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

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

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 22 Chemical Coordination and Integration.

VERY SHORT ANSWER QUESTIONS

Question 1.
Which of the two adrenocortial layers, zona glomerulosa and zona reticularis lies outside enveloping the other?
Solution:
Zona glomerulosa envelops zona reticularis from the outside.

Question 2.
What is erythropoiesis? Which hormone stimulates it?
Solution:
The process of formation of RBC is Erythropoiesis. Erythropoietin, a Peptide hormone secreted from the juxtaglomerular cells of kidney stimulates erythropoiesis.

Question 3.
Name the only hormone secreted by pars intermedia of the pituitary gland.
Solution:
The only hormone secreted by Pars intermedia ^ of pituitary gland is Melanocyte Stimulating hormone (MSH). This hormone causes dispersal of pigment granules in the pigment cells, which darken the colour in certain animals like fishes and amphibians.

Question 4.
Name the endocrine gland that produces calcitonin and mention the role played by this ‘ hormone.
Solution:
Calcitonin/thyrocalcitonin linear polypeptide hormone comprising of 32 amino acids that is produced in humans primarily by the parafollicular cells of the thyroid gland. It checks excess Ca2+ and phosphate in plasma by • decreasing mobilization from bones.

Question 5.
Name the hormone that helps in cell-mediated immunity.
Solution:
The hormone thymosin plays a maj or role in the development and differentiation of T-lymphocytes, which provide cell-mediated immunity.

Question 6.
A patient complains of constant thirst, excessive 3. passing of urine and low blood pressure. When the doctor checked the patients’ blood glucose and blood insulin level, the level were normal or 4. slightly low. The doctor diagnosed the condition a diabetes insipid us. But he decided to measure one more hormone in patients blood. Which hormones does the doctor intend to measure?
Solution:
The doctor intends to measure the hyperglycaemia hormone, and its action is opposite to that of insulin Excess of glucose in blood suppresses the secretion of glucose, whereas fall in glucose level enhance glucose production.

Question 7.
Correct the following statements by replacing the term underlined.
(a) Insulin is a steroid hormone.
(b) TSH is secreted from the corpus leteum.
(c) Tetraiodothyronine is an emergency hormone.
(d) the pineal gland is located on the anterior part of the kidney.
Solution:
(a) Insulin is a peptide hormone
(b) TSH is secreted from the pars distalis region ofpitutary.
(c) Adrenaline is an emergency hormone.
(d) The adrenal gland is located on the anterior part of the kidney.

Question 8.
Match the following columns.
Column I                           Column II
A. Oxytocin                     1. Amino acid derivative
B. Epinephrine                2. Steroid
C. Progesterone             3. Protein
D. Growth hormone       4. Peptide
Solution:
The correct matching is
Column I                             Column II
A. Oxytocin                 –       Peptide
B. Epinephrine            –       Amino acid derivative
C. Progesterone         –        Steroid
D. Growth hormone   –       Protein

SHORT ANSWER QUESTIONS

Question 1.
What is the role-played by luteinising hormones in males and females respectively.
Solution:

  1. LH and FSH stimulate activity of gonads and hence are called gonadotropins.
  2. Luteinising hormone (LH) in males stimulates the synthesis and secretion of hormones called androgens from testis. Androgens along with FSH (Follicle Stimulating Hormone) regulate spermatogenesis.
  3. LH induces ovulation of fully mature follicles in females and maintains the corpus luteum, formed from the remnants of the graafian follicles after ovulation. This secretes progesterone.

Question 2.
George comes on a vacation to India from US. The long journey disturbs his biological system and he suffers from jet lag. What is the cause of his discomfort?
Solution:

  1. The melatonin hormone secreted by the pineal gland is also called as ‘sleep hormone’ as it promotes sleep-wake cycle.
  2. The disruption of the body clock as it is out of synchronisation because of the unfamiliar time zone of the destination causes Jet lag. The body experiences different patterns of light and dark conditions than it is normally used to, this disrupts the natural sleep-wake cycle.
  3. A hormone that plays a key role in body rhythms and causes jet lag is melatonin. Eyes perceive darkness after the sun sets and alert the hypothalamus to begin releasing melatonin, which promotes sleep. Conversely, when the eyes perceive sunlight, they induce the hypothalamus to with hold melatonin prodtiction.
  4. The hypothalamus however cannot readjust its schedule instantly and it may take several days, to overcome this problem.

Question 3.
Inflammatory responses can be controlled by a certain steroid. Name the steroid, its source and also its other important functions.
Solution:

  1. Glucocorticoids cortisol in particular, produce anti-inflammatory reactions and suppress the immune response.
  2. The middle zone, in adrenal cortex which is the widest of three zones called zona fasciculata is the source for glucocorticoids.
  3. The glucocorticoids as the name suggests affect carbohydrate metabolism and metabolism of proteins and fats.
  4. They stimulate gluconeogenesis, lipolysis and proteolysis. They also inhibit utilization of amino acid and cellular uptake. Cortisol is also called stress hormone as it copes with stress.

Question 4.
Old people have weak immune system. What could be the reasons?
Solution:

  1. A major role in the development of the immune system is played by thymus.
  2. The thymus gland is a lobular structure located on the dorsal side of the heart and the aorta. It is derived from the endoderm of the embryo. Thymus secretes a hormone named thymosin which stimulates the development of white blood cells (WBCs), involved in producing immunity.
  3. In old individuals, thymus is degenerated which results in decreased production of thymosin. The immune system as a result becomes weak, in old people.

LONG ANSWER QUESTIONS

Question 1.
Calcium plays a very important role in the formation of bones. Write on the role of endocrine glands and hormones responsible for maintaining calcium homeostasis.
Solution:
The hormones and endocrine glands that are responsible for maintaining calcium
homeostasis, are thyroid and parathyroid glands j and their associated hormones are calcitonin and Parathyroid Hormone (PTH).
(i) Parathyroid glands – These glands developed from the endoderm of the embryo. The cells of parathyroid glands are
of two types – chief cells and oxyphil cells. The chief cells of the parathyroid glands secrete parathyroid hormone (PTH).
This hormone (PTH) is involved in regulation of calcium and phosphate balance between the blood and other tissue. It mobilises the release of calcium into the blood from bones. PTH increases reabsorption of calcium by the body organs like intestine and kidneys.
(ii) Thyroid gland – It is the largest endocrine gland located anterior to the thyroid cartilage of the larynx in the neck.

This gland plays a major role in maintaining calcium homeostasis. It releases thyrocalcitonin hormone produced by the parafollicular cells, also called, ‘C’ cells. This hormone is secreted when the calcium level in blood gets high.

It is a 32 amino acid peptide hormone that lowers the calcium level by suppressing release of calcium ions from the bones. Calcitonin thus has an action opposite to that of the parathyroid hormone in calcium homeostasis.
NCERT Exemplar Solutions for Class 11 Biology Chapter 22 Chemical Coordination and Integration 1

Question 2.
Hypothalamus is a supper master endocrine gland. Elaborate.
Solution:
Hypothalamus is a minor but extremely important part of the diencephalon that is involved in the mediation of endocrine, autonomic and behavioural fiinction.

It consists of several groups of neuro secretory cells called nuclei which produce hormones. Hypothalamus provides anatomical connection between the nervous and endocrine system. It controls the release of major hormones by

hypophysis which include :
(i) Growth Hormone Releasing Hormone stimulates the anterior lobe of the pituitary gland to release growth hormone or somatostatin.
(ii) MSH Releasing Hormone stimulates the intermediate lobe of the pituitary gland to secrete Melanocyte Stimulating Hormone (MSH)
The hormones released from hypothalamus are involved in the processes like temperature regulation, control of water balance in body, sexual behaviour and reproduction, control of daily cycles in physiological state, behaviour and mediation’ of emotional response hypothalamus is thus called as super master endocrine gland of body.
(iii) Prolactin Releasing Hormone (PRH) stimulates the anterior lobe of the pituitary gland to secret prolactin.
(iv) Gonadotropin Releasing Hormone stimulates the anterior lobe of the pituitary gland to release gonadotropic hormones (FSHandlH).
(v) Thyrotropm Releasing Hormone (TRH) stimulates the anterior lobe of pituitary gland to release Thyroid Stimulating Hormone (TSH).
(vi) Adrenocorticotrophic releasing Hormone (ARH) stimulates the anterior lobe of pituitary gland to secrete Adrenocorticotropic Hormone (ACTH). ACTH stimulates the synthesis and secretion of steroid hormones called glucocorticoids by adrenal glands.
The hormones released from hypothalamus are involved in the processes like temperature regulation, control of water balance in body, sexual behaviour and reproduction, control of daily cycles in physiological state, behaviour and mediation of emotional responses. Hypothalamus is thus called as super master endocrine gland of body.

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 22 Chemical Coordination and Integration, help you. If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 22 Chemical Coordination and Integration, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination

NCERT Exemplar Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination.

VERY SHORT ANSWER QUESTIONS

Question 1.
Rearrange the following in the correct order of involvement in electrical impulse movement.
Solution:
The correct order of involvement in electrical impulse movement is as follows:
(i) Dendrites
(ii) Cell body
(iii) Axon
(iv) Axon terminal (vi) Synaptic knob
NCERT Exemplar Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination 1.1

Question 2.
Which cells of the retina enable us to see coloured objects around us?
Solution:
Cone cells present in unable us to see the colours. There are three types of cones which possess their own characteristic photopigments that respond to red, green and blue light.

Question 3.
Arrange the following in the order of reception and transmission of sound wave from the ear drum. Cochlear nerve, external auditory canal, ear drum, stapes, incus, malleus, cochlea.
Solution:
The reception and transmission of sound waves occurs in following order – External Auditory canal —» Eardrum —» Malleus —> Incus —> Stapes —>• Cochlea —> Cochlear nerve

Question 4.
During resting potential, the axonal membrane is polarized, indicate the movement of H-ve and -ve ions leading to polarisation diagrammatically.
Solution:
NCERT Exemplar Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination 4.1

Question 5.
Our reaction like aggressive behaviour, use of abusive words, restlessness etc. are regulated by brain, name the parts involved.
Solution:
Functions as aggressive behaviour, use or abusive words, restlessness, etc. The inner part of cerebral hemispheres and a group of associated deep structures called limbic lobe or limbic system along with hypothalamus are involved.

Question 6.
What do grey and white matter in the brain represent?
Solution:
A major component of CNS is Grey matter consisting of neutronal cell bodies, dendrite, unmyelinated axons, glial cells and capillaries. White matter is also a component of CNS and consists mostly of gilal cell and myelinated axons.

Question 7.
Where is the hunger centre located in human brain?
Solution:
Hypothalamus in human brain contains many centres which control urge for eating and drinking.

Question 8.
Complete the statement by choosing appropriate match among the following.
NCERT Exemplar Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination 8.1

Solution:
A. -> (3), B. -> (4), C. -> (2), D. -> (1)
NCERT Exemplar Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination 8.2

SHORT ANSWER QUESTIONS

Question 1.
The major parts of the human neural system is depicted below. Fill in the empty boxes with appropriate
words.
NCERT Exemplar Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination s1.1

Solution:
The major parts of the human neural system is filled in the boxes with appropriate words
NCERT Exemplar Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination s1.2

Question 2.
Neuron system and computers share certain common features. Comment in five lines.
Solution:
In various organs the sensory neurons is present to sense the environment and extend the message to the brain. So, it is equivalent to input device of computers.
Brain acts as the CPU, or Central Processing Unit. The information gathered by sensory neurons is processed by brain and it gives command to the concerned organ to act accordingly. This message is taken or conveyed by motor neurons which act as output devices.

Question 3.
What is the function described to Eustachian tube?
Solution:
The eustachian tube forms connection between the middle ear cavity with the pharynx. It helps in equalising the pressure on either sides of the ear drum. At the pharyngeal opening of the eustachian tube there is a valve which normally remains closed.
The valve opens during yawning, swallowing and during an abrupt change in altitude, when air enters or leaves the tympanic cavity to v equalise the pressure of air on the two sides of the tympanic membrane.

LONG ANSWER QUESTIONS

Question 1.
Explain the process of the transport and release of neurotransmitter with the help of a labelled diagram showing a complete neuron, axon terminal and synapse.
Solution:
The three main parts of a neuron include the
(i) Cell body
(ii) Axon
(iii) Dendrites
Stimulus or nerve impulse of any kind passes from one neuron to another via axon. This nerve impulse is wave of bioelectric/electrochemical disturbance that passes along the neuron during conduction of an excitation.
NCERT Exemplar Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination l1.1

  • Within a synapse transport and release of a neuro transmiter occurs.
  • 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.
  • Upon arrival of an impulse (action potential) at the axon terminal, it stimulates the movement of the synaptic vesciles towards the membrane, where they fuse with the plasma membrane and release their neurotransmitters in the synaptic cleft.
  • The released neurotransmitters bind to their specific receptors, present on the post-synaptic membrane. This binding opens ion channels allowing the entry of ions, that can generate a new action potential in the
    NCERT Exemplar Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination l1.2

Question 2.
Explain the structure of middle and internal ear with the help of diagram.
Solution:
Ears are a part of statoacoustic organ meant for balancing and hearing the external ear in most mammals is a heap of tissue also called pinna. It is a part of auditory system.

The human ear consists of three main parts external ear, middle, ear and internal ear.

Structure of Middle Ear

  • The middle ear consists of three bones or ossicles-the malleus (hammer), incus (anvil and stapes (stir-up).
  • These bones are attached to one another in a chain-like manner.
  • The malleus is attached to the tympanic membrane and the stapes is attached to the oval window (a membrane beneath the stapes) of cochlea.
  • These three ossicles increase the efficiency of transmission of sound waves to the inner ear.
  • The middle ear also opens into the eustachian tube, which connects with the pharynx and maintains the pressure between the middle ear and the outside atmosphere.

Structure of Internal Ear

  • Thd inner ear consists of a labyrinth of chambers filled with fluid within temporal bone of the skull. The labyrinth consists of two parts the bony and membranous labyrinth. The bony labyrinth is a series of channels.
  • Membranous labyrinth lies inside these channels which is surrounded by a fluid called perilymph. The membranous labyrinth is filled with a fluid called endolymph. The coiled portion of the labyrinth is called cochlea.
  • The cochlea has two large canal separated by a small cochlear duct (scala media). An upper vestibular canal (scala vestibuli) and a lower tympanic canal (scala tympani). The vestibular and tympanic canals contain perilymph and the cochlear duct is filled with endolymph.
  • The wall of membranous labyrinth comes in contact with the fenestra ovalis at the base of scale vestibuli while the fenestra rotunda.
    NCERT Exemplar Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination l2.1

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 21 Neural control and co-ordination, help you. If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 21 Neural control and co-ordination, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 11 Biology Chapter 20 Locomotion and Movement

NCERT Exemplar Solutions for Class 11 Biology Chapter 20 Locomotion and Movement

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 20 Locomotion and Movement.

VERY SHORT ANSWER QUESTIONS

Question 1.
Name the cells/ tissues in human body which
(a) exhibit amoeboid movement
(b) exhibit ciliary movement
Solution:
(a) Macrophages and leucocytes in blood exhibit amoeboid movement. Cytoskeletal elements like microfilaments are also involved in amoeboid movement.
(b) Ciliary Movement occurs mostly in the internal organs, lined by the ciliated epithelium, e.g., cilia in trachea helps in removing dust particle and foreign substances inhaled along with atmospheric air.
Passage of ova through the female reproductive tract is also facilitated by the ciliary movement. This is due to the presence of ciliated epithelium in the Fallopian tube.

Question 2.
Locomotion requires a perfect coordinated activity of muscular …… systems.
Solution:
Locomotion requires a prefect coordinated activity of muscular, skeletal and neural systems.

Question 3.
Sarcolemma, sarcoplasm and sarcoplasmic reticulum refer to particular type of cell in our body. Which is this cell and to what parts of that cell do these names refer to?
Solution:
Muscle fibre is lined by the plasma membrane called sarcolemma. Muscle fibre is a syncitium because sarcoplasm (the cytoplasm) of muscle fibre contains number of nuclei and sarcoplasmic reticulum is the endoplasmic reticulum of the muscle fibre and is the store house of calcium ions.

Question 4.
Label the different components of actin filament in the diagram given below
NCERT Exemplar Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 4.1v
Solution:
Each actin filament is made of two ‘F (filamentous) actins helically wound to each other and each ‘F’ actin is a polymer of monomeric ‘G’ (globular) actins.
The different components of action filament can be represented as
NCERT Exemplar Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 4.2v

Question 5.
What is the difference between the matrix of bones and cartilage?
Solution:
Difference between the matrix of ones and cartilage
NCERT Exemplar Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 5.1v

Question 6.
Which tissue is affected by mysthenia gravis? What is the underlying cause.
Solution:
Myasthenia gravis is autoimmune disorder of skeletal muscle, which affects neuromuscular junction, that leads to fatigue, weakening and paralysis of the skeletal muscle.

Question 7.
How do our bone joints function without grinding noise and pain?
Solution:
The presence of synovial fluid, between articulating surface of the two bones enclosed within synovial cavity of synovial joints to enables out joints to function without grinding noise and pain.

Question 8.
Give the location of a ball and socket joint in a human body.
Solution:
In human body Ball and socket joint are present between humerus and pectoral girdle. These joints allows free movement of bone in all direction. E.g., shoulder joints (humerus bone in socket of pectoral girdle) and

SHORT ANSWER QUESTIONS

Question 1.
With respect to rib cage, explain the following
(a) bicephalic ribs
(b) true ribs
(c) floating ribs
Solution:
There are 12 pairs of ribs. Each rib consist of a thin flat bone dorsally connected to the vertebral column and ventrally to the sternum.
(a) Bicephalic ribs each rib has two articulating surfaces on its dorsal end hence, are called as bicephatic ribs.
(b) The first seven pairs of ribs are true ribs. These ribs are dorsally attached to the thoracic vertebrae and ventrally connected to the sternum with the help of hyaline cartilage.
(c) The last two pair (11th and 12th) of ribs are not connected ventrally to the sternum therefore, called as floating ribs.
NCERT Exemplar Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 1.1s

Question 2.
Exchange of calcium between bone and ^extracellular fluid takes place under the
influence of certain hormones
(a) What will happen if of Ca2+ is in extracellular fluid?
(b) What will happen if very less amount of Ca2+ is in the extracellular fluid?
Solution:
Parathyroid and thyroid glands, function under the feed back control of blood calcium
(a) More Ca2+ concentration in extracellular fluid is associated with hyperparathyroidism. It causes demineralisation, resulting in softening and bending of the bones. This condition leads to osteoprosis.
(b) Very less amount of Ca2+ in extracellular fluid is associated with hypoparathyroidism. This increases the excitability or nerves and muscles, causing cramps, sustained contraction of the muscles of larynx, face, hands and feet. This disorder called parathyroid tetany or hypercalcemic tetany.

Question 3.
Rahul exercises regularly by visiting a gymnasium. Of late he is gaining weight. What could be the reasons? Choose the correct answer and elaborate.
(a) Rahul has gained weight due to accumulation of fats in body
(b) Rahul has gained weight due to increased muscle and less of fat
(c) Rahul has gained weight because his muscle shape has improved
(d) Rahul has gained weight because he is accumulating water in the body
Solution:
(b) Rahul has gained weight because the shape of his muscle has changed. Regular exercise increases the body muscle. There is an enlargement of muscles due to increase in the amount of sarcoplasm and mitochondria and the strength he to developed led him to gain the mass and size of body muscle and reduction in fat content.

Question 4.
Radha was running on a treadmill at a great speed for 15 minutes continuously. She stopped the treadmill and abruptly came out. For the next few minutes, she was breathing heavily/fast. Answer the following questions.
(a) What happened to her muscles when she did strenuously exercised?
(b) How did her breathing rate change?
Solution:
(a) Her muscles got fatigues due to continuous exercise because of the accumulation of lactic acid within skeletal muscles. Pain is also often experienced in the fatigued muscles.
(b) Her breathing rate changes from normal to
high as during as her body muscles require thus oxygen for the ATP production, than the normal value, her breathing thus enhances, to take most oxygen from the atmosphere.

Question 5.
Write a few lines about gout.
Solution:
Gout is a disease caused due to improper purine metabolism. It causes accumulation of uric acid and its crystals in the joints. The level of uric acid and crystals of its salts get raised in blood causing their accumulation in the joint to which causes gouty arthritis. The excess of urates in blood can also lead to the formation stones in the kidneys.

Question 6.
What are the points for articulation of pelvic and pectoral girdles?
Solution:

  1. Each half of the pectoral girdle consist of a clavicle and a scapula.
  2. The dorsal flat, triangular body of scapula has a slightly elevated ridge called the spine that, projects flat expanded process called the acromion and the clavicle articulating with it.
  3. There a depression below the acromion is called the glenoid cavity which articulates with the head of the humerous to form the shoulder joint.
  4. Pelvic girdle consist of two coxal bones, each formed by the fusioin of three bones, ilium, ischium and pubis. It articulates with femur through a cavity called acetabulum forming thigh joint.

LONG ANSWER QUESTIONS

Question 1.
How does a muscle shorten during its contracting and return to its original form during relaxation?
Solution:
Muscles contract due to formation of cross-bridge between the actin and myosin filament
(i) An ATP molecule j oins the active site on the head of myosin myofilament. These heads contains an enzyme, myosin ATPase along with Ca2+ and Mg2+ ions that catalyses the break down of ATP.
NCERT Exemplar Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 1.3l
(ii) The energy is transferred to myosin head which straightens to join an active site on actin myofilament, forming a across-
NCERT Exemplar Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 1.1l
(iii) The energised cross-bridges move, causing the attached actin filaments to move towards the centre of A-band. The Z-line is also pulled inwards causing shortening of sarcomere, contraction. During contraction A-bands retain the length, while I-bands get reduced.
(iv) The myosin head releases ADP and Pi where relaxes to its low energy state. The head detaches from actin myofilaments when new ATP molecule joins it and cross-bridge are broken.
(v) In the next cycle, the free head cleaves the new ATP. The cycles of cross-bridge formation and breakage is repeated causing further sliding.
NCERT Exemplar Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 1.2l
(vi) After contraction muscle relaxation occurs when the calcium ions are pumped back to the sarcoplasmic cistemae, thus, blocking the sites on actin myofilaments. The Z-line returns to original positions or relaxation.

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 20 Locomotion and Movement, help you. If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 20 Locomotion and Movement, drop a comment below and we will get back to you at the earliest.

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

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

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 19 Excretory Products and their Elimination.

VERY SHORT ANSWER QUESTIONS

Question 1.
Where does the selective reabsorption of glomerular filtrate take place?
Solution:
The selective reabsorption of glomerular filtrate takes place in Proximal Convoluted Tubules (PCT) and Distal Convoluted Tubules (DCT).

Question 2.
What is the excretory product from kidneys of reptiles?
Solution:
The excretory product from the kidney of reptile is uric acid.

Question 3.
What is the composition of sweat produced by sweat glands?
Solution:
Sweat produced by sweat glands is a watery fluid that containing NaCl, small amounts of urea, lactic acid, etc. It’s primary function is to facilitate a cooling effect on the body surface and also to helps in removal of water.

Question 4.
Identify the glands that perform the excretory function in prawns.
Solution:
In prawns, the excretory organs are known as antennary glands or green glands. These glands are white pea sized structures and opaque enclosed in the coxa of each 2nd antenna. They mainly excrete ammonia.

Question 5.
What is the excretory structure in Amoeba?
Solution:
Conractile vacuole performs the function of excretion as well as osmoregulation in amoeba.

Question 6.
The following abbreviations are used in the context of excretory functions, what do they stand for?
(a) ANF
(b) ADH
(c) GFR
(d) DCT
Solution:
(a) ANF Stands for Atrial Natriuretic Factor
(b) ADH Stands for Antidiuretic Hormone
(c) GFR Stands for Glomerular Filtration Rate
(d) DCT Stands for Distal Convoluted Tubule

Question 7.
Differentiate glycosuria from ketonuria.
Solution:
Difference between glycosuria and ketonuria is as follows
NCERT Exemplar Solutions for Class 11 Biology Chapter 19 Excretory Products and their Elimination 1

Question 8.
Mention any two metabolic disorders, which can be diagnosed by analysis of urine.
Solution:
Metabolic disorders that can be diagnosed by analysis of urine are
(i) Hematuria- It is a disorder in which blood cells are present in the urine, which could be a sign of kidney stone or a tumor in urinary tract.
(ii) Albuminuria- It is a disorder in which albumin is present in urine and occurs in nephritis i.e., inflammaton of glomeruli. In this condition the size of filtering slits becomes enlarged.

Question 9.
What are the main processes of urine formation?
Solution:
Urine formation includes glomerular filtration (ultra Alteration), selective reabsorption and tubular secretion that occurs in different parts of the nephron.
Glomerular filteration is carried out by glomerulus and is involve the filteration of blood.
Selective reabsorption is the absorption of filtrate through renal tubules either activity or passively.
Tubular secretin involves secretion through tubular cells in urine in order to maintain ionic and acid-base balance of body fluids.

Question 10.
Fill in the blanks appropriately
Organ Excretory wastes
(a) Kidneys ………..
(b) Lungs ………..
(c) Liver ………..
(d) Skin ………..
Solution:
Organ                             Excretory wastes
(a) Kidneys     ——>     Urine
(b) Lungs        ——>    C02
(c) Liver           ——>    Urea
(d) Skin           ——->   Sweat

SHORT ANSWER QUESTIONS

Question 1.
Show the structure of a renal corpuscle with the help of a diagram.
Solution:
The structure of a renal corpuscle is shown below.
NCERT Exemplar Solutions for Class 11 Biology Chapter 19 Excretory Products and their Elimination 2

Question 2.
What is the role played by renin-angiotensin in the regulation of kidney frictions?
Solution:
On activation by fall in the glomerular blood pressure/flow renin is released from the Juxta-Glomerular Apparatus (JGA).

It 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 Glomerular Filteration Rate (GFR). Angiotensin II also activates the adrenal cortex to release aldosterone.

This Aldosterone causes reabsprption of Na+ and water from the distal parts of the tubule. Which leads to an increase in blood pressure and GFR. This complex mechanism is generally known as Renin Angiotensin Aldosterone System of RAAS.

Question 3.
The composition of glomerular filtrate and urine is not same. Comment.
Solution:
Glomerular filtrate contains all the content of the blood plasma except proteins. About 180 litres of glomerular filtrate like water, glucose, nutrients ions etc. occurs. As a result, now the composition of urine is quite different from that of the glomerular filtrate. Some ions are also added to this fluid by tubules i.e. tubular secretion to maintain ionic and acid base balance of body fluids. Thus the composition of glomerular filtrate andd urine is not same.

Question 4.
What is the procedure advised for the correction of extreme renal failure? Give a brief account of it.
Solution:
The ultimate method for the correction of acute/ extreme renal failure (kidney failure) is, Kidney transplantation it is to minimise chances of rejection by the immune system of the host, functional kidney is used as a transplant from a donor, preferably close relative modem clinical procedures have increased the success rate of such a complicated technique.

Question 5.
Explain, why a haemodialysing unit called artificial kidney?
Solution:

  1. Haemodialysis is a method that become a boon for thousands of uremic (accumulation of urea in blood) patients all over the world. Haemodialysing unit act as artificial kidney by removing urea from patients blood due to kidney failure.
  2. In this process blood is drained from artery and pumped into a dialysing unit after the addition of an anticoagulant named heparin.
  3. The unit contains a coiled cellophane tube which is surrounded by a dialysing fluid having the same composition as that of plasma except nitrogenous waste.
  4. The porous cellophane mfcmbrane of the tube allows the passage of molecules that is based on concentration gradient.
  5. Absence of nitrogenous water in dialysing fluid these substances freely move out thereby clearing the blood.
  6. In the end the cleared blood is pumped back to the body through a vein after the addition of anti¬heparin to it thereby completing the process.
    NCERT Exemplar Solutions for Class 11 Biology Chapter 19 Excretory Products and their Elimination 3

LONG ANSWER QUESTIONS

Question 1.
Explain the mechanism of formation of concentrated urine in mammals.
Solution:
Mammals have the ability to produce concentrated urine. The loop of Henle and vasa recta play important role in it which is discussed as follows:
(i) the proximity between the Henle’s loop and vasa recta as well as the counter current that is formed due to the fiow of filtrate in two limp’s of Henle’s loop in opposite direction and help in opposite direction and help in maintaining an increasing osmolality towards the inner medullary interstitium, i.e., from 300 mOsmoL-1 in the cortex to about 1200 mOsmol-1 in the inner medulla.
NCERT Exemplar Solutions for Class 11 Biology Chapter 19 Excretory Products and their Elimination 4
NCERT Exemplar Solutions for Class 11 Biology Chapter 19 Excretory Products and their Elimination 6
(ii) This gradient is caused mainly due to NaCl and urea ascending limb of Henle’s loop transports NaCl, that is exchanged with the descending limb of vasa recta.
(iii) Through the ascending portion of vasa recta. NaCl is returned to the interstitium.
(iv) Similarly, a small amount of urea enters the thin segment of the ascending limb of Henle’s loop, which is transported back to the intersitium by the collecting tubule.
(v) This special arrangement of Henle’s loop, and vasa recta, is called the counter current mechanism.
(vi) The rate of dissipation is reduced by the counter current exchange. This in turn, reduces the rate at which the current must pump Na+ to maintain any given gradient.
(vii) Presence of such interstitial garden helps in an easy passage of water from the collecting tubule thereby concentrating the filtrate (urine).
(viii) Human kidneys produces urine nearly four times concentrated than the initial filtrate formed.

Question 2.
Describe the structure of a human kidney with the help of a labelled diagram.
Solution:

  • Human kidney are reddish-brown, bean-shaped structures that is situated between the last thoracic and third lumbar vertebra, which is closer to the dorsal inner wall of the abdominal cavity.
  • Each kidney of an adult human measures 10-12 cm in length, 5-7 cm in width, 2-3 cm in thickness with an average weight of 120-170 gm
  • The kidney is covered by a fibrous connective tissue i.e., the renal capsula, that protects the kidney.
  • Internally, it consists of outer dark cortex and an inner light medulla, both containing nephron, nephron is the structural and functional units of kidney.
  • The median concave border of a kidney contains a notch called hilum, that functions as route entry and exit of blood vessels, nerves and ureter.
  • The renal cortex is granular in appearance that contains convoluted tubules that malpighian corpuscles. The renal medulla contains loop of henle, collecting ducts and tubules and ducts ofBertini.
  • Medulla is divided into conical masses, the medullary pyramids that further form papillae.
  • The papillae form calyces, which join to renal pelvis leading to ureter. Between the medullary pyramids, cortex extends into medulla and forms renal columns which are called as column of Bertini.
    NCERT Exemplar Solutions for Class 11 Biology Chapter 19 Excretory Products and their Elimination 5

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

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

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation.

VERY SHORT ANSWER QUESTIONS

Question 1.
Name the blood component which is viscous and straw coloured fluid.
Solution:
Blood is a special connective tissue consisting of a fluid matrix, plasma and cells.

Question 2.
Complete the missing word in the statement given below.
(a) Plasma without …….. factors is called serum.
(b)………and monocytes are phagocytic cells.
(c) Eosinophils are associated with …….. reactions. .
(d) ……. ions play a significant role in clotting.
(e) One can determine the heart beat rate by counting the number of ……….. in an ECG.
Solution:
(a) Clotting
(b) Neutrophils
(c) allergic
(d) Calcium
(e) QRS complex

Question 3.
Name the vascular connection that exists between the digestive tract and liver.
Solution:
Hepatic portal system is the vascular connection that exists between the digestive tract and liver.

Question 4.
Given below are the abnormal conditions related to blood circulation. Name the disorders.
(a) Acute chest pain due to failure of 02 supply to heart muscles.
(b) Increased systolic pressure.
Solution:
(a) Angina
(b) High Blood Pressure

Question 5.
Which coronary artery disease is caused due to narrowing of the lumen of arteries?
Solution:
Atherosclerosis is the coronary artery disease caused due to the narrowing of the lumen of arteries due to deposition of calcium, fat, cholesterol and fibrous tissue the arteries become narrow, affecting vessels that supply blood to the heart muscles.

Question 6.
Define the following terms and give their location.
(a) Purkinje fibre
(b) Bundle of His
Solution:
(a) Purkinje Fibres are the fibres that conduct impulse, and the contraction impulses from AV node into the walls of ventricles.
(b) Bundle of His are mass of specialised fibres that originates from the AV node.

Question 7.
State the functions of the following in blood
(a) fibrinogen
(b) globulin
(c) neutrophils
(d) lymphocytes
Solution:
(a) Fibrinogens are the components of blood plasma that are inactive. In the presence of enzyme thrombin, they form a clot or coagulum of a network of threads called fibrin, in which dead and damaged elements of blood are trapped.
(b) Globulins are primarily involved in immunity,
i. e., defence mechanisms of the body.
(c) Neutrophils are phagocytic cells, that destroy foreign organisms entering the body,
(d) Lymphocytes are specialised cells which are responsible for the immune responses in the body. There are two major types of lymphocytes, that are involved in this process are B and T-lymphocytes.

Question 8.
How will you interpret an electocardiagram (ECG) in which time taken in QRS complex is higher?
Solution:
Electrocardiograph (ECG) is a graphical representation of the electrical activity of the heart during a cardiac cycle. A patient is connected to the machine having three electrical leads (one to each wrist and one to the left ankle) that continuously monitor the activity of heart.
Multiple leads are attached to the chest regions, for a detailed evaluation of the heart functions The QRS complex represent the depolarisation of the ventricles, that initiates the ventricular contraction. The contraction starts shortly after Q and marks the beginning of the systole. The time taken in QRS complex is 0.12 second in normal ECG
NCERT Exemplar Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation 1
The larger Q and R wave indicate a myocardial infarction (heart attack). The S-T segment is elevated in acute myocardial infarction and depressed when the heart muscle receives insufficient oxygen.

SHORT ANSWER QUESTIONS

Question 1.
The walls of ventricles are much thicker than atria. Explain.
Solution:
The walls of ventricles are thicker than the atria. It is due to the greater pressure exerted by pumping out of blood through ventricles of heart in compare to atria. Ventricles need to pump blood further and with much force.

Question 2.
Differentiate between
(a) blood and lymph
(b) basophilsand eosinophils
(c) tricuspid and bicuspid valve
Solution:
(a) Difference between blood and lymph are as follows
NCERT Exemplar Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation 2
(b) Difference between basophils and eosinophils are as follows
NCERT Exemplar Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation 3
(c)Difference between tricuspid valve and bicuspid valve are as follows
NCERT Exemplar Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation 4

Question 3.
Briefly describe the followings
(a) anaemia
(b) angina pectoris
(c) atherosclerosis
(d) hypertension
(e) heart failure
(f) erythroblastosis foetalis
Solution:
(a) Anaemia This is the most common disorder
of the blood, which is caused due to decrease in the number of RBC than the normal amount and also due to less quantity of haemoglobin than the normal value in blood. This is the most common disorder of the blood.
(b) Angina Pectoris When there is blockage in coronary artery, thus insufficient supply of blood reaches to heart muscles. That results in chest pain, fear, anxiety, pale skin, profuse
v sweating and vomiting. The pain usually starts in the centre of the chest spreads down to the left arm which last for only few second.
(c) Atherosclerosis It refers the deposition of cholesterol or fatty substance in the inner lining of arteries called atherosclerotic plaque. Sometimes arteries get completely blocked, this may result in stroke of heart attack.
(d) Hypertension It is sometimes also known as arterial hypertension. The blood pressure in the arteries gets elevated. It could be primary or secondary hypertension which
are caused by various conditions which affect kidneys, arteries heart or endocrine system.
(e) Heart Failure It is the state of heart, causes in when heart does not pump blood effectively enough to meet the requirement of the body.
(f) Erythroblastosis foetalis It is a haemolytic disease causes in new boms, which is an allo-immune condition that develops in foetus, when igG molecules produced by mother pass through placenta and attack RBC. It causes reticulocytosis and anaemia. It develops due to Rh incompatibility between the couples.
In a man with RH+ blood and women with Rlr blood, the second pregnancy foetus may have this problem due to IgG accumulation in women during first child development and delivery.

Question 4.
Explain the functional significance of lymphatic system?
Solution:
Lymphatic system comprises blood vessels that carries a fluid called lymph. It contains white blood cells, which are responsible for fighting against any diseases. It removes and filter the interstitial fluid from tissues, later absorbs and transports fatty acids and fats as chyle from digestive system and also transport cells to immune system.

LONG ANSWER QUESTIONS

Question 1.
Explain Rh-Incompatibility in humans.
Solution:
In nearly 80% of human Rh antigen is observed on the surface of RBCs. Such individuals are called Rh positive (Rh+) and those individuals where this antigen is not present are called Rh negative (Rh).
Both Rh+ and Rh individuals are phenotypically normal. The problem in them arises during blood transfusion and pregnancy.
(i) Incompatibility During Blood Transfusion The first blood transfusion of Rh+ blood to the person with Rh” blood causes no harm because the Rh person develops anti Rh factors or antibodies in his/ her blood, but second transfusion of Rh+ blood to the Rh person because anti Rh factors are already formed it destroys the red blood corpuscles of the donor. .
(ii) Incompatibility During Pregnancy If father’s blood is Rh+, mother blood is Rh and the foetus blood is Rh+. It will lead to a serious problem. Rh antigens of the foetus do not get exposed to the Rh’ve blood of the mother in the first pregnancy as the two bloods are well separated by the placenta.
But in the subsequent Rh+ foetus, the anti Rh factors (antibodies) present in causes mother destroys the foetal RBC due to mixing of blood. This causes the Haemolytic Disease of the New Born (HDN), called as erythroblastosis foetalis. In some cases new born may survive but will be anaemic and may also suffer with jaundice.
NCERT Exemplar Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation 5
This condition can be avoided by administering anti-Rh antibodies to the mother immediately after the delivery of the first child.

Question 2.
Explain different types of blood groups and donor compatibility by making a table.
Solution:
There are more than 30 surface antigens in blood cells which give rise to different blood groups.
ABO Grouping on basis of the presence or absence of two surface antigens on the RBCs namely, A and B. The plasma of different individuals contain two natural antibodies. The distribution of antigen and antibody in divided four groups into. A, AB, B and O.
NCERT Exemplar Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation 6
Form the given table it is evident that group ‘O’ blood can donate to persons with any other blood group and hence ‘O’ group individuals are called ‘Universal donors’. Person with ‘AB’ blood can accept blood from persons with AB, as well as the other groups of blood. Hence, such persons are called ‘Universal recipients’.

Question 3.
In the diagrammatic presentation of heart given below, mark and label. SAN, AVN, AV bundles, bundle of his and Purkinje fibres.
Solution:
The diagrammatic presentation of heart with labelled SAN, AVN, AV bundles bundle of His and purkinje fibres in heart is shown below.
NCERT Exemplar Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation 7

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

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

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 17 Breathing and Exchange of Gases.

VARY SHORT ANSWER QUESTIONS

Question 1.
Define the following terms?
(a) Tidal volume
(b) Residual volume
(c) Asthma
Solution:
(a) Tidal Volume (TV) is volume of air inspired or expired during a normal respiration. It is approx. 500 ml. i. e., a healthy man can inspire or expire approximately 6000 to 8000 mL of air per minute.
(b) Residual Volume : (RV) is volume of air remaining in the lungs even after a forcible expiration. This averages 1100 mL to 1200 mL.
(c) Asthma is an allergic reaction that causes constriction of the bronchiole muscles, thereby reducing the air passage thus the amount of the air that can get to the alveoli.

Question 2.
A fluid filled double membranous layer surrounds the lungs. Name it and mention its important function.
Solution:
Pleural membrane is a fluid filled double membranous layer surrounds the lung. It protects the lung and provides lubrication to it.

Question 3.
Cigarette smoking causes emphysema. Give reason.
Solution:
Emphysema is a chronic disorder of respiratory system, in which inflation or abnormal distension of alveolar wall occurs. Cigarette smoking and the inhalation of smoke or toxic substances over a time period causes the damaging of septa present between the alveoli, and its elastic tissue is replaced by the connective tissue in lungs.
Hence, decreases the respiratory surface and causes emphysema. It causes shortness of breath, production of sputum, chronic bronchitis, etc.

Question 4.
What is the amount of 02 supplied to tissues through every 100 mL of oxygenated blood under normal physiological conditions?
Solution:
Every 100 mL of oxygenated blood can deliver around 5 mL of 02 to the tissue under normal physiological conditions.

Question 5.
A major percentage (97%) of 02 is transported by RBCs in the blood. How does the remaining percentage (3%) of 0transported?
Solution:
About 97% of 02 is transported by RBCs in the blood. The remaining 3% of 02 is carried in a dissolved state through the plasma.

Question 6.
Complete the missing terms
(a) Inspiratory Capacity (IC) =…..+ IRV
(b) …. = TV + ERV
(c) Functional Residual Capacity (FRC) = ERV+ …..
Solution:
(a) Inspiratory Capacity (IC) = (TV) + (IRV) Tidal Volume. Inspiratory Reserve Volume
(b) Expiratory Capacity (EC) = (TV + (ERV) Tidal Volume. Expiratory Reserve Volume.
(c) Functional Residual Capacity (FRC) = (ERV) Expiratory + (RV) Reserve Volume. Residual Volume.

Question 7.
Name the organs of respiration in the following organisms.
(a) Flatworm …..
(b) Birds ……
(c) Frog …..
(d) Cockroach …….
Solution:
(a) Flatworm General body surface
(b) Birds Lungs
(c) Frog Lungs and moist skin
(d) Cockroach Tracheal tubes.

SHORT ANSWER QUESTIONS

Question 1.
State the different modes of CO2 transport in blood.
Solution:
The blood carries carbon dioxide in three forms.
(i) In dissolved State About 7% of C02 is carried by physical solution. Under normal temperature and pressure.
(ii) As carbamino Compounds – Carbon dioxide binds directly with Hb to form an unstable compound carbaminocompounds (COzHb). About 23% CO2 is transported in this form. When pC02 is high and p02 is low as in the tissues, more binding of CO2 occurs whereas, when pCO2 is low and pO2 is high as in alveoli as tissue dissociation of CO2 from carbamino-haemoglobin takes place.
NCERT Exemplar Solutions for Class 11 Biology Chapter 17 Breathing and Exchange of Gases 1
(iii) As bicarbonate Ions C02 reacts with water in the presence of carbonic anhydrase to form carbonic acid (H2C03) in RBC H2C03– dissociates into hydrogen and bicarbonate ions (HCO-).
The whole reaction proceeds as follows
NCERT Exemplar Solutions for Class 11 Biology Chapter 17 Breathing and Exchange of Gases 2
The carbonic anhydrase reaction mainly occur in RBC as it contain high concentration of enzyme carbonic anhydrase and minute quantity of it is present in plasma too.

Question 2.
For completion of respiration process, write the given steps in sequential manner.
(a) Diffusion of gases (O2 and CO2) across alveolar membrane.
(b) Transport of gases by blood.
(c) Utilisation of O2 by the cells for catabolic reactions and resultant release of CO2.
(d) Pulmonary ventilation by which atmospheric air is drawn in and CO2 rich alveolar air is released out.
(e) Diffusion of O2 and CO2 between blood and tissues.
Solution:
NCERT Exemplar Solutions for Class 11 Biology Chapter 17 Breathing and Exchange of Gases 3

LONG ANSWER QUESTIONS

Question 1.
Explain the mechanism of breathing with neat labelled sketches.
Solution:
Mechanism of breathing involves two stages:
Inspiration is the process, during which atmospheric air is drawn in expiration is the process by which the alveolar air is released out.
The movement of air into and out ofa the lungs is carried out by creating a pressure gradient between the lungs and the atmosphere, with the help of diaphragm and inter costal muscles.
NCERT Exemplar Solutions for Class 11 Biology Chapter 17 Breathing and Exchange of Gases 4

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 17 Breathing and Exchange of Gases, help you. If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 17 Breathing and Exchange of Gases, drop a comment below and we will get back to you at the earliest.

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

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

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption.

VERY SHORT ANSWER QUESTIONS

Question 1.
The food mixes thoroughly with the acidic gastric juice of the stomach by the churning movements of its muscular wall. What do we call the food then?
Solution:
For 4-5 hours, the food is stored in stomach and gets thoroughly mixed with the acidic gastric juice of stomach by the churning movements of its muscular wall. The food at this stage is called as chyme.

Question 2.
Trypsinogen is an inactive enzyme of pancreatic juice. An enzyme, enterokinase, activates it. Which tissue/cells secrete this enzyme?/ How is it activated?
Solution:
Trypsinogen is activated to trypsin in the presence of which enzyme enterkinase is secreted by the intestinal mucosa.

Question 3.
In which part of alimentary, canal does absorption of water, simple sugars and alcohol takes place?
Solution:
The absorption of water, simple sugars, alcohol and some lipid soluble drugs take place by the stomach wall.

Question 4.
Name the enzyme involved in the breakdown of nucleotides into sugars and bases?
Solution:
The enzymes nucleotidases and nucleosidases are involved in the breakdown of nucleotides into sugars and bases.
NCERT Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 14

Question 5.
What do we call the type of teeth attachment to jaw bones in which each tooth is embedded in a socket of jaws bones?
Solution:
The type of attachment where teeth are embedded in the socket of jaw bone is called thecodont.

Question 6.
Stomach is located in upper left portion of the abdominal cavity and has three major parts. Name these three parts.
Solution:
The three major of stomach are cardio, fundus and pylorus.
NCERT Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 1

Question 7.
Does gall bladder make bile?
Solution:
Gall bladder is involved in the storage of bile and not associated with the bile formation rather, bile is secreted and from the hepatic cells of the liver.

Question 8.
Correct the following statements by deleting one of entries (given in bold).
(a) Goblet cells are located in the intestinal mucosal epithelium and secrete chymot- rypsin/mucus.
(b) Fats are broken down into di-and monog-lycerides with the help of amylase/lipases.
(c) Gastric glands of stomach mucosa have oxyntic cell/chief which secrete HC1.
(d) Saliva contains enzymes that digest starch/ protein.
Solution:
(a) Goblet cells are located in the intestinal
mucosal epithelium and secrete mucus.
(b) Fats are broken down into di and monoglycerides with the help of lipases. Fats —Llp‘’e’ > Diglycerides —> Monoglycerides.
NCERT Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 2
(c) Gastric glands of stomach mucosa have oxyntic cells which secrete HCl
(d) Saliva contains enzymes that digest starch
NCERT Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 3

SHORT ANSWER QUESTIONS

Question 1.
What is pancreas? Mention the major secretions of pancreas that are helpful in digestion.
Solution:
The pancreas is both exocrine and as well as gland endocrine situated between the limbs of ‘U’ shaped duodenum.
NCERT Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 4
Internal structure of pancreas consist of two parts, i.e., the exocrine and endocrine part.
(i) Exocrine part consists of rounded lobules called acini, that secretes alkaline pancreatic juice of pH 8.4 and is mainly involved in the digestion of starch, proteins, fats and nucleic acids.
(ii) Endocrine part secretes hormones like, insulin and glucagon that regulate glucose metabolism.

Question 2.
Name the part of the alimentary canal where major absorption of digested food takes place. What are the absorbed forms of different kinds of food materials?
Solution:
The principle organ for the absorption of nutrients small intestine is the proces’s of digestion complete here and the final products of digestion are absorbed through the mucosa into the blood stream.
The absorbed form of different food materials are
NCERT Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 5

Question 3.
List the organs of human alimentary canal and name the major digestive glands with their location.
Solution:
Humn digestive system consists of two main parts: alimentary canal and digestive glands. The organ of human alimentary canal are mouth, pharynx oesophagus, stomach, small intestine, large intestine, rectum and anus. Major digestive glands with their locations are as follows:
NCERT Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 6

Question 4.
What are three major types of cells found in the gastric glands? Name their secretions.
Solution:
Following are the three major types of cells found in gastric glands. –
(i) Mucous neck cells (Goblet cells) are involved in the secretin of mucus and are present throughout the epithelium of gastrointestinal tract.
(ii) Peptic of Chief cells (Zymogenic cells) are Involved in the secretion of gastric enzymes such as proenzymes pepsinogen and prorenin and usually basal in location.
(iii) Parietal or oxyntic cells are large and most numerous present on the side walls of the gastric glands. They are involved in the secretion of HC1 and Castlis intrinsci Factor (CIF).

Question 5.
How is the intestinal mucosa protected from the acidic food entering from stomach?
Solution:
The intestinal mucosal epithelium has goblet cell which secrete mucus. The mucus along with bicarbonate present in the gastric juice help in lubrication and protection of mucosal epithelium from the acidic food entering from the stomach.

LONG ANSWER QUESTIONS

Question 1.
A person had roti and dal for his lunch. Trace the changes in those during its passage through the alimentary canal.
NCERT Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 7
Solution:
Digestion of Roti (Carbohydrates)
(a) Digestion of Carbohydrates in the Oral Cavity
In oral cavity, the roti get mixed with saliva that contains an enzyme salivary amylase (ptyalin), which converts starch of roti into maltose, isomaltose and small dextrins called a – dextrin. 30% of starch is hydrolysed in the oral cavity.
NCERT Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 8
(b) Digestion of Carbohydrates in the Small Intestine
The partially digested roti passes from oral cavity to oesophagus and then reaches to stomach by peristalsis. The stomach stores the food for 4-5 hours. The gastric juice does not contain carbohydrate digesting enzyme. The partially digested food is now called as chyme. In intestine, following action occurs.
(i) Action of Pancreatic Juice – Carbohydrates in the chyme are hydrolysed by pancreatic amylase into disaccharides.
Polysaccharides(starch) —Amylas’: > Disaccharides
NCERT Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 9
(ii) Action of Intestinal Juice – Intestinal Juice contain maltase, isomaltase, sucrase (invertase), lactase and a – dextrinase. hi the presence of these enzymes food is converted into simpler compounds like glucose, fructose, galactose, etc.
NCERT Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 10
Digestion of Protein
Proteins are made up of amino acids. So proteins are broken down to amino acid during the process of digestion.
Saliva lacks any protein digesting enzyme so, digestion starts further in stomach.
(a) Digestion of Protein in Stomach. The stomach normally stores food for 4-5 hours. The gastric glands of the stomach secrete gastric juice that contains HCI, proenzymes like-pepsinogen and prorennin. Various reactions in stomach are discussed bwlow.
NCERT Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 11
(b) Digestion of Protein in Small Intestine (i) Action of Pancreatic Juice – The enzymes , trypsinogen, chymotrypsinogen and procarboxypeptidase in pancreatic juice are all concerned with the protein digestion.
NCERT Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 12
(ii) Action of Intestinal Juice – Intestinal juice contain enzymes enterokinase, amino peptidase and dipeptidase
NCERT Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 13
The macromolecules are broken down into simpler components are the products of roti and dal (carbohydrates and proteins) which are further absorbed by the villi in small intestine and the rest undigested food is removed in the form of faeces by large intestine.

Question 2.
Discuss mechanisms of absorption.
Solution:

  • Absorption is a process by which the end product of digestion passes through the intestinal mucosa into the blood or lymph.
  • It is carried out by passive, active or facilitated transport mechanism. Small amount of monosaccharide like glucose, amino acids and some electrolytes like chloride ions are absorbed by simple diffusion.
  • Some of the substance like fructose and some amino acids are absorbed with the help of the carrier ions like Na+ are absorbed by the active transport.
  • Fatty acid and glycerol are insoluble, thus they cannot be absorbed by the blood. They are first incorporated into small droplets called micelles which move into the intestinal mucosa.
  • They are reformed into very small protein coated fat globules called chylomicrons which are transported into the lacteals of the villi. The lacteals ultimately release the absorbed substance into the blood stream.
  • The maximum absorption of food takes place in small intestine.

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 16 Digestion and Absorption, help you. If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 16 Digestion and Absorption, drop a comment below and we will get back to you at the earliest.

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

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

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 15 Plant Growth and Development.

VERY SHORT ANSWER QUESTIONS

Question 1.
Fill the places with appropriate word/words.
(a) A phase of growth which is maximum and fastest is ………
(b) Apical dominance as expressed in dicotyledonous plants is due to the presence of more ……. in the apical bud than in the lateral ones.
(c) In addition to auxin, a ……. must be supplied to culture medium obtain a good callus in plant tissue culture.
(d)…….. of a vegetative plants are the sites of photoperiodic perception.
Solution:
(a) A phase of growth which is maximum and latest is exponential phase.
(b) Apical dominance as expressed in dicotyledonous plants is due to the presence or more auxins in the apical bud than in the lateral ones.
(c) In addition to auxin, a cytokinin must be g supplied to culture medium to obtain a good callus in plant tissue culture.
(d) Leaves of vegetative plants are the sites of photoperiodic perception.

Question 2.
Plant Growth Substances (PGS) have innumerable practical applications. Name the PGS you should use to
(a) increase yield of sugarcane
(b) promote lateral shoot growth
(c) cause sprouting of potato tuber
(d) inhibit seed germination
Solution:
(a) Spraying gibberellins
(b) Application of auxins
(c) Ethylene
(d) ABA

Question 3.
A primary root grows from 5 cm to 19 cm in a week. Calculate the growth rate and relative growth rate over the period.
Solution:
Growth depends upon three factors – initial size (WQ), rate of growth (r) and time interval (+) for which the rate of growth is retained.
NCERT Exemplar Solutions for Class 11 Biology Chapter 15 Plant Growth and Development 1

Thus absolute growth rate is 0.1907 while relative growth rate is 3.8 cm.

Question 4.
Gibberellins were first discovered in Japan when rice plants were suffering from bakane (the foolish seedling disease) caused by a fungus Gibberella fujikuroi.
(a) Give two functions of this phytohormone.
(b) Which property of gibberellin caused foolish seedling disease in rice?
Solution:
Following are the two functions of gibberellin:
(i) It produces the phenomenon of bolting i. e., the growth of the intemodal region of stem in rosette plants.
(ii) It induces seed germination and break bud and seed domancy.
(b) The rice seeding/plant show excessive growth in their intemodal region when gets infected by fungus Gibberellafujikuroi. This fungus produces excessive amount of plant hormone GA that makes plants taller in comparison to the normal plant foolishly and many results into death of the plant.

Question 5.
Classify the following plants into Long Day Plants (LDP), Short Day plants (SDP) and Day Neutral Plants (DNP) Xanthium, henbane (Hyoscyamus niger), spinach, rich, strawberry, Bryophyllum, sunflower, tomato, maize.
Solution:
Long Day Plant (LDP) The plants that requires the exposure light for a longer period exceeding a well defined critical duration of light are long day plants.

Among the above given plant LDP are for henbane, Bryophyllum and spinach. Short Day Plants (SDP) The Plants that requires light for a period less than well defined critical duration of light, e.g., Xanthium, rice, strawberry.

Day Natural Plants (DNP) The exposure to light does not affect the flowering in certain plants, e.g, DNP, sunflower, tomato, maize.

Question 6.
A farmer grows cucumber plants in his field. He wants to increase the number of female flowers in them. Which plant growth regulator can be applied to achieve this?
Solution:
Ethylene, is a plant growth regulator that has feminizing effect on sex expression. Ethylene promotes formation of female flowers in monoecious plants like cucumber.

Question 7.
Where are the following hormones synthesised in plants?
(a) IAA
(b) Gibberellins
(c) Cytokinins
Solution:
(a) IAA i.e., Indole acetic acid. It is synthesised at the growing apices of the plant, e.g., shoot tip, leaf primordia and developing seeds.
(b) Gibberellins It is synthesised in the apical shoot buds, young leaves, root tips and developing seeds.
(c) Cytokinins are synthesised mainly in routs, but synthes also occurs in the endosperm of seeds, growing embryo etc.

Question 8.
Growth is one of the charactristic of all living organism? Do unicellular organism also grow? If so, what are the parameters?
Solution:
1. Lag phase- In this phase growth is slow.
2. Exponential phase- It shows rapid growth and maintains maximum growth for sometime.
3. Stationary phase- In this phase Growth diminishes and ultimately stops.

Question 9.
In the figure of sigmoid growth curve given below, label segments 1,2 and 3.
NCERT Exemplar Solutions for Class 11 Biology Chapter 15 Plant Growth and Development 2
Solution:
Growth is the main characteristic that distinguish living organisms from non-living. All living organism grow in number and then accumulate biomass and grow in size as well.

Increase in number of cells as well as increase in size and length of each cell, exhibits growth of all living organism.
In unicellular organism, the growth is synchronized with reproduction.

These organism when divide they produce offspring (reproduction) i.g., each cell accumulate (synthesise) protoplasm and increase in size but at a certain a limit and divide to from two cells.

Question 10.
The rice seedlings infected with fungus Gibberellafujikuroi is called foolish seedlings? What was the reason behind it?
Solution:
The rice seedling infected with fungus Gibberella fujikuroi are called foolish seedling because infected plants grow excessively taller than rest of the non infected rice plants in the field fall over and be unharvestable.

SHORT ANSWER QUESTIONS

Question 1.
Nicotiana tobacum, a short day plant, when exposed to more than critical period of light fails to flower. Explain.
Solution:
Short day plants are those plants that flower only when the exposure to duration of light is below critical period. Tobacco, being a short day plant is unable to show flowering when it is exposed to light above than the critical period.

Question 2.
Explain in 2-3 lines each of the following terms with the help of examples taken from different plant tissues.
(a) Differentiation
(b) De-differentiation
(c) Re-differentiation
Solution:
(a) Differentiation is permanent in composition
structure size and function of cells, tissue or organs. For example the meristematic tissues in plants gives rise to new cells which then mature and get differentiated into tissue or an organ of the plant, e.g., cells, distal to root apical meristem form root cap, cell of the periphery form epiblema, followed by cortex, etc.
(b) De-differentiation is the process of regain of differentiated cells so that they again become differentiated and able to divide, e.g., in dicot stem, the cortical cells get de-differentiate and become meristematic to form cambium (interfascicular cambium, and fascicular cambiums).
(c) Re-differentiation The cambium cells thus formed, again re-differentiate to form secondary cortex cells, secondary xylem and phloem elements and phelloderm in case of secondary growth of woody dicot plants.

Question 3.
The role of ethylene and abscissic acid is both positive and negative. Justify the statement.
Solution:
NCERT Exemplar Solutions for Class 11 Biology Chapter 15 Plant Growth and Development 3

Question 4.
In animals, there are special glands secreting
hormones, whereas there are no glands in plants. Where are plant hormones formed? How are the hormones translocated to the site of activity?
Solution:
The plant hormones are synthesised by the plant cells needed. Few hormones are specifically synthesised at a particular part of the plant like auxin synthesised in growing shoot apices and ethylene is secretes by ripened fruits.

Cytokinin is found in dividing cells. Unlike plants animal 1 being more advanced, and organised they have proper hormone secreting glands and organs.

These are transported through the transport system of their body in both plant and animals. In plants, hormone are translocated via xylem and phloem to the site of activity.

Question 5.
In a slide showing different types of cells can you identify which type of the cell may be meristematic and the one which is incapable of dividing and how?
Solution:
On the basis of the following characteristics the meristemtic cells can be identified.
(i) Cell consist thin cellulose wall and dense cytoplasm with large nucleus.
(ii) Among meristematic cells, plasmodesmal connections are more numerous.
(iii) Cell division, i. e., mitosis arid its various stages are distinctly visible.
(iv) Chromosomes of cells replicate and divide into two homologous chromatids.
All these features contribute to open ended growth where structure is in complete in meristematic regions.
Whereas, cells incapable of dividing show features such as
(i) Attains particular shape, size and thickening.
(ii) Undergoes structural and physiological differentiation.
(iii) Different types of cell are formed such as epidermis, cortex, vascular tissues.

Question 6.
A rubber band stretches and reverts back to its original position. Bubble gum stretches, but it would not return to its original position.
Is there any difference between the two processes? Discuss it with respect to plant growth (hint elasticity (reversible) plasticity (irreversible).
Solution:
NCERT Exemplar Solutions for Class 11 Biology Chapter 15 Plant Growth and Development 4

Question 7.
Label the diagram.
A. This is which part of a dicotyledonous plants?
B. If we remove part 1 from the plant, what will happen?
NCERT Exemplar Solutions for Class 11 Biology Chapter 15 Plant Growth and Development 5
Solution:
Representation the labelling of the given diagram is as follows
NCERT Exemplar Solutions for Class 11 Biology Chapter 15 Plant Growth and Development 6

A. the plant part in the given diagram is growing shoot apex.
B. Removal of shoot apex will help to overcome the apical dominance. Thus, the lateral buds grow faster, giving rise to branches and give the plant a bushy appearance.

Question 8.
Both animals and plants grow. Why do we say that growth and differentiation in plants is open and not so in animals? Does this statement hold true for sponges also?
Solution:

  1. Growth in plant is totally different from the animal growth as growth in plant is unlimited and indefinite.
  2. Root and shoot in tips in the plants are open ended i.e. they always grow and form new organ to replace the older and senescent one due to presence of meristem cells, which are capable to grow and divide.
  3. Thus, the plant growth continues throughout the life. On the contrary, animal growth is limited as growth /stops as soon as they mature. Sponges are those animals which show cellular level of organisation.
  4. These animals posses totipotent cells which are capable of giving rise to all other cells in sponges.
  5. A small part detached from a sponge can regenerate into a whole new sponge.
  6. However growth in sponges cannot be called open or indefinite as they cannot grow beyond a certain size.
    Practically, they do not show open ended growth.

Question 9.
Define parthenocarpy. Name the plant hormone used to induce parthenocarpy.
Solution:
Parthenocarpy is the process where fruit develop without fertilisation and so, it lacks contain seed. Seedless fruits are developed in some plants. Certain phytohormone induce development of fruit without fertilisation. This can also be induced artificially by spraying auxin an*i gibberellins in certain plants like, grapes, papaya, etc. ,

Question 10.
While eating watermelons, all of us wish it was seedless. As a plant physiologist can you suggest any method by which this can be achieve.
Solution:
The seedless, fruits can be produced by the process of parthenocarpy. In this fruits are
developed without fertilisation, so, seeds are not formed in the fruit. Artifically parthenocarpy can be induced by spraying auxin and gibberellin to produce seedless watermelons.

Question 11.
On germination a seed first produces shots with leaves, flowers appear later,
A. Why do you think this happens?
B. How is this advantageous to the plant?
Solution:
A. With the germination of seeds the plant
enters into vegetative growth period. This period tabes light stimulus i. e., photoperiod) and synthesise the florigen (a flowering hormone) that flowering.
B. The vegetative growth period prepares the plant to bear reproductive structure like flower, fruits and seeds, and allows it to grow, mature and reproduce.

Question 12.
Fill in the blanks
A. Maximum growth is observed in …… phase.
B. Apical dominance isdue to ………. .
C. …… initiate rooting.
D. pigment involved in phote photoperception in fiowering plants in ……….
Solution:
A. Exponential
B. Auxin
C. Cytokinin
D. Phytochrome.

LONG ANSWER QUESTIONS

Question 1.
Some varieties of wheat are known as spring wheat while others are called winter wheat. Former variety is sown and planted in spring and is harvested by the end of the same season. However, winter varieties, if planted in spring, fail to flower or produce mature grains within a span of a flowering season. Explain, why?
Solution:

  • Some annual plants such as wheat do not flower, unless they experience a low temperature during spring they remain vegetative but after receiving low temperature (in winter) they grow further to bear flowers and fruits.
  • During winter the low temperature prevents precocious reproductive development in autumn, thus enabling the plant to reach vegetative maturity before reproductive phase.
  • Thus, in spring when spring varieties are planted they flower and bear fruits prior to end of growing season.
  • But, if the winter varieties are planted in spring, they fail to flower and produce mature grains before the end of growing season, as they could not perceive low temperature of winters.

Question 2.
Name a hormone which
A. is gaseous in nature
B. Is responsible for phototropism
C. induces femaleness in flowers of cucumber
D. is used for killing weeds (dicots)
E . induces flowering in long day plants.
Solution:
A. Ethylene is a hormone which is gaseous in nature.
B. Auxin (synthetic auxin 2-4D) is responsible for phototropism.
C. Ethylene induces ferminising effect. External supply of very small quantity of ethylene can increase the number of female flowers and hence fruits as in cucumber.
D. Synthetic auxin (2-4D) that kills broad leaved dicot weeds and is used as weedicides.
E. Gibberellins induces flowering in long-day plants.

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 15 Plant Growth and Development, help you. If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 15 Plant Growth and Development, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants

NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants.

VERY SHORT ANSWER QUESTIONS

Question 1.
Energy is released during the oxidation of 3. compounds in respiration. How is this energy stored and released as and when it is needed?
Solution:
The energy currency of every living cell Adenosine Triphosphate (ATP).
Complex organic food molecules such as sugars, fats and proteins are rich sources of energy for cell because much of the energy used to form these molecules is stored within the chemical bonds that hold them together. So, the cells release the stored energy through a series of oxidation reactions.

During oxidation of food, the product of reaction has a lower energy content than the donor molecule. At the same time, electron acceptor molecules capture some of the energy lost during oxidation and store it for later use.

Cells convert the energy from oxidation reactions to energy-rich molecules such as ATP that can be used through the cell for metabolism and construct new cellular components.

Question 2.
How does Respiratory Quotient 5. (RQ) indicate which type of substrate, i.e., carbohydrate, fat or protein is getting oxidised?
R.Q=A/B
What do A and B stand for?
What type of substrates have R.Q. of 1,< 1 or > 1?
Solution:
The ratio of C02 evolved and 02 consumed in respiration is called the Respiratory Quotient (RQ) or respiratory ratio.
NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants 1
Example:
(i) During aerobic respiration carbohydrates have RQ = 1
(ii) During germination of seeds proteins and fats have RQ of < 1. (iii) Under aerobic conditions substrates like organic acids have RQ of > 1

Question 3.
F0 – F1 particles participate in the synthesis of …… .
Solution:
F0 – F1 particles present in the inner mitochondrial membrane are involved in the Adenosine Triphosphate synthesis. It is known as the energy currency of the cell.
NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants 2

Question 4.
When does anaerobic respiration occur in man and yeast?
Solution:
In animals anaerobic respiration occurs in the situation of oxygen deficiency during heavy exercise, when pyruvic acid is reduced to lactic acid by the enzyme lactate dehydrogenase.

In yeast, the incomplete oxidation of glucose occurs in anaerobic conditions, during which pyruvic acid is converted to COz and ethanol by the action of enzyme pyruvic acid decarboxylase and alcohol dehydrogenase.

Question 5.
Which of the following will release more energy on oxidation? Arrange them in ascending order.
(a) 1 gm of fat
(b) 1 gm of protein
(c) 1 gm of glucose
(d) 0.5 gm of protein + 0.5 gm glucose
Solution:
The ascending order of substrate that will release more energy on oxidation will be as follows
1 gm protein < 0.5 gm in protein < 1 gm glucose < 1 gm fat + 0.5 gm glucose

SHORT ANSWER QUESTIONS

Question 1.
If a person is feeling dizzy, glucose or fruit juice is given immediately but not a cheese sandwich, which might have more energy. Explain.
Solution:
The glucose is absorbed and reaches blood quickly and gives instant energy. Whereas, cheese sandwich require time for digestion, and absorption. Sick person needs immediate energy supply, so glucose or fruit juices containing glucose are given to them.

Question 2.
Pyruvic acid is the end product of glycolysis. What are the three metabolic fats of pyruvic acid under aerobic and anaerobic conditions? Write in the space provided in the diagram.
NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants 3

Solution:
The three metabolic products formed under 3. aerobic and anaerobic conditions are Lactic acid, Ethanol and Acetyl Co-A Lactic acid is formed under anaerobic condition in skeletal muscles by the oxidation of pyruvic acid.

Ethanol is formed under anaerobic condition by the oxidation of pyruvic acid in yeast.

Acetyl Co-A is formed by the oxidation of pyruvic acid that take place within the mitochondria under aerobic condition.
NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants 4
NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants 5

Question 3.
Oxygen is an essential requirement for aerobic respiration by it enters the respiratory process at the end? Discuss.
Solution:
Aerobic respiration needs oxygen in order to generate ATP. Oxygen acts as final acceptor in respiratory process.

In pulse e (electrons) that energy from the electron transport chain ETC and take up protons from medium to form water.

It plays a vital role in respiration. 02 enters in the respiratory process at the end. It drives the process of aerobic respiration by removing hydrogen from the system. Thus, acting as final hydrogen acceptor.

By the process of oxidative phosphorylation the energy is produced, utilising the energy of oxidation reduction reactions.

Question 4.
The figure given below shows the steps in glycolysis. Fill in the missing steps A, B, C, D and also indicate whether ATP is being used up or released at step E?
NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants 6

Solution:
Process ofglycolysis is summarised as follow
NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants 7

Question 5.
Do you know any step in the TCA cycle where there is substrate level phosphorylation.
Which one?
Solution:
In an intermediate reaction TCA cycle,
succinyl Co-A is converted succinic acid and one GTP molecule is synthesised through substrate level
NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants 8
GTP formed in this reaction gives rise to ATP as follows
GTP+ADP GDP + ATP

Question 6.
In a way green plants and cyanobacteria have synthesised all the food on the earth. Comment.
Solution:
Cyanobacteria are unicellular prokaryotic organisms. Besides, some primitive cellular cell organelles, they have photosynthetic lamellae where photosynthetic pigments like chlorophyll-a c, phycocyanin and phycoerythrin, are present.

These coloured pigments confer typical blue green colour to the bacteria and enable them to manufacture food for themselves and aquatic animals.

Green plants are multicellular are organisms, which is capable of making food by using C02, H20 and light energy in specialized cell organelles called chlorcplast. So bacteria and green plants make food for living organisms on earth.

Question 7.
When a substrate is being metabolised, why does not all the energy that is produced get released in one step. It is released in multiple steps. What is the advantage of step-wise release?
Solution:
This is an enzyme that has dual nature. When C02 concentration is good enough in atmosphere. It acts as carboxylase. But if concentration of O2 increase, its nature changes and it binds with O2 and acts as oxygenase enzyme that forces CO2 to enter in C2 cycle that leads to photorespiration and loss of CO2.

Question 8.
Respiration requires 02. How did the first cells on the earth manage to survive in an atmosphere that lacked O2?
Solution:
Respiration always does not require O2. There are some organisms which respire in anaerobic condition i.e. in the absence of O2.
The first cells of earth i.g., chemosynthetic bacteria, which are the primitive organisms found earlier on earth. They obtain energy by breaking down inorganic molecules like H2S, NO2 – etc.
12H4S + 6 CO2 -> C6H12O6 + 6H2O + 12S

Question 9.
It is known that red muscle fibres in animals can work for longer periods of time continuously. How is this possible?
Solution:
There are basically two kinds of muscle fibres red muscles and white muscles
Red muscles work continuously for a longer time because
(i) These muscle fibres are dark red, due to the presence of red haemoprotein called myoglobin. It binds and stores oxygen as oxymyoglobin in the red fibres. Oxymyoglobin liberates oxygen for utilisation during muscle contraction.
(ii) Mitochondria are more in numbers, hence they work for long periods of time.
(iii) Red muscles possesses less sarcoplasmic reticulum.
(iv) They carry out considerable aerobic oxidation without accumulating much lactic acid. Thus without fatigue red muscle fibres can contract for a longer period.
(v) These muscle fibre have slow rate of contraction for long periods, e.g., extensor muscles of the human back.

Question 10.
RuBP carboxylase, PEPcase, pyruvate dehydrogenase, ATPase, cytochrome oxidase, hexokinase, lactate dehydrogenase, Select/ choose enzymes from the list above which are involved in
(a) Photosynthesis
(b) Respiration
(c) Both in photosynthesis and respiration
Solution:
RuBP Carboxylase is an enzyme that takes part in dark reaction of photosynthesis. It catalyses the fixing of C02 in C3 cycle
PEPcase an enzymes that takes part in photosynthesis of C4 plants. It catalyses the reaction of fixing of C02 to form first stable product oxaloacetate. 4 carbon compound.

Pyruvate dehydrogenase is an enzyme involved in aerobic respiration and catalyses the reaction ( of formation of acetyle Co-A from pyruvic acid.

It requires the participation of NAD and Co- enzyme-A.
NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants 9
ATPase is a part of both respiration and photosynthesis. Both these processes uses etc, associated proton pump and ATP synthase.
These all play a key part in the process is used by ETC pump hydrogen ions across a membrane.
The protons flows back through ATP synthase, driving the production of ATP.

Cytochrome Oxidase is involved in both respiration and photosynthesis. It acts as electron carrier in the electron transport chain. Hexokinase is an enzymes which is also involved in, respiration. In glycolysis, it catalyses the first reaction, i.e., formation of glucose -6- phosphate from glucose molecule.

It uses one ATP molecule which transfers P04 group to glucose molecules.
NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants 10
Lactate Dehydrogenase is an enzyme which is involved in anaerobic respiration in bacteria Lactobacillus.

Pyruvic acid formed at the end of glycolysis is converted to lactic acid by the help of homo- fermentative lactic acid bacteria. Hydrogen from NADH molecule is transferred to pyruvate is then transferred to pyruvate molecule lactic acid molecule leading to the formation of acid.
NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants 11

Question 11.
How does a tree trunk exchange gases with the environment although it lacks stomata?
Solution:
The old tree trunk is covered by dead woody tissue called cork. The epidermal layers of such tree get ruptured and outer cortical cells are loosely arranged. These structures are called as lenticels.
These are the sites of gaseous exchange and transpiration.
NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants 12

Question 12.
Mention the important series of events of aerobic respiration that occur in the matrix of the mitochondrion as well as one that take place in inner membrane of the mitochondrion.
Solution:
Kreb ’ s cycle occurs in the matrix of mitochondria. It is given in the following series of reactions
NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants 13
Electron transport chain is carried out in the inner mitochondria membrane
The inner mitochondrial membrane is specific about possessing proton (H+) and electron (e ) acceptors in a particular sequence called electron transport chain. It consists four enzyme complexes.
The electrons either follow the pathway of complexes I, III and IV or II, III and IV that depends that upon the substrates from Kreb’s cycle.
Following are the ways through which the transfer of electrons and hydrogen atoms takes place.
Complex I It consists of flavoproteins of NADH dehydrogenase (FPN), of which FMN is the prosthetic group. It is combined with the flavoprotein is non-heme iron of NADH dehydrogenase. This complex spans inner mitochondrial membrane and is also able to translocate protons across it form matrix side to outer side.
Complex II It consists of flavoprotein of succinate dehydrogenase, of which FAD is the prosthetic group. It is combined with the flavoprotein is non-heme iron of succinate dehydrogenase.
Between complexes II and in the mobile carrier coenzyme-Q (Co-Q) or ubiquinone (UQ) is present
Complex III It consists of cytochrome-/) and cytochrome-c that is associated with cytochrome-h is non-heme iron of complex III. Between complexes III and IV is the mobile carrier cytochrome-c.
Complex IV It consists of cytochrome-a and cytochrome-a3, and bound copper that are required for this complex reaction to occur. This cytochrome also called cytochrome oxidase. It is the only electron carrier in which the heme iron has a free ligand that can react directly with molecular oxygen.
NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants 14

  1. Thus, hydride ions are transferred from the substance to be oxidised to NAD+. From NAD+ the hydrogen atoms are transferred to FMN of flavor protein 1 (Fp’N). After FMN the hydrogen atom splits into an electron and a proton.
  2. In further stages transfer of e~s occur but there is no longer a transfer of hydrogens. The electron passes to co-enzyme-Q, and from co-enzyme Q to cytochromes- b, cp c, a and ay The proton is released free.
  3. As the hydrogen atom or electron passes down by F0 – F1 particle at the same time oxidation of one coenzyme and reduction occurs at another steps. Oxygen is able to diffuse inside the mitochondria.
  4. It is converted to anionic form 02-, combines with 2H+ and forms metabolic water reduced co-enzyme NADH + H+ that helps in pushing out three pairs of H+ to outer chamber while FADH2 sends two pairs of H+ to outer chamber.
  5. Oxidative phosphorylation is the synthesis of ATP molecules, with the help of energy liberated during oxidation of reduced co-enzyme (NADH2, FADH2) produced in respiration.
  6. The enzyme required for this synthesis is called ATP synthase present in inner mitochondria membrane.

The following figures shows this process
NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants 15

LONG ANSWER QUESTIONS

Question 1.
Oxygen is critical for aerobic respiration. Explain its role with respect to ETS.
Solution:
The oxidation of glucose starts with glycolysis in cytoplasm which followed by Krebs’ cycle and finally Electron transport Chain (ETC) in inner mitochondrial membrane. The end of ETC 02 is required.
Where, it acts as final hydrogen acceptor. 02 is responsible for removing electrons from the system. In the absence of oxygen, electrons could not be passed through the co-enzymes, intum proton pump will not be established and ATP will not be produced via oxidative phosphorylation. Thus oxygen plays an important role in aerobic respiration in mitochondrial matrix.

Question 2.
Enumerate the assumptions that we undertake in making the respiratory balance sheet. Are these assumptions valid for a living system? Compare fermentation and aerobic respiration in this context.
Solution:
The assumption that we undertake is making the respiratory balance sheet one as follows:
(i) Respiratory substrate is glucose
(ii) There is sequential pathway i.e., glycolysis in cytoplasm, TCA cycle in mitochondrial matrix and ETS in inner mitochondriol membrane.
(iii) NADH synthesised in glycolysis enters into ETC for phosphorylation.
(iv) None of the intermediates in the pathway are utilised to synthesise any other compound.
These assumptions are not valid for a living system because of following reasons:
(i) Glycolysis, TCA and ETC work simultaneously and do not take place one after the other.
(ii) ATP is uutilised when needed.
(iii) Rate of enzyme actions are controlled by multiple means.
Comparison between fermentation and aerobic respiration in this context is as follows:
(i) Fermentation is partial breakdown of glucose whereas aerobic respiration is complete breakdown of glucose.
(ii) Net gain of only 2 ATP in fermentation whereas in aerobic respiration 38 ATP is produced.

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 14 Respiration in Plants, help you. If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 14 Respiration in Plants, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 11 Biology Chapter 13 Photosynthesis

NCERT Exemplar Solutions for Class 11 Biology Chapter 13 Photosynthesis

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 13 Photosynthesis.

VERY SHORT ANSWER QUESTIONS

Question 1.
Examine the figure.
NCERT Exemplar Solutions for Class 11 Biology Chapter 13 Photosynthesis 1
(b) Can these be passed on to the progeny? How?
(c) Name the metabolic process taking place in the places marked (A) and (B).
Solution:
(a) Figure shows the chloroplast, which is green
in colour and performs photosynthesis in plants. The structure is present in plant cell.
(b) Yes, chloroplast has the power of self replication because of presence of extra nuclear DNA. Hence, known as semi- autonomous organelle.
(c) The metabolic processes that occurs in the marked places are as follows.
A-It is the stroma of chloroplast, where dark reaction of photosynthesis takes place.
B-It is the structure of extra nuclear DNA that is responsible for replication of chloroplast, when it is required in the photo synthesising cells.

Question 2.
2H2O—>4H+ + O2 + 4e
Based on the above equation, answer the following questions
(a) Where does this reaction take place in plants?
(b) What is the significance of this reaction?
Solution:
(a) This reaction takes place in reaction centre
PS II, that is located on the inner surface of thylakoid membrane. It is known as water splitting centre, where electrons are extracted from water and the reaction is catalysed by Mn+ and Cl ions.
(b) Spliting of water is an important event in photosynthesis are
(i) It liberates molecular oxygen as byproduct of photosynthesis and is the significant source of oxygen in air, or is essential for all living beings on earth.
(ii) Hydrogen ions produced takes part in reducing NADP to NADPH. It is a strong reducing agent.
(iii) The electrons released are transferred from PS II to PS I through a series of electron carriers thus, creating a gradient for the ATP synthesis.

Question 3.
Cyanobacteria and some other photosynthetic bacteria don’t have chloroplasts. How do they conduct photosynthesis?
Solution:

  1. In Cyanobacteria complex lamellar system (thylakoids) are present instead of chloroplast. These thylakoids are functionally analogous to the plastids of eukaryotic cells.
  2. Pigment like chlorophyll-a, C-phycocyanin, C-phycoerythrin embedded in these lamellar system and they trap solar energy and perform photosynthesis. They perform oxygenic photosynthesis.
  3. Photosynthetic bacteria possess related pigments called bacterichlorophyll which are of different types (a,b,c,d,e,f and g). Groups that contain chlorophyll, perform photosynthesis, but do not evolve oxygen.
  4. Bacteriochlorophyll, perform photosynthesis but do not evolve oxygen.
  5. Bacteriochlorophylls are photoreceptors similar to chlorophylls except for the reduction of an additional pyrrole ring and other minor differences that shift their absorption maxima to near infrared, to wavelength as long as 1000 nm.
  6. Thus, they utilize light wavelengths not used by green plants or cyanobacteria.
  7. Bacteriopheophytin is a variant of bacteriochlorophyll that has two protons are present instead of magnesium ion at its centre.

Question 4.
(a) NADP reductase enzyme is located on ……..
(b) Breakdown of proton gradient leads to release of ……….. .
(b) ATP molecules
Solution:
(a) NADP reductase enzyme is located on the outer side of thylakoid membrane.
(b) ATP molecules

Question 5.
Can girdling experiments be done in monocots? If yes, how? If no, why note?
Solution:
The girdling experiment cannot be done in monocots. The monocots vascular bundles are scattered all over the width of stem, so we cannot get the specific band of the phloem tissue which we get in dicot.

Question 6.
3CO2, + 9ATP + 6NADPH + water –>Glyceraldehyde 3-phosphate + 9ADP + 6NADP+ + 8Pi.
Analyse the above reaction and answer the following questions
(a) How many molecules of ATP and NADPH are required to fix one molecule of CO2?
(b) Where in the chloroplast does this process occur?
Solution:
(a) 2 molecules of ATP for phosphorylation and
two molecules ofNADPH for reduction are required to fix one molecule of CO2 (b) The calvin cycle occurs in the stroma of the chloroplast.

Question 7.
Does moonlight support photosynthesis? Find out.
Solution:
Moonlight does not carry enough energy to excite chlorophyll molecules, i.e; reaction centre PSI and PSII, so light dependent reactions are not initiated. Thus, photosynthesis cannot occur in moonlight.

Question 8.
ATPase enzyme consists of two parts. What are those parts? How are they arranged in the thylakoid membrane? Conformational change occur in which part of the enzyme?
Solution:
ATP synthase enzyme consists of two parts:
(a) F1– head piece is a peripheral membrane protein complex and contain the site for synthesis of ATP from ADP + pi (inorganic phosphate).
(b) F0-integral membrane protein complex that form the channel through which proton cross the inner membrane.
The arrangement of F1 and F0 in thylakoid membrane is as follows.
F0– is a portion present within the thylakoid membrane.
F1 is a portion of ATP synthase enzyme present in the stroma of chloroplast.
The conformational change occurs in F1 portion of ATP synthase thus, it facilitates the ATP synthesis.

SHORT ANSWER QUESTIONS

Question 1.
Succulents are known to keep their stomata closed during the day to check transpiration. How do they meet their photosynthetic C02 requirements?
Solution:
Succulent plants grow in dry and xeric conditions so, to prevent water loss through transpiration the stomata remains closed during day time. So that the gaseous exchange does not take place.

Thus plants have developed the mechanism to fix C02 during night in the form of malic acid, which is a 4 carbon compound and are released during the day, inside the photosynthetic cells.

Question 2.
Chlorophyll-‘a’ is the primary pigment for the light reaction. What are accessory pigments? What is their role in photosynthesis?
Solution:
Accessory pigments are also photosynthetic pigments, like chlorophyll-/), xanthophyll and carotenoids which are not directly involved in emission of excited electron, but they help in harvesting solar radiation and pass it on to chlorophyll-o.

This pigment itself absorbs maximum radiation at blue and red region. So the chief pigment of photosynthesis is chlorophyll and others (i.e, chlorophyll-/; xanthophyll and carotenoid) are accessory pigments.

Question 3.
Do reactions of photosynthesis called, as ‘Dark Reaction’ need light? Explain.
Solution:
Dark reactions is a type of independent reactions. Through various biochemical reactions CO2 is reduced to produce C6H12O6 (glucose) which does not need light. But they depend on the products formed during light reactions, i.e., NADPH2 and ATP.

Question 4.
How are photosynthesis and respiration related to each other?
Solution:
Photosynthesis and respiration are related to each other as in both mechanisms, the plants gain energy.
In photosynthesis, plants gain energy from solar radiations whereas, in respiration, they break down glucose molecule to get energy in the form of ATP molecules.

The product of photosynthesis i.e., glucose (food) is utilised in respiration to yield energy in the form of ATP. While doing so, it release many other simple molecules (CO2 + H2O) that are utilised in photosynthesis to produce more sugar.

Question 5.
If a green plant is kept in dark with proper ventilation, can this plant carry out photosynthesis? Can anything be given as supplement to maintain its growth or survival?
Solution:
The plant in given conditions cannot carry out photosynthesis. Light is necessary for any green plant to make its own food. The plant should be watered properly for its survival.

Question 6.
Photosynthetic organisms occur at different depths in the ocean. Do they receive qualitatively and quantitatively the same light? How do they adapt to carry out photosynthesis under these conditions.
Solution:
Mostly algae are present at various depth in ocean. These show great variations in its photosynthetic pigment. These can absorb different wave lengths of light and could perform photosynthesis.
Green algae-chlorophyll-o, (absorbs red) and b(absorbs blue violet).
Brown algae-chlorophyll-o, c and fucoxanthin (absorbs yellow).
Rhodophyceae-chlorophyll-o, d and phyocoerythrin.

Question 7.
What conditions enable RuBisCO to function as an oxygenase? Explain the ensuing process.
(a) Can we artificially induce the property of nitrogen-fixation in a plant, leguminous or non leguminous?
(b) What kind of relationship is observed between mycorrhiza and pine trees?
(c) Is it necessary for a microbe to be in close association with a plant to provide mineral nutrition? Explain with the help of one example.
Solution:
This is an enzyme that has dual nature. When CO2 concentration is good enough in atmosphere. It acts as carboxylase. But if concentration of O2 increase, its nature changes and it binds with O2 and acts as oxygenase enzyme that forces CO2 to enter in C2 cycle that leads to photorespiration and loss of CO2.

Question 8.
Why does the rate of photosynthesis decrease at higher temperatures?
Solution:
Photosynthesis is an enzyme specific process. All enzymes works at an optimum temperature (/. e., 25-35°C). As temperature increases, enzyme gets denatured thus leading to fall in the rate of photosynthesis.

Question 9.
Explain how during light reaction of photosynthesis, ATP Synthesis is a chemiosmotic phenomenon.
Solution:

  1. In light reaction plants solar radiation is trapped by photosynthetic pigments, which converts light energy into chemical energy.
  2. Photophosphorylation is the main event of light reaction i.e., formation of ATP from ADP + Pi by using energy of excited electron movement through electron transport chain, that is present in thylakoid membrane.
  3. The movement of ions across a selectively permeable membrane, down the electrochemical/ proton gradient is known as chemiosmosis.
  4. Chemiosmosis hypothesis of ATP formation was first proposed by Mitchell (1961), according to ATP generated by enzyme ATP synthase via a membrane, proton pump and proton gradient.
  5. ATP synthase allows ions 02 protons to pass through membrane and proton pump.
  6. Which creates a high concentration of protons (H+) in the lumen and hence diffuses across the membrane to activate ATPase, releasing ATP molecules. One molecule of ATP is released for every two (H+) ions passing through ATPase.
    NCERT Exemplar Solutions for Class 11 Biology Chapter 13 Photosynthesis 2

Question 10.
In What kind of plants do you come across ‘Kranz anatomy’? To which conditions are those plants better adapted? How are these plants better adapted than the plants, which lack this anatomy?
Solution:

    1. Kranz anatomy how dimorphism in the chloroplast structure. It is found in C4 plants. The cells of leaves consists two types of chloroplast in them.
    2. Granal Chloroplast is found in the mesophyll cells of leaves.
    3. Chloroplast have well developed grana in them. These chloroplast fixes C02 effectively even if it is present in lower concentrations.
    4. PEP carboxylase fixes CO2 to form oxaloacetic acid (4 carbon compound).
    5. Agranal Chloroplast is found in bundle sheath cells of the leaves. C3 cycle occurs in these cells in the presence of RuBisCo enzyme.
    6. The C4 plants are well adapted to high O2 concentrations and high temperature.
      C4 plants can absorb CO2 even when CO2 concentration in much low thus C4 plants can perform high rate of photosynthesis even the stomata are closed or there is the shortage of water thus, they can conserve water.
    7. Since, PEP-carboxylase is insensitive to O2 thus excess O2 has inhibitory effect in C4 pathway and no photosynthesis occurs in C4 plant.
    8. Thus, C4 plants are better adapted to tropical and desert (hot acid habitats) areas than the plants, that lack kranz anatomy.

Question 11.
Tomatoes, carrots and chilies are red in colour due to the presence of one pigment. Name the pigment. Is it a photosynthetic pigment?
Solution:
The pigments are chromoplasts, these are fat soluble carotinoid pigments like carotenes and xanthophylls. These are called accessory pigments, they absorb light and transfer energy to Chlorophyll a.

Question 12.
Observe the diagram and answer the following.
NCERT Exemplar Solutions for Class 11 Biology Chapter 13 Photosynthesis 3
(a) Which group of plants exhibit these two types of cells?
(b) What is the first product of C4 cycle?
(c) Which enzyme is there in bundle sheath cells and mesophyll cells?
Solution:
(a) Monocot plants that belongs to Graminae/
Poaceae family, e.g., sugarcane, maize etc., possess these two types of cells, i.e., bundle sheath and mesophyll cell (in kranz anatomy).
(b) First product of C4 cycle is 4-carbon compound oxaloacetic acid.
(c) Mesophyll cells consists PEP carboxylase enzyme to fix atmospheric C02 to form 4-carbon compound oxalo acetic acid, whereas bundle sheath cells consists RuBP carboxylase that fixes C02 to form 3-carbon compound 3 PGA (3 phosphoglyceric acid).

LONG ANSWER QUESTIONS

Question 1.
In the figure given below, the back line (upper) indicates action spectrum for photosynthesis and the lighter line (lower) indicates the absorption spectrum of chlorophyll-a, answer the following
NCERT Exemplar Solutions for Class 11 Biology Chapter 13 Photosynthesis 4
(a) What does the action spectrum indicate? How can we plot an action spectrum? Explain with an example.
(b) How can we derive an absorption spectrum for any substances?
(c) If chlorophyll-a is responsible for light reaction of photosynthesis, why do the action spectrum and absorption spectrum not overlap?
Solution:
(a) Action spectrum depicts the relative rates of
photosynthesis at different wavelenghths of light. Action spectrum of photosynthesis can be plotted by measurement of oxygen evolution at different wavelength. Englemann (1882) by using a green algae plotted action spectrum.
(b) Absorption spectrum of a substance can be derived by calculating amount of energy of different wavelength of light absorbed.
(c) Chlorophyll a is responsible for light reaction of photosynthesis, but the action spectrum and absorption spectrum do not overlap because, though chlorophyll a is the main pigment responsible for the absorption of light, other thylakoids pigment like chlorophyll b, xanthophylls, carotenoids,

which are accessory pigments also absorb and transfer the energy to chlorophyll a. Indeed they not only enable a wider range of wavelength of incoming light to be utilized for photosynthesis but also protect chlorophyll from photo-oxidation.

Question 2.
What are the important events and end products of the light reaction?
Solution:
Following are important events of light reaction:
(i) Excitation of chlorophyll molecule to release a pair of electrons and use their energy in the formation of ATP from ADP + Pi. This process is known as photophosphorylation.
(ii) Splitting of water molecule
(a) 2H2O —» 4H+ + 4e + 0+
(b) NADP + 2H+ ->NADPH2
End products of light reaction are NADPH and ATP. Reducing power is produced in the light reaction i.e., ATP and NADPH2 molecules that are used up in dark reaction and O2 is evolved as a by product by the splitting of water.

Question 3.
Why does not photorespiration take place in C4 plants?
Solution:

  • Photorespiration is associated with C3 cycle, where plant lose CO2 fixation due to the increase in concentrate ion of O2 change in the nature of activity of RuBP carboxylase-oxygenase.
  • While C4 plants have evolved a mechanism to avoid loss of CO2.
  • There is not a direct involvement of RuBP carboxylase-oxygenase as C3 cycle operates in bundle sheath cells, where both temperature and oxygen level low.
  • CO2 fixation occurs by another enzyme PEP carboxylase in mesophyll cells andoxaloacetate is formed, that is later converted into malic acid and transported to bundle sheath cells.
  • There, it liberates CO2, which is used in Calvin cycle, operating in bundle sheath cells of C4 plants.
  • We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 13 Photosynthesis, help you.
  • If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 13 Photosynthesis, drop a comment below and we will get back to you at the earliest.

 

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 13 Photosynthesis, help you. If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 13 Photosynthesis, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition

NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition.

VERY SHORT ANSWER QUESTIONS

Question 1.
Name a plant, which accumulate silicon.
Solution:
Oryza sativa and Triticum aestivum are the plants that accumulates silicon. These plants absorbs silicon actively and accumulate them in their biomass.

Question 2.
Mycorrhiza is a mutualistic association. How do the organisms involved in this association gain from each other?
Solution:
Mycorrhiza is a mutualistic (symbiotic) association between fungus and roots of plants. The roots provide shelter and food to the fungus and the fungus helps plants in absorption of minerals, water uptake and protection against fungus.

Question 3.
Nitrogen fixation is shown by prokaryotes and not eukaryotes. Comment.
Solution:
Prokaryotes like Rhizobium and Anabaena are capable of nitrogen fixation as they contain enzyme nitrogenase but eukaryotes lack this enzyme.

Question 4.
A farmer adds Azotobacter culture to soil before sowing maize. Which mineral element is being replenished?
Solution:
Azotobacter provides nitrogen fixing bacteria which converts free nitrogen into nitrate and nitrites. It increases soil fertility.

Question 5.
What type of conditions are created by leghaemoglobin in the root nodule of a legume?
Solution:
Leghaemoglobin present in the root nodules of leguminous plants is responsible for creating anaerobic conditions and hence acts as an oxygen scavenger, protecting enzyme nitrogenase to come in contact with oxygen and help in the proper functioning of enzyme, i. e., conversion of atmospheric nitrogen to ammonia (NHj).

Question 6.
Yellowish edges appear in leaves deficient in.
Solution:
Yellowish edges or chlorosis appears in the leaves due to the deficiency of nitrogen. Its deficiency also causes delaying of flowering, interference in protein synthesis and dormancy of lateral buds.

Question 7.
Name the macronutrient which is a component of all organic compounds but it not obtained from soil.
Solution:
Carbon is an essential macronutrient, which is a component of all organic compounds but is not obtained by soil. Plant take it from atmosphere in the form of C02. Its concentration in atmosphere is about 0.03%. Plants use C02 for photosynthesis (as a source of carbon) to synthesises glucose.

Question 8.
Name one non-symbiotic nitrogen fixing prokaryote.
Solution:
Azotobacter is a non-symbotic nitrogen fixing prokaryote. It flourishs in the rice fields.

Question 9.
Complete the equation for reductive amination …….. .
NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 1
Solution:
NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 2

Question 10
Excess of Mn in soil leads to deficiency of Ca, Mg and Fe. Justify.
Solution:
When higher amounts of Mn2+ is absorbed by plants. The toxicity expressed in the form of brown sports surrounded by chlorotic vein.
It is due to the following reasons
(i) Reduction in uptake of Fe3+ and Mn2+.
(ii) Inhibition of binding of Mn2+ to specific enzymes.
(iii) Inhibition of Ca2+ translocation in shoot apex.
Thus, excess of Mn2+ causes deficiency of iron, magnesium and calcium.

SHORT ANSWER QUESTIONS

Question 1.
How is sulphur important for plants? Name the amino acids in which it is present.
Solution:
Sulphur is a macronutrient that is important for normal plant growth and development. It is also an integral part of some amino acids, proteins and helps in deciding the secondary structure of proteins as it forms disulphide bonds.

It is absorbed by the plants as SO42- ion. It is present in vitamins (biotin, thiamine), proteins, coenzyme-A, amino acid (cystein and methionine) etc. It is also an essential component of plants like (onion, garlic) and mustard.

Its deficiency causes chlorosis in young leaves, extensive root growth, formation of hard and woody stem. It also causes the reduction in juice content of citrus fruit and tea yellow disease of tea.

Question 2.
How are organisms like Pseudomonas and Thiobacillus of great significance in nitrogen cycle?
Solution:
In biological nitrogen fixation, the atmospheric N2 gets reduced to NH3 by the help of enzyme nitrogenase reductase present in some prokaryotes. NH3 is then oxidised in to N02 and NO3 by some other bacteria (Nitrosomonas and Nitrobacter) following are the various steps involved in nitrogen fixation.
NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 3
Pseudomonas and Thiobacillus are involved in the process of denitrification. They convert nitrate (NO3) and nitrite (NO2) into free nitrogen (N2), that is released into the atmosphere.

Question 3.
Carefully observe the following figure
NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 4
(a) Name the technique shown in the figure and the scientist who demonstrated this technique for the first time.
(b) Name atleast three plants for which this technique can be employed for their commercial production.
(c) What is the significance of aerating tube and feeding funnel in this setup?
Solution:
(a) Hydroponics,’Julius Von Sachs (1860)
(b) (i) Solanum lycopersicum (tomato)
(ii) Hibiscus asculentus (ladiesfinger)
(iii) Solanum melongena (brinjal)
(c) Aerating tube provides oxygen for the normal growth and development of the roots growing in the liquid solution. Feeding funnel is used to add water and nutrients in the hydroponic system when required.

Question 4.
Name of most crucial enzyme found in root nodules for N2-fixation? Does it require a special pink coloured pigment for its functioning? Elaborate.
Solution:
The most crucial enzyme found in the root nodules for N2-fixation is nitrogenase. It is a Mo – Fe protein that catalyses the conversion of atmospheric nitrogen to ammonia. Pink colour 8.
pigment present is root nodules of leguminous plants is called leghaemoglobin creates * anaerobic conditions for the functioning of nitrogenase enzyme.

Question 5.
Carnivorous plants exhibit nutritional adaption. Citing an example explain this fact.
Solution:
Carnivorous plants fulfill their nutritional requirements by feeding on small animals, like insects or protozoans, e.g. Nepenthes, Venus fly trap, Utricularia etc. Carnivorous plants grow in soil deficient in nitrogen.

In pitcher plant leaves are modified into pitcher which stores the juice to lure an insect. When the insect come to suck this juice, chemicals present in nectar dissolve the skin of the prey and the plant obtains nutrients (mainly nitrogen) from its skin.

Question 6.
A farmer adds/supplied Na, Ca, Mg and Fe regularly to his field and yet he observes that the plants show deficiency of Ca, Mg and Fe. Give a valid reason and suggest a way to help the farmer improve the growth of plants.
Solution:
Plant can tolerate a specific amount of micronutrients. A lesser amount of micro- nutrient can cause deficiency symptoms and higher amount can cause toxicity.

The concentration of mineral ion which reduces the dry weight of the tissues by 10% is called toxic concentration. This concentration is different for different micronutrients as well as for different plants e.g., Mn2+ is toxic beyond 600 mgg -1; (for soyabean) and (for sunflower) and beyond 5300 μgg-1.

It has also been observed that the toxicity of one micronutrient causes the deficiency of other nutrients.
To overcome such problems, farmers should use these nutrients in prescribed concentration so that the excess uptake of one element do not reduce the uptake of the element.

Question 7.
We find that Rhizobium forms nodules on the roots of leguminous plants. Also Frankia another microbe forms nitrogen fixing nodules on the roots of non-leguminous plant Alnus.
(a) Can we artificially induce the property of nitrogen-fixation in a plant, leguminous or non leguminous?
(b) What kind of relationship is observed between mycorrhiza and pine trees?
(c) Is it necessary for a microbe to be in close association with a plant to provide mineral nutrition? Explain with the help of one example.
Solution:
(a) Artificial induction in leguminous and non- leguminous plants have been tried by scientists. It’s success rate is very low because expression of gene is highly specific phenomenon. When it desired gene is introduced that may not work because conditions for its expressions are very specific.
(b) Symbiotic mutualistic relationship (mutualism) is observed between the pine roots and mycorrhiza as both are benefitted mutually.
(c) Yes it is necessary for a microbe to be in close association with plant to provide mineral nutrition, to develop a physical relationship for example Rhizobium gets into the root and involve root tissues, then only helps in nitrogen-fixation.

Question 8.
With the help of examples describe the classification of essential elements based on the function they perform.
Solution:
Based on the diverse functions of essential elements, these are categorised into following categories given below:
(i) Constituent of biomolecules: These are the essential component of biomolecules. Hence, known as structural elements of cells, e.g., carbon, hydrogen, oxygen and nitrogen.
(ii) Energy related Chemical compound: Some elements also function in providing energy to the cell e.g. phosphorus is a component of ATP which function as energy currency -of all the living system in which magne¬sium is a component of chlorophyll, which is involved in the conversion of light en¬ergy to chemical energy.
(iii) Enzyme showing catalytic effects: Many of the essential elements are required in the form of cofactors by enzymes. They function as the activator or inhibitor of enzymes, e.g., Mg2+ acts as an activator of several enzymes in both photosynthesis e.g., Ribulose bisphosphate(RuBP), Car boxylase , Phosphoenol pymvate carboxy¬lase and respiration (e.g., hexokinase and phosphofructokinase). While Zn2+ acts as an activator of alcohol dehydrogenase while Mo of nitrogenase during the course of nitrogen fixation.

Question 9.
Trace the events starting from the coming in contact of Rhizobium to a leguminous root till nodule formation. Add a note on importance of leghaemoglobin.
Solution:
Formation of Root Nodule The coordinated activities of the Rhizobiam bacteria regume depends on the chemical interaction between these symbiotic partners.
In the following diagram the principle stages in the nodule formation are summarised.
NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 5
Leghaemoglobin is an oxygen scavenger, that protects nitrogen enzyme from 02 and also creates anaerobic conditions for the reduction of N2 to NH3 by Rhizobium bacteria.

Question 10.
Give the biochemical events occurring in the root nodule of a pulse plant. What is the end product? What is its fate?
Solution:
Formation of root nodule in pulse plant is the result of infection of roots by Rhizobium. The following figure shows the process of nodule formation.
NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 6
(b) Successful infection of the root hair causes it to curl
NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 7
(c) Infected thread carries the bacteria to enter the cortex. Bacteria cause cortical and pericycle cells to divide, lead to nodule formation
NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 8
(d) Mature nodule with vascular tissues continuous with those of the roots.
The chemical reaction is as follows
N2 + 8e+ 8H++ 16 ATP ->2NH3 + H2 + 16ADP + P1i
The reaction takes place in the presence of enzyme nitrogenase that acts in anaerobic conditions, which is created by leghaemoglobin.
Fate of Ammonia
There are two ways by which ammonia is further used.
NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 9
This reaction and transfer of NH2 group take place for amino acid to other amino acid catalysed by enzyme transaminase.

Question 11.
Hydroponics have been shown to be a successful technique for growing of plants. Yet most of the crops are still grown on land. Why?
Solution:
Hydroponics is a soil less culture and successful technique for plants, still many crops are grown on land because
(i) The major concern is its cost. The setting and handling of hydroponics requires much more investment than that of the soil based production.
(ii) Sanitization is extremely important, because especially with indoor hydroponic environments. Water borne disease can spread quickly through some methods of hydroponic production.
(iii) It is relatively a new technique and not used by the traditional farmers due to lack of knowledge.
(iv) Plants are less adaptable to the surrounding atmosphere. However weather and narrow oxygenation may minimise the production and quality of plant yield.

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 12 Mineral Nutrition, help you. If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 12 Mineral Nutrition, drop a comment below and we will get back to you at the earliest.