NCERT Solutions For Class 11 Biology Biomolecules

NCERT Solutions For Class 11 Biology Biomolecules

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules:

Section Name Topic Name
9 Biomolecules
9.1 How to Analyse Chemical Composition?
9.2 Primary and Secondary Metabolites
9.3 Biomacromolecules
9.4 Proteins
9.5 Polysaccharides
9.6 Nucleic Acids
9.7 Structure of Proteins
9.8 Nature of Bond Linking Monomers in a Polymer
9.9 Dynamic State of Body Constituents – Concept of Metabolism
9.10 Metabolic Basis for Living
9.11 The Living State
9.12 Enzymes
9.13 Summary

NCERT Solutions Class 11 BiologyBiology Sample Papers

NCRT TEXTBOOK QUESTIONS SOLVED

1. What are macromolecules? Give examples.
Solution: Macromolecules are large high molecular weight substances with complex molecular structure and occur in colloidal state (being insoluble) in intracellular fluid. These are formed by polymerization of large number of micromolecules. Polysaccharides, proteins and nucleic acids are few examples.

2. Illustrate a glycosidic, peptide, and a phospho- diester bond.
Solution. (i) Glycosidic bond is the type of chemical linkage between the monosaccharide units of disaccharides, oligosaccharides and polysaccharides, which is formed by the removal of a molecule of water.
NCERT Solutions For Class 11 Biology Biomolecules Q2
(ii)Peptide bonds are formed by the reaction between carboxyl (- COOH) of one amino acid and amino (- NH2) group of other amino acid with the elimination of water.
NCERT Solutions For Class 11 Biology Biomolecules Q2.1
(iii) In a polynucleotide chain, adjacent nucleotides are joined together by a bond called phosphodiester bond. This bond links a phosphate group and sugar group of two adjacent nucleotides by means of an oxygen bridge.
NCERT Solutions For Class 11 Biology Biomolecules Q2.2

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3. What is meant by tertiary structure of proteins?
Solution: The helical polypeptide molecule may fold on itself and assume a complex but specific form-spherical, rod-like or any form in between these. These geometrical shapes,are known as tertiary (3°) structure of protein molecules. The coils and folds of the polypeptide molecules are so arranged as to hide the non-polar amino acid chains inside and to expose the polar side chains. The tertiary structure of a protein brings distant amino acid side chains nearer to form active sites of enzymatic proteins. The tertiary structure is maintained by weak bonds such as hydrogen, ionic, disulphide and hydrophilic – hydrophobic bonds, formed between one part of a polypeptide and another. This structure is easily disrupted by pH, temperature and chemicals stopping the function of proteins.

The protein molecular weight calculator is a simple tool for computing the molecular mass of a protein using a list of its amino acids.

4. Find and write down structures ©f 10 interesting small molecular weight biomolecules.
Solution: Interesfing small molecular weight biomolecules are minerals (like sodium, potassium, calcium, zinc, iodine etc), gases (like Oz, N2, C02, NH3) sugars – (ribose, deoxyribose, glucose, fructose), lipids, amino acids, nucleotides (pyrimidines & purine). Structures of 10 interesting small molecular weight biomolecules are as follows:
NCERT Solutions For Class 11 Biology Biomolecules Q4

NCERT Solutions For Class 11 Biology Biomolecules Q4.1

NCERT Solutions For Class 11 Biology Biomolecules Q4.2

NCERT Solutions For Class 11 Biology Biomolecules Q4.3

5. Proteins have primary structure. If you are given a method to know which amino acid is at either of two termini (ends) of a protein, can you connect this information to purity or homogeneity of a protein?
Solution: There are several methods provided by several scientists to find out the sequence of amino acids. Frederick Sanger proposed Sanger’s reagent to know the amino acid sequence in a polypeptide chain.
Sanger used 1-fluoro 2, 4 dinitrobenzene (FD NB) to determine insulin structure. FDNB specifically binds with N-terminal amino acid to form a dinitrophenyl (DNP) derivative of peptide. This DNP- derivative peptide can be identified by chromatography. The identified sequence of amino acids shows the homogeneity of a protein molecule.

6. Find out and make a list of proteins used as therapeutic agents. Find other applications of proteins.
Solution: Proteins used as therapeutic agents are: thrombin, fibrinogen, enkephalins, antigens, antibodies, streptokinase, protein tyrosine kinase, diastase, renin, insulin, oxytocin, vasopressin etc. Proteins are also used in cosmetics, dairy industries, textile industries, research techniques, biological buffers etc.

7. Explain the composition of triglycerides. jSfflTriacylglycerols (triglycerides) are the esters of glycerol with fatty acids.
Solution: They are insoluble in water and non-polar in character and commonly known as neutral fats. The neutral or depot fats are composed of carbon, hydrogen and oxygen like carbohydrates but have far fewer oxygen atoms than carbon atoms unlike the carbohydrates.
NCERT Solutions For Class 11 Biology Biomolecules Q7
(i) Glycerol – A glycerol molecule has 3
carbons, each bearing a hydroxyl (-OH) group. .
(ii) Fatty acids – A fatty acid molecule is an unbranched chain of carbon atoms with each carbon atom (C) forming four bonds to other atoms. It has a carboxyl group- COOH at one end and hydrogen atom (H) bonded to all or most carbon atoms forming a hydrogen chain. The carbon- hydrogen bonds are non-polar. Therefore, the hydrocarbon chain does not dissolve in water. Because the carboxyl group contains the polar C = O and OFI groups. It tends to dissolve in water even though the rest of fatty acid molecule will not. Triacylglycerols of plants, in general, have higher content of unsaturated fatty acids as compared to that of animals.

8. Can you describe what happens when milk is converted into curd or yoghurt, from your understanding of proteins.
Solution: Milk is converted into curd or yoghurt due to denaturation of proteins. In denaturation, disruption of bonds that maintains secondary and tertiary structure leads to the conversion of globular proteins into fibrous proteins. This involves a change in physical, chemical and biological properties of protein molecules.

9. Can you attempt building models of biomolecules using commercially available atomic models (Ball and stick models).
Solution: Yes, models of biomolecules can be prepared using commercially available atomic models.
Ball and stick models and space filling models are 3D or spatial molecular models which serve to display the structure of chemical products and substances or biomolecules. With ball and stick models, the centers of the atoms are connected by straight lines which represent the covalent bonds. Double and triple bonds are often represented by springs which form curved connections between the balls. The bond angles and bond lengths reflect the actual relationships, while the space occupied by the atoms is either not represented at all or only denoted essentially by the relative sizes of the spheres.

10. Attempt titrating an amino acid against a weak base and discover the number of dissociating (ionizable) functional groups in the amino acid.
Solution: The existence of different ionic forms of amino acids can be easily understood by the titration curves. The number of dissociating functional group is one in case of neutral and basic amino acids and two in case of acidic amino acids.

11. Draw the structure of the amino acid, alanine.
Solution:
NCERT Solutions For Class 11 Biology Biomolecules Q11

12. What are gums made of ? Is fevicol different ?
Solution: Gums are hetero-polysaccharides (poly-mers) of large number of different monosac-charide units. Yes, fevicol is a different kind of polymer. It is a synthetic sticky substance called resin which is manufactured by esteri-fication of organic compounds.

13. Find out a qualitative test for proteins, fats and oils, amino acids and test, any fruit juice, saliva, and urine for them.
Solution: Biuret test for protein : The biuret test is a chemical test used for determining the presence of peptide bonds. In a positive test, a copper II ion (Cu2+ ion) is reduced to copper I (Cu+) which forms a complex with the nitrogen and carbon of peptide bonds in an alkaline solution. A violet colour indicates the presence of proteins.
Ninhydrin test for amino acid: Ninhydrin (2,2 Dihydroxy indane-l,3-dione) is a chemical used to detect ammonia or primary and secondary amines. When reacting with these free amines, a deep blue or purple colour known as Ruhemann’s purple is evolved. Amino acid analysis of proteins is also done by ninhydrine. Most of the amino acids (including a-amino acids) are hydrolysed and reacted with ninhydrin except proline (a secondary amine). Amino acid containing a free amino group and a free carboxylic acid group reacts together with ninhydrin to produce coloured product. When the amino group is secondary, the condensation product is yellow.
NCERT Solutions For Class 11 Biology Biomolecules Q13
Solubility test for fats and oils : A positive solubility test for fats is that the fat dissolves in lighter fluid and not in water. In this test, 5 drops of fat or oil are added in two test tubes containing 10 drops of lighter fluid and 10 drops cold water respectively.
Fruit juice contains sugar so it cannot be tested by the above-mentioned tests. Saliva contains proteins, mineral salts, amylase etc., so it can be tested for protein and amino acids. Urine contains proteins, so it can be tested for it.

14. Find out how much cellulose is made by all the plants in the biosphere.
Solution: About 100 billion tonnes of cellulose is prepared per year by the plants of the world.

15. Describe the important properties of enzymes.
Solution: The important properties of enzymes are as follows:
(i) The enzymes are generally proteins which are high molecular weight complex globular proteins. They can associate with non-protein substance for their activity.
(ii) The enzymes do not start a chemical reaction but only accelerate it. They combine temporarily with the substrate molecules and are not consumed or changed permanently in the reaction which they catalyse.
(iii) The enzyme controlled reactions are reversible.
(iv) The enzymes are specific in action. An enzyme catalyses only a particular kind of reaction or acts on a particular substrate only.
(v) The enzymes are thermolabile i.e., heat sensitive and can function best at an optimum temperature. Similarly, enzymes show maximum activity at optimum pH.
(vi) The enzymes are inactivated by poisons and radiation.

NCERT Exemplar Class 11 Biology Chapter 22 Chemical Coordination and Integration

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NCERT Exemplar Class 11 Biology Chapter 22 Chemical Coordination and Integration

Multiple Choice Questions

Q1. Select the right match of endocrine gland and their hormones among the options given below.

A. Pineal (i) Epinephrine
B. Thyroid (ii) Melatonin
c: Ovary (iii) Estrogen
D. Adrenal medulla (iv) Tetraiodothyronine

Options:
(a) A—(iv), B—(ii), C—(iii), D—(i)
(b) A—(ii), B—(iv), C—(i), D—(iii)
(c) A—(iv), B—(ii), C—(i), D—(iii)
(d) A—(ii), B—(iv), C—(iii), D—(i)

Ans:(d)

A. Pineal (ii) Melatonin
B. Thyroid (iv) Tetraiodothyronine
C. Ovary (iii) Estrogen
D. Adrenal medulla (i) Epinephrine

Q2. Which of the following hormones is not secreted by anterior pituitary?
(a) Growth hormone (b) Follicle stimulating hormone
(c) Oxytocin (d) Adrenocorticotrophic hormone
Ans: (c) The pars distalis region of pituitary, commonly called anterior pituitary, produces 6 Growth Hormone (GH), Prolactin (PRL), Thyroid Stimulating Hormone (TSH), Adrenocorticotrophic Hormone (ACTH), Luteinizing Hormone (LH) and Follicle Stimulating Hormone (FSH). Neurohypophysis (pars nervosa) also known as posterior pituitary, stores and releases two
hormones called oxytocin and vasopressin, which are actually synthesised by the hypothalamus and are transported axonally to neurohypophysis.

Q3. Mary is about to face an interview. But during the first five minutes before the interview she experiences sweating, increased rate of heart beat, respiration, etc. Which hormone is responsible for her restlessness?
(a) Estrogen and progesterone (b) Oxytocin and vasopressin
(c) Adrenaline and noradrenaline (d) Insulin and glucagon
Ans:(c) Mary is about to face an interview. But during the first five minutes before the interview she experiences sweating, increased rate of heart beat, respiration, etc. Adrenaline and non-adrenaline hormone are responsible for her restlessness.

Q4. The steroid responsible for balance of water and electrolytes in our body is
(a) Insulin (b) Melatonin (c) Testosterone (d) Aldosterone
Ans: (d) Vasopressin acts mainly at the kidney and stimulates resorption of water and electrolytes by the distal tubules and thereby reduces loss of water through urine (diuresis). Hence, it is also called as anti-diuretic hormone (ADH).

Q5. Thymosin is responsible for
(a) Raising the blood sugar level (b) Raising the blood calcium level
(c) Differentiation of T lyipphocytes (d) Decrease in blood RBC
Ans: (c) Thymosin is responsible for differentiation of T-lymphocytes.

Q6. In the mechanism of action of a protein hormone, one of the second messengers is
(a)    Cyclic AMP  
(b)   Insulin             
(c)    T3                   
(d)  Gastrin
Ans: (a) Hormones which interact with membrane-bound receptors normally do not enter the target cell, but generate secondary messengers (e.g., cyclic AMP, cGMP, DAG, IP3, Ca++ etc.) which in turn regulate cellular metabolism.

Q7. Leydig cells produce a group of hormones called
(a) Androgens (b) Estrogens
(c) Aldosterone (d) Gonadotropins
Ans: (a) Leydig cells produce a group of hormones called androgens.

Q8. Corpus luteum secretes a
(a) Prolactin (b) Progesterone (c) Aldosterone (d) Testosterone
Ans: (b) Corpus luteum secretes progesterone.

Q9. Cortisol is secreted from gland called
(a) Pancreas (b) Thyroid (c) Adrenal (d) Thymus
Ans: (c) Cortisol is secreted from gland called adrenal.

Q10. A hormone responsible for normal sleep-wake cycle is
(a) Epinephrine (b) Gastrin (c) Melatonin (d) Insulin
Ans: (c) A hormone responsible for normal sleep-wake cycle is melatonin.

Q11. Hormones are called chemical signals that stimulate specific target tissues. Which is the correct location of these receptors in case of protein hormones?
(a) Extra cellular matrix
(b) Blood
(c) Plasma membrane
(d) Nucleus
Ans: (c) Protein hormone receptors present on the plasma membrane of the target cells are called membrane-bound receptors.

Q12. Choose the correct option among the following:

Column A Column B
A. Epinephrine (i) Stimulates in muscle growth
B. Testosterone (ii) Decrease in blood pressure
C. Glucagon (iii) Decrease in liver glycogen content
D. Atrial natriuretic factor (iv) Increases heart beat

Options:
(a) A—(ii), B—(i), C—(iii), D—(iv)
(b) A—(iv), B—(i)„ C—(iii), D—(ii)
(c) A—(i), B—(ii), C—(iii), D—(iv)
(d) A—(i), B—(iv), C—(ii), D—(iii)

Ans. (b)

A. Epinephrine (iv) Increases heart beat
B. Testosterone (0 Stimulates in muscle growth
C. Glucagon (iii) Decrease in liver glycogen content
D. Atrial natriuretic factor (ii) Decrease in blood pressure

Very Short Answer Type Questibns
Q1.There are many endocrine glands in human body. Name the glands which is absent in male and the one absent in female.
Ans: In Males—Ovary and in Females—Testis.

Q2. Which of the two adrenocortical layers, zona glomerulosa and zona reticularis lies outside enveloping the other?
Ans: Outer layer—Zona glomerulosa
Inner layer—Zona reticularis

Q3. What is erythropoiesis? Which hormone stimulates it?
Ans: Formation of RBC is known as erythropoiesis and the hormone erythropoietin stimulates the process.

Q4. Name the only hormone secreted by pars intermedia of the pituitary gland.
Ans: MSH or Intermedin

Q5. Name the endocrine gland that produces calcitonin and mention the role played by this hormone.
Ans: Thyroid gland also secretes a protein hormone called thyrocalcitoni’n (TCT) which regulates the blood calcium levels. TCT is secreted by ‘C’ cells of thyroid glands. TCT is a hypocalcaemic hormone which lower the blood calcium level by increasing calcium deposition in the bones, so checks osteoporosis.

Q6. Name the hormone that helps in cell-mediated immunity.
Ans: Thymosin.

Short Answer Type Questions
Q1. What is the role-played by luteinizing hormones in males and females respectively?
Ans: LH stimulates the synthesis and secretion of androgens called male hormones. In females, LH is essential for ovulation. In females, LH induces ovulation of fully mature follicles (graafian follicles) and maintains the corpus luteum formed from the remnants of the graafian follicles after ovulation.

Q2. What is the role of second messenger in hormone action?
Ans: Hormones which do not enter the target cell, interact with specific receptors located on the surface of the target cell membranes and generates second messengers (e.g., cAMP) on the inner side of plasma membrane. The second messenger, in turn, carries out all the hormonal functions.

Q3. On an educational trip to Uttaranchal, Ketki and her friends observe that many local people were having swollen necks. Please help Ketki and her friends to find out the solutions to the following questions.
a. Which probable disease are these people suffering from?
b. How is it caused?
c. What effect does this condition have on pregnancy?
Ans: a. Goitre ‘
b. Iodine deficiency in diet
c. Hypothyroidism during pregnancy causes defective development and maturation of the growing baby leading to stunted growth (cretinism), mental retardation, low intelligence quotient, abnormal skin, deaf- mutism, etc.

Q4. 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?
Ans: George comes on a vacation to India from US. The long journey disturbs his biological system and he suffers from jet lag. It is due to the disturbance in diurnal rhythm. Melatonin plays a very important role in the regulation of a 24-hour (diurnal) rhythm of our body. For example, it helps in maintaining the normal rhythms of sleep-wake cycle, body temperature.

Q5. Inflammatory responses can be controlled by a certain steroid. Name the steroid, its source and also its other important functions.
Ans: Glucocorticoids, particularly cortisol, produces anti-inflammatory reactions and suppresses the immune response. Cortisol stimulates the RBC production. Glucocorticoids stimulate gluconeogenesis, lipolysis and proteolysis; and inhibit cellular uptake and utilisation of amino acids. Cortisol is also involved in maintaining the cardio-vascular system as well as the kidney functions.

Q6. Old people have weak immune system. What could be the reason?
Ans: Thymus is degenerated in old individual resulting in a decreased production of thymosins. As a result the immune responses of old persons become weak.

Q7. What are the effects of hypothyroidism (observed during pregnancy) on the development and maturation of a growing baby?
Ans: Hypothyroidism during pregnancy causes defective development and maturation of the growing baby leading to stunted growth (cretinism), mental retardation, low intelligence quotient, abnormal skin, deaf-mutism, etc.

Q8. Mention the difference between hypothyroidism and hyperthyroidism.
Ans: Iodine is essential for the normal rate of hormone synthesis in the thyroid.
Deficiency of iodine in our diet results in hypothyroidism and enlargement of the thyroid gland, commonly called goitre. Hypothyroidism during pregnancy causes defective development and maturation of the growing baby leading to stunted growth (cretinism), mental retardation, low intelligence quotient, abnormal skin, deaf-mutism, etc. In adult women, hypothyroidism may cause menstrual cycle to become irregular. Due to cancer of the thyroid gland or due to development of nodules of the thyroid glands, the rate of synthesis and secretion of the thyroid hormones is increased to abnormal high levels leading to a condition called hyperthyroidism which adversely affects the body physiology.

Q9. You have learnt that a characteristic feature of endocrine system is the presence of feedback loops. By this what is meant if hormone A stimulates gland ‘X to secrete hormone B, the production of ‘A ’ could be modified when the level of B changes in our blood. An example is the relation between hormones LH and estrogen (E2). An old woman exhibits the following features. High levels of LH in blood but low levels of E2 in the blood. Another woman exhibits high level of LH in blood and also high level of E2 in the blood. Where is the defect in both these women? Provide suitable diagram to support this answer.
Ans: If an old woman exhibits the high levels of LH in blood but low levels of E2 in the blood then it may be due to any abnormality in the anterior pituitary.
If an old woman exhibits the high levels of LH in blood and high levels of E2 in the blood then it may be due to any abnormality in the anterior pituitary or ovary.

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NCERT Exemplar Class 11 Biology Chapter 19 Excretory Products and Their Elimination

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NCERT Exemplar Class 11 Biology Chapter 19 Excretory Products and Their Elimination

Multiple Choice Questions

Q1. The following substances are the excretory products in animals. Choose the least toxic from among them.
(a) Urea (b) Uric acid
(c) Ammonia (d) Carbon dioxide
Ans: (b) Ammonia is the most toxic form and requires large amount of water for its elimination, whereas uric acid, being the least toxic, can be removed with a minimum loss of water.

Q2. Filtration of the blood takes place at
(a) PCT (b) DCT
(c) Collecting ducts (d) Malpighian body
Ans: (d) Filtration of the blood takes place at malpighian body.

Q3. Which of the following statements is incorrect?
a. ADH—prevents conversion of angiotensinogen in blood to angiotensin
b. Aldosterone—facilitates water reabsorption
c. ANF—enhances sodium reabsorption
d. Renin—causes vasodilation
Ans: (a) ADH is a hormone released from the posterior pituitary gland that causes an increase in blood pressure through reabsorption of water.

Q4. A large quantity of one of the following is removed from our body by lungs
(a) C02 only (b) H20 only
(c) C02 and H20 (d) Ammonia
Ans: (a) A large quantity of C02 is removed from our body by lungs.

Q5. The pH of human urine is approximately
(a) 6.5 (b) 7 (c) 6 (d) 7.5
Ans: (c) The pH of human urine is approximately 6.

Q6. Different types of excretory structures and animals are given below. Match them appropriately and mark the correct answer from among those given below:

Excretory structure/ organ Animals
A. Protonephridia (i) Prawn
B. Nephridia (ii) Cockroach
C. Malpighian tubules (iii) Earthworm
D. Green gland or Antennal gland (iv) Flatworms

Options:
(a) D—(i), (C)—(ii), B—(iii), A—(iv)
(b) B— (i), (C)—(ii), A—(iii), B—(iv)
(c) D—(i), (C)—(ii), A—(iii), B-(iv)
(d) B—(i), (C)—(ii), B—(iii), D—(iv)

Ans. (a)

Excretory structure/ organ Animals
A. Protonephridia (iv) Flatworms
B. Nephridia (iii) Earthworm
C. Malpighian tubules (ii) Cockroach                      •
D. Green gland or Antennal gland (i) Prawn

 

Q7. Which one of the following statements is incorrect?
(a) Birds and land snails are uricotelic animals.
(b) Mammals and frogs are ureotelic animals.
(c) Aquatic amphibians and aquatic insects are ammonotelic animals.
(d) Birds and reptiles are ureotelic.
Ans: (d) Reptiles (snakes and lizards), birds, land snails and insects excrete nitrogenous wastes as uric acid in the form of pellet or paste with a minimum loss of water and are called uricotelic animals.

Q8. Which of the following pairs is wrong?
(a) Uricotelic — Birds (b) Ureotelic — Insects
(c) Ammonotelic — Tadpole (d) Ureotelic — Elephant
Ans: (b) Insects — Uricotelic

Q9. Which one of the following statements is incorrect?
(a) The medullary zone of kidney is divided into a few conical masses called medullary pyramids projecting into the calyces.
(b) Inside the kidney the cortical region extends in between the medullary pyramids as renal pelvis
(c) Glomerulus along with Bowman’s capsule is called the renal corpuscle
(d) Renal corpuscle, proximal convoluted tubule (PCT) and distal convoluted tubule (DCT) of the nephron are situated in the cortical region of kidney
Ans: (b) The cortex extends in between the medullary pyramids as renal columns called Columns of Bertini.

Q10. The condition of accumulation of urea in the blood is termed as
(a) Renal Calculi (b) Glomerulonephritis
(c) Uremia (d) Ketonuria
Ans: (c) The condition of accumulation of urea in the blood is termed as uremia.

Q11. Which one of the following is also known as antidiuretic hormone?
(a) Oxytocin (b) Vasopressin (c) Adrenaline (d) Calcitonin
Ans: (b) Vasopressin is also known as antidiuretic hormone (ADH).

Q12. Match the terms given in Column I with their physiological processes given in Column II and choose the correct answer

Column I Column II
A. Proximal convoluted tubule (i) Formation of concentrated urine
B. Distal convoluted tubule (ii) Filtration of blood
C. Henle’s loop (iii) Reabsorption of 70-80% of electrolytes
D. Counter-current mechanism (iv) Ionic balance
E. Renal corpuscle (v) Maintenance of concentration gradient in medulla

Options:

(a) A—(iii), B—(v), C—(iv), D—(ii), E—(i)
(b) A—(iii), B—(iv), C—(i), D—(v), E—(ii)
(c) A—(i), B—(iii), C—(ii), D—(v), E—(iv)
(d) A—(iii), B—(i), C—(iv), D—(v), E—(ii)

Ans: (b)

A. Proximal convoluted tubule (iii) Reabsorption of 70-80% of electrolytes
B. Distal convoluted tubule (iv) Ionic balance
C. Henle’s loop (i) Formation of concentrated urine
D. Counter-current mechanism (v) Maintenance of concentration gradient in medulla
E. Renal corpuscle (ii) Filtration of blood

Q13. Match the abnormal conditions given in Column A with their explanations given in Column B and choose the correct option.

Column A Column B
A. Glycosuria (i) Accumulation of uric acid in joints
B. Renal calculi (ii) Inflammation in glomeruli
C. Glomerulonephritis (iii) Mass of crystallised salts within the kidney
D. Gout (iv) Presence of glucose in urine

Options:
(a) A—(i), B—(iii), C—(ii), D—(iv)
(b) A—(iii), B—(ii), C—(iv), D—(i)
(c) A—(iv), B—(iii), C—(ii), D—(i)
(d) A—(iv), B—(ii), C—(iii), D—(i)

Ans. (c)

A. Glycosuria (iv) Presence of glucose in urine
B. Renal calculi (iii) Mass of crystallised salts within the kidney
C. Glomerulonephritis (ii) Inflammation in glomeruli
D. Gout (i) Accumulation of uric acid in joints

 

Q14. We can produce concentrated/dilute urine. This is facilitated by a special mechanism. Identify the mechanism.
(a) Reabsorption from PCT
(b) Reabsorption from collecting duct
(c) Reabsorption/secretion in DCT
(d) Counter current mechanism in Henle’s loop/Vasa recta
Ans: (d) We can produce concentrated/dilute urine. This is facilitated by a special mechanism called counter current mechanism in Henle’s loop/Vasa recta.

Q15. Dialysing unit (artificial kidney) contains a fluid which is almost same as plasma except that it has
(a) High glucose
(b) High urea
(c) No urea
(d) High uric acid
Ans: (c) Dialysing fluid = Plasma – nitrogenous wastes (urea)

Very Short Answer Type Questions
Q1. Where does the selective reabsorption of Glomerular filtrate take place?
Ans: DCT

Q2. What is the excretory product from kidneys of reptiles?
Ans: Uric acid

Q3. What is the composition of sweat produced by sweat glands?
Ans: Water, minerals, lactic acid and urea.

Q4. Identify the glands that perform the excretory function in prawns.
Ans: Antennal glands or green glands

Q5. What is the excretory structure in amoeba?
Ans: Contractile vacuole

Q6. The following abbreviations are used in the context of excretory functions, what do they stand for?
a.ANF
b. ADH
c. GFR
d. DCT
Ans: a. ANF—Atrial Natriuretic factor
b. ADH—Antidiuretic hormone
c. GFR—Glomerular Filtration Rate
d. DCT—Distal Convoluted Tubule
Q7. Differentiate Glycosuria from Ketonuria.
Ans: Glycosuria—Presence of glucose in urine.
Ketonuria—Presence of ketone bodies in urine.

Q8. What is the role of sebaceous glands?
Ans: Sebaceous glands eliminate certain substances like sterols, hydrocarbons and waxes through sebum. This secretion provides a protective oily covering for the skin.

Q9. Name two actively transported substances in Glomerular filtrate.
Ans: Glucose and aminoacids

Q10. Mention any two metabolic disorders, which can be diagnosed by analysis of urine.
Ans: Glycosuria and ketonuria

Q11. What are the main processes of urine formation?
Ans: The main processes are filtration, reabsorption, secretion and concentration/ dilution

Q12. Sort the following into actively or passively transported substances during reabsorption of GFR:
glucose, aminoacids, nitrogenous wastes, Na+, water
Ans: Actively transported—Glucose, aminoacids and Na+
Passively transported—Nitrogenous wastes and water

Q13. Complete the following:
a. Urinary excretion = Tubular reabsorption + tubular secretion
b. Dialysis fluid = Plasma
Ans: a. Urinary excretion = Glomerular filtration – tubular reabsorption + tubular secretion
b. Dialysis fluid = Plasma – nitrogenous wastes

Q14. Mention the substances that exit from the tubules in order to maintain a concentration gradient in the medullary interstitium.
Ans: NaCl and Urea.

Q15. Fill in the blanks appropriately                  .

       Organ                              Excretory wastes

  1. Kidneys _____________________
  2. Lungs _____________________
  3. Liver _____________________
  4. Skin _____________________

Ans: a. Kidneys — Urea
b. Lungs — C02 and HzO
c. Liver — Bilirubin, biliverdin, cholesterol, degraded steroid hormones, vitamins and drugs
d. Skin — Sweat (NaCl, urea, lactic acid) and sebum (sterols, hydrocarbons and waxes).

Short Answer Type Questions
Q1. Show the structure of a renal corpuscle with the help of a diagram.
Ans:
NCERT Exemplar Class 11 Biology Chapter 19 Excretory Products and Their Elimination Img 1

Q2. What is the role played by Renin-Angiotensin in the regulation of kidney function?
Ans: Renin is released from JGA on activation due to fall in the glomerular blood pressure/flow. Renin converts angiotensinogen in blood to angiotensin-I and further to angiotensin-II. Angiotensin-II being a powerful vasoconstrictor, increase the glomerular blood pressure and thereby GFR. Angiotensin-II also activates the adrenal cortex to release aldosterone. Aldosterone causes reabsorption of Na+ and water from the distal parts of the tubule. This also leads to.an increase in blood pressure and thereby GFR. This is generally known as the Renin-Angiotensin mechanism.

Q3. Aquatic animals generally are ammonotelic in nature whereas terrestrial forms are not. Comment.
Ans: Ammonia is the most toxic form and requires large amount of water for its elimination, terrestrial adaptation necessitated the production of lesser toxic nitrogenous wastes like urea and uric acid for conservation of water. Mammals, many terrestrial amphibians and marine fishes mainly excrete urea and are called ureotelic animals. Ammonia produced by metabolism is converted into urea in the liver of these animals and released into the blood which is filtered and excreted out by the kidneys.

Q4. The composition of glomerular filtrate and urine is not same. Comment.
Ans: A comparison of the volume of the filtrate formed per day (180 litres per
day) with that of the urine released (1.5 litres), suggest that nearly 99% of the filtrate has to be reabsorbed by the renal tubules. This process is called reabsorption. For example, substances like glucose, amino acids, Na+, etc.,
in the filtrate are reabsorbed actively so, these substances are not present in urine.

Q5. What is the procedure advised for the correction of extreme renal failure? Give a brief account of it.
Ans: Kidney transplantation is the ultimate method in the correction of acute renal failures (kidney failure). A functioning kidney is used in transplantation from a donor, preferably a close relative, to minimise its chances of rejection by the immune system of the host. Modem clinical procedures have increased the success rate of such a complicated technique.

Q6. How have the terrestrial organisms adapted themselves for conservation of water?
Ans: Terrestrial adaptation necessitated the production of lesser toxic nitrogenous wastes like urea and uric acid for conservation of water. Mammals, many terrestrial amphibians and marine fishes mainly excrete urea and are called ureotelic animals. Ammonia produced by metabolism is converted into urea in the liver of these animals and released into the blood which is filtered and excreted out by the kidneys. Some amount of urea may be retained in the kidney matrix of some of these animals to maintain a desired osmolarity. Reptiles, birds, land snails and insects excrete nitrogenous wastes as uric acid ‘ in the form of pellet or paste with a minimum loss of water and are called uricotelic animals.

Q7. Label the parts in the following diagram
NCERT Exemplar Class 11 Biology Chapter 19 Excretory Products and Their Elimination Img 2

Q8. Explain, why a haemodialysing uhit called’artificial kidney?
Ans: Malfunctioning of kidneys can lead to accumulation of urea in blood, a condition called uremia, which is highly harmful and may lead to kidney failure. In such patients, urea can be removed by a process called hemodialysis. Blood drained from a convenient artery is pumped into a dialysing unit (also called artificial kidney) after adding an anticoagulant like heparin. The unit contains a coiled cellophane tube surrounded by a fluid (dialysing fluid) having the same composition as that of plasma except the nitrogenous wastes. The porous cellophane membrane of the tube allows the passage of molecules based on concentration gradient. As nitrogenous wastes are absent in the dialysing fluid, these substances freely move out, thereby clearing the blood. The cleared blood is pumped back to the body through a vein after adding anti-heparin to it.

Q9. Comment upon the hormonal regulation of selective reabsorption.
Ans: Osmoreceptors in the body are activated by changes in blood volume, body fluid volume and ionic concentration. An excessive loss of fluid from the body can activate these receptors which stimulate the hypothalamus to release antidiuretic hormone (ADH) or vasopressin from the neurohypophysis. ADH facilitates water reabsorption from latter parts of the tubule, thereby preventing diuresis.

Long Answer Type Questions

Q1. Explain the mechanism of formation of concentrated urine in mammals.
Ans: Mechanism of Concentration of the Filtrate:
Mammals have the ability to produce a concentrated urine. The Henle’s loop and vasa recta play a significant role in this. The flow of filtrate in the two limbs of Henle’s loop is in opposite directions and thus forms a counter current. The flow of blood through the two limbs of vasa recta is also in a counter current pattern. The proximity between the Henle’s loop and vasa recta, as well as the counter current in them help in maintaining an increasing osmolarity towards the inner medullary interstitium, i.e., from 300 mOsmo1L in the cortex to about 1200 mOsmolLin the inner medulla. This gradient is mainly caused by NaC1 and urea. NaC1 is transported by the ascending limb of Henle’s loop which is exchanged with the descending limb of vasa recta. NaCl is returned to the interstitium by the ascending portion of vasa recta. Similarly, small amounts of urea enter the thin segment of the ascending limb of Henle’s ioop which is transported back to the interstitium by the collecting tubule. The above described transport of substances facilitated by the special arrangement of Henle’s loop and vasa recta is called the counter current mechanism. This mechanism helps to maintain a concentration gradient in the medullary interstitium. Presence of such interstitial gradient helps in an easy passage of water from the collecting tubule thereby concentrating the filtrate (urine). Human kidneys can produce urine nearly four times concentrated than the initial filtrate formed.

Q2. Draw a labelled diagram shewing reabsorption and secretion of major substances at different parts of the nephron.
Ans:
NCERT Exemplar Class 11 Biology Chapter 19 Excretory Products and Their Elimination Img 3

Q3. Explain briefly, micturition and disorders of the excretory system.
Ans:
• Micturition: Urine formed by the nephrons is ultimately carried to the urinary bladder where it is stored till a voluntary signal is given by the central nervous system (CNS). This signal is initiated by the stretching of the urinary bladder as it gets filled with urine. In response, the stretch receptors on the walls of the bladder send signals to the CNS. The CNS passes on motor messages to initiate the contraction of smooth muscles of the bladder and simultaneous relaxation of the urethral sphincter causing the release of urine. The process of release of urine is called micturition and the neural mechanisms causing it is called the micturition reflex.
• Disorders of the Excretory System: Malfunctioning of kidneys can lead to accumulation of urea in blood, a condition called uremia, which is highly harmful and may lead to kidney failure. In such patients, urea can be removed by a process called hemodialysis. Kidney transplantation is the ultimate method in the correction of acute renal failures (kidney failure).
• Renal calculi: Stone or insoluble mass of crystallised salts (oxalates, etc.) formed within the kidney.
• Glomerulonephritis: Inflammation of glomeruli of kidney.

Q4. How does tubular secretion help in maintaining ionic and acid-base balance in body fluids?
Ans: During urine formation, the tubular cells secrete substances like H+, K+ and ammonia into the filtrate. Tubular secretion is also an important step in urine formation as it helps in the maintenance of ionic and acid base balance of body fluids.              .

  • PCT helps to maintain the pH and ionic balance of the body fluids by selective secretion of hydrogen ions, ammonia and potassium ions into the filtrate
  • DCT is also capable of selective secretion of hydrogen and potassium ions and NH3 to maintain the pH and sodium-potassium balance in blood.
  • Collecting duct also plays a role in the maintenance of pH and ionic balance of blood by the selective secretion of H+ and K+

Q5. The” glomerular filtrate in the loop of Henle gets concentrated in the descending and then gets diluted in the ascending limbs. Explain.
Ans: A hairpin shaped Henle’s loop has a descending and an ascending limb. Reabsorption is minimum in its ascending limb. However, this region plays a significant role in the maintenance of high osmolarity of medullary interstitial fluid. The descending limb of loop of Henle is permeable to water but almost impermeable to electrolytes. This concentrates the filtrate as it moves down. The ascending limb is impermeable to water but allows transport of electrolytes actively or passively. Therefore, as the concentrated filtrate pass upward, it gets diluted due to the passage of electrolytes to the medullary fluid.

Q6. Describe the structure of a human kidney with the help of a labelled diagram.
Ans: In humans, the excretory system consists of a pair of kidneys, one pair of
ureters, a urinary bladder and a urethra. Kidneys are reddish brown, bean shaped structures situated between the levels of last thoracic and third lumbar vertebra close 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 g. Towards the centre of the inner concave surface of the kidney is a notch called hilum through which ureter, blood vessels and nerves enter. Inner to the hilum is a broad funnel shaped space called the renal pelvis with projections called calyces. The outer layer of kidney is a tough capsule. Inside the kidney, there are two zones, an outer cortex and an inner meditlla. The medulla is divided into a few conical masses (medullary pyramids) projecting into the calyces (sing.: calyx). The cortex extends in between the medullary pyramids as renal columns called Columns of Bertini. Each kidney has nearly one million complex tubular structures called nephrons, which are the functional units.

NCERT Exemplar Class 11 Biology Chapter 19 Excretory Products and Their Elimination Img 4

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NCERT Exemplar Class 11 Biology Chapter 21 Neural Control and Coordination

NCERT Exemplar Class 11 Biology Chapter 21 Neural Control and Coordination are part of NCERT Exemplar Class 11 Biology. Here we have given NCERT Exemplar Class 11 Biology Chapter 21 Neural Control and Coordination. https://www.cbselabs.com/ncert-exemplar-problems-class-11-chapter-21-neural-control-coordination/

NCERT Exemplar Class 11 Biology Chapter 21 Neural Control and Coordination

Multiple Choice Questions

Q1. Chemicals which are released at the synaptic junction are called
(a) Hormones
(b) Neurotransmitters
(c) Cerebrospinal fluid
(d) Lymph
Ans: (b) Chemicals released at the synaptic junction are called neurotransmitters.

Q2. Potential difference across resting membrane is negatively charged. This is due to differential distribution of the following ions
(a)Na+ and K+ ions
(b) C03++and Cl“ ions
(c) Ca*+ and Mg++ ions                         
(d) Ca+4 and CL ions
Ans: (a) Potential difference across resting membrane is negatively charged. This is due to differential distribution of Na+ and K+ ions.

Q3. Resting membrane potential is maintained by
(a) Hormones (b) Neurotransmitters
(c) Ion pumps (d) None of the above
Ans: (c) Resting membrane potential is maintained by ion pumps.

Q4. The function of our visceral organs is controlled by
(a) Sympathetic and somatic neural system
(b) Sympathetic and parasympathetic neural system
(c) Central and somatic neural system.
(d) None of the above
Ans: (b) The function of our visceral organs is controlled by sympathetic and parasympathetic neural systems.

Q5. Which of the following is not involved in knee-jerk reflex?
(a) Muscle spindle (b) Motor neuron
(c) Brain (d) Intemeurons
Ans: (c) Brain is not involved in knee-jerk reflex.

Q6. An area in the brain which is associated with strong emotions is
(a) Cerebral cortex (b) Cerebellum
(c) Limbic system (d) Medulla
Ans: (c) An area in the brain is associated with strong emotions is limbic system.

Q7. Mark the vitamin present in rhodopsin.
(a) VitA (b) Vit B (c) VitC (d) VitD
Ans: (a) Vit A is present in rhodopsin.

Q8. Human eyeball consists of three layers and it encloses
(a) Lens, iris, optic nerve
(b) Lens, aqueous humor and vitreous humor
(c) Cornea, lens, iris
(d) Cornea, lens, optic nerve
Ans: (b) Human eyeball consists of three layers and it encloses lens, aqueous humor and vitreous humor.

Q9. Wax gland present in the ear canal is called
(a) Sweat gland
(b) Prostate gland
(c) Cowper’s gland –
(d) Sebaceous gland/ceruminous gland
Ans: (d) Wax gland present in the ear canal is called sebaceous gland/ceruminous gland.

Q10. The “part of internal ear responsible for hearing is
(a) Cochlea (b) Semicircular canal
(c) Utriculus (d) Sacculus
Ans: (a) The part of internal ear responsible for hearing is cochlea.

Q11. The organ of Corti is a structure present in
(a) External ear (b) Middle ear
(c) Semicircular canal (d) Cochlea
Ans: (d) The organ of Corti is a structure present in cochlea.

Q12. While travelling to higher altitudes, people can feel pain in the ear and dizziness. Which part, among the following is involved?
(a) Cochlea, ear ossicles
(b) Tympanic membrane
(c) Eustachian tube, utricle, saccule and semicircular canals
(d) None of the above
Ans: (c)

Very Short Answer Type Questions
Q1. Rearrange the following in the correct order of involvement in electrical impulse movement: Synaptic knob, dendrites, cell body, Axon terminal, Axon
Ans: Dendrites—Cell body—Axon—Axon terminal—Synaptic knob.

Q2. Comment upon the role of ear in maintaining the balance of the body and posture.
Ans: The crista and macula are the specific receptors of the vestibular apparatus responsible for maintenance of balance of the body and posture.

Q3. Which cells of the retina enable us to see coloured objects around us?
Ans: Cone cells of the retina enable us to see the coloured objects around us.

Q4. 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.
Ans: Ear drum, malleus, incus, stapes, cochlea, chochlear nerve.

Q5. During resting potential, the axonal membrane is polarised, indicate the movement of +ve and -ve ions leading to polarisation diagrammatically.
Ans: Neurons are excitable cells because their membranes are in a polarised state. Different types of ion channels are present on the neural membrane. These ion channels are selectively permeable to different ions. When a neuron is not conducting any impulse, i.e., resting, the axonal membrane is comparatively more permeable to potassium ions (K+) and nearly impermeable to sodium ions (Na+). Similarly, the membrane is impermeable to negatively charged proteins present in the axoplasm. Consequently, the axoplasm inside the axon contains high concentration of K+ and negatively charged proteins and low concentration of Na+. In contrast, the fluid outside the axon contains a “low concentration of K+, a high concentration of Na+ and thus form a concentration gradient. These ionic gradients across the resting membrane are maintained by the active transport of ions by the sodium-potassium pump which transports 3 Na+ outwards for 2 K+ into the cell. As a result, the outer surface of the axonal membrane possesses a positive charge while its inner surface becomes negatively charged and therefore is polarised. The electrical potential difference across the resting plasma membrane is called as the resting potential.
NCERT Exemplar Class 11 Biology Chapter 21 Neural Control and Coordination Img 1

Q6. Name the structures involved in fhe protection of the brain.
Ans: The human brain is well protected by the skull. Inside the skull, the brain is covered by cranial meninges consisting of an outer layer called dura mater, a very thin middle layer called arachnoid and an inner layer (which is in contact with the brain tissue) called pia mater. Piamater is a vascular membrane which is richly supplied with blood capillaries. Space between the duramater and arachnoid is called subdural space. Space between the arachnoid and pia mater is called subarachnoid space. Subarachnoid space is filled with the cerebrospinal fluid (CSF) which acts as a cushion for CNS from shocks.

Q7. Our reactions like aggressive behaviour, use of abusive words, restlessness etc. are regulated by brain, name the parts involved.
Ans: Limbic system and hypothalamus

Q8. What do grey and white matter in the brain represent?
Ans: The layer of cells which covers the cerebral hemisphere is called cerebral cortex and is thrown into prominent folds. The cerebral cortex is referred to as the grey matter due to its greyish appearance. The neuron cell bodies are concentrated here giving the colour. Fibres of the tracts are covered with the myelin sheath, which constitute the inner part of cerebral hemisphere. They give an opaque white appearance to the layer and, hence, is called the white matter.

Q9. Where is the hunger centre located in human brain?
Ans: Hypothalamus

Q10. Which sensory organ is involved in vertigo (sensation of oneself or objects spinning around)?
Ans: Vestibular apparatus of inner ear

Q11. While travelling at a higher altitude, a person complains of dizziness and vomiting sensation. Which part of the inner ear is disturbed during the journey?
Ans: Vestibular apparatus (saccule, utricle and semicircular canals)

Q12. Complete the statement by choosing appropriate match among the following:

(a) Resting potential (i) Chemicals involved in the transmission of impulses at synapses.
(b) Nerve impulse («) Gap between the presynaptic and postsynaptic neurons.
(c) Synaptic cleft (iii) Electrical potential difference across the resting neural membrane.
(d) N euro transmitters (iv) An electrical wave-like response of a neuron to a stimulation.

Ans:

(a) Resting potential (iii) Electrical potential difference across the resting neural membrane.
(b) Nerve impulse (iv) An electrical wave-like response of a neuron to a stimulation.
(c) Synaptic cleft (ii) Gap between the presynaptic and postsynaptic neurons.
(d) N euro transmitters (i) Chemicals involved in the transmission of impulses at synapses.

Short Answer Type Questions
Q1. The major parts of the human neural system is depicted below. Fill in the empty boxes with appropriate words.
NCERT Exemplar Class 11 Biology Chapter 21 Neural Control and Coordination Img 2

Q2. What is the difference between electrical transmission and chemical transmission?
Ans:

Electrical transmission Chemical transmission
1. Mediated by electrical synapses. 1. Mediated through neuro transmitters.
2. The membranes of pre- and post-synaptic neurons are in very close proximity. 2. The membranes of pre- and post- synaptic neurons are separated by synaptic cleft.
3. Electrical current can flow directly from one neuron into the other across the synapses. 3. Neurotransmitters are involved in the transmission of impulses at the synapses.
4. This transmission is faster. 4. This transmission is slower.
5. These are rare in our system. 5. These are common in our system.

 

Q3. Neural system and computers share certain common features. Comment in five lines. (Hint: CPU, input-output devices).
Ans: Neural system and computers share certain common features. The neural system has brain as command and control centre similar to the computer that has CPU (Central processing unit). Sensory organs are input devices of neural system like the mouse and keyboard of the computer. Responses of the body are the output of the neural system.like the data analysis and typed material of the computer. Nerves are comparable to the wires of the computers.

Q4. If someone receives a blow on the back of neck, what would be the effect on the person’s CNS?
Ans: If someone receives a blow on the back of neck, it may result in the dislocation of the cervical vertebrae that may lead to the injury of the spinal cord passes through neural canal. Injury of spinal cord may lead to paralysis.

Q5. What is the function ascribed to Eustachian tube?
Ans: A Eustachian tube connects the middle ear cavity with the pharynx. The Eustachian tube helps in equalising the pressures on either sides of the ear drum.

Q6. Label the following parts in the given diagram using arrow.
NCERT Exemplar Class 11 Biology Chapter 21 Neural Control and Coordination Img 3
a. Aqueous chamber
b. Cornea
c. Lens
d. Retina
e. Vitreous chamber
f. Blind spot
NCERT Exemplar Class 11 Biology Chapter 21 Neural Control and Coordination Img 4

Long Answer Type Questions
Q1. Explain the process of the transport and release of a neurotransmitter with the help of a labelled diagram showing a complete neuron, axon terminal and synapse.
Ans: A nerve impulse is transmitted from one neuron to another through junctions called synapses. A synapse is formed by the membranes of a presynaptic neuron and a postsynaptic neuron, which may or may not be separated by a gap called synaptic cleft. At a chemical synapse, the membranes of the pre- and post-synaptic neurons are separated by a fluid-filled space called synaptic cleft. Chemicals called neurotransmitters are involved in the transmission of impulses at these synapses. The axon terminals contain vesicles filled with these neurotransmitters. When an impulse (action potential) arrives at the axon terminal, it stimulates the movement of the synaptic vesicles towards the membrane where they fuse with the plasma membrane and release their neurotransmitters in the synaptic cleft. The released neurotransmitters bind to their specific receptors, present on the postsynaptic membrane. This binding opens ion channels allowing the entry of ions which can generate a new potential in the postsynaptic neuron. The new potential developed may be either excitatory or inhibitory.

NCERT Exemplar Class 11 Biology Chapter 21 Neural Control and Coordination Img 5

Q2. Name the parts of human forebrain indicating their respective functions.
Ans: Forebrain:
The forebrain consists of cerebrum, thalamus and hypothalamus. Cerebrum forms_ the major part of the human brain. A deep cleft divides the cerebrum longitudinally into two halves, which are termed as the left and right cerebral hemispheres. The hemispheres are connected by a tract of nerve fibres called corpus callosum. The layer of cells which covers the cerebral hemisphere is called cerebral cortex and is thrown into prominent folds. The cerebral cortex is referred to as the grey matter due to its greyish appearance. The neuron cell bodies are concentrated here giving the colour. The cerebral cortex contains motor areas, sensory areas and large regions that are neither clearly sensory nor motor in function. These regions called as the association areas are responsible for complex functions like intersensory associations, memory and communication. Fibres of the tracts are covered with the myelin sheath, which constitute the inner part of cerebral hemisphere. They give an opaque white appearance to the layer and, hence, is called the white matter. The cerebrum wraps around a structure called thalamus, which is a major coordinating centre for sensory and motor signaling. Another very important part of the brain called hypothalamus lies at the base of the thalamus. The hypothalamus contains a number of centres which control body temperature, urge for eating and drinking. It also contains several groups of neurosecretory cells, which secrete hormones called hypothalamic hormones. The inner parts of cerebral hemispheres and a group of associated deep structures like amygdala, hippocampus, etc., form a complex structure called the limbic lobe or limbic system. Along with the hypothalamus, it is involved in the regulation of sexual behaviour, expression of emotional reactions (e.g., excitement, pleasure, rage and fear), and motivation.

Q3. Explain the structure of middle and internal ear with the help of diagram.
Ans: The middle ear contains three ossicles called malleus, incus and stapes which are attached to one another in a chainlike fashion. The malleus is attached to the tympanic membrane and the stapes is attached to the oval window of the cochlea. The ear ossicles increase the efficiency of transmission of sound waves to the inner ear. A Eustachian tube connects the middle ear cavity with the pharynx. The Eustachian tube helps in equalising the pressures on either sides of the ear drum.
NCERT Exemplar Class 11 Biology Chapter 21 Neural Control and Coordination Img 6

• The fluid-filled inner ear called labyrinth consists of two parts, the bony and the membranous labyrinths. The bony labyrinth is a series of channels. Inside these channels lies the membranous labyrinth, 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 membranes constituting cochlea, the Reissner’s and basilar, divide the surrounding perilymph filled bony labyrinth into an upper scala vestibuli and a lower scala tympani. The space within cochlea called scala media is filled with endolymph. At the base of the cochlea, the scala vestibuli ends at the oval window, while the scala tympani terminates at the round window which opens to the middle ear. The organ of Corti is a structure located on the basilar membrane which contains hair cells that act as auditory receptors. The hair cells are present in rows on the internal side of the organ of Corti. The basal end of the hair cell is in close contact with the afferent nerve fibres. A large number of processes called stereo cilia are projected from the apical part of each hair cell. Above the rows of the hair cells is a thin elastic membrane called tectorial membrane.
• The inner ear also contains a complex system called vestibular apparatus, located above the cochlea. The vestibular apparatus is composed of three semi-circular canals and the otolith (macula is the sensory part of saccule and utricle). Each semi-circular canal lies in a different plane at right angles to each other. The membranous canals are suspended in the perilymph of the bony canals. The base of canals is swollen and is called ampulla, which contains a projecting ridge called crista ampullaris which has hair cells. The saccule and utricle contain a projecting ridge called macula. The crista and macula are the specific receptors of the vestibular apparatus responsible for maintenance of balance of the body and posture.

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

NCERT Exemplar Class 11 Biology Chapter 17 Breathing and Exchange of Gases are part of NCERT Exemplar Class 11 Biology. Here we have given NCERT Exemplar Class 11 Biology Chapter 17 Breathing and Exchange of Gases. https://www.cbselabs.com/ncert-exemplar-problems-class-11-chapter-17-breathing-exchange-gases/

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

Multiple Choice Questions

Q1. Respiration in insects is called direct because ‘

(a) The cells exchange 02/C02 directly with the air in the tubes
(b) The tissues exchange 02/C02 directly with coelomic fluid
(c) The tissues exchange 02/C02 directly with the air outside through body surface
(d) “ Tracheal tubes exchange 02/C02 directly with the haemocoel which then exchange with tissues
Ans: (d) Respiration in insects is called direct because tracheal tubes exchange 02/C02 directly with the haemocoel which then exchange with tissues.

Q2.A person suffers punctures in his chest cavity in an accident, without any damage to the lungs, its effect could be
(a) Reduced breathing rate
(b) Rapid increase in breathing rate
(c) No change in respiration                 
(d) Cessation of breathing
Ans: (d)A person suffers punctures in his chest cavity in an accident, without any damage to the lungs, its effect could be cessation of breathing.

Q3. It is known that exposure to carbon monoxide is harmful to animals because
(a) It reduces C02 transport
(b) It reduces 02 transport
(c) It increases C02 transport               
(d) It increases 02 transport
Ans: (b) CO is a poisonous gas which binds with Hb more rapidly than 02 to form carboxyhaemoglobin. CO makes .the most stable combination with the Hb of blood. CO has 200-250 times more affinity for Hb as compared to 02. When the inhaled air contains CO gas then a person suffers from suffocation because product cannot dissociate so decreases free oxygen. So it reduces 02 transport.

Q4. Mark the true statement among the following with reference to normal breathing.
(a) Inspiration is a passive process whereas expiration is active
(b) Inspiration is an active process whereas expiration is passive
(c) Inspiration and expiration are active processes
(d) Inspiration and expiration are passive processes
Ans: (b) Inspiration is an active process whereas expiration is passive.

Q5. Mark the incorrect statement in context to 02 binding to Hb.
(a) Lower pH
(b) Lower temperature
(c) Lower pC02                                       
(d) Higher p02
Ans: (a) 02 binding to Hb occurs in the following conditions: lower temperature, lower pC02 and higher p02.

Q6.Mark the correct pair of muscles involved in the normal breathing in humans
(a)External and internal intercostal muscles
(b)Diaphragm and abdominal muscles
(c)Diaphragm and external intercostal muscles
(d) Diaphragm and intercostal muscles
Ans: (d) Diaphragm and intercostal muscles involved in the normal breathing in humans. ,

Q7. Incidence of Emphysema—a respiratory disorder is high in cigarette smokers. In such cases
(a) The bronchioles are found damaged
(b) The alveolar walls are found damaged
(c) The plasma membrane is found damaged
(d) The respiratory muscles are found damaged
Ans: (b) Emphysema is a chronic disorderin which alveolar walls are damaged due to which respiratory surface is decreased.

Q8. Respiratory process is regulated by certain specialised centres in the brain. One of the following listed centres can reduce the inspiratory duration upon stimulation
(a) Medullary inspiratory centre (b) Pneumotaxic centre
(c) Apneustic centre (d) Chemosensitive centre
Ans: (b) Pneumotaxic centre can reduce the inspiratory duration upon stimulation.

Q9. C02 dissociates from carbaminohaemoglobin when
(a)    pC02 is high and p02 is low           
(b)     p02 is high and pC02 is low
(c)     pC02 and p02 are equal                 
(d)     None of the above
Ans: (b) C02 dissociates from carbaminohaemoglobin when p02 is high and pC02 is low,

Q10. In breathing movements, air volume can be estimated by .
(a)    Stethoscope                                     
(b)     Hygrometer
(c)    Sphygmomanometer                    
(d)     Spirometer
Ans: (d) In breathing movements, air volume can be estimated by spirometer.

Q11.From the following relationships between respiratory volumes and capacities, mark the correct option.
NCERT Exemplar Class 11 Biology Chapter 17 Breathing and Exchange of Gases Img 1

i. Inspiratory Capacity (IC) = Tidal Volume + Inspiratory Residual Volume (IRV) ,
iii. Residual Volume (RV) = TLC – VC

Q12. The oxygen-haemoglobin dissociation curve will show a right shift in case of
(a) High pC02                                          
(b) Highp02
(c) Low pC02                                           
(d) Less H+ concentration
Ans: (a) Curve shift is right in following conditions: (1) Decrease in p02, (2) Increase in pC02 (Bohr effect), (3) Increase in body temperature, (4) Increase in H+ ion concentration, (5) Decrease in pH, (6) Increase in 2, 3 diphosphoglycerate.

Q13. Match the following and mark the correct options

Animal Respiratory organ
A. Earthworm (0 Moist cuticle
B. Aquatic Arthropods (ii) Gills
C. Fishes (iii) Lungs
D. Birds/Reptiles (iv) Trachea

(a) A—(ii), B—(i), C—(iv), D—(iii)
(b) A—(i), B—(iv), C—(ii), D—(iii)
(c) A—(i), B—(iii), C—(ii), D—(iv)
(d) A—(i), B—(iv), C—(ii), D—(iii)

Ans: (d)

Animal Respiratory organ
A. Earthworm 0) Moist cuticle
B. Aquatic Arthropods (iv) Trachea
C. Fishes (ii) Gills
D. Birds/Reptiles (iii) Lungs

 

Very Short Answer Type Questions
Q1. Define the following terms
a. Tidal volume
b. Residual volume
c. Asthma
Ans: a. Tidal volume: 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: Volume of air remaining in the lungs even after a forcible expiration. This averages 1100 mL to 1200 mL. Residual air mainly occurs in alveoli.
c. Asthma: Asthma is a difficulty in breathing causing wheezing due to inflammation of bronchi and bronchioles. In asthma, due to flattening of
tracheal vessels, alveoli are deprived of oxygen. Asthma is characterised by spasm in bronchial muscle.

Q2. A fluid-filled double membranous layer surrounds the lungs. Name it and mention its important function.
Ans: Pleural fluid is found in between the two membranes of lung and it reduces the friction on the lung surface.

Q3. Name the primary site of exchange of gases in our body?
Ans: Alveoli

Q4. Cigarette smoking causes emphysema. Give reason.
Ans: Cigarette smoking causes damage of the alveolar walls leading to decreased respiratory surfaces for exchange of gases.

Q5. What is the amount of 02 supplied to tissues through every 100 mL of oxygenated blood under normal physiological conditions?
Ans: 5 mL of oxygen/100 mL of oxygenated blood.

Q6. A major percentage (97%) of 02 is transported by RBCs in the blood. How does the remaining percentage (3%) of 02 transported?
Ans: Through Plasma

 

Q7. Arrange the following terms based on their volumes in an ascending order
a. Tidal Volume (TV)
b. Residual Volume (RV)
c. Inspiratory Reserve Volume (IRV)
d. Expiratory Capacity (EC)
Ans: a. Tidal Volume (TV): 500 mL
b. Residual Volume (RV): 1100 mL-200 mL
c. Inspiratory Reserve Volume (IRV): 2500 mL-3000 mL
d. Expiratory Capacity (EC): 1500 mL-1600 mL

Q8. Complete the missing terms
a. Inspiratory Capacity (IC) = _____+ IRV
b. _____ = TV + ERV
c. Functional Residual Capacity (FRC) = ERV + _____
Ans. a. Inspiratory Capacity (IC) = TV + IRV
b. EC = TV + ERV
c. Functional Residual Capacity (FRC) = ERV + RV

Q9. Name the organs of respiration in the following organisms:
a. Flatworm
b.Birds
c. Frog
d. Cockroach
Ans: a. Flatworm—Entire body surface
b. Birds—Lung
c. Frog—Lung and moist skin
d. Cockroach—Tracheal tubes

Short Answer Type Questions

Q1. State the different modes of C02 transport in blood.
Ans: Nearly 20-25% of C02 by RBCs
Nearly 70% of C02 as bicarbonates Nearly 7% of C02 as dissolved state in plasma

Q2. Compared to 02, the diffusion rate of C02 through the diffusion membrane per unit difference in partial pressure is much higher. Explain.
Ans: Solubility is an important factor deciding diffusion rate. As the solubility of C02 is 20-25 times higher than 02, diffusion of C02 through the diffusion membrane per unit difference in partial pressure is much higher.

Q3. For completion of respiration process, write the given steps in sequential manner.                                                                        .

  1. Diffusion of gases (02 and C02) across alveolar membrane.
  2. Transport of gases by blood.
  3. Utilisation of 02 by the cells for catabolic reactions and resultant release of C02.
  4. Pulmonary ventilation by which atmospheric air is drawn in and C02 rich alveolar air is released out.
  5. Diffusion of 02 and C02 between blood and tissues.

Ans: Respiration involves the following steps:

  1. Breathing or pulmonary ventilation by which atmospheric air is drawn in and C02 rich alveolar air is released out.
  2. Diffusion of gases (02 and C02) across alveolar membrane.
  3. Transport of gases by the blood.
  4. Diffusion of 02 and C02 between blood and tissues.
  5. Utilisation of 02 by the cells for catabolic reactions and resultant release of C02.

Long Answer Type Questions

Q1. Explain the transport of 02 and C02 between alveoli and tissue with a diagram.
Ans:

  • Transport of gases: Blood is the medium of transport for 02 and C02. 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. Nearly 20-25% of C02 is transported by RBCs whereas 70% of it is carried as bicarbonate. About 7% of C02 is carried in a dissolved state through plasma.
  • Transport of oxygen: Haemoglobin is a red coloured iron containing pigment present in the RBCs. 02 can bind with haemoglobin in a reversible manner to form oxyhaemoglobin. Each haemoglobin molecule can carry a maximum of four molecules of 02. Binding of oxygen with haemoglobin is primarily related to partial pressure of 02. Partial pressure of C02, hydrogen ion concentration and temperature are the other factors which can interfere with this binding. A sigmoid curve is obtained when percentage saturation of haemoglobin with 02 is plotted against the p02. This curve is called the Oxygen dissociation curve and is highly useful in studying the effect of factors like pC02, H+ concentration, etc., on binding of 02 with haemoglobin. In the alveoli, where there is high p02, low pC02, lesser H+ concentration and lower temperature, the factors are all favourable for the formation of oxyhaemoglobin, whereas in the tissues, where low p02, high pC02, high H+ concentration and higher temperature exist, the conditions are favourable for dissociation of oxygen from the oxyhaemoglobin. This clearly indicates that 02 gets bound to haemoglobin in the lung surface and gets dissociated at the tissues. Every 100 mL of oxygenated blood can deliver around 5 mL of 02 to the tissues under normal physiological conditions.
  • Transport of carbon dioxide: C02 is carried by haemoglobin as carbamino-haemoglobin (about 20-25%). This binding is related to the partial pressure of C02. p02 is a major factor which could affect this binding. When pC02 is high and p02 is low as in the tissues, more binding of carbon dioxide occurs whereas, when the pC02 is low and p02 is high as in the alveoli, dissociation of C02 from carbamino-haemoglobin takes place, i.e., C02 which is bound to haemoglobin from the tissues is delivered at the alveoli. RBCs contain a very high concentration of the enzyme, carbonic anhydrase and minute quantities of the same is present in the plasma too. This enzyme facilitates the following reaction in both directions

NCERT Exemplar Class 11 Biology Chapter 17 Breathing and Exchange of Gases Img 2

At the tissue site where partial pressure of C02 is high due to catabolism, C02 diffuses into blood (RBCs and plasma) and forms HC02 and H+. At the alveolar site where pC02 is low, the reaction proceeds in the oppositedirection leading to the formation of C02 and H20. Thus, C02 trapped as bicarbonate at the tissue level and transported to the alveoli is released out as C02. Every 100 mL of deoxygenated blood delivers approximately 4 mL of C02 to the alveoli.

NCERT Exemplar Class 11 Biology Chapter 17 Breathing and Exchange of Gases Img 3

Q2. Explain the mechanism of breathing with neat labelled sketches.
Ans: Breathing involves two stages:
a. Inspiration: Inspiration is initiated by the contraction of diaphragm, which increases the volume of thoracic chamber in the anteroposterior axis. The contraction of external inter-costal muscles lifts up the ribs and the sternum causing an increase in the volume of thoracic chamber in the dorso-ventral axis also. Such an increase in thoracic volume leads to a similar increase in pulmonary volume resulting in decreased intra- pulmonary pressure to less than atmospheric pressure. This causes the movement of external air into the lungs, i.e., inspiration.
b. Expiration: The inter-costal muscles return the diaphragm and sternum to their normal positions with relaxation of the diaphragm. This reduces the thoracic volume and thereby the pulmonary volume. As a result an increase in intra-pulmonary pressure to slightly above the atmospheric pressure causes the expulsion of air from the lungs i.e., expiration.

NCERT Exemplar Class 11 Biology Chapter 17 Breathing and Exchange of Gases Img 4

Q3. Explain the role of neural system in regulation of respiration.
Ans: Human beings have a significant ability to maintain and moderate the respiratory rhythm to suit the demands of the body tissues. This is done by the neural system. A specialised centre present in the medulla region of the brain called respiratory rhythm centre is primarily responsible for this regulation. Another centre present in the pons region of the brain called pneumotaxic centre can moderate the functions of the respiratory rhythm centre. Neural signal from this centre “can reduce the duration of inspiration and thereby alter the respiratory rate. A chemosensitive area is situated adjacent to the rhythm centre which is highly sensitive to C02 and hydrogen ions. Increase in these substances can activate this centre, which in turn can signal the rhythm centre to make necessary adjustments in the respiratory process by which these substances can be eliminated. Receptors associated with aortic arch and carotid artery also can recognise changes in C02 and H+ concentration and send necessary’ signals to the rhythm centre for remedial actions. The role of oxygen in the regulation of respiratory rhythm is quite insignificant.

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