by the central nervous system (CNS). This signal is initiated by the stretching of the urinary bladder as it gets filled with urine. In response, the stretch receptors on the walls of the bladder send signals to the CNS. The CNS passes on motor messages to initiate the contraction of smooth muscles of the bladder and simultaneous relaxation of the urethral sphincter causing the release of urine. An adult human excretes, on an average, 1 to 1.5 litres of urine per day.

7. Match the items of column Iwith those of column II:
Column I
Column II
(a)Ammonotelism(i)Birds
(b)Bowman’s capsule(ii)Water reabsorption
(c)Micturition(iii)Bony fish
(d)Uricotelism(iv)Urinary bladder
(d)ADH(v)Renal tubule

Answer

Column I
Column II
(a)Ammonotelism(i)Bony fish
(b)Bowman’s capsule(ii)Renal tubule
(c)Micturition(iii)Urinary bladder
(d)Uricotelism(iv)Birds
(d)ADH(v)Water reabsorption         

Page No: 308

8. What is meant by the term osmoregulation?
 
Answer
 
Osmoregulation is the regulation of blood volume, body fluid volume and ionic concentration.
 
9. Terrestrial animals are generally either ureotelic or uricotelic, not ammonotelic, why?
 
Answer
 
Terrestrial animals are generally either ureotelic or uricotelic, not ammonotelic because:
→ Ammonia is the most toxic form and requires large amount of water for its elimination, whereas uric acid, being the least toxic, can be removed with a minimum loss of water. Therefore, it needs to be converted into a less toxic form such as urea or uric acid.
→ As ammonia is readily soluble, is generally excreted by diffusion across body surfaces or through gill surfaces (in fish) as ammonium ions. 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.

10. What is the significance of juxta glomerular apparatus (JGA) in kidney function?
 
Answer
 
The JGA plays a complex regulatory role. A fall in glomerular blood flow/glomerular blood pressure/GFR can activate the JG cells to release renin which converts angiotensinogen in blood to angiotensin I and further to angiotensin II. Angiotensin II, being a powerful vasoconstrictor, increases the glomerular blood pressure and thereby GFR. Angiotensin II also activates the adrenal cortex to release Aldosterone. Aldosterone causes reabsorption of Na+ and water from the distal parts of the tubule. This also leads to an increase in blood pressure and GFR. This complex mechanism is generally known as the Renin-Angiotensin mechanism.

11. Name the following:
 
(a) A chordate animal having flame cells as excretory structures
► Flatworms

(b) Cortical portions projecting between the medullary pyramids in the human kidney
► Columns of Bertini

(c) A loop of capillary running parallel to the Henle’s loop.
► Vasa Recta
 
12. Fill in the gaps:

(a) Ascending limb of Henle’s loop is ____________to water whereas the descending limb is___________to it.
► impermeable, permeable

(b) Reabsorption of water from distal parts of the tubules is facilitated by hormone____________.
► vassopressin

(c) Dialysis fluid contains all the constituents as in plasma except________.
► nitrogenous waste

(d) A healthy adult human excretes (on an average) _______ gm of urea/day.
► 25 to 30
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CLASS 11th CHAPTER -18 Body Fluids and Circulation |Biology | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Biology Chapter -18 Body Fluids and Circulation  includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Biology exams must go through NCERT Solutions for Class 11 Chapter -18 Body Fluids and Circulation. NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

Class 11th Chapter -18 Body Fluids and Circulation | NCERT BIOLOGY SOLUTION |

Exercises 
 
Page No: 289
 
1. Name the components of the formed elements in the blood and mention one major function of each of them.
 
Answer
 
The components of the formed elements in the blood with their major function are:
→ Erythrocytes (RBC): The erythrocytes play a significant role in transport of respiratory gases.
→ Leucocytes (WBC): The leucocytes play an important role to fight against infections.
→ Thrombocytes (Platelets): Platelets are involve in the coagulation or clotting of blood.
 
2. What is the importance of plasma proteins?
 
Answer
 
The major plasma proteins are fibrinogen, globulins and albumins.
• Fibrinogen play important role in blood coagulation. 
• Globulins protects the body against infecting agents
• Albumins helps in maintaining the fluid volume within the vascular space.
 
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

Answer

Column I
  
Column II
(a)
Eosinophils
(iii)
Resist Infections
(b)
RBC
(v)
Gas transport
(c)
AB Group
(ii)
Universal Recipient
(d)
Platelets
(i)
Coagulation
(e)
Systole
(iv)
Gas transport

4. Why do we consider blood as a connective tissue?
 
Answer
 
Blood as a connective tissue because:
→ Blood serves the purpose of connecting the body systems by transporting substances.
→ Blood is mesodermal in origin like any other connective tissues.
 
5. What is the difference between lymph and blood?

Answer

Lymph
Blood
It is red-coloured fluid It is a colourless fluid.
It contains plasma and lesser number of WBCs and platelets. It contains plasma, RBCs, WBCs, and platelets.
Its plasma lacks proteins. Its plasma has proteins, calcium, and phosphorus.
It transports nutrients from the tissue cells to the blood, through lymphatic vessels. It transports nutrients and oxygen from one organ to another.
It helps in body defence and is a part of the immune system. It helps in the circulation of oxygen and carbon dioxide.
The flow of lymph is slow. The flow of blood in the blood vessels is fast.

6. What is meant by double circulation? What is its significance?
 
Answer
 
Double circulation is a process during which blood passes twice through the heart during one complete cycle. It consist of two separate circulation:

→ Systemic circulation: In this circulation the oxygenated blood from the left ventricle of heart is pumped to all the body parts (except lungs) through aorta. After this, deoxygenated blood from various parts comes back to heart by superior and inferior vena cava into the right atria. This completes the 1st circulation.

→ Pulmonary circulation: In pulmonary circulation the deoxygenated blood brought back from body parts is pumped to lungs by the right ventricle through the pulmonary artery. In lungs deoxygenated blood is again converted back to oxygenated blood and sent back to the heart in left atria by pulmonary veins. This completes the 2nd circulation.

The most important significance of this system is that the oxygenated and deoxygenated blood does not mix anywhere in the body thus improves the oxygen supplying capacity of the heart.
 
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
 
Answer

(a) Blood and Lymph
Lymph
Blood
It is a colourless fluid. It is red-coloured fluid.
It contains plasma and lesser number of WBCs and platelets. It contains plasma, RBCs, WBCs, and platelets.
It transports nutrients from the tissue cells to the blood, through lymphatic vessels. It transports nutrients and oxygen from one organ to another.
It helps in body defence and is a part of the immune system. It helps in the circulation of oxygen and carbon dioxide.

(b) Open and Closed system of circulation


Open system of circulation Closed system of circulation
In this system, blood is pumped by the heart, through large vessels, into body cavities called sinuses. In this system, blood is pumped by the heart, through a closed network of vessels.
The body tissues are in direct contact with blood.
The body tissues are in indirect contact with blood.
Blood flows at low pressure. Blood flows at high pressure.
Blood flow cannot be regulated. 
Blood flow can be regulated
This is present in arthropods and molluscs. This is present in annelids, echinoderms, and vertebrates.
 
(c) Systole and Diastole

Systole
Diastole
It is the contraction of the heart chambers to drive blood into the aorta and the pulmonary artery. It is the relaxation of the heart chambers between two contractions.
Systole decreases the volume of the heart chambers and forces the blood out of them. Diastole brings the heart chambers back into their original sizes to receive more blood

(d) P-wave and T-wave

Systole
Diastole
In an electrocardiogram (ECG), the P-wave indicates the activation of the Sino-atrial node. In an electrocardiogram (ECG), the T-wave represents ventricular relaxation.
During this phase, the impulse of contraction is generated by the SA node, causing atrial depolarisation. During this phase, the ventricles relax and return to their normal state.
It is of atrial origin. It is of ventricular origin.

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

Answer

The heart of vertebrates evolved from the simple 2-chambered heart of fishes to complex multi-chambered hearts.
Fishes have a 2-chambered heart with an atrium and a ventricle. Amphibians and the reptiles except crocodiles have a 3-chambered heart with two atria and a single ventricle, whereas crocodiles, birds and mammals possess a 4-chambered heart with two atria and two ventricles.
In fishes the heart pumps out deoxygenated blood which is oxygenated by the gills and supplied to the body parts from where deoxygenated blood is returned to the heart (single circulation).
In amphibians and reptiles, the left atrium receives oxygenated blood from the gills/lungs/skin and the right atrium gets the deoxygenated blood from other body parts. However, they get mixed up in the single ventricle which pumps out mixed blood (incomplete double circulation).
In birds and mammals, oxygenated and deoxygenated blood received by the left and right atria respectively passes on to the ventricles of the same sides. The ventricles pump it out without any mixing up, i.e., two separate circulatory pathways are present in these organisms, hence, these animals have double circulation.

9. Why do we call our heart myogenic?
 
Answer
 
Heart is called myogenic because all the normal activities of the heart are regulated intrinsically or the activities are auto regulated by specialised muscles known as nodal tissues without the need for an external stimulus to be delivered by the nervous system.
 
10. Sino-atrial node is called the pacemaker of our heart. Why?
 
Answer
 
The sino-atrial (SA) node is a node of specialised cardiac muscle fibres located in the upper part of the right atrium of the heart. The cardiac impulse originates from the SA node triggers a sequence of electrical events in the heart and is responsible for initiating and maintaining the rhythmic contractile activity of the heart. Therefore, it is called the pacemaker.
 
11. What is the significance of atrio-ventricular node and atrio-ventricular bundle in the functioning of heart?
 
Answer
 
The atrio-ventricular node and atrio-ventricular bundle has the ability to generate action potentials without any external stimuli. Their main function is to get excited by the action potential initiated by the Sino- atrial node and conduct the stimulus to the remaining part of the heart through which they branch thus helping in the rhythmic contraction and relaxation of the heart.
 
12. Define a cardiac cycle and the cardiac output.
 
Answer

The sequential events taking place in the heart which is the contraction or systole and relaxation or diastole of both the atria and ventricles is called cardiac cycle.
The volume of blood pumped out by the ventricles in one minute is called the cardiac output. 
 
13. Explain heart sounds.

Answer
 
Two prominent sounds are produced during each cardiac cycle. The first heart sound (lub) is associated with the closure of the tricuspid and bicuspid valves whereas the second heart sound (dub) is associated with the closure of the semilunar valves. These sounds are of clinical diagnostic significance.
 
14. Draw a standard ECG and explain the different segments in it.
 
Answer
 
Standard ECG
 
Each peak in the ECG is identified with a letter from P to T that corresponds to a specific electrical activity of the heart.
The P-wave represents the electrical excitation or depolarization of the atria. Depolarisation of atria leads to atricular systole.
The QRS complex represents the depolarization of the ventricles which initiates ventricular systole.
The T-wave represents the return of the ventricles from excited to normal state (repolarisation). The end of T-wave marks the end of systole.
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CLASS 11th CHAPTER -17 Breathing and Exchange of gases |Biology | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Biology Chapter -17 Breathing and Exchange of Gases  includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Biology exams must go through NCERT Solutions for Class 11 Chapter -17 Breathing and Exchange of Gases. NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

Class 11th Chapter -17 Breathing and Exchange of Gases | NCERT BIOLOGY SOLUTION |

Exercises
 
Page No: 277
 
1. Define vital capacity. What is its significance?
 
Answer
The maximum volume of air a person can breathe in after a forced expiration is called vital capacity. It helps in finding differentiate causes of lung disease.
 
2. State the volume of air remaining in the lungs after a normal breathing.

Answer
The volume of air remaining in the lungs after a normal breathing is called Functional residual capacity (FRC). This includes expiratory reserve volume (ERV) and residual volume (RV). ERV=1000 to 1100 ml
RV = 1100 to 1200 ml
Thus, FRC = 2100 to 2300 ml

3. Diffusion of gases occurs in the alveolar region only and not in the other parts of respiratory system. Why?
 
Answer
Alveoli are the primary sites of exchange of gases. Exchange of gases also occur between blood and tissues. O2 and CO2 are exchanged in these sites by simple diffusion mainly based on pressure/concentration gradient. Alveolar region is having enough pressure gradient to facilitate diffusion of gases while other regions of the respiratory system don’t have the required pressure gradient. Solubility of the gases as well as the thickness of the membranes involved in diffusion are also some important factors that can affect the rate of diffusion.

4. What are the major transport mechanisms for CO2? Explain.

Answer
CO2 is carried by haemoglobin as carbamino-haemoglobin (about 20-25 per cent). This binding is related to the partial pressure of CO2 . pO2 is a major factor which could affect this binding. When pCO2 is high and pO2 is low as in the tissues, more binding of carbon dioxide occurs whereas, when the pCO2 is low and pO2 is high as in the alveoli, dissociation of CO2 from carbamino-haemoglobin takes place, i.e., CO2 which is bound to haemoglobin from the tissues is delivered at the alveoli. RBCs contain a very high concentration of the enzyme, carbonic anhydrase and minute quantities of the same is present in the plasma too. This enzyme facilitates the following reaction in both directions.
Transport of CO2
At the tissue site where partial pressure of CO2 is high due to catabolism, CO2 diffuses into blood (RBCs and plasma) and forms HCO3– and H+, . At the alveolar site where pCO2 is low, the reaction proceeds in the opposite direction leading to the formation of CO2 and H2O. Thus, CO2 trapped as bicarbonate at the tissue level and transported to the alveoli is released out as CO2 as shown in above figure. Every 100 ml of deoxygenated blood delivers approximately 4 ml of CO2 to the alveoli.
 
5. What will be the pO2 and pCO2 in the atmospheric air compared to those in the alveolar air?
(i) pO2 lesser, pCO2 higher
(ii) pO2 higher, pCO2 lesser
(iii) pO2 higher, pCO2 higher
(iv) pO2 lesser, pCO2 lesser
 
Answer
(ii) pO2 higher, pCO2 lesser

6. Explain the process of inspiration under normal conditions.

Answer
Inspiration is initiated by the contraction of diaphragm which increases the volume of thoracic chamber in the antero-posterior axis. The contraction of external inter-costal muscles lifts up the ribs and the sternum causing an increase in the volume of the thoracic chamber in the dorso-ventral axis. The overall increase in the thoracic volume causes a similar increase in pulmonary volume. An increase in pulmonary volume decreases the intra-pulmonary pressure to less than the atmospheric pressure. This pressure gradient forces the air from outside to move into the lungs and inspiration takes place.

7. How is respiration regulated?
 
Answer
The respiration is regulated bu neural system. A specialised centre present in the medulla region of the brain called respiratory rhythm centre is primarily responsible for this regulation. Another centre present in the pons region of the brain called pneumotaxic centre can moderate the functions of the respiratory rhythm centre. Neural signal from this centre can reduce the duration of inspiration and thereby alter the respiratory rate. A chemosensitive area is situated adjacent to the rhythm centre which is highly sensitive to CO2 and hydrogen ions. Increase in these substances can activate this centre, which in turn can signal the rhythm centre to make necessary adjustments in the respiratory process by which these substances can be eliminated. Receptors associated with aortic arch and carotid artery also can recognise changes in CO2 and H+ concentration and send necessary signals to the rhythm centre for remedial actions. The role of oxygen in the regulation of respiratory rhythm is quite insignificant.
 
8. What is the effect of pCO2 on oxygen transport?
 
Answer
pCO2 plays a major role in transportation of oxygen. In the alveoli, the low pCO2 and high pO2 favours the formation of haemoglobin. In the tissues, the high pCO2 and low pO2 favours the dissociation of oxygen from oxyhaemoglobin. Hence, the affinity of haemoglobin for oxygen is enhanced by the decrease of pCO2 in blood. Therefore, oxygen is transported in blood as oxyhaemoglobin and oxygen dissociates from it at the tissues.

9. What happens to the respiratory process in a man going up a hill?
 
Answer
When a man going up a hill he has to exert more effort to climb which increases the consumption of oxygen. As a result, the partial pressure of oxygen in haemoglobin decreases which creates more demand for oxygen. Thus, the breathing rate increases to fill this gap.

10. What is the site of gaseous exchange in an insect?
 
Answer
 
Insects have a network of tubes known as tracheal tubes to transport atmospheric air within the body. The tracheae open on the lateral surface of the animal through minute pores called spiracles.
 
11. Define oxygen dissociation curve. Can you suggest any reason for its sigmoidal pattern?
 
Answer
 
When percentage saturation of haemoglobin with O2 is plotted against the pO2 a sigmoid curve is obtained which is called Oxygen dissociation curve.
The dissociation curve is sigmoidal pattern 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.
 
12. Have you heard about hypoxia? Try to gather information about it, and discuss with your friends.
 
Answer
Hypoxia is defined as a condition of the body in which the tissue have shortage of oxygen. It generally happens because of a mismatch between oxygen demand and supply.

13. Distinguish between
(a) IRV and ERV
(b) Inspiratory capacity and Expiratory capacity
(c) Vital capacity and Total lung capacity
 
Answer
 
(a) IRV and ERV
 
Inspiratory reserve volume(IRV)
Expiratory reserve volume (ERV)
It is the maximum volume of air that can be inhaled after a normal inspiration. It is the maximum volume of air that can be exhaled after a normal expiration.
It is about 2500-3500 mL in the human lungs. It is about 1000-1100 mL in the human lungs.

(b) Inspiratory capacity and Expiratory capacity

Inspiratory capacity (IC)
Expiratory capacity (EC)
It is the volume of air that can be inhaled after a normal expiration. It is the volume of air that can be exhaled after a normal inspiration.
It includes tidal volume and inspiratory reserve volume. (IC = TV + IRV) It includes tidal volume and expiratory reserve volume. (EC = TV + ERV)

(c) Vital capacity and Total lung capacity

Vital capacity (VC)
Total lung capacity (TLC)
It is the maximum volume of air that can be exhaled after a maximum inspiration.  It is the volume of air in the lungs after maximum inspiration. It includes IC, ERV, and residual volume.
It includes IC and ERV It includes IC, ERV, and residual volume.
It is about 4000 mL in the human lungs. It is about 5000-6000 mL in the human lungs.

14. What is Tidal volume? Find out the Tidal volume (approximate value) for a healthy human in an hour.
 
Answer
 
Tidal volume is the volume of air inspired or expired during normal respiration. It is approximately 500 ml in a healthy man.
The hourly tidal volume for a healthy human (taking 12 breathes/min) can be calculated as:
= 500 ml × 12 × 60 minute = 360000 ml
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CLASS 11th CHAPTER -16 Digestion and Absorption |Biology | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Biology Chapter – 16 Digestion and Absorption includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Biology exams must go through NCERT Solutions for Class 11 Chapter -16 Digestion and Absorption. NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

Class 11th Chapter -16 Digestion and Absorption | NCERT BIOLOGY SOLUTION |

Exercises
 
Page No: 267
 
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
► (i) pepsin, lipase and rennin

(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
► (ii) intestinal juice
 
2. Match column I with column II
Column I
Column II
(a)
Bilirubin and biliverdin
(i)
Parotid
(b)
Hydrolysis of starch
(ii)
Bile
(c)
Digestion of fat
(iii)
Lipases
(d)
Salivary gland
(iv)
Amylases

Answer

Column I
Column II
(a)
Bilirubin and biliverdin
(ii)
Bile
(b)
Hydrolysis of starch
(iv)
Amylases
(c)
Digestion of fat
(iii)
Lipases
(d)
Salivary gland
(i)
Parotid

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?
 
Answer
 
(a) Villi are finger like structures that absorb the nutrients from digested food in the intestine. The main function of villi is to absorption of nutrients so they are present in small intestine as this is the place where absorption of food takes place. Villi is not present in stomach because the food is still yet to be broken down therefore, similar finger like structures called “rugae” is present which secrete pepsin and the gastric juices for the digestion to take place in the stomach.
 
(b) Pepsinogen changes into its active form by the action of hydrochloric acid.


(c) The walls of the alimentary canal are made up of four layers. These are:
• Serosa: It is the outermost layer and is made up of a thin mesothelium with some connective tissues.
• Muscularis: It is a thin layer of smooth muscles which are usually arranged into an inner circular and outer longitudinal layer.
• Sub-mucosa: It is a layer of loose connective tissues, containing nerves, blood, and lymph vessels which supports mucosa.
• Mucosa: It is the innermost lining of the lumen of the alimentary canal which is mainly involved in absorption and secretion.
 
(d) Bile helps in emulsification of fats and also activates lipases. Therefore, bile helps in digestion of fats.
 
4. State the role of pancreatic juice in digestion of proteins.
 
Answer
 
Pancreatic juice contains a variety of inactive enzymes such as trypsinogen, chymotrypsinogen, and carboxypeptidases which help in the digestion of proteins and the partially hydrolysed proteins, like proteoses and peptones, into dipeptides.
 
5. Describe the process of digestion of protein in stomach.
 
Answer
The food that enters the stomach becomes acidic on mixing with this gastric juice which is secreted by gastric glands present on the wall of stomach. The main components of gastric juice are hydrochloric acid, pepsinogen, mucus, and rennin. The inactive proenzyme, pepsinogen when acted upon by hydrochloric acid gets converted into active enzyme pepsin which coverts protein into proteoses and peptides.
Renin which is present in gastric juice of infants helps in the digestion of milk protein.
 
6. Give the dental formula of human beings.
 
Answer
The dental formula in humans is represented as I 2/2, C 1/1, PM 2/2, M 3/3 in which
I = Incisors
C = Canine
PM = Pre molar
M = Molar
Thus, the dental formula of human beings is 2123/2123.
 
7. Bile juice contains no digestive enzymes, yet it is important for digestion. Why?
 
Answer
Bile juice not contain any digestive enzymes yet it plays an important role in the digestion of fats. as it 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. It also makes the medium of food from acidic to alkaline and activates lipase.
 
8. Describe the digestive role of chymotrypsin. What two other digestive enzymes of the same category are secreted by its source gland?

Answer
Chymotrypsin is an active enzyme formed by the action of trypsin in the pancreatic juice. This helps in the digestion of proteins peptones and proteoses converting it into dipeptides.
The other two digestive enzyme secreted by its source gland is amylase and lipases.
 
9. How are polysaccharides and disaccharides digested?

Answer
The polysaccharides and disaccharides are partially digested by the amylase enzyme present in the pancreatic juice. The remaining digestion takes place by enzymes in the succus entericus. The enzyme maltase converts maltose into two molecules of glucose, lactase converts lactose into glucose and galactose, sucrase converts sucrose into glucose and fructose.
 
10. What would happen if HCl were not secreted in the stomach?
 
Answer
If HCl were not secreted in the stomach then it would affect protein digestion. The HCl secreted by glands present on stomach walls provides acidic medium to food. 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 and affects its digestion.
 
11. How does butter in your food gets digested and absorbed in the body?
 
Answer
Digestion of butter
Butter is mainly composed of fat. Bile juice secreted by the liver contains bile salts break down large fat globules into smaller globules which increases the surface area for the action of lipase. This is known 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 fatty aicds and glycerol.
 
Absorption of butter
The ultimate products of fat i.e., fatty acids and glycerol is not water soluble so they can’t absorbed the blood directly. Therefore, they are first incorporated into small droplets called micelles and then transported into the villi of the intestinal mucosa. They are re-formed into very small protein coated fat globules called the chylomicrons which are transported into the lymph vessels in the villi. These lymph vessels ultimately release the absorbed substances into the blood stream.

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

Answer
 
The digestion of protein starts in stomach and completes in small intestine.
In stomach, active pepsin is formed by the action of HCl on inactive pepsinogen  converts proteins into proteases and peptones.
In small intestine, the pancreatic juice contains inactive enzymes – trypsinogen, chymotrypsinogen and procarboxypeptidases in the inactive form. Trypsinogen is activated by an enzyme, enterokinase, secreted by the intestinal mucosa into active trypsin, which in turn activates the other enzymes in the pancreatic juice. Chymotrypsinogen and procarboxypeptidases get converted into chymotrypsin and carboxypeptidase in the presence of trypsin. Chymotrypsin converts the proteins into peptides, and carboxypeptidase further converts peptides into smaller peptide chains and amino acids.
 
13. Explain the term thecodont and diphyodont.
 
Answer
 
The type of attachment in which each tooth is embedded in a socket of jawbone is called thecodont.
Majority of mammals including human beings form two sets of teeth during their life, a set of temporary milk or deciduous teeth, replaced by a set of permanent or adult teeth. This type of dentition is called diphyodont.
 
14. Name different types of teeth and their number in an adult human.
 
Answer
Different types of teeth and their number in an adult human are:
→ Incisors: Total number is 8
→ Canine: Total number is 4
→ Premolars: Total number is 8
→ Molars: Total number is 12
 
15. What are the functions of liver?
 
Answer
The functions of liver are:
→ The liver secretes bile juice which helps in the digestion of fats.
→ It secretes an anticoagulant called heparin which prevent clotting of blood inside blood vessels.
→ It produces a protein, angiotensinogen, which helps the kidneys in maintaining body fluid osmoregulation.
→ It is an important place of lymph formation.
→ It produces red blood cells in the embryo.
→ It also helps in breakdown of insulin and other hormones, haemogloin, some toxic substances and conversion of ammonia into urea takes place in liver.
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CLASS 11th CHAPTER -15 Plant Growth and Development |Biology | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Biology Chapter – 15 Plant Growth and Development includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Biology exams must go through NCERT Solutions for Class 11 Chapter -15 Plant Growth and Development. NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

Class 11th Chapter -15 Plant Growth and Development | NCERT BIOLOGY SOLUTION |

Exercises

Page No: 253

1. Define growth, differentiation, development, dedifferentiation, redifferentiation, determinate growth, meristem and growth rate.
 
Answer
 • Growth is an irreversible permanent increase in size of an organ or its parts or even of an individual cell.

• Differentiation is the process in which the cells derived from root apical and shoot-apical meristems and cambium differentiate and mature to perform specific functions.
 
• Development is a term that includes all changes that an organism goes through during its life cycle from germination of the seed to senescence.
 
• Dedifferentiation is the process in which permanent plant cells regain the power to divide under certain conditions.
 
• Redifferentiation is the process in which de-differentiated cells become mature again and lose their capacity to divide.
 
• Determinate growth is the type of growth in which growth stops after a certain phase.
 
• Meristem are the specialised regions in the plants where active cell divisions take place.
 
• The increased growth per unit time is termed as growth rate.
 
2. Why is not any one parameter good enough to demonstrate growth throughout the life of a flowering plant?
 
Answer
Growth, at a cellular level, is principally a consequence of increase in the amount of protoplasm. Measuring the growth of protoplasm involves many parameters such as the weight of the fresh tissue sample, the weight of the dry tissue sample, the differences in length, area, volume, and cell number measured during the growth period. Hence, there cannot be one parameter good enough to demonstrate growth throughout the life of a flowering plant.
 
3. Describe briefly:
(a) Arithmetic growth
(b) Geometric growth
(c) Sigmoid growth curve
(d) Absolute and relative growth rates
 
Answer
(a) In arithmetic growth, following mitotic cell division, only one daughter cell continues to divide while the other differentiates and matures. The elongation of roots at a constant rate is an example of arithmetic growth. On plotting length of the organ against time, a linear curve is obtained. Mathematically, it is expressed as:
Lt = L0 + rt
Here, Lt is length at time ‘t’, L0 is length at time 0 and r is the rate per unit time.

(b) In most systems, the initial growth is slow and called lag phase, and it increases rapidly thereafter at an exponential rate and called log or exponential phase. Here, both the progeny cells following mitotic cell division retain the ability to divide and continue to do so. However, with limited nutrient supply, the growth slows down leading to a stationary phase. The graph of the geometric growth gives a sigmoid curve.
 
(c) A sigmoid curve is a characteristic of living organism growing in a natural environment. This curve is divided into three phases – lag phase, log phase or exponential phase of rapid growth, and stationary phase.
Sigmoid curve Graph
Exponential growth can be expressed as:
W1= W0ert
W1 = final size (weight, height, number etc.)
W0 = initial size at the beginning of the period
r = growth rate
t = time of growth
e = base of natural logarithms
 
(d) The measurement and the comparison of total growth per unit time is called the absolute growth rate. 
The growth of the given system per unit time expressed on a common basis, e.g., per unit initial parameter is called the relative growth rate.
 
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.
 
Answer
The five main groups of natural plant growth regulators are:
(i) Auxins
(ii) Gibberellic acid
(iii) Cytokinins
(iv) Ethylene
(v) Abscisic acid
A note on discovery, physiological functions and agricultural/horticultural applications of Auxins are:
 
→ Discovery: The first observations regarding the effects of auxins were made by Charles Darwin and Francis Darwin when they observed that the coleoptiles of canary grass responded to unilateral illumination by growing towards the light source (phototropism).
After a series of experiments, it was concluded that the tip of coleoptile was the site of transmittable influence that caused the bending of the entire coleoptile. Auxin was isolated by F.W. Went from tips of coleoptiles of oat seedlings.

→ Physiological Functions:
• They control plant cell-growth.
• They cause the phenomenon of apical dominance.
• They control division in the vascular cambium and xylem differentiation.
• They induce parthenocarpy and prevent abscission of leaves and fruits.

→ Horticulture Application:
• They help to initiate rooting in stem cuttings, an application widely used for plant propagation.
• 2-4 D is used weedicide to kill broadleaf, dicotyledonous weeds.
• They induce parthenocarpy in tomatoes.
• It promote flowering e.g. in pineapples.

Page No: 254

5. What do you understand by photoperiodism and vernalisation? Describe their significance.
 
Answer
 
The response of plants to periods of day/night is termed as photoperiodism. It is hypothesised that the hormonal substance responsible for flowering is formed in the leaves, subsequently migrating to the shoot apices and modifying them into flowering apices. Photoperiodism helps in studying the response of flowering in various crop plants with respect to the duration of exposure to light.

There are plants for which flowering is either quantitatively or qualitatively dependent on exposure to low temperature. This phenomenon is termed vernalisation. It refers specially to the promotion of flowering by a period of low temperature. It prevents precocious reproductive development late in the growing season, and enables the plant to have sufficient time to reach maturity.

6. Why is Abscisic acid also known as stress hormone?
 
Answer
Abscisic acid stimulates the closure of stomata in the epidermis and increases the tolerance of plants to various kinds of stresses. Therefore, it is also called the stress hormone. It promotes seed dormancy and ensures seed germination during favourable conditions. It helps seeds withstand desiccation. It also helps in inducing dormancy in plants at the end of the growing season and promotes abscission of leaves, fruits, and flowers.

7. ‘Both growth and differentiation in higher plants are open’. Comment.
 
Answer
The higher plants retain the capacity for unlimited growth throughout their life. This ability of the plants is due to the presence of meristems at certain locations in their body. The cells of such meristems have the capacity to divide and self-perpetuate. Therefore, growth in higher plants are open. Also, some of these cells always undergo differentiation after some rounds of cell division. Hence, the differentiation is also open.
 
8. ‘Both a short day plant and a long day plant can flower simultaneously in a given place’. Explain.
 
Answer
Flowering in some plants depends on relative durations of light and dark periods. The short-day plant and long-day plant can flower at the same place, provided they have been given an adequate photoperiod.
 
9. Which one of the plant growth regulators would you use if you are asked to:
(a) Induce rooting in a twig
► Auxins
 
(b) Quickly ripen a fruit
► Ethylene

(c) Delay leaf senescence
► Cytokinins

(d) Induce growth in axillary buds
► Cytokinins

(e) ‘Bolt’ a rosette plant
► Gibberellins

(f) Induce immediate stomatal closure in leaves.
► Abscisic acid
 
10. Would a defoliated plant respond to photoperiodic cycle? Why?

Answer
No, a defoliated plant will not respond to the photoperiodic cycle as leaves are the sites of perception of light/dark duration. Therefore, in the absence of leaves, the plant would not respond to light.
 
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.
 
Answer
 
(a) If GA3 is applied to rice seedlings, then the rice seedlings will show internode-elongation and increase in height.

(b) If dividing cells stop differentiating, then the plant organs such as leaves and stem will not be formed. The mass of undifferentiated cell is called callus.

(c) If a rotten fruit gets mixed with unripe fruits, then the ethylene produced from the rotten fruits will hasten the ripening of the unripe fruits.

(d) If you forget to add cytokinin to the culture medium, then cell division, growth, and differentiation will be slower.
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CLASS 11th CHAPTER -14 Respiration in Plants |Biology | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Biology Chapter – 14 Respiration in Plants includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Biology exams must go through NCERT Solutions for Class 11 Chapter -14 Respiration in Plants. NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

Class 11th Chapter -14 Respiration in Plants | NCERT BIOLOGY SOLUTION |

Page No: 238
 
Exercises
 
1. Differentiate between
(a) Respiration and Combustion
(b) Glycolysis and Krebs’cycle
(c) Aerobic respiration and Fermentation
 
Answer

(a) Respiration and Combustion

Respiration
Combustion
It is a biochemical process.It is a physiochemical process.
It occurs inside the cells.It does not occur inside cells.
It is a biologically-controlled process.It is an uncontrolled process.
No light is producedLight may be produced
ATP is generated.ATP is not generated.
Enzymes are requiredEnzymes are not required

(b) Glycolysis and Krebs’cycle

Glycolysis
Krebs’cycle
The breakdown of glucose to pyruvic acid is called glycolysis.The further processing of pyruvic acid through aerobic route is called Krebs’ cycle.
It is a linear pathway.It is a cyclic pathway.
It occurs in the cell cytoplasm.It occurs in the mitochondrial matrix.
It occurs in all living beings.It occurs in aerobic respiration.
It generates 2 NADH2 and 2 ATP molecules on the breakdown of one glucose molecule.It produces 6 NADH2, 2 FADH2, and 2 ATP molecules on the breakdown of two acetyl-CoA molecules.
No carbon dioxide evolvedCarbon dioxide evolved

(c) Aerobic respiration and Fermentation

Aerobic respiration
Fermentation
The complete oxidation of organic substances in the presence of oxygen.The incomplete oxidation of glucose in the absence of oxygen.
It occurs in the cytoplasm and mitochondriaIt occurs in the cytoplasm
The end products are carbon dioxide and waterThe end products are ethyl alcohol and carbon dioxide
Complete oxidation of the respiratory substrate takes placeIncomplete oxidation of the respiratory substrate takes place
About 36 ATP molecules are producedOnly 2 ATP molecules are produced


2. What are respiratory substrates? Name the most common respiratory substrate.
 
Answer
A compound which is oxidized during respiration is called respiratory substrate. Glucose is the most common respiratory substrate.
 
3. Give the schematic representation of glycolysis?
Answer
 
schematic representation of glycolysis
 
4. What are the main steps in aerobic respiration? Where does it take place?
 
Answer
 
The main steps in aerobic respiration and places of their occurrence are:
→ Glycolytic breakdown of glucose in to pyruvic acid in cytoplasm.
→ Krebs cycle in matrix of mitochondria
→ Electron transport system in inner mitochondrial membrane
→ Oxidative phosphorylation in the inner mitochondrial membrane.
 
5. Give the schematic representation of an overall view of Krebs cycle.
 
Answer
 
Krebs Cycle
 
6. Explain ETS.
 
Answer
The metabolic pathway through which the electron passes from one carrier to another is called the electron transport system (ETS). It is located in the inner mitochondrial membrane. Electrons from NADH produced in the mitochondrial matrix during citric acid cycle are oxidised by an NADH dehydrogenase (complex I), and electrons are then transferred to ubiquinone located within the inner membrane. Ubiquinone also receives reducing equivalents via FADH2 (complex II) that is generated during oxidation of succinate in the citric acid cycle. The reduced ubiquinone (ubiquinol) is then oxidised with the transfer of electrons to cytochrome c via cytochrome bc 1 complex (complex III). The cytochrome c acts as a mobile carrier between complex III and cytochrome c oxidase complex, containing cytochrome a and a3, along with copper centres (complex IV).
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 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 FADH2, on oxidation, gives 3 ATP molecules.
 
7. Distinguish between the following:
(a) Aerobic respiration and Anaerobic respiration
(b) Glycolysis and Fermentation
(c) Glycolysis and Citric acid Cycle
 
Answer

(a) Aerobic respiration and Anaerobic respiration

Aerobic respiration
Anaerobic respiration
It takes place in the presence of oxygen.It takes place in the absence of oxygen.
It occurs in cytoplasm and mitochondria.It occurs in cytoplasm
The end products of aerobic respiration are carbon dioxide and water.The end products of fermentation are ethyl alcohol and carbon-dioxide.
Complete oxidation of respiratory substrate takes place.Incomplete oxidation of respiratory substrate takes place.
36-38 ATP molecules are produced.Only 2 ATP molecules are produced.

(b) Glycolysis and Fermentation

Glycolysis
Fermentation
Glycolysis occurs during aerobic and anaerobic respiration.Fermentation is a type of anaerobic respiration.
The end products of aerobic respiration is pyruvic acidThe end products of aerobic respiration is ethanol or lactic acid

(c) Glycolysis and Citric acid Cycle

Glycolysis
Citric acid Cycle
It is a linear pathway.It is a cyclic pathway.
It occurs in the cell cytoplasm.It occurs in the mitochondrial matrix.
It occurs in all living beings.It occurs in aerobic respiration.
It generates 2 NADH2 and 2 ATP molecules on the breakdown of one glucose molecule.It generates 6 NADH2, 2 FADH2, and 2 ATP molecules on the breakdown of two acetyl-CoA molecules.

8. What are the assumptions made during the calculation of net gain of ATP?
 
Answer
 
The following assumptions are made during the calculation of net gain of ATP:
• There is a sequential, orderly pathway functioning, with one substrate forming the next and with glycolysis, TCA cycle and ETS pathway following one after another.
• The NADH synthesised in glycolysis is transferred into the mitochondria and undergoes oxidative phosphorylation.
• None of the intermediates in the pathway are utilised to synthesise any other compound.
• Only glucose is being respired – no other alternative substrates are entering in the pathway at any of the intermediary stages.
 
9. Discuss “The respiratory pathway is an amphibolic pathway.”
Answer
The breaking down process occurring within the organism is known as catabolic process and the synthesis process is known as anabolism. The respiratory pathway involves both the catabolic and anabolic process, so it is referred as amphibolic pathway.
 
10. Define RQ. What is its value for fats?
 
Answer
The ratio of the volume of CO2 evolved to the volume of O2 consumed in respiration is called the respiratory quotient (RQ). The RQ for carbohydrates is 1. The RQ for fat and protein is less than 1.
For example: Calculations for a fatty acid, tripalmitin
2(C5H98O6) + 145O2 → 102 CO2 + 98 H2O
RQ = Volume of CO2 evolved/Volume of O2 consumed
      = 102 CO2/145O2 = 0.7

11. What is oxidative phosphorylation?
 
Answer
 
The generation of ATP from ADP during electron transport system by utilizing the energy obtained during oxidative reaction is called oxidative phosphorylation.
 
12. What is the significance of step-wise release of energy in respiration?
 
Answer
 
The significance of step-wise release of energy in respiration are:
→ It facilitates the utilization of the relatively higher proportion of the energy in ATP synthesis.
→ The activities of enzymes for the different steps may be enhanced or inhibited by specific compounds. This provides a mean of controlling the rate of the pathway and the energy output according to the need of the cell.
→ The same pathway may be utilized for forming intermediates used in the synthesis of other biomolecules like amino acids.
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CLASS 11th CHAPTER – 13 Photosynthesis in Higher Plants |Biology | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Biology Chapter – 13 Photosynthesis in Higher Plants includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Biology exams must go through NCERT Solutions for Class 11 Chapter -13 Photosynthesis in Higher Plants. NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

Class 11th Chapter -13 Photosynthesis in Higher Plants | NCERT BIOLOGY SOLUTION |

Page No: 224
 
Exercises
 
1. By looking at a plant externally can you tell whether a plant is C3 or C4? Why and how?
 
Answer
 
We can’t tell whether a plant is C3 or C4 by looking at a plant externally. However, plants which which are adapted to dry climates follow the C4 pathway. Unlike C3 plants, the leaves of C4 plants have a special anatomy but this difference can only be observed at the cellular level.

2. By looking at which internal structure of a plant can you tell whether a plant is C3 or C4? Explain.
 
Answer
 
As leaves of C4 plants have a special anatomy called Kranz anatomy. This makes them different from C3 plants. Special cells, known as bundle-sheath cells, surround the vascular bundles. These cells have a large number of chloroplasts. They are thick-walled and have no intercellular spaces. Therefore, we can tell whether a plant is C3 or C4 through internal structure.
 
3. Even though a very few cells in a C4 plant carry out the biosynthetic – Calvin pathway, yet they are highly productive. Can you discuss why?
 
Answer

In C4 plants photorespiration does not occur because they have a mechanism that increases the concentration of CO2 at the enzyme site. This takes place when the C4 acid from the mesophyll is broken down in the bundle sheath cells to release CO2 that results in increasing the intracellular concentration of CO2. In turn, this ensures that the RuBisCO functions as a carboxylase minimising the oxygenase activity. Thus, the photosynthesis rate increases and make C4 plants more productive.

Page No: 225

4. RuBisCo is an enzyme that acts both as a carboxylase and oxygenase. Why do you think RuBisCo carries out more carboxylation in C4 plants?
 
Answer
 
The enzyme RuBisCo is absent from the mesophyll cells of C4 plants. It is present in the bundle-sheath cells surrounding the vascular bundles. In C4 plants, the Calvin cycle occurs in the bundle-sheath cells. The primary CO2 acceptor in the mesophyll cells is phosphoenol pyruvate -a three-carbon compound. It is converted into the four-carbon compound oxaloacetic acid (OAA). OAA is further converted into malic acid. Malic acid is transported to the bundle-sheath cells, where it undergoes decarboxylation and CO2 fixation occurs by the Calvin cycle. This prevents the enzyme RuBisCo from acting as an oxygenase.
 
5. Suppose there were plants that had a high concentration of Chlorophyll b, but lacked chlorophyll a, would it carry out photosynthesis? Then why do plants have chlorophyll b and other accessory pigments?
 
Answer
 
If there were complete absence of chlorophyll a in a plant, it would not carry out photosynthetic activity at all because chlorophyll a is the chief pigment associated with photosynthesis as it traps light. Other accessory pigments like chlorophyll b, santhophylls and carotenoids are equally essential as they also absorb light and transfer energy to chlorophyll a. They also enable a wider range of wavelength of incoming light to be utilised for photosynthesis and protect chlorophyll a from photo- oxidation.
 
6. Why is the colour of a leaf kept in the dark frequently yellow, or pale green? Which pigment do you think is more stable?
 
Answer
 
Chlorophyll or green pigment is unable to absorb energy in the absence of light therefore loses its stability. Thus, the colour of leaf changes to yellow or pale green. This shows that Carotenoids and Xanthophyll pigments are more stable.
 
7. Look at leaves of the same plant on the shady side and compare it with the leaves on the sunny side. Or, compare the potted plants kept in the sunlight with those in the shade. Which of them has leaves that are darker green? Why?
 
Answer
The plants placed in light will have darker leaves as compared to leaves of a plant placed in shade. As leaves in shade get lesser light for photosynthesis so they perform lesser photosynthesis as compared to the leaves or plants kept in sunlight. To increase the rate of photosynthesis, the leaves present in shade have more chlorophyll pigments. This increase in chlorophyll content increases the amount of light absorbed by the leaves, which in turn increases the rate of photosynthesis which makes the leaves or plants in shade greener than the leaves or plants kept in the sun.
 
8. Figure 13.10 shows the effect of light on the rate of photosynthesis. Based on the graph, answer the following questions:
(a) At which point/s (A, B or C) in the curve is light a limiting factor?

(b) What could be the limiting factor/s in region A?
(c) What do C and D represent on the curve?
 
Answer
 
Figure 13.10
 
(a) At point A
(b) Light is a limiting factor also, water, temperature, and the concentration of carbon dioxide could also be limiting factors in the region A.
(c) C represents the stage beyond which light is not a limiting factor. D represents the stage beyond which intensity of light has no effect on the rate of photosynthesis.
 
9. Give comparison between the following:
(a) C3 and C4 pathways
(b) Cyclic and non-cyclic photophosphorylation
(c) Anatomy of leaf in C3 and C4 plants
 
Answer
 
(a) C3 and C4 pathways

C3 pathways
C4 pathways
The primary acceptor of CO2  is RUBP – a five-carbon compound.The primary acceptor of CO2 is phosphoenol pyruvate – a three-carbon compound.
The first stable product is 3 phosphoglycerate.The first stable product is oxaloacetic acid.
It occurs only in the mesophyll cells of the leaves.It occurs in the mesophyll and bundle-sheath cells of the leaves.
It is a slower process of carbon fixation and photo-respiratory losses are high.It is a faster process of carbon fixation and photo-respiratory losses are low.

(b) Cyclic and non-cyclic photophosphorylation

Cyclic photophosphorylation
Non-cyclic photophosphorylation
It occurs only in photosystem I.It occurs both in photosystems I and II.
It involves only the synthesis of ATP.It involves the synthesis of ATP and NADPH2.
In this process, photolysis of water does not occur. Therefore, oxygen is not produced.In this process, photolysis of water takes place and oxygen is liberated.
In this process, electrons move in a closed circle.In this process, electrons do not move in a closed circle.

(c) Anatomy of leaf in C3 and C4 plants

Anatomy of leaf in C3
Anatomy of leaf in C4
Bundle-sheath cells are absentBundle-sheath cells are present
RuBisCo is present in the mesophyll cellsRuBisCo is present in the bundle-sheath cells.
The first stable compound produced is 3-phosphoglycerate – a three-carbon compound.The first stable compound produced is oxaloacetic acid – a four-carbon compound.
Photorespiration occursPhotorespiration does not occur
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