NCERT Solutions for Class 11

From the following table, students can find NCERT Solutions for Class 11 Maths, Physics, Chemistry, English Commerce, Business Studies, Computer Science, etc., Solving NCERT Solutions for class 11 will help you to solve the NCERT Class 12 questions. NCERT Solutions for class 11 is provided by the best teachers in order to help students of class 11 to ace the class 11 exam with decent grades. Class 11 NCERT solutions will not only help students to secure a good score in class 11 board exams but also helps to crack the engineering and medical competitive entrance exams.

NCERT Solutions for Class 11 Biology Chapter 22 Chemical Coordination and Integration

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 22 Chemical Coordination and Integration:

Section NameTopic Name
22Chemical Coordination and Integration
22.1Endocrine Glands and Hormones
22.2Human Endocrine System
22.3Hormones of Heart, Kidney and Gastrointestinal Tract
22.4Mechanism of Hormone Action
22.5Summary

NCERT Solutions Class 11 BiologyBiology Sample Papers

NCERT TEXTBOOK QUESTIONS FROM SOLVED

1. Define the following:
(a) Exocrine gland,
(b) Endocrine gland,
(c) Hormone.
Solution:
(a) Exocrine gland is a gland that pours its secretion on the surface or into a particular region by means of ducts for performing a metabolic activity, e.g., sebaceous glands, sweat glands, salivary glands and intestinal glands.
(b) Endocrine gland is an isolated gland (separates even from epithelium forming it) which secretes informational molecules or hormones that are poured into venous blood or lymph for reaching the target organ because the gland is not connected with the target organ by any duct. Therefore endocrine gland is also called ductless gland e.g. thyroid gland.
(c) Hormone is a substance that is manu-factured and secreted in very small quantities into the blood stream by an endocrine gland or a specialized nerve cell and regulates the growth or functioning of a specific tissue organ in a distant part of the body e.g insulin.

2. Diagrammatically indicate the location of the various endocrine glands in our body.
Solution:
NCERT Solutions For Class 11 Biology Chemical Coordination and Integration Q2

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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-l Tract.
Solution:
(a) Two types of hormones are produced by hypothalamus : releasing hormones (that stimulate secretion of pituitary hormones) and inhibiting hormones (that inhibit secretion of pituitary hormones).
These hormones are:

  1. Thyrotrophin-releasing hormone Adreno-
  2. corticotrophin-releasing hormone
  3. Follicle-stimulating hormone-releasing hormone
  4. Luteinizing hormone-releasing hormone
  5. Growth hormone-releasing hormone
  6. Growth inhibiting hormone
  7. Prolactin releasing hormone
  8. Prolactin inhibiting hormone
  9. Melanocyte stimulating hormone¬releasing hormone
  10. Melanocyte stimulating hormone- inhibiting hormone.

(b) Different parts of pituitary secrete different hormones.
Hormones secreted by anterior lobe of pituitary are:

  1. Follicle stimulating hormone
  2. Luteinizing hormone
  3. Thyroid stimulating hormone
  4. Adrenocorticotrophic hormone
  5. Somatotrophic or Growth hormone
  6. Prolactin hormone or Luteotrophic hormone.
    Middle (intermediate) lobe of pituitary : Melanocyte stimulating hormone.
    Posterior lobe of pituitary:
    (i) Oxytocin
    (ii) Vasopressin or antidiuretic hormone.

(c) Thyroid secretes 3 hormones:

  1. Thyroxine or tetraiodothyronine
  2. Triiodothyronine
  3. Calcitonin.

(d) Parathyroid gland secretes a single hormone called parathormone (PTH) or Collip’s hormone.

(e) Adrenal glands have two regions, namely, outer adrenal cortex and inner adrenal medulla. Both these regions secrete different hormones.
Hormones of adrenal cortex are grouped into three categories:

  1. Glucocorticoids, e.g., cortisol
  2. Mineralocorticoids, e.g., aldosterone
  3. Sexcorticoids e.g testosterone. Adrenal medulla secretes two hormones
    (i) Epinephrine (adrenaline)
    (ii)Nor-epinephrine (nor-adrenaline).

(f) Pancreas secretes following hormones:

  1. Insulin
  2. Glucagon
  3. Somatostatin.

(g) Testis secretes androgens such as testosterone.

(h) Ovary secretes:

  1. Estrogens such as estradiol
  2. Progesterone
  3. Relaxin.

(i) Thymus secretes thymosin hormone.

(j) Atrium secretes atrial natriuretic factor (ANF).

(k) Kidney secretes:
(i) Renin (ii) Erythropoetin

(l) G.I. tract secretes :

  1. Gastrin
  2. Secretin
  3. Cholecystokinin
  4. Enterocrinin
  5. Duocrinin
  6. Villikinin.

4. Fill in the blanks:
Hormones                                           Target gland
(a) Hypothalamic hormones        ………………..
(b) Thyrotrophin (TSH)                 ………………..
(c) Corticotrophin (ACH)              ………………..
(d) Gonadotrophins (LH, FSH)   ………………..
(e) Melanotrophin (MSH)              ………………..
Solution:
(a) Pituitary
(b) Thyroid
(c) Adrenal cortex
(d) Gonads -Testes in male and ovaries in female
(e) Skin.

5. Write short notes on the functions of the following hormones:
(a) Parathyroid hormones (PTH)
(b) Thyroid hormones
(c) Thymosin
(d) Androgens
(e) Estrogens
(f) Insulin and Glucagon.
Solution:
(a) Parathyroid hormone increases the level of calcium and decreases the level of phosphate in the blood.
(b) Thyroid gland secretes three hormones: thyroxine, triiodothyronin and calcitonin. Thyroxine and triiodothyronin control the general metabolism of the body, promote growth of body tissues and stimulates tissue differentiation. Calcitonin regulates the concentration of calcium in the blood.
(c) Thymosin is secreted by thymus. It accelerates cell division, stimulates the development and differentiation of T-lymphocytes and also hastens attainment of sexual maturity.
(d) Androgens are secreted by testis. They stimulate the development of male reproductive system, formation of sperms, development of male accessory sex characters and also determines the male sexual behaviour and the sex urge.
(e) Estrogens are secreted by ovaries. They stimulate the female reproductive tract to grow to full size and become functional, differentiation of ova and development of accessory sex characters.
(f) Insulin is secreted by the |3-cells of the pancreas. It lowers blood glucose level, and promotes synthesis of proteins and fats. Glucagon is secreted by the a-cells of the pancreas. It increases the level of glucose in the blood.

6. Give example(s) of
(a) Hyperglycemic hormone and hypoglyce-mic hormone
(b) Hypercalcemic hormone
(c) Gonadotrophic hormones
(d) Progestational hormone
(e) Blood pressure lowering hormone
(f) Androgens and estrogens.
Solution:
(a)Glucagon, Insulin
(b) Parathormone (PTH)
(c) Follicle stimulating hormone (FSH) and Luteinizing hormone (LH)
(d) Progesterone
(e) Atrial natriuretic factor
(f) Testosterone and Estradiol.

7. Which hormonal deficiency is responsible for the following:
(a) Diabetes meilitus
(b) Goitre
(c) Cretinism.
Solution:
(a) Insulin
(b) Thyroxine and Triiodothyronine
(c) Thyroxine and Triiodothyronine.

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

9. Match the following :
Column I         Column II
(a) T4              (i) Hypothalamus
(b) PTH          (ii)Thyroid
(c) GnRH      (iii)Pituitary
(d) LH            (iv) Parathyroid.
Solution:
(a) – (ii); (b) – (iv); (c) – (i); (d) – (iii)

NCERT Solutions for Class 11 Biology Chapter 21 Neural Control and Coordination:

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 21 Neural Control and Coordination:

Section NameTopic Name
21Neural Control and Coordination
21.1Neural System
21.2Human Neural System
21.3Neuron as Structural and Functional Unit of Neural System
21.4Central Neural System
21.5Reflex Action and Reflex Arc
21.6Sensory Reception and Processing
21.7Summary

NCERT Solutions Class 11 BiologyBiology Sample Papers

NCERT TEXTBOOK QUESTIONS FROM SOLVED

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

(b) Eye: Eye is a hollow spherical structure composed of three coats:
– Outer fibrous coat
– Middle vascular coat
– Inner nervous coat
(i) Fibrous coat: It is thick and protects the eyeball. It has two distinct regions – sclera and cornea. Sclera covers most of the eye ball. The sclera or white of the eye contains many collagen fibres. Cornea is a transparent portion that forms the anterior one – sixth of the eyeball. The cornea is avascular (i.e., lacks blood supply).
(ii)Vascular coat: It comprises of 3 regions : choroid, iris, ciliary body.
(a) Choroid : It lies adjacent to sclera and contains numerous blood vessels and pigmented cells.
(b) Iris: The iris is a circular muscular diaphragm containing the pigment giving eye its colour. It extends from the ciliary body across the eyeball in front of the lens. It 2. has an opening in the centre called the pupil.
It contains two types of smooth muscles, circular muscles (sphincters) and radial muscles (dilators), of ectodermal origin.
(c) Ciliary body: Behind the peripheral margin of the iris, the vascular coat is thickened to form the ciliary body. It is composed of the ciliary muscles and the ciliary processes.
(iii) Nervous coat: It consists of retina which is neural and sensory layer of an eye ball. It consists of three layers; ganglion cells, bipolar cells and photoreceptor cells (rods and cones).
Lens: It is a transparent, biconvex, elastic structure that bends light waves as they pass through its surface. It is composed of epithelial cells that have large amounts of clear cytoplasm in the form of fibres.
Chambers of eyeball: The lens, suspensory ligament and ciliary body divide the eye into an anterior aqueous chamber and a posterior vitreous chamber which are filled with aqueous humour and vitreous humour respectively.

(c) Ear: There are three portions in an ear:
(i) External ear: It further has 2 regions: pinna and external auditory canal or meatus.
(a) Pinna: The pinna is a projecting elastic cartilage covered with skin. Its most prominent outer ridge is called the helix. The lobule is the soft pliable part at its lower end composed of fibrous and adipose tissue richly supplied with blood capillaries. It is sensitive as well as effective in collecting sound waves.
(b) External auditory canal: It is an S-shaped tube leading inward from the pinna. It is a tubular passage supported by cartilage in its exterior part and by bone in its interior part.
(ii) Middle ear: It consists of 3 small bones called ear ossicles – malleus, incus and stapes, which are attached to one another and increase efficiency of transmission of sound waves to inner ear.
(iii) Internal ear: It consists of bony and

2. Compare the following:
(a) Central neural system (CNS) and Peripheral neural system (PNS).
(b) Resting potential and action potential.
(c) Choroid and retina.
Solution: (a) CNS: It lies along the mid-dorsal axis of the body. It is a hollow, dorsally placed structure and comprises of brain and spinal cord. It is a centre of information processing and control.
PNS: Nerves arising from the central nervous system constitute the peripheral nervous system. It carries information to and from the CNS. It includes spinal nerves and cranial nerves.
(b) Resting potential: Outside the plasma membrane of a nerve fibre is the extracellular fluid which is positively charged with respect to the cell contents inside the plasma membrane. A resting nerve fibre shows a potential difference between inside and outside of this plasma membrane. This difference in the electrical charges across the plasma membrane is called the ‘resting potential’. A membrane with resting potential across it, is said to be electrically polarized. Action potential : Action potential is another name of nerve impulse. The contents inside a cell at the excited state becomes positively charged with respect to extracellular fluid outside it. This change in polarity across the plasma membrane is known as an action potential. The membrane with reversed polarity across it is said to be depolarized.
(c) Choroid: Choroid lies adjacent to the sclera and contains numerous blood vessels that supply nutrients and oxygen to the other tissues especially of retina. It contains abundant pigment cells and is dark brown in colour.
Retina: It is the neural and sensory layer of the eye ball. It is a very delicate coat and lines the whole of the vascular coat. Its external surface is in contact with the choroid and its internal surface with vitreous humour. It contains ganglion cells, bipolar cells and photoreceptor cells. membranous labyrinth. Membranous labyrinth consists of three semicircular ducts, utricle, saccule and cochlea.

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3. Explain the following processes:
(a) Polarisation of the membrane of a nerve fibre.
(b) Depolarisation of the membrane of a nerve fibre.
(c) Conduction of a nerve impulse along a nerve fibre.
(d) Transmission of a nerve impulse across a chemical synapse.
Solution: (a) Polarisation of the membrane of a nerve fibre : In the resting (not conducting impulse) nerve fibre the plasma membrane separates two solution of different chemical composition but having approximately the same total number of ions. In the external medium (tissue fluid), sodium ions (Na+) and Cl ions predominate, whereas within the fibre (intracellular fluid) potassium ions (K+) predominate. The differential flow of the positively charged ions and the inability of the negatively charged organic (protein) ions within the nerve fibre to pass out cause an increasing positive charge on the outside of the membrane and negative charge on the inside of the membrane. This makes the membrane of the resting nerve fibre polarized, extracellular fluid outside being electropositive (positively charged) with respect to the cell contents inside it.
(b) Depolarisation of the membrane of a nerve fibre: During depolarisation, the activation gates of Na channels open, and the K channels remain closed. Na+ rush into the axon. Entry of sodium ions leads to depolarisation (reversal of polarity) of the nerve membrane, so that the nerve fibre contents become electropositive with respect to the extracellular fluid.
(c) Conduction of a nerve impulse along a nerve fibre: Nervous system transmits information as a series of nerve impulses. A nerve impulse is the movement of an action potential as a wave through a nerve fibre. Action potentials are propagated, that is, self-generated along the axon. The events that set up an action potential at one spot on the nerve fibre also transmit it along the entire length of the nerve fibre. The action potential then moves to the neighbouring region of the nerve fibre till it covers the whole length of the fibre.
(d) Transmission of a nerve impulse across a chemical synapse: 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. The axon terminals contain vesicles filled with these neurotransmitters. When an impulse (action potential) arrives at the axon terminal, it stimulates the movement of the synaptic vesicles towards the membrane where they fuse with the plasma membrane and burst to 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
Solution: (a)
NCERT Solutions For Class 11 Biology Neural Control and Coordination Q4
(b)
NCERT Solutions For Class 11 Biology Neural Control and Coordination Q4.1
(c)
NCERT Solutions For Class 11 Biology Neural Control and Coordination Q4.2
(d)
NCERT Solutions For Class 11 Biology Neural Control and Coordination Q4.3

5. Write short notes on the following:
(a) Neural coordination (b) Forebrain
(c) Midbrain                       (d) Hindbrain
(e) Retina                             (f) Ear ossicles
(g) Cochlea                          (h) Organ of Corti
(i) Synapse
Solution: (a) Neural coordination : When higher animals respond to various stimuli, each response to a specific stimulus generally involves many organs (parts) of their bodies. Therefore, it is necessary that all the concerned organs (parts) of the body should work in a systematic manner to produce the response. The working together of various organs (parts) of the body of multicelullar organism in a proper manner to complement the functions of each other is called coordination. This is achieved by three overlapping processes of nervous system-sensory input, integration and motor output.
(b) Forebrain: It consists of: Olfactory lobes, the paired structures concerned with the sense of smell. Cerebrum which is the largest and most complex of all the parts of the human brain. It is divided by a cleft into left and right cerebral hemispheres which are connected by a large bundle of myelinated fibres the. corpus callosum. The outer cover of cerebral hemisphere is called cerebral cortex. It consists of sensory and motor areas. Hypothalamus region of forebrain contains centres which control body temperature, hunger and also contains group of neurosecretory cells.
(c) Midbrain: 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) Hindbrain: 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 of 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: Retina is the inner layer of an eye and it contains three layers of cells-from inside to outside – ganglion cells, bipolar cells and photoreceptor cells. There are two types of photoreceptor cells, namely, rods and cones. These cells contain the light-sensitive proteins called the photopigments. The daylight (photopic) vision and colour vision are functions of cones and the twilight (scotopic) vision is the function of the rods. The rods contain a purplish-red protein called the rhodopsin or visual purple, which contains a derivative of Vitamin A. In the human eye, there are three types of cones which possess their own characteristic photopigments that respond to red, green and blue lights. The sensations of different colours are produced by various combinations of these cones and their photopigments. When these cones are stimulated equally, a sensation of white light is produced.
(f) Ear ossicles : There is a small flexible chain of three small bones called as ear ossicles – the malleus (hammer shaped), the incus (anvil shaped) and the stapes (stirrup shaped) in the middle ear. Malleus is attached to the tympanic membrane on one side and incus on the other side. Incus in turn is connected with the stapes. Malleus is the largest ossicle, however stapes is the smallest ossicle.
(g) Cochlea : It is the main hearing organ which is connected with saccule. It is a spirally coiled tube that resembles a snail shell in appearance. It tapers from a broad base to an almost pointed apex.
(h) Organ of Corti: It 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) Synapse : It is the junction between the axon of one neuron and the dendrite or cyton of another neuron for transmission of nerve impulse.

6. Give a brief account of
(a) Mechanism of synaptic transmission.
(b) Mechanism of vision.
(c) Mechanism of hearing.
Solution: (a) Mechanism of synaptic transmission: Refer answer 3 (d)
(a) Mechanism of vision: The light rays in visible wavelength focused on the retina through the cornea and lens generate potentials (impulses) in rods and cones. Light induces
dissociation of the retinal from opsin resulting in changes in the structure of the opsin. This causes membrane permeability changes. As a result, 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 action potentials (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.
(b) Mechanism of hearing : 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 (malleus, incus and stapes) 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. As a result, 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?
Solution: (a)In humans, colour vision results from the activity of cone cells, a type of photoreceptor cells. In the human eye, there are three types of cones which possess their own characteristic photopigments that respond to red, green and blue lights. The sensations of different colours are produced by various combinations of these cones and their photopigments. When these cones are stimulated equally, sensation of white light is produced. Yellow light, for instance, stimulates green’and red cones approximately to equal extent, and this is interpreted by the brain as yellow colour.
(b) Ears (cristae and maculae present in internal ears).
(c) The iris contains two sets of smooth muscles – sphincters and dilators. These muscles regulate the amount of light entering the eyeball by varying the size of pupil. Contraction of sphincter muscles makes the pupil smaller in bright light so that less light enters the eye. Contraction of dilator muscles widens the pupil in dim light so that more light goes in eye to fall on 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.
Solution: (a) The action potential is largely determined by Na+ ions. The action potential results from the following sequential events
(i) Disturbance caused to the membrane of a nerve fibre by a stimulus results in leakage of  Na+ into the nerve fibre.
(ii) Entry of Na+ lowers the trans-membrane potential difference.
(iii) Decrease in potential difference makes the membrane more permeable to Na+ than to K+ ions so that more Na+ enter the fibre than K+ leave it.
(iv) Accumulation of Na+ in the nerve fibre initiates depolarisation (action potential), making the axonic contents positively charged relative to the extracellular fluid.
(v) With continued addition of Na+ the potential reaches zero and then plus 40-50 millivolts. This is the peak of action potential.
(vi) Permeability of a depolarised membrane to Na+ then rapidly drops, there are now as many Na+ on the inside of the membrane as on the outside.
(b) Refer answer 6 (b)
(c) Refer answer 6 (c)

9. Differentiate between
(a) Myelinated and non-myelinated axons
(b) Dendrites and axons
(c) Rods and cones
(d) Thalamus and Hypothalamus
(e) Cerebrum and Cerebellum
Solution: (a) Differences between myelinated and non-myelinated axons are as follows:
NCERT Solutions For Class 11 Biology Neural Control and Coordination Q9
(b) Axon and dendrites can be differentiated as follows:
NCERT Solutions For Class 11 Biology Neural Control and Coordination Q9.1

NCERT Solutions For Class 11 Biology Neural Control and Coordination Q9.2
(c) The differences between rods and cones are as follows:
NCERT Solutions For Class 11 Biology Neural Control and Coordination Q9.3
(d) Thalamus and hypothalamus can be differentiated as follows:
NCERT Solutions For Class 11 Biology Neural Control and Coordination Q9.4
(e) Cerebrum and cerebellum can be differentiated as follows:
NCERT Solutions For Class 11 Biology Neural Control and Coordination Q9.5

10. Answer the following.
(a) Which part of the ear determines the pitch ofa sound?
(b) Which part of the human brain is the most developed?
(c) Which part of our central neural system acts as a master clock?
Solution: (a) The receptor cells in the organ of Corti (Internal ear).
(b) Cerebrum (cerebral hemispheres).
(c) Pineal gland present in diencephalon of forebrain acts as a master clock, which maintains biological rhythm.

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

12. Distinguish between
(a) Afferent neurons and efferent neurons
(b) Impulse conduction in myelinated nerve fibre and unmyelinated nerve fibre
(c) Aqueous humour and vitreous humour
(d) Blind spot and yellow spot
(e) Cranial nerves and spinal nerves
Solution: (a)
NCERT Solutions For Class 11 Biology Neural Control and Coordination Q12
(b) Refer answer 9(a)
(c)
NCERT Solutions For Class 11 Biology Neural Control and Coordination Q12.1

NCERT Solutions For Class 11 Biology Neural Control and Coordination Q12.2
(d)
NCERT Solutions For Class 11 Biology Neural Control and Coordination Q12.3
(e)

NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement:

Section NameTopic Name
20Locomotion and Movement
20.1Types of Movement
20.2Muscle
20.3Skeletal System
20.4Joints
20.5Disorders of Muscular and Skeletal System
20.1Summary

NCERT Solutions Class 11 BiologyBiology Sample Papers

NCERT TEXTBOOK QUESTIONS FROM SOLVED

1. Draw the diagram of a sarcomere of skeletal muscle showing different regions.
Solution: 
NCERT Solutions For Class 11 Biology Locomotion and Movement Q1

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

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3. Describe the important steps in muscle contraction.
Solution: Mechanism of muscle contraction is explainei by sliding filament theory which states that contraction of a muscle fibre takes place by the sliding of the thin filaments over the th’ ck filaments. As a nerve impulse reaches the terminal end of the axon, synaptic vesicles fuse with the axon membrane and release a chemical transmitter, acetylcholine and binds to receptor sites of the motor end plate. When depolarization of the motor end plate reaches a certain level, it creates an action potential. An action potential (impulse) passes from the motor end plate over the sarcolemma and then into the T-tubules and sarcoplasmic reticulum and stimulates the sarcoplasmic reticulum to release calcium ions into the sarcoplasm. The calcium ions bind to troponin causing a change in its shape and position. This in turn alters shape and the position of tropomyosin, to which troponin binds. This shift exposes the active sites on the F-actin molecules. Myosin cross-bridges are then able to bind to these active sites. The heads of myosin molecules project laterally from thick myofilaments towards the surrounding thin myofilaments. These heads are called cross bridges. The head of each myosin molecule contains an enzyme mysoin ATPase. In the presence of myosin ATPase,Ca++ and Mg++ ions, ATP breaks down into ADP and inorganic phosphate, releasing energy in the head.
Energy from ATP causes energized myosin cross bridges to bind to actin.
NCERT Solutions For Class 11 Biology Locomotion and Movement Q3
The energized cross-bridges move, causing thin myofilaments to slide along the thick myofilaments.

4. Write true or false. If false change the statement so that it is true.
(a) Actin is present in thin filament.
(b) H-zone of striated muscle fibre represents both thick and thin filaments.
(c) Human skeleton has 206 bones.
(d) There are 11 pairs of ribs in man.
(e) Sternum is present on the ventral side of the body.
Solution: (a) True
(b) False – H-Zone of striated muscle fibres represents only thick filaments.
(c) True
(d) False – There are 12 pairs of ribs in man.
(e) True

5. Write the differences between:
(a) Actin and Myosin
(b) Red and White muscles
(c) Pectoral and Pelvic girdle
Solution: (a) Actin filaments and myosin filaments can be differentiated as follows:
NCERT Solutions For Class 11 Biology Locomotion and Movement Q5
(b) Differences between red muscle fibres and white muscle fibres are given in the following table:
NCERT Solutions For Class 11 Biology Locomotion and Movement Q5.1

NCERT Solutions For Class 11 Biology Locomotion and Movement Q5.2

NCERT Solutions For Class 11 Biology Locomotion and Movement Q5.3
(c) Differences between pectoral and pelvic girdles are given in the following table:
NCERT Solutions For Class 11 Biology Locomotion and Movement Q5.4

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

7. What are the different types of movements exhibited by the cells of human body?
Solution: The cells of human body show three types of movements: amoeboid, ciliary and muscular.
Amoeboid movements: These are found in leucocytes of blood and phagocytes of certain body organs. In such cells, movements are brought with the help of temporary finger-like cytoplasmic projections, called pseudopodia or false feet. So it is also called pseudopodial movement. These pseudopodia are formed by flow of cytoplasm, called cyclosis (simplest form of movement), and cytoskeletal structures like microfilaments.
Ciliary movements: Large number of our internal tubular organs are lined by ciliated epithelium. For instance, the cilia of the cells lining the trachea, oviducts and vasa efferentia propel dust particles, eggs and sperms respectively by their coordinated movements in specific directions in these organs. Muscular movements: These are brought about by the action of skeleton, joints and muscles. These are of two types: movements of body parts and locomotion.

8. How do you distinguish between a skeletal muscle and a cardiac muscle?
Solution: We can distinguish between a skeletal muscle and a cardiac muscle on the basis of the features discussed in the following table:
NCERT Solutions For Class 11 Biology Locomotion and Movement Q8

9. Name the type of joint between the following:
(a) atlas/axis
(b) carpal/metacarpal of thumb
(c) between phalanges
(d) femur/acetabulum
(e) between cranial bones
(f) between pubic bones in the pelvic girdle
Solution: (a) Pivot joint
(b) Saddle joint
(c) Hinge joint
(d) Ball and socket joint
(e) Fibrous joint
(f) Cartilaginous joint

10. Fill in the blank spaces:
(a) All mammals (except a few) have……. cervical vertebra.
(b) The number of phalanges in each limb of human is…….
(c) Thin filament of myofibril contains two ‘F’ actins and two other proteins namely…….and…….
(d) In a muscle fibre Ca++ is stored in …….
(e)…….and…….pairs of ribs are called floating ribs.
(f) The human cranium is made of……. bones.
Solution: (a) 7
(b) 14
(c) tropomyosin, troponin
(d) sarcoplasmic reticulum
(e) 11th and 12th
(f) 8

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

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 19 Excretory Products and their Elimination:

Section NameTopic Name
19Excretory Products and their Elimination
19.1Human Excretory System
19.2Urine Formation
19.3Function of the Tubules
19.4Mechanism of Concentration of the Filtrate
19.5Regulation of Kidney Function
19.6Micturition
19.7Role of other Organs in Excretion
19.8Disorders of the Excretory System
19.9Summary

NCERT Solutions Class 11 BiologyBiology Sample Papers

NCERT TEXTBOOK QUESTIONS FROM SOLVED

1.Define Glomerular Filtration Rate (GFR).
Solution. The amount of filtrate formed by the kidneys per minute is called glomerular filtration rate (GFR). It is approximately 125 mL/min. in a healthy person.

2.Explain the autoregulatory mechanism of GFR.
Solution. The kidneys have built-in mechanisms for the regulation of glomerular filtration rate. One such efficient mechanism is carried out by juxta glomerular apparatus (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.

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

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

5.Describe the role of liver, lungs and skin in excretion.
Solution. Other than the kidneys, lungs, liver and skin also help in the elimination of excretory wastes. Lungs remove large amounts of C02 (18 litres/day) and also significant quantities of water every day. Liver secretes bile which contains substances like bilirubin, biliverdin, cholesterol, degraded steroid hormones, vitamins and drugs. Most of these substances ultimately pass out along with digestive wastes. The sweat and sebaceous glands in the skin can eliminate certain substances through their secretions. Sweat produced by the sweat glands is a watery fluid containing NaCl, small amounts of urea, lactic acid etc. Sebaceous glands eliminate certain substances like sterols, hydrocarbons and waxes through sebum.

6.Explain micturition.
Solution. The process of passing out urine from the urinary bladder is called micturition. Urine formed by the nephrons is ultimately carried to the urinary bladder where it is stored. This causes stretching of the wall of bladder that leads to the stimulation of stretch receptors on the walls of the bladder. This sends signal to the CNS. The CNS passes on motor messages to initiate the contraction of smooth muscles of the bladder and simultaneous relaxation of the urethral sphincter causing the release of urine.

7.Match the items of column I with 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
(e) ADH                                       (v)Renal tubule
Solution. (a) – (iii), (b) – (v), (c) – (iv), (d) – (i), (e) – (ii)

8.What is meant by the term osmoregulation?
Solution. The regulation of water and solute contents of the body fluids by the kidney is called osmoregualtion.

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

10. What is the significance of juxta glomerular apparatus (JGA) in kidney function?
Solution. Juxta glomerular apparatus (JGA) is a special sensitive region formed by cellular modifications in the distal convoluted tubule and the afferent arteriole at the location of their contact. The 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.

11 .Name the following.
(a) A chordate animal having flame cells as excretory structures.
(b) Cortical portions projecting between the medullary pyramids in the human kidney.
(c) A loop of capillary running parallel to the Henle’s loop.
Solution. (a) Cephalochordate – Amphioxus
(b) Columns of Bertini
(c) Vasa recta

12.Fill in the gaps.
(a) Ascending limb of Henle’s loop is________to water whereas the descending limb is________to it.
(b) Reabsorption of water from distal parts of the tubules is facilitated by hormone________
(c) Dialysis fluid contains all the constituents as in plasma except________
(d) A healthy adult human excretes (on an average)________gm of urea/day.
Solution.
(a) Ascending limb of Henle’s loop is impermeable to water whereas the descending limb is permeable to it.
(b) Reabsorption of water from distal parts of the tubules is facilitated by hormone ADH.
(c) Dialysis fluid contains all the constituents as in plasma except nitrogenous wastes.
(d) A healthy adult human excretes (on an average) 25 – 30 gm of urea/day.

NCERT Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation:

Section NameTopic Name
18Body Fluids and Circulation
18.1Blood
18.2Lymph (Tissue Fluid)
18.3Circulatory Pathways
18.4Double Circulation
18.5Regulation of Cardiac Activity
18.6Disorders of Circulatory System
18.7Summary

NCERT TEXTBOOK QUESTIONS FROM SOLVED

1. Name the components of the formed elements in the blood and mention one major function of each of them.
Solution: Blood corpuscles are the formed ele-ments in the blood, they constitute 45% of the blood. Formed elements are – (erythrocytes, RBCs or red blood corpuscles), (leucocytes, WBCs or white blood corpuscles) and throm¬bocytes or blood platelets. The major function of RBCs is to transport oxygen from lungs to body tissues and COz from body tissues to the lungs. White blood cells provide immunity to the body. Blood platelets play important role in blood clotting.

2. What is the importance of plasma proteins?
Solution: Plasma proteins constitute about 7 to 8% of plasma. These mainly include albumin, globulin, prothrombin and fibrinogen. Prothrombin and fibrinogen are needed for blood clotting. Albumins and globulins retain water in blood plasma and helps in maintaining osmotic balance. Certain globulins

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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) Systol                         (v) Gas transport
Solutlion.(a) – (iii); (b) – (v); (c) – (ii); (d) – (i); (e) – (iv).

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

5. What is the difference between lymph and blood?
Solution: The differences between blood and lymph are given below:
NCERT Solutions For Class 11 Biology Body Fluids and Circulation Q5

6. What is meant by double circulation? What is its significance?
Solution: The type of blood circulation in which oxygenated blood and deoxygenated blood do not get mixed is termed double circulation. It includes systemic circulation and pulmonary circulation. The circulatory pathway of double circulation is given in the following flow chart.
NCERT Solutions For Class 11 Biology Body Fluids and Circulation Q6
Flow chart: Double blood circulation Double circulation or separation of systemic and pulmonary circulations provides a higher metabolic rate to the body and also allows the two circulations to have different blood pressures according to the need of the organs they supply.

7. Write the differences between:
(a) Blood and lymph
(b) Open and closed system of circulation
(c) Systole and diastole
(d) P-wave and T-wave
Solution: (a) Refer answer 5.
(b) The differences between open and closed circulatory system are given below:
NCERT Solutions For Class 11 Biology Body Fluids and Circulation Q7

NCERT Solutions For Class 11 Biology Body Fluids and Circulation Q7.1
(c) Systole is contraction of heart chambers in order to pump out blood while diastole is relaxation of heart chambers to receive blood. The contraction of a chamber or systole decreases its volume and forces the blood out of it, whereas its relaxation or diastole brings it back to its original size to receive more blood.
(d) P wave is a small upward wave of elec-trocardiograph that indicates the atrial depolarisation (contraction of atria). It is caused by the activation of SA node. T-wave is a dome shaped wave of electro-cardiograph which represents ventricular repolarisation (ventricular relaxation).

8. Describe the evolutionary change in the pattern of heart among the vertebrates.
Solution: Vertebrates have a single heart. It is a hollow, muscular organ composed of cardiac muscle fibres. Two types of chambers in heart are atria and ventricles. The heart of lower vertebrates have additional chambers, namely sinus venosus and conus arteriosus or bulbus arteriosus or truncus arteriosus. During the course of development, in higher vertebrates, the persistent portions viz, auricles and ventricles are retained. However, these get complicated by incorporating several valves inside them and becoming compartmentali sed.
In fishes, heart is two chambered (1 auricle and 1 ventricle). Both the accessory chambers, sinus venosus and conus arteriosus are present. The heart pumps out deoxygenated blood which is oxygenated by the gills and sent to the body parts from where deoxygenated blood is carried to the heart. It is called single circulation and heart is called venous heart. Lung fish, amphibians and reptiles have three chambered heart, (2 auricles and 1 ventricle). The left atrium gets oxygenated blood from the gills/lungs/skin/buccopharyngeal cavity and the right atrium receives the deoxygenated blood from other body parts. But both oxygenated and deoxygenated blood get mixed up in single ventricle which pumps out mixed blood. This is called incomplete double circulation.
Crocodiles, birds and mammals have a complete four chambered heart (right and left auricles; right and left ventricles). Oxygenated and deoxygenated blood never get mixed. Right parts of the heart receive deoxygenated blood from all other body parts and send it to lungs for oxygenation whereas left parts of heart receive oxygenated blood from lungs and send it to other body parts. This mode of circulation is termed as complete double circulation which includes systemic and pulmonary circulation. There are no accessory chambers in heart of birds and mammals.

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

10. Sino-atrial node is called the pacemaker of our heart. Why?
Solution: Sino-atrial node (SAN) is a mass of neuromuscular tissue which lies in the wall of right atrium. It is responsible for initiating and maintaining the rhythmic contractile activity of the heart. Therefore, it is called the pacemaker.

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

12. Define a cardiac cycle and the cardiac output.
Solution: The sequential events in the heart which are repeated cyclically is called cardiac cycle and it consists of systole (contraction) and diastole (relaxation) of both the atria and ventricles. The duration of a cardiac cycle is 0.8 seconds. Periods of cardiac cycle are atrial systole (0.1 second), ventricular systole (0.3 second) and complete cardiac diastole (0.4 second).
The amount of blood pumped by heart per minute is called cardiac output. It is calculated by multiplying stroke volume (volume of blood pumped by each ventricle per minute) with heart rate (number of beats per minute). The heart of normal person beats 72 times per minute and pumps out about 70 mL of blood per beat. Therefore, cardiac output averages 5000 mL or 5 litres.

13. Explain heart sounds.
Solution: The beating of heart produces characteristic sounds which can be heard by using stethoscope. In a normal person, two sounds are produced per heart beat. The first heart sound Tubb’ is low pitched, not very loud and of long duration. It is caused partly by the closure of the bicuspid and tricuspid valves and partly by the contraction of muscles in the ventricles.
The second heart sound ‘dubb’ is high pitched, louder, sharper and shorter in duration. It is caused by the closure of the semilunar valves and marks the end of ventricular systole.

14. Draw a standard ECG and explain the different segments in it.
Solution: ECG is graphic record of the electric current produced by the excitation of the cardiac muscles. The instrument used to record the changes is an electrocardiograph. A normal electrogram (ECG) is composed of a P wave, a QRS wave (complex) and a T wave. The P Wave is a small upward wave that represents electrical excitation or the atrial depolarisation which leads to contraction of both the atria (atrial contraction). It is caused by the activation of SA node. The impulses of contraction start from the SAnode and spread throughout the artia.
The QRS Wave (complex) represents ventricular depolarisation (ventricular contraction). It is caused by the impulses of the contraction from AV node through the bundle of His and Purkinje fibres and the contraction of the ventricular muscles. Thus this wave is due to the spread of electrical impulse through the ventricles.
The T Wave represents ventricular repolarisation (ventricular relaxation). The potential generated by the recovery of the ventricle from the depolarisation state is called the repolarisation wave. The end of the T-wave marks the end of systole.
ECG gives accurate information about the heart. Therefore, ECG is of great diagnostic value in cardiac diseases.

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

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 17 Breathing and Exchange of Gases:

Section NameTopic Name
17Breathing and Exchange of Gases
17.1Respiratory Organs
17.2Mechanism of Breathing
17.3Exchange of Gases
17.4Transport of Gases
17.5Regulation of Respiration
17.6Disorders of Respiratory System
17.7Summary

NCERT TEXTBOOK QUESTIONS FROM SOLVED

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

2. State the volume of air remaining in the lungs after a normal breathing.
Solution: When a person breathes normally, the amount which remains in the lung after normal expiration, is called functional residual capacity. It is the sum of residual volume and the expiratory reserve volume (FRC = RV + ERV). It is about 2100 – 2300 mL of air.

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3. Diffusion of gases occurs in the alveolar region only and not in the other parts of respiratory system. Why?
Solution: For efficient exchange of gases, respi: atory surface must have certain characteristics such as (i) it must be thin, me ist and permeable to respiratory gases (ii) it must have large surface area, (iii) it must be highly vascular. Only alveolar region has these characteristics. Thus, diffusion of gases occurs in this region only.

4. What are the major transport mechanisms for CO2? Explain.
Solution: Nearly 20-25 percent of CO2 is transported by haemoglobin of RBCs, 70 percent of it is carried as bicarbonate ion in
plasma and about 7 percent of CO2 is carried in a dissolved state through plasma. CO2 is carried by haemoglobin as carbamino- haemoglobin. This binding is related to the partial pressure of CO2.

5. What will be the p02 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
Solution: (ii) Air that has entered the alveoli through the bronchioles is called alveolar air. It has the same partial pressure of CO2 and 02 as is in the atmospheric air. Then, there occurs gaseous exchange between the adjacent blood capillaries and the alveoli. CO2 diffuses from blood into the alveolar air and O2 diffuses from alveolar air to the blood. As a result, new alveolar air has higher pCO2and lesser pO2, than the atmospheric air.

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

7. How is respiration regulated?
Solution: Respiration is under both nervous and chemical regulation.
The respiratory centre in brain is composed of groups of neurons located in the medulla oblongata and pons varolii. The respiratory centre regulates the rate and depth of the breathing.
Dorsal respiratory group of neurons are located in the dorsal portion of the medulla oblongata. This group of neurons mainly causes inspiration.
Ventral group of neurons are located in the ventrolateral part of the medulla oblongata. These can cause either inspiration or expiration.
Pneumotaxic centre is located in the dorsal part of pons varolii. It sends signals to all the neurons of dorsal respiratory group and only to inspiratory neurons of ventral respiratory group. Its job is primarily to limit inspiration. Chemically, respiration is regulated by the large numbers of chemoreceptors located in the carotid bodies and in the aortic bodies. Excess carbon dioxide or hydrogen ions mainly stimulate the respiratory centre of the brain and increases the inspiratory and expiratory-signals to the respiratory muscles. Increased C02 lowers the pH resulting in acidosis. The role of oxygen in the regulation of respiratory rhythm is quite insignificant.

8. What is the effect of pCO2on oxygen transport?
Solution: Increase in pCO2 tension in blood brings rightward shift of the oxygen dissociation curve of haemoglobin thereby decreasing the affinity of haemoglobin for oxygen. This effect is called Bohr’s effect. It plays an important role in the release of oxygen in the tissues.

9. What happens to the respiratory process in a man going up a hill?
Solution: Rate of breathing will increase in order to supply sufficient oxygen to blood because air in mountainous region is deficient in oxygen.

10. What is the site of gaseous exchange in an insect?
Solution: Tracheae (Tracheal respiration) is the site of gaseous exchange in an insect. .

11. Define oxygen dissociation curve. Can you suggest any reason for its sigmoidal pattern?
Solution: The relationship between the partial pressure of oxygen (pO2) and percentage saturation of the haemoglobin with oxygen (O2)is graphically illustrated by a curve called oxygen haemoglobin dissociation curve (also called oxygen dissociation curve).
The sigmoidal pattern of oxygen haemoglobin dissociation curve is the result of two properties which play significant role in the transport of oxygen. These two properties are:
(i) Minimal loss of oxygen from haemoglobin occurs above p02 of 70-80 mm Hg despite significant changes in tension of oxygen beyond this. This is depicted by relatively flat portion of the curve.
(ii)Any further decline in p02 from 40 mm Hg causes a disproportionately greater release of oxygen from the haemoglobin. It results in the steeper portion of the curve and causes the curve to be sigmoid.

12. Have you heard about hypoxia? Try to gather information about it, and discuss with your friends.
Solution: Hypoxia is a condition of oxygen shortage in the tissues. It is of two types:
(i) Artificial hypoxia: It results from shortage of oxygen in the air as at high altitude. It causes mountain sickness characterised by breathelessnes, headache, dizziness and bluish tinge on skin.
(ii) Anaemic hypoxia: It results from the reduced oxygen carrying capacity of the blood due to anaemia or carbon monoxide poisoning. In both cases, less haemoglobin is available for carrying 02.

13. Distinguish between
(a) IRV and ERV
(b) Inspiratory capacity and expiratory capacity.
(c) Vital capacity and total lung capacity.
Solution:
(a) Differences between IRV and ERV are as follows:
NCERT Solutions For Class 11 Biology Breathing and Exchange of Gases Q13
(b)Differences between inspiratory capacity and expiratory capacity are as follows:
NCERT Solutions For Class 11 Biology Breathing and Exchange of Gases Q13.1
(c) Differences between vital capacity and total lung capacity are as follows:
NCERT Solutions For Class 11 Biology Breathing and Exchange of Gases Q13.2

14. What is tidal volume? Find out the tidal volume (approximate value) for a healthy human in an hour.
Solution: Tidal volume is the volume of air inspired or expired with each normal breath. This is about 500 mL in an adult person. It is composed of about 350 mL of alveolar volume and about 150 mL of dead space volume. The alveolar volume consists of air that reaches the respiratory surfaces of the alveoli and engages in gas exchange. The dead space volume consists of air that does not reach the respiratory surfaces.
A healthy man can inspire or expire approximately 6000 to 8000 mL of air per minute. Therefore, tidal volume for a healthy human in an hour is 360 – 480 mL of air.

NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption:

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption:

Section NameTopic Name
16Digestion and Absorption
16.1Digestive System
16.2Digestion of Food
16.3Absorption of Digested Products
16.4Disorders of Digestive System
16.5Summary

NCERT TEXTBOOK QUESTIONS FROM SOLVED

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 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.
Solution: (a) (i) Pepsin, lipase and rennin
(b) (ii) Intestinal juice

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

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3. Answer briefly:
(a) Why are villi present in the intestine and not in the stomach?
(b) How does pepsinogen change into its active form ?
(c) What are the basic layers of the wall of alimentary canal?
(d) How does bile help in the digestion of fats ?
Solution: (a) The absorptive surface area of small intestine is enormously increased by microvilli and as maximum absorption
of digested food takes place in small intestine as compared to other organs, therefore, villi are present in small intestine and not in stomach. Moreover, stomach is primarily associated with temporary storage of food.
(b) The proenzyme pepsinogen, on exposure to hydrochloric acid, secreted by oxyntic cells of gastric glands gets converted into the active enzyme pepsin, the proteolytic enzyme of the stomach.
(c) The wall of alimentary canal from oesophagus to rectum possesses four layers, namely serosa, muscularis, sub-mucosa and mucosa. Serosa is the outermost layer and is made up of a thin mesothelium with some connective tissues. Muscularis is formed by smooth muscles. The sub-mucosal layer is formed of loose connective tissues containing nerves, blood and lymph vessels. In duodenum, glands are also present in sub-mucosa. The innermost layer lining the lumen of the alimentary canal is the mucosa. This layer forms irregular folds (rugae) in the stomach and small finger¬like foldings called villi in the small intestine.
(d) Bile has no enzymes but contains bile salts, namely, sodium bicarbonate, sodium glycocholate and sodium taurocholate that reduce the surface tension of large fat droplets and break them into many small droplets by a process known as emulsification. These small fat droplets present large surface area for lipase (fat digesting enzyme) to act upon them. Moreover, bile also activates lipases.

4. State the role of pancreatic juice in digestion of proteins.
Solution: The pancreatic juice contains inactive enzymes – trypsinogen, chymotrypsinogen, procarboxypeptidases. Trypsinogen is acti¬vated by an enzyme enterokinase, (secreted by the intestinal mucosa) into active trypsin, which in turn activates the other enzymes of the pancreatic juice. Proteins, proteoses and peptones (partially hydrolysed proteins) in the chyme reaching the intestine are acted upon by these proteolytic enzymes of pancre¬atic juice.
NCERT Solutions For Class 11 Biology Digestion and Absorption Q4

5. Describe the process of digestion of protein in stomach.
Solution: The gastric glands of the stomach secrete gastric juice that contains HCl and proenzymes – pepsinogen and prorennin. The proenzyme pepsinogen, on exposure to HCl gets converted into the active enzyme pepsin, the proteolytic enzyme of stomach. The pepsin converts proteins into proteoses and peptones (peptides). Prorennin is found in gastric juice of infants and is activated by pepsin into active rennin. It helps in digestion of milk protein casein.
NCERT Solutions For Class 11 Biology Digestion and Absorption Q5

6. Give the dental formula of human beings.
Solution: The dental formula of human beings is
NCERT Solutions For Class 11 Biology Digestion and Absorption Q6
It shows arrangement of teeth in each half of
the upper and lower jaw.
NCERT Solutions For Class 11 Biology Digestion and Absorption Q6.1

7. Bile juice contains no digestive enzymes, yet it is important for digestion. Why ?
Solution: Bile has no enzymes but contains bile salts, namely, sodium bicarbonate, sodium glycocholate and sodium taurocholate that reduce the surface tension of large fat drop¬lets and break them into many small droplets by a process known as emulsification. These small fat droplets present large surface area for lipase (fat digesting enzyme) to act upon them. Moreover, bile also activates lipases.
NCERT Solutions For Class 11 Biology Digestion and Absorption Q7

8. Describe the digestive role of chymotrypsin. Which two other digestive enzymes of the same category are secreted by its source gland ?
Solution: Chymotrypsin is a proteolytic enzyme of pancreatic juice secreted by exocrine part of pancreas. It helps in digestion of proteins. It converts proteins, peptones and proteoses into oligopeptides and dipeptides. Two other proteolytic enzymes present in pancreatic juice are trypsinogen and procarboxypeptidase.

9. How are polysaccharides and disaccharides digested ?
Solution: Digestion of polysaccharides (starch and glycogen) starts from buccal cavity. In buccal cavity, polysaccharides are acted upon by salivary amylase or ptyalin which splits starch and glycogen into disaccharides and small dextrins called ‘a’ dextrin.
NCERT Solutions For Class 11 Biology Digestion and Absorption Q9
The digestion of carbohydrates does not occur in stomach because gastric juice itself has no carbohydrase.
In small intestine, the food mixes with two juices, pancreatic juice and intestinal juice. Pancreatic juice contains a carbohydrase named pancreatic amylase. This enzyme hydrolyses more starch and glycogen.
NCERT Solutions For Class 11 Biology Digestion and Absorption Q9.1
Intestinal juice contains carbohydrases; maltase, isomaltase, a-dextrinase, sucrase and lactase which act on disaccharides as follows:
NCERT Solutions For Class 11 Biology Digestion and Absorption Q9.2
fructose and galactose are monomers of carbohydrates. These are absorbed by intestinal mucosa.

10. What would happen if HCl were not secreted in the stomach?
Solution: HCl is secreted by parietal or oxyntic cells of gastric glands. It serves the following functions:

  1. It activates the pepsinogen and prorennin into their active form pepsin and rennin.
  2. It provides the acidic pH (pH 1.8) optimal for pepsin.
  3. It kills the harmful bacteria present in the food.
  4. It stops the action of saliva on food. Pepsin and rennin are the principle proteolytic enzymes of stomach. If these enzymes are not activated by HCl then digestion of protein will not take place in stomach, and also the harmful bacteria can cause various diseases.

11. How does butter in your food get digested and absorbed in the body ?
Solution: Butter is a saturated fat. Fats and oils of the ingested food are triglycerides.
They are digested by lipases. Small intestine is the principal organ for fat digestion.
In the small intestine food meets three secretions, bile, pancreatic juice and intestinal juice, all alkaline in nature.
Bile contains no enzyme but it contains bile salts which reduces the surface tension of large fat droplets and breaks them into smaller ones (emulsification).
NCERT Solutions For Class 11 Biology Digestion and Absorption Q11
Emulsified triglycerides Pancreatic juice contains pancreatic lipase, which is the principal fat digesting enzyme. It is activated by bile.
NCERT Solutions For Class 11 Biology Digestion and Absorption Q11.1
Fatty acid + Glycerol Intestinal lipase found in intestinal juice hydrolyses some triglycerides, diglycerides and monoglycerides to fatty acids and glycerol like pancreatic lipase.
Fatty acids, glycerol and monoglycerides are the end products of fat digestion and being insoluble in water cannot be directly absorbed from the intestinal contents. So they combine with the bile salts and phospholipids to form micelles (water soluble). From the micelles fatty acids, glycerides, sterols and fat soluble vitamins are absorbed into the intestinal cells by diffusion where they are resynthesised in the ER and are converted into very small protein coated fat molecules (droplets) called chylomicrons. The latter are released from the intestinal cells into the lymph present in the lymphatic capillaries, the lacteals. These lacteals 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.
Solution: Proteins of ingested food are broken down into amino acids by proteases (peptidases). Proteases are secreted in inactive forms called proenzymes which are converted into active forms at site of their action. Protein digestion starts in the stomach and is completed in the small intestine. Saliva contains no protease.
Digestion of proteins in stomach : Chief cells of gastric gland secrete pepsinogen and prorennin, which act as follows:
NCERT Solutions For Class 11 Biology Digestion and Absorption Q12

NCERT Solutions For Class 11 Biology Digestion and Absorption Q12.1
Digestion of proteins in small intestine: In small intestine, peptones and proteoses are acted upon by enzymes of pancreatic juice and intestinal juice.
Pancreatic juice contains 3 inactive proteases; trypsinogen, chymotrypsinogen and pro-carboxypeptidase. Their action is as follows:
NCERT Solutions For Class 11 Biology Digestion and Absorption Q12.2
Dipeptides + Amino acids Intestinal juice contains two digestive pro-teases; aminopeptidases and dipeptidases and a nondigestive enterokinase (enteropep- tidase).
NCERT Solutions For Class 11 Biology Digestion and Absorption Q12.3
Amino acids are the end products of protein digestion which are absorbed by intestinal cells.

13. Explain the term thecodont and diphyodont.
Solution: Thecodont: In human, each tooth is embedded in a socket of jaw bone. Such teeth are described as thecodont.
Diphyodont: 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.
Solution: Adult human has 32 teeth with the
NCERT Solutions For Class 11 Biology Digestion and Absorption Q14
Human has heterodont dentition i.e., having four different types of teeth. The number of different types of teeth in human are as follows:incisors = 8, canines = 4, premolars = 8, molars = 12

15. What are the functions of liver?
Solution: Liver is the largest gland of the body and consists of hepatic cells. Besides being a digestive gland, the liver performs a number of functions for the welfare of body. Its varied functions are as follows

  1. Secretion of bile.
  2. Glycogenesis, gluconeogenesis and glycogenolysis.
  3. Storage of fat, glycogen, vitamins like A, D, E, K and B12, blood, water, etc.
  4. Deamination of amino acids.
  5. Synthesis of urea.
  6. Elimination of excretory substances.
  7. Detoxification of harmful substances.
  8. Formation and breakdown of blood
    corpuscles, i.e., in embryos, liver is haemopoietic (produces red blood corpuscles) and in adults its Kupffer cells phagocytise and destroy worn out and dead RBCs.
  9. Secretion of blood proteins, i.e., prothrombin and fibrinogen.
  10. Secretion of anticoagulant heparin.
  11. Production of heat.
  12. Secretion of enzymes.

NCERT Solutions for Class 11 Biology Chapter 15 Plant Growth and Development:

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 15 Plant Growth and Development:

Section NameTopic Name
15Plant Growth and Development
15.1Growth
15.2Differentiation, Dedifferentiation and Redifferentiation
15.3Development
15.4Plant Growth Regulators
15.5Photoperiodism
15.6Vernalisation
15.7Summary

NCERT Solutions Class 11 BiologyBiology Sample Papers

NCERT TEXTBOOK QUESTIONS FROM SOLVED

1. Define growth, differentiation, development, dedifferentiation, redifferentiation, determinate growth, meristem and growth rate.
Solution: Growth is defined as a vital process which brings about an irreversible and permanent change in the shape, size, form, weight and volume of a cell, organ or whole organism, accompanied with increase in dry matter.
Differentiation is a localised qualitative change in size, biochemistry, structure and function of cells, tissues or organs, e.g., fibre, vessel, tracheid, sieve tube, mesophyll, leaf etc. Thus it is a change in form and physiological activity. It results in specialisation for particular functions.
Development may be defined as a process which includes growth, differentiation and maturation in a regular sequence in the life history of a cell, organ or organism viz., seed germination, growth, differentiation, flowering, seed formation and senescence. Dedifferentiation is the process by which the differentiated cells which have lost the ability to divide under certain circumstances, become meristematic and regain the divisibility. Redifferentiation is defined as maturation or differentiation of dedifferentiated cells to form cells which are unable to divide e.g., secondary xylem elements, cork cells etc., are formed by redifferentiation of secondary cambial cells.
Determinate growth is the ability of a cell, tissue or the organism to grow for a limited period of time. Meristem is a tissue consisting of unspecialised immature cells, possessing the power of continuous cell division and adding new cells to the body. Growth rate is defined as the increased growth per unit time.

2. Why is not any one parameter good enough to demonstrate growth throughout the life of a flowering plant?
Solution: A flowering plant consists of a number of organs viz., roots, stem, leaves, flowers, fruits etc. growing differently under different stages of life cycle. These plant organs require different parameters to demonstrate their growth. In plant organs like fruits, bulbs, corms etc. fresh weight is used for measuring their growth. In case of fruits, increase in volume, diameter etc., are also used as other parameters for the measurement of their growth. For flat organs like leaves, increase in surface area is used as the parameter. Stem and roots primarily grow in length and then in girth, thus increase in length and diameter are used for measuring their growth. Consequently, the flowering plants exhibit several parameters to demonstrate growth.

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3. Describe briefly
(a) Arithmetic growth
(b) Geometric growth
(c) Sigmoid growth curve
(d) Absolute and relative growth rates
Solution: (a) Arithmetic growth: If the length of a plant organ is plotted against time it shows a linear curve, the growth is called arithmetic growth. In this growth, the rate of growth is constant and increase in growth occurs in arithmetic progression e.g., length of a plant is measured as 2,4, 6, 8,10,12 cms at a definite interval of 24 hrs. It is found in root or shoot elongating at constant rate. Arithmetic growth is expressed as Lt = L0 + rHere, L= length after time t. L0 = length at the beginning, r = growth rate.
NCERT Solutions For Class 11 Biology Plant Growth and Development Q3
(b) Geometric growth: Geometric growth is the growth where both the progeny cells following mitosis retain the ability to divide and continue to do so. It occurs in many higher plants and in unicellular organisms when grown in nutrient rich medium. Number of cells is initially small so that initial growth is slow which is called lag phase. Later on, there is rapid growth at exponential rate. It is called log or exponential phase.
NCERT Solutions For Class 11 Biology Plant Growth and Development Q3.1
(c) Sigmoid growth curve: Geometric growth cannot be sustained for long. Some cells die. Limited nutrient availability causes slowing down of growth. It leads to stationary phase. There may be actually a decline. Plotting the growth against time will give a typical sigmoid or S-curve.
NCERT Solutions For Class 11 Biology Plant Growth and Development Q3.2
S-curve of growth is typical of most living organisms in their natural environment. It also occurs in cells, tissues and organs of plants.
(d) Absolute growth rate is the measurement of total growth per unit time. Relative growth rate is growth per unit time per unit initial growth.
Growth in given time period/ Measurement at start of time period
Suppose two leaves have grown by 5 cm2 in one day. Initial size of leaf A was 5 cm2 while that of leaf B was 50 cm2. Though their absolute growth is the same (5 cm2/day), relative rate of growth is faster in leaf A(5/5) because of initial small size than in leaf B(5/50).

4. List five main groups of natural plant growth regulators. Write a note on discovery, physiological functions and agricultural/ horticultural applications of any one of them.
Solution: There are five main groups of natural plant growth regulators which are very much recognised as natural hormones in plants. These are:

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

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

Physiological functions of auxins:

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

Agricultural/horticultural application of auxins:

  1. Application of auxins like IAA, IBA, NAA induce rooting in stem cuttings of many plants. This method is widely used to multiply several economically useful plants.
  2. Normally, auxins inhibit flowering however in litchi and pineapple, application of auxin promotes flowering thus used in orchards.
  3. Auxin induces parthenocarpy in some plants including tomato, pepper, cucumber and Citrus, thus, produces seedless fruits of more economic value.
  4. Auxins like 2, 4-D and 2, 4, 5-T are commercially used as weedicides, due to their low cost and greater chemical stability. They are selective herbicides (killing broad-leaved plants, but not grasses).
  5. For checking premature fruit drop, auxins are applied which prevent the formation of abscission zone in the petiole or just below the fruit. Auxin regulates maturing fruit on the trees of apples, oranges and grape fruit. High doses of auxins can
    cause fruit drop. Thus, heavy applications of synthetic auxins are used commercially to promote a coordinated abscission of various fruits to facilitate harvesting.
  6. Auxin, produced in the apical bud, suppresses the development of lateral buds, i.e., apical dominance. Thus practically used in prolonging the dormancy period of potato tubers.
  7. Naphthalene acetamide is used to prevent the lodging (excessive elongation and development of weak plants, specially in gramineae) or falling of crops.
  8. Auxin (2,4-D) promotes callus formation in tissue culture. Complete plantlets are regenerated from callus tissue, using auxins and cytokinin which are then transplanted into the soil. Now-a-days, this is a widely practised method of propagation in the field of agriculture and horticulture.

5. What do you understand by photoperiodism and vernalisation? Describe their significance.
Solution: The physiological mechanism for flower-ing is controlled by two factors: photoperiod or light period, i.e., photoperiodism and low temperature, i.e., vernalisation. Photoperiodism is defined as the flowering response of a plant to relative lengths of light/ dark period. Significance of photoperiodism is as follows:

  1. Photoperiodism determines the season in which a particular plant shall flower. For example, short day plants develop flowers in autumn-spring period (e.g., Dahlia, Xanthium) while long day plants produce flowers in summer (e.g., Amaranthus).
  2. Knowledge of photoperiodic effect is useful in keeping some plants in vegetative growth (many vegetables) to obtain higher yield of tubers, rhizomes etc. or keep the plant in reproductive stage to yield more flowers and fruits.
  3. A plant can be made to flower throughout the year by providing favourable photoperiod.
  4. Helps the plant breeders in effective cross-breeding in plants.
  5.  Enable a plant to flower in different seasons.
    Vernalisation is promotion or induction of flowering by exposing a plant to low temperature for some time. Significance of vernalisation is as follows :
    (i) Crops can be grown earlier.
    (ii)Plants can be grown in such regions where normally they do not grow.
    (iii)Yield of the plant is increased.
    (iv)Resistance to cold and frost is increased.
    (v) Resistance to fungal diseases is increased.

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

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

8. ‘Both a short day plant and a long day plant can produce flower simultaneously in a given place’. Explain.
Solution: A short day plant (SDP) flowers only when it receives a long dark period and short photoperiod, e.g., Xanthium, Dahlia etc. On the other hand, a long day plant (LDP) will flower only when it receives a long photoperiod and short dark period, e.g., wheat, oat etc. Thus critical photoperiod is that continuous duration of light which must not be exceeded in SDP and should always be exceeded in LDP in order to bring them to flower. Xanthium requires light for less than 15.6 hrs and Henbane requires light for more than 11 hrs. Xanthium (a SDP) and Henbane (DP) will flower simultaneously in light period between 11 to 15.6 hrs.

9. Which one of the plant growth regulators would you use if you are asked to
(a) induce rooting in a twig
(b) quickly ripen a fruit
(c) delay leaf senescence
(d) induce growth in axillary buds
(e) ‘bolt’ a rosette plant
(f) induce immediate stomatal closure in leaves.
Solution: (a) Auxins like IBA, NAA.
(b) Ethylene
(c) Cytokinins
(d) Cytokinins
(e) Gibberellins
(f) Abscisic acid (ABA)

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

11. What would be expected to happen if:
(a) GA3 is applied to rice seedlings
(b) dividing cells stop differentiating
(c) a rotten fruit gets mixed with unripe fruits
(d) you forget to add cytokinin to the culture medium.
Solution:
(a) The coleoptile will elongate rapidly, as GA3 helps in cell growth.
(b) The development of callus (mass of undifferentiated cells) will take place.
(c) The unripe fruits will ripe quickly because of the increased rate of respiration due to emission of ethylene from rotten fruit.
(d) Cell division will retard and shoot will not initiate from the callus.

NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants:

Section NameTopic Name
14Respiration in Plants
14.1Do Plants Breathe?
14.2Glycolysis
14.3Fermentation
14.4Aerobic Respiration
14.5The Respiratory Balance Sheet
14.6Amphibolic Pathway
14.7Respiratory Quotient
14.8Summary

NCERT Solutions Class 11 BiologyBiology Sample Papers

NCERT TEXTBOOK QUESTIONS FROM SOLVED

1. Give the schematic representation of an overall view of Krebs’ cycle.
Solution: 
NCERT Solutions For Class 11 Biology Respiration in Plants Q1

2. Differentiate between
(a) Respiration and Combustion
(b) Glycolysis and Krebs’cycle
(c) Aerobic respiration and Fermentation
Solution: (a) Differences between respiration and combustion are as follows :
NCERT Solutions For Class 11 Biology Respiration in Plants Q2
(b) Differences between glycolysis and Krebs’ cycle are as follows:
NCERT Solutions For Class 11 Biology Respiration in Plants Q2.1
NCERT Solutions For Class 11 Biology Respiration in Plants Q2.2
(C)Differences between aerobic respiration and fermentation are as follows:
NCERT Solutions For Class 11 Biology Respiration in Plants Q2.3

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

4. Give the schematic representation of glycolysis.
Solution: 
NCERT Solutions For Class 11 Biology Respiration in Plants Q4

NCERT Solutions For Class 11 Biology Respiration in Plants Q4.1
5. Explain ETS.
Solution: An electron transport chain or system (ETS) is a series of coenzymes and cytochromes that take part in the passage of electrons from
a chemical to its ultimate acceptor. Reduced coenzymes participate in electron transport chain. Electron transport takes place on cristae of mitochondria [oxysomes ( F0 -F1 , particles) found on the inner surface of the membrane of mitochondria]. NADH formed in glycolysis and citric acid cycle are oxidised by NADH dehydrogenase (complex I) and the electrons are transferred to ubiquinone. Ubiquinone also receives reducing equivalents via FADH2 through the activity of succinate dehydrogenase (complex II). The reduced ubiquinone is then oxidised by transfer of electrons of cytochrome c via cytochrome Fc, complex (complex III). Cytochrome c acts as a mobile carrier between complex III and complex IV. Complex IV refers to cytochrome c oxidase complex containing cytochromes a and a3and two copper centres. When the electrons are shunted over the carriers via complex I to IV in the electron transport chain, they are coupled to ATP synthetase (complex V) for the formation of ATP from ADP and Pi. Oxygen functions as the terminal acceptor of electrons and is reduced to water along with the hydrogen atoms. Reduced coenzymes (coenzyme I, II and FAD) do not combine directly with the molecular O2. Only their hydrogen or electrons are transferred through various substances and finally reach O2. The substances useful for the transfer of electron are called electron carriers. Only electrons are transferred through cytochromes (Cyt F1 Cyt c,,C2, a, a3) and finally reach molecular O2. Both cytochrome a and a3 form a system called cytochrome oxidase. Copper is also present in Cyt a3 in addition to iron. The molecular oxygen that has accepted electrons now receives the protons that were liberated into the surrounding medium to give rise to a molecule of water. The liberated energy is utilised for the synthesis of ATP from ADP and Pi.

6. What are the main steps in aerobic respiration? Where does it take place?
Solution: Aerobic respiration is an enzymatically controlled release of energy in a stepwise catabolic process of complete oxidation of organic food into carbon dioxide and water with oxygen acting as terminal oxidant. It
occurs by two methods, common pathway and pentose phosphate pathway. Common pathway is known so because its first step, called glycolysis, is common to both aerobic and anaerobic modes of respiration. The common pathway of aerobic respiration consists of three steps – glycolysis, Krebs’ cycle and terminal oxidation. Aerobic respiration takes place within mitochondria. The final product of glycolysis, pyruvate is transported from the cytoplasm into the mitochondria.

7. What are the assumptions made during the calculation of net gain of ATP?
Solution: It is possible to make calculations of the net gain of ATP for every glucose molecule oxidised; but in reality this can remain only a theoretical exercise. These calculations can be made only on certain assumptions that:

  • There is a sequential, orderly pathway functioning, with one substrate forming the next and with glycolysis, TCA cycle and ETS pathway following one after another.
    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.

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

8. Distinguish between the following:
(a) Aerobic respiration and Anaerobic respira¬tion.
(b) Glycolysis and Fermentation.
(c) Glycolysis and Citric acid cycle.
Solution: (a) Differences between aerobic and anaerobic respiration are as follows:
NCERT Solutions For Class 11 Biology Respiration in Plants Q8
(b) Differences between glycolysis and fermentation are as follows:
NCERT Solutions For Class 11 Biology Respiration in Plants Q8.1

9. Discuss “The respiratory pathway is an amphibolic pathway”.
Solution: Amphibolic pathway is the one which is used for both breakdown (catabolism) and build-up (anabolism) reactions. Respiratory pathway is mainly a catabolic process which serves to run the living system by providing energy. The pathway produces a number of intermediates. Many of them are raw materials for building up both primary and secondary metabolites. Acetyl CoA is helpful not only in Krebs’ cycle but is also raw material for synthesis of fatty acids, steroids, terpenes, aromatic compounds and carotenoids, a-ketoglutarate is organic acid which forms glutamate (an important amino acid) on amination. OAA (Oxaloacetic acid) on amination produces asparate. Both aspartate and glutamate are components of proteins. Pyrimidines and alkaloids are other products. Succinyl CoA forms cytochromes and chlorophyll.
Hence, fatty acids would be broken down to acetyl CoA before entering the respiratory pathway when it is used as a substrate. But when the organism needs to synthesise fatty acids, acetyl CoA would be withdrawn from the respiratory pathway for it. Hence, the respiratory pathway comes into the picture both during breakdown and synthesis of fatty acids. Similarly, during breakdown and synthesis of proteins too, respiratory intermediates form the link. Breaking down processes within the living organism is catabolism, and synthesis is anabolism. Because the respiratory pathway is involved in both anabolism and catabolism, it would hence be better to consider the respiratory pathway as an amphibolic pathway rather than as a catabolic one.

10. Define RQ. What is its value for fats?
Solution: Respiratory quotient (RQ) is the ratio of the volume of carbon dioxide produced to the volume of oxygen consumed in respiration over a period of time. Its value can be one, zero, more than 1 or less than one.
NCERT Solutions For Class 11 Biology Respiration in Plants Q10
Volume of C02 evolved Volume of 02 consumed
RQ is less than one when the respiratory substrate is either fat or protein.
C57 H104O6 + 80 O2-» 57 CO2+ 52H2O
RQ = 57CO2/80O2 = 0.71
RQ is about 0.7 for most of the common fats.

11. What is oxidative phosphorylation?
Solution: Oxidative phosphorylation is the synthesis of energy rich ATP molecules with the help of energy liberated during oxidation of reduced co-enzymes (NADH, FADH2) produced in respiration. The enzyme required for this synthesis is called ATP synthase. It is considered to be the fifth complex of electron transport chain. ATP synthase is located in FT or head piece of F0 -F1 or elementary particles. The particles are present in the inner mitochondrial membrane. ATP synthase becomes active in ATP formation only where there is a proton gradient having higher concentration of H+ or protons on the F0 side as compared to F x side (chemiosmotic hypothesis of Peter Mitchell).
Increased proton concentration is produced in the outer chamber or outer surface of inner mitochondrial membrane by the pushing of proton with the help of energy liberated by passage of electrons from one carrier to another. Transport of the electrons from NADH over ETC helps in pushing three pairs of protons to the outer chamber while two pairs of protons are sent outwardly during electron flow from FADH2. The flow of protons through the F0 channel induces F1 particle to function as ATP-synthase. The energy of the proton gradient is used in attaching a phosphate radical to ADP by high energy bond. This produces ATP. Oxidation of one molecule of NADH2 produces 3 ATP molecules while a similar oxidation of FADH2 forms 2 ATP molecules.

12. What is the significance of step-wise release of energy in respiration?
Solution: The utility of step-wise release of energy in respiration are given as follows :
(i) There is a step-wise release of chemical bond energy which is very easily trapped in forming ATP molecules.
(ii) Cellular temperature is not allowed to rise.
(iii) Wastage of energy is reduced.
(iv) There are several intermediates which can be used in production of a number of biochemicals.
(v) Through their metabolic intermediates different substances can undergo respiratory catabolism.
(vi) Each step of respiration is controlled by its own enzyme. The activity of different enzymes can be enhanced or inhibited by specific compounds.
This helps in controlling the rate of respiration and the amount of energy liberated by it.

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NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis in Higher Plants

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis in Higher Plants:

Section NameTopic Name
13Photosynthesis in Higher Plants
13.1What do we Know?
13.2Early Experiments
13.3Where does Photosynthesis take place?
13.4How many Pigments are involved in Photosynthesis?
13.5What is Light Reaction?
13.6The Electron Transport
13.7Where are the ATP and NADPH Used?
13.8The C4 Pathway
13.9Photorespiration
13.10Factors affecting Photosynthesis
13.11Summary

NCRT TEXTBOOK QUESTIONS SOLVED

1. By looking at a plant externally can you tell whether a plant is C3 or C? Why and how?
Solution: It is not possible to distinguish externally between a C3 and C4 plant, but generally tropical plants are adapted for C cycle.

2. By looking at which internal structure of a plant can you tell whether a plant is C3 or C? Explain.
Solution: C plants live in hot moist or arid and nonsaline or saline habitats. Internally the leaves show kranz anatomy. In kranz anatomy, the mesophyll is undifferentiated and its cells occur in concentric layers around vascular bundles. Vascular bundles are surrounded by large sized bundle sheath cells which are arranged in a wreath-like manner (kranz – wreath). The mesophyll and bundle sheath cells are connected by plasmodesmata or cytoplasmic bridges. The chloroplasts of the mesophyll cells are smaller. They have well developed grana and a peripheral reticulum but no starch. Mesophyll cells are specialised to perform light reaction, evolve 02 and produce assimilatory power (ATP and NADPH). They also possess enzyme PEPcase for initial fixation of CO2 The chloroplasts of the bundle sheath cells are agranal.

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3. Even though very few cells in a C plant carry out the biosynthetic – Calvin pathway, yet they are highly productive. Can you discuss why?
Solution: Since, through C cycle, a plant can photosynthesise even in presence of very low concentration of CO2 (upto 10 parts per million), the partial closure of stomata due to xeric conditions would not bring much
effect. Therefore, the plants can adapt to grow at low water content, high temperature and bright light intensities. This cycle is specially suited to such plants which grow in dry climates of tropics and subtropics. Besides, the photosynthetic rate remains higher due to absence of photorespiration in these plants. It can be visualised that both C cycle and photorespiration are the result of evolution or might have been one of the reasons of evolution for the adaptation of plants to different environments. C plants are about twice to efficient as Cplants in converting solar energy into production of dry matter.

4. RuBisCO is an enzyme that acts both as a carboxylase and oxygenase. Why do you think RuBisCO carries out more carboxylation in C plants?
Solution: RuBisCO is an enzyme which acts both as carboxylase (carboxylation during photosynthesis) and oxygenase (during photorespiration). But RuBisCO carries out more carboxylation in C4 plants. In C plants, initial fixation of carbon dioxide occurs in mesophyll cells. The primary acceptor of C02 is phosphoenol pyruvate or PER It combines with carbon dioxide in the presence of PEP carboxylase or PEPcase to form oxaloacetic acid or oxaloacetate. Malic acid or aspartic acid is translocated to bundle sheath cells through plasmodesmata. Inside the bundle sheath cells they are decarboxylated (and deaminated in case of aspartic acid) to form pyruvate and CO2 . CO2 is again fixed inside the bundle sheath Cells through Calvin cycle. RuBP of Calvin cycle is called secondary or final acceptor of CO2 in C plants. Pyruvate is sent back to mesophyll cells.

5. Suppose there were plants that had a high concentration of chlorophyll b, but lacked chlorophyll a, would it carry out photosynthesis? Then why do plants have chlorophyll b and other accessory pigments?
Solution: Plants that do not possess chlorophyll a will not carry out photosynthesis because it is the primary pigment and act as the reaction centre. It performs the primary reactions of photosynthesis or conversion of light into chemical or electrical energy. Other photosynthetic pigments are called accessory pigments. They absorb light energy of different wavelengths and hence broaden the spectrum of light absorbed by photosynthetic pigments. These pigments hand over the absorbed energy to chlorophylla.

6. Give comparison between the following:
(a) C3 andC pathways
(b) Cyclic and non-cydic photophosphorylation
(c) Anatomy of leaf in C3 and C plants.
Solution: (a) The differences between C3 and C
NCERT Solutions For Class 11 Biology Photosynthesis in Higher Plants Q6

NCERT Solutions For Class 11 Biology Photosynthesis in Higher Plants Q6.1
(b) The differences between cyclic and non- cyclic photophosphorylation are as follows :
NCERT Solutions For Class 11 Biology Photosynthesis in Higher Plants Q6.2

NCERT Solutions For Class 11 Biology Photosynthesis in Higher Plants Q6.3
(c) Differences between the leaf anatomy of C3 and C4plants are as follows :
NCERT Solutions For Class 11 Biology Photosynthesis in Higher Plants Q6.4

7. Look at leaves of the same plant on the shady side and compare it with the leaves on the sunny side. Or ompare the potted plants kept in the sunlight with those in the shade. Which of them has leaves that are darker green? Why?
Solution: The leaves of the shaded side are darker green than those kept in sunlight due to two reasons:
(i) The chloroplasts occur mostly in the mesophyll cells along their walls for receiving optimum quantity of incident light.
(ii)The chloroplasts align themselves in vertical position along the lateral walls of high light intensity and along tangential wails in moderate light.

8. The given figure shows the effect of light on the rate of photosynthesis. Based on the graph, answer the following questions.
NCERT Solutions For Class 11 Biology Photosynthesis in Higher Plants Q8
(a) At which point/s (A, B or C) in the curve is light limiting factor?
(b) What could be the limiting factor/s in region A?
(c) What do C and D represent on the curve?
Solution: (a) At regions A and B light is the limiting factor.
(b) In the region A’, light can be a limiting factor.
(c) C is the region where the rate of photosynthesis is not increased when light intensity is increased. D is the point where some other factors become limiting.

9. Why is the colour of a leaf kept in the dark frequently becomes yellow, or pale green? Which pigment do you think is more stable?
Solution: Carotenoid pigments are found in all photosynthetic cells. They are accessory pigments also found in roots, petals etc. These pigments do not breakdown easily thus temporarily reveal their colour due to unmasking, following breakdown of chlorophylls. Thus the colour of leaf kept in dark is yellow or pale green.