NCERT Exemplar Class 11 Biology Chapter 9 Biomolecules img

NCERT Exemplar Class 11 Biology Chapter 9 Biomolecules

NCERT Exemplar Class 11 Biology Chapter 9 Biomolecules are part of NCERT Exemplar Class 11 Biology. Here we have given NCERT Exemplar Class 11 Biology Chapter 9 Biomolecules.

NCERT Exemplar Class 11 Biology Chapter 9 Biomolecules

Multiple Choice Questions

Q1. It is said that elemental composition of living organisms and that of inanimate objects (like earth’s crust) are similar in the sense that all the major elements are present in both. Then what would be the difference between these two . groups? Choose a correct answer from the following.
(a) Living organisms have more gold in them than inanimate objects
(b) Living organisms have more water in their body than inanimate objects
(c) Living organisms have more carbon, oxygen and hydrogen per unit mass than inanimate objects
(d) Living organisms have more calcium in them than inanimate objects.
Ans: (c)

Element % Weight of
Earth’s Crust Human Body
Hydrogen (H) 0.14 0.5
Carbon (C) 0.03 18.5
Oxygen(0) 46.6 65.0
Nitrogen (N) very little 3.3
Sulphur (S) 0.03 0.3
Sodium (Na) 2.8 0.2
Calcium (Ca) 3.6 1.5
Magnesium (Mg) 2.1 0.1
Silicon (Si) 27.7 negligible

Q2. Many elements are found in living organisms either free or in the form of compounds. One of the following is not found in living organisms.
(a) Silicon (b) Magnesium (c) Iron (d) Sodium
Ans: (a) See Answer 2.
Q3. Aminoacids, as the name suggests, have both an amino group and a carboxyl group in their structure. In addition, all naturally occurring aminoacids (those which are found in proteins) are called L-aminoacids. From this, can you guess from which compound can the simplest aminoacid be made?
(a) Formic acid (b) Methane (c) Phenol acid (d) Glycine
Ans: (d) Glycine is an amino acid (which have both an amino group and a carboxyl group in their structure).
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Q4. Many organic substances are negatively charged, e.g., acetic acid, while others are positively charged e.g., ammonium ion. An aminoacid under certain conditions would have both positive and negative charges simultaneously in the same molecule. Such a form of aminoacid is called
(a) Positively charged form (b) Negatively charged form
(c) Neutral form (d) Zwitterionic form
Ans: (d) In aqueous solution, the carboxyl group can lose a proton and amino group can accept a proton, giving rise to a dipolar ion called Zwitter ion. Zwitter ion is neutral but contains both positive and negative charges.
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Q5. Sugars are technically called carbohydrates, referring to the fact that their formulae are only multiple of C(H20). Hexoses therefore have six carbons, twelve hydrogens and six oxygen atoms. Glucose is a hexose. Choose from among the following another hexose.
(a) Fructose (b) Erythrose ~(c) Ribulose (d) Ribose
Ans: (a) Sugars are technically called carbohydrates, referring to the fact that their formulae are only multiple of C(H20). Hexoses therefore have six carbons, twelve hydrogens and six oxygen atoms. E.g., glucose and fructose.

Q6. When you take cells or tissue pieces and grind them with an acid in a mortar and pestle, all the small biomolecules dissolve in the acid. Proteins polysaccharides and nucleic acids are insoluble in mineral acid and get precipitated. The acid soluble compounds include amino acids, nucleosides, small sugars etc. When one adds a phosphate group to a nucleoside one gets another acid soluble biomolecule called
(a) Nitrogen base
(b) Adenine
(c) Sugar phosphate
(d) Nucleotide
Ans: (d) Neucliotide = base + sugar + phosphate

Q7. When we homogenise any tissue in an acid, the acid soluble pool represents
(a) Cytoplasm (b) Cell membrane
(c) Nucleus (d) Mitochondria
Ans: (a) When we homogenise any tissue in an acid, the acid soluble pool represents cytoplasm.

Q8. The most abundant chemical in living organisms could be
(a) Protein (b) Water (c) Sugar (d) Nucleic acid
Ans: (b) Most abundant component of cell is water.

Component % of the total Cellular Mass
Water 70-90
Proteins 10-15
Nucleic acids 5-7
Carbohydrates 3
Lipids 2
Ions 1

 

Q9. A homopolymer has only one type of building block called monomer repeated V number of times. A heteropolymer has more than one type of monomer. Proteins are heteropolymers usually made of aminoacids. While a nucleic acid like DNA or RNA is made up of only 4 types of nucleotide monomers, proteins are made of
(a) 20 types of monomers (b) 40 types of monomers
(c) 30 types of monomers (d) only one type of monomer
Ans: (a) A homopolymer has only one type of building block called monomer repeated V number of times. A heteropolymer has more than one type of monomer. Proteins are heteropolymers usually made of amino acids. While a nucleic acid like DNA or RNA is made of of only 4 types of nucleotide monomers, proteins are made of 20 types of monomers.

Q10. Proteins perform many physiological functions. For example, some proteins function as enzymes. One of the following represents an additional function that some proteins discharge
(a) Antibiotics
(b) Pigment conferring colour to skin
(c) Pigment making colours of flowers
(d) Hormones
Ans: (d) Proteins perform many physiological functions. For example, some proteins function as enzymes. Hormones represents an additional function that some proteins discharge (like insulin).

Q11. Glycogen is a homopolymer made of
(a) Glucose units (b) Galactose units
(c) Ribose units (d) Amino acids
Ans: (a) Glycogen is a homopolymer made of glucose units.

Q12. The number of ‘ends’ in a glycogen molecule would be
(a) Equal to the number of branches plus one
(b) Equal to the number of branch points
(c) One ‘
(d) Two, one on the left side and another on the right side
Ans: (d) In a polysaccharide chain (say glycogen), the right end is called the reducing end and the left end is called the non-reducing end.

Q13. The primary structure of a protein molecule has
(a) Two ends (b) One end (c) Three ends (d) No ends
Ans: (a) The primary structure of a protein molecule has two ends.
A protein is imagined as a line, the left end represented by the first amino acid and the right end is represented by the last amino acid. The first amino acid is also called as N-terminal amino acid. The last amino acid is called the C-terminal amino acid.

Q14. Enzymes are biocatalysts. They catalyse biochemical reactions. In general they reduce activation energy of reactions. Many physico-chemical processes are enzyme mediated.’ Some examples of enzyme mediated reactions are given below. Tick the wrong entry.
(a) Dissolving C02 in water
(b) Unwinding the two strands of DNA .
(c) Hydrolysis of sucrose
(d) Formation of peptide bond
Ans: (a) Dissolving C02 in water is a physical process.

Very Short Answer Type Questions
Q1. Medicines are either man made (i.e., synthetic) or obtained from living organisms like plants, bacteria, animals etc. and hence the latter are called natural products. Sometimes natural products are chemically altered by man to reduce toxicity or side effects. Write against each of the following whether they were initially obtained as a natural product or as a synthetic chemical.
a. Penicillin
b. Sulfonamide
c. Vitamin C
d. Growth Hormone
Ans: a. Penicillin: Natural product
b. Sulfonamide: Synthetic chemical
c. Vitamin C: Natural product
d. Growth Hormone: Natural product

Q2. Select an appropriate chemical bond among ester bond, glycosidic bond, peptide bond and hydrogen bond and write against each of the following.
a. Polysaccharide
b. Protein
c. Fat
d. Water
Ans: a. Polysaccharide: Glycosidic bond
b. Protein: Peptide bond
c. Fat: Ester bond
d. Water: Hydrogen bond

Q3. Write the name of any one aminoacid, sugar, nucleotide and fatty acid.
Ans: Glycine (amino acid), Ribose (sugar), Cytidylic acid (nucleotide) and
Arachidonic acid (fatty acid).

Q4. Reaction given below is catalysed by oxidoreductase between two substrates A and A’, complete the reaction. A reduced + A‘ oxidised -»
Ans: A reduced + A’ oxidised —»A oxidised + A’ reduced

Q5. How are prosthetic groups different from co-factors?
Ans: Prosthetic groups are organic compounds and are distinguished from other cofactors in that they are tightly bound to the apoenzyme. For example, in peroxidase and catalase, which catalyze the breakdown of hydrogen peroxide to water and oxygen, haem is the prosthetic group and it is a part of the active site of the enzyme.
Cofactor may be organic or inorganic (metal ions).

Q6. Glycine and Alanine are different with respect to one substituent on the a-carbon. What are the other common substituent groups?
Ans: The R-group in these proteinaceous amino acids could be a hydrogen (the amino acid is called glycine), a methyl group (alanine), hydroxy methyl (serine), etc.
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Q7. Starch, Cellulose, Glycogen, Chitin are polysaccharides found among the following. Choose the one appropriate and write against each.
Cotton fibre __________
Exoskeleton of Cockroach __________
Liver __________
Peeled potato __________
Ans: Cotton fibre : Cellulose
Exoskeleton of Cockroach : Chitin
Liver: Glycogen
Peeled potato: Starch

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Q4. Nucleic acids exhibit secondary structure, justify with example.
Ans: Nucleic acids exhibit a wide variety of secondary structures. For example, one of the secondary structures exhibited by DNA is the famous Watson—Crick model. This model says that DNA exists as a double helix. The two strands of polynucleotides are antiparallel, i.e. run in the opposite direction. The backbone is formed by the sugar-phosphate-sugar chain. The nitrogen bases are projected more or less perpendicular to this backbone but face inside. A and G of one strand compulsorily base pairs with T and C, respectively, on the other strand. There are two hydrogen bonds between A and T and three hydrogen bonds between G and C. Each strand appears like a helical staircase. Each step of ascent is represented by a pair of bases. At each step of ascent, the strand turns 36°. One full turn of the helical strand would involve ten steps or ten base pairs. Attempt drawing a line diagram. The pitch would be 34 A. The rise per base pair would be 3.4 A. This form of DNA with the above mentioned salient features is called B-DNA.

Q5. Comment on the statement “living state is a non-equilibrium steady state to be able to perform work”.
Ans: The most important fact of biological systems is that all living organisms exist in a steady-state characterised by concentrations of each of these biomolecules. These biomolecules are in a metabolic flux. Any chemical or physical process moves spontaneously to equilibrium. The steady state is a non-equilibrium state. One should remember from physics that systems at equilibrium cannot perform work. As living organisms work continuously, they cannot afford to reach equilibrium. Hence the living state is a non-equilibrium steady-state to be able to perform work; living process is a constant effort to prevent falling into equilibrium. This is achieved by energy input. Metabolism provides a mechanism for the production of energy. Hence the living state and metabolism are synonymous. Without metabolism there cannot be a living state.

Long Answer Type Questions
Q1. Formation of enzyme-substrate complex (ES) is the first step in catalysed reactions. Describe the other steps till the formation of product.
Ans: The catalytic cycle of an enzyme action can be described in the following steps:
(1) First, the substrate binds to the active site of the enzyme, fitting into the active site.
(2) The binding of the substrate induces the enzyme to alter its shape, fitting more tightly around the substrate.
(3) The active site of the enzyme, now in close proximity of the substrate breaks the chemical bonds of the substrate and the new enzyme-product complex is formed.
(4) The enzyme releases the products of the reaction and the free enzyme is ready to bind to another molecule of the substrate and run through the catalytic cycle once again.

Q2. What are different classes of enzymes? Explain any two with the type of reaction they catalyse.
Ans: Enzymes are divided into 6 classes each with 4—13 subclasses and named accordingly by a four-digit number.
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Q3. Nucleic acids exhibit secondary structure. Describe through Watson-Crick Model.
Ans: Nucleic acids exhibit a wide .variety of secondary structures. For example, one of the secondary structures exhibited by DNA is the famous Watson- Crick model. This model says that DNA exists as a double helix. The two strands of polynucleotides are antiparallel, i.e. run in the opposite direction. The backbone is formed by the sugar—phosphate—sugar chain. The nitrogen bases are projected more or less perpendicular to this backbone but face inside. A and G of one strand compulsorily base pairs with T and C, respectively, on the other strand. There are two hydrogen bonds between A
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and T and three hydrogen bonds between G and C. Each strand appears like a helical staircase. Each step of ascent is represented by a pair of bases. At each step of ascent, the strand turns 36°. One full turn of the helical strand would involve ten steps or ten base pairs. Attempt drawing a line diagram. The pitch would be 34 A. The rise per base pair would be 3.4 A. This form of DNA with the above mentioned salient features is called B-DNA.

Q4. What is the difference between a nucleotide and nucleoside? Give two examples of each with their structure.
Ans: Living organisms have a number of carbon compounds in which heterocyclic rings can be found. Some of these are nitrogen bases—adenine, guanine, cytosine, uracil and thymine. When found attached to a sugar, they are called nucleosides. If a phosphate group is also found esterified to the sugar they are called nucleotides. Adenosine, guanosine, thymidine, uridine and cytidine are nucleosides. Adenylic acid, thymidylic acid, guanylic acid, uridylic acid and cytidylic acid are nucleotides.
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Q5. Describe various forms of lipid with a few examples.
Ans: Lipids are generally water insoluble. They could be simple fatty acids. A fatty acid has a carboxyl group attached to an R-group. The R-group could be a methyl (-CH3), or ethyl (—C2H5) or higher number of-CH2 groups (1 carbon to 19 carbons). For example, palmitic acid has 16 carbons including carboxyl carbon. Arachidonic acid has 20 carbon atoms including the carboxyl carbon. Fatty acids could be saturated (without double bond) or unsaturated (with one or more C=C double bonds). Another simple lipid is glycerol which is trihydroxy propane.
• Many lipids have both glycerol and fatty acids. Here the fatty acids are found esterified with glycerol. They can be then monoglycerides,
diglycerides and triglycerides. These are also called fats and oils based on melting point. Oils have lower melting point (e.g., gingely oil) and hence remain as oil in winters.
• Some lipids have phosphorous and a phosphorylated organic compound in them. These are phospholipids. They are found in cell membrane. Lecithin is one example. Some tissues especially the neural tissues have lipids with more complex structures.

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NCERT Exemplar Class 11 Biology Chapter 14 Respiration in Plants

NCERT Exemplar Class 11 Biology Chapter 14 Respiration in Plants

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

NCERT Exemplar Class 11 Biology Chapter 14 Respiration in Plants

Multiple Choice Questions

Q1. The ultimate electron acceptor of respiration in an aerobic organism is
(a) Cytochrome (b) Oxygen
(c) Hydrogen (d) Glucose
Ans: (b) The ultimate electron acceptor of respiration in an aerobic organism is oxygen.

Q2. Phosphorylation of glucose during glycolysis is catalyzed by
(a) Phosphoglucomutase (b) Phosphoglucoisomerase
(c) Hexokinase (d) Phosphorylase
Ans: (c) Phosphorylation of glucose during glycolysis is catalyzed by hexokinase.

Q3. Pyruvic acid, the key product of glycolysis can have many metabolic fates. Under aerobic condition it forms
(a) Lactic acid
(b) C02 + H20
(c) Acetyl CoA + C02
(d) Ethanol + CO
Ans: (c) Pyruvic acid, the key product of glycolysis can have many metabolic fates. Under aerobic condition it forms Acetyl CoA + CO2.

Q4. Electron Transport System (ETS) is located in mitochondrial
(a) Outer membrane
(b) Inter membrane space
(c) Inner membrane
(d) Matrix
Ans: (c) Electron Transport System (ETS) is located in mitochondrial inner membrane.

Q5. Which of the following exhibits the highest rate of respiration? .
(a) Growing shoot apex
(b) Germinating seed
(c) Root tip
(d) Leaf bud
Ans: (b) Germinating seed exhibits the highest rate of respiration.

Q6. Mitochondria are called powerhouses of the cell. Which of the following observations support this statement?
(a) Mitochondria synthesise ATP
(b) Mitochondria have a double membrane .
(c) The en2ymes of the Krebs’ cycle and the cytochromes are found in mitochondria
(d) Mitochondria are found in almost all plant and animal cells
Ans: (a) Mitochondria are called powerhouses of the cell because mitochondria synthesise ATP.

Q7. The end product of oxidative phosphorylation is
(a) NADH
(b) Oxygen
(c) ADP
(d) ATP + H20
Ans: (d) The end product of oxidative phosphorylation isATP+H2o.

Q8. Match the following and choose the correct option from those given below.

Column A Column B
A. Molecular oxygen (i) a-Ketoglutaric acid
B. Electron acceptor (ii) Hydrogen acceptor
C. Pyruvate dehydrogenase (iii) Cytochrome C
D. Decarboxylation (iv) Acetyl Co A

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

Ans. (a)

A. Molecular oxygen (ii) Hydrogen acceptor
B. Electron acceptor (iii) Cytochrome C
C. Pyruvate dehydrogenase (iv) Acetyl Co A
D. Decarboxylation (i) a-Ketoglutaric acid

Very Short Answer Type Questions

Q1. Energy is released during the oxidation of compounds in respiration. How is this energy stored and released as and when it is needed?
Ans: Energy released during the oxidation of compounds in respiration is immediately stored in ATP in the form of chemical bonds.
ADP + iP + energy —> ATP
As and when needed, this bond energy is broken and utilised
ATP —> ADP + iP + energy

Q2. Explain the term “Energy Currency”. Which substance acts as energy currency in plants and animals?
Ans: Every function of the cell requires energy. Energy currency stores and releases the energy as and when needed in the cell. ATP is called energy currency in both plants and animals.

Q3. Different substrates get oxidized during respiration. How does Respiratory Quotient (RQ) indicate which type of substrate, i.e., carbohydrate, fat or protein is getting oxidized?
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Q5. When does anaerobic respiration occur in man and yeast?
Ans: In alcoholic fermentation (by yeast), incomplete oxidation of glucose occurs under anaerobic conditions by sets of reactions where pyruvic acid is converted to ethanol and C02.
PA —> Ethanol + C02
In animal cells also, like muscles during exercise, when oxygen is inadequate for cellular respiration pyruvic acid is reduced to lactic acid.

Q6. Which of the following will release more energy on oxidation? Arrange them in ascending order?
a. 1 g of fat b. 1 g of protein
c. 1 g of glucose
d. 0.5 g of protein + 0.5 g glucose
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Q7. The product of aerobic glycolysis in skeletal muscle and anaerobic fermentation in yeast are respectively ________ and ______.
Ans. The product of aerobic glycolysis in skeletal muscle and anaerobic fermentation in yeast are respectively pyruvic acid and ethanol + CO2.

Short Answer Type Questions

Q1. If a person is feeling dizzy, glucose or fruit juice is given immediately but not a cheese sandwich, which might have more energy. Explain.
Ans: Glucose or fruit juice is absorbed easily through the alimentary canal. In the cells glucose is oxidised and energy is released immediately. A cheese sandwich provides energy after digestion and absorption which takes long time.

Q2. What is meant by the statement “aerobic respiration is more efficient”?
Ans: Aerobic respiration is more efficient because fermentation accounts for
only a partial breakdown of glucose, whereas in aerobic respiration it is completely degraded to C02 and H20. Also, in fermentation there is a net gain of only two molecules of ATP for each molecule of glucose degraded to pyruvic acid, whereas many more molecules of ATP are generated under aerobic conditions.

Q3. Pyruvic acid is the end product of glycolysis. What are the three metabolic fates of pyruvic acid under aerobic and anaerobic conditions? Write in the space provided in the diagram.
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Q4. The energy yield in terms of ATP is higher in aerobic respiration than during anaerobic respiration. Why is there anaerobic respiration even in organisms that live in aerobic condition like human beings and angiosperms?
Ans: Anaerobic respiration even in organisms that live in aerobic condition like human beings and angiosperms due to the oxygen deficiency. In our skeletal muscles during strenuous exercise oxygen deficiency leads to anaerobic respiration. In germinating seeds anaerobic respiration leads to release of energy for emerging the seedling from the soil.

Q5. Oxygen is an essential requirement for aerobic respiration but it enters the respiratory process at the end? Discuss.
Ans: Although the aerobic process of respiration takes place only in the presence of oxygen, the role of oxygen is limited to the terminal stage of the process. Yet, the presence of oxygen is vital, since it drives the whole process by removing hydrogen from the system. Oxygen acts as the final hydrogen acceptor.

Q6. Respiration is an energy releasing and enzymatically controlled catabolic process which involves a step-wise oxidative breakdown of organic substances inside living cells. In this statement about respiration explain the meaning of
1. Step-wise oxidative breakdown and
2. Organic substances (used as substrates).
Ans: 1. Step-wise oxidative breakdown: During oxidation within a cell, all the energy contained in respiratory substrates is not released free into
the cell, or in a single step. It is released in a series of slow step-wise reactions controlled by enzymes, and it is trapped as chemical energy in the form of ATP. Hence, it is important to understand that the energy released by oxidation in respiration is not (or rather cannot be) used directly but is used to synthesise ATP, which is broken down whenever (and wherever) energy needs to be utilised.
2. Organic substances (used as substrates): The compounds that are oxidised during this process are known as respiratory substrates. Usually carbohydrates are oxidised to release energy, but proteins, fats and even organic acids can be used as respiratory substances in some plants, under certain conditions.

Q7. Comment on the statement – Respiration is an energy producing process but ATP is being used in some steps of the process.
Ans: In the respiration pathway, there are some steps where energy is utilised for phosphorylation. For example, conversion of glucose to glucose-6-phosphate consume one ATP. But at the end of the respiratory process many more ATP are produced.

Q8. 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?
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Q9. Why is respiratory pathway referred to as an amphibolic pathway? Explain.
Ans: During breakdown and synthesis of protein, respiratory intermediates form the link. Breaking down processes within the living organism is catabolism, and synthesis is anabolism. Because the respiratory pathway is involved in both anabolism and catabolism, it would hence be better to consider the respiratory pathway as an amphibolic pathway rather than as a catabolic one.

Q10. We commonly call ATP as the energy currency of the cell. Can you think of some other energy carriers present in a cell? Name any two.
Ans: Yes, some other energy carriers are also present in a cell like GTP (guanosine triphosphate), ADP (adenosine diphosphate) and creatine phosphate.

Q11. ATP produced during glycolysis is a result of substrate level phosphorylation. Explain.
Ans: ATP produced during glycolysis is a result of substrate level phosphorylation because these ATP are produced without the electron transport system (ETS) and chemiosmosis.
During substrate level phosphorylation ATP is directly synthesised from ADP and inorganic phosphate (iP).

Q12. Do you know any step in the TCA cycle where there is substrate level phosphorylation. Which one?
Ans: During TCA cycle there is one step where substrate level phosphorylation takes place. This occurs during conversion of succinyl-CoA to succinic acid.

Q13. In a way green plants and cyanobacteria have synthesised all the food on the earth. Comment.
Ans: All the energy required for ‘life’ processes is obtained by oxidation of some macromolecules that we call ‘food’. Only green plants and cyanobacteria can prepare their own food; by the process of photosynthesis they trap light energy and convert it into chemical energy that is stored in the bonds of carbohydrates like glucose, sucrose and starch.

Q14.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?
Ans: The complete combustion of glucose, which produces C02 and H20 as end products, yields energy most of which is given out as heat.
C6H12O6  +  6O2  —> 6CO2+ 6H2O + Energy
If this energy is to be useful to the cell, it should be able to utilise it to synthesise other molecules that the cell requires. The strategy that the plant cell uses is to catabolise the glucose molecule in such a way that not all the liberated energy goes out as heat. The key is to oxidise glucose not in one step but. in several small steps enabling some steps to be just large enough such that the energy released can be coupled to ATP synthesis.

Q15. Respiration requires 02. How did the first cells on the earth manage to survive in an atmosphere that lacked 02?
Ans: During the process of respiration, oxygen is utilised, and carbon dioxide, water and energy are released as products. The combustion reaction requires oxygen. But some cells live where oxygen may or may not be available. There are sufficient reasons to believe that the first cells on this planet lived in an atmosphere that lacked oxygen. Even among present-day living organisms, we know of several that are adapted to anaerobic conditions.

Q16. It is known that red muscle fibres in animals can work for longer periods of time continuously. How is this possible?
Ans: Muscle contains a red coloured oxygen storing pigment called myoglobin. Myoglobin content is high in some of the muscles which gives a reddish appearance. Such muscles are called the Red fibres. These muscles also contain plenty of mitochondria which can utilise the large amount of oxygen stored in them for ATP production. These muscles, therefore, can also be called aerobic muscles.

Q17. The energy yield in terms of ATP is higher in aerobic respiration than during anaerobic respiration. Explain.
Ans: The energy yield in terms of ATP is higher in aerobic respiration than during anaerobic respiration because fermentation (anaerobic respiration) accounts
for only a partial breakdown erf glucose, whereas in aerobic respiration it is completely degraded to CO2 and H2O. So, in fermentation there is a net gain of only two molecules of ATP for each molecule of glucose degraded to pyruvic acid, whereas many more molecules of ATP are generated under aerobic conditions.

Q18. 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
Ans: a. Photosynthesis: RuBP carboxylase, PEPcase, ATPase
b. Respiration: Hexokinase, ATPase, Pyruvate dehydrogenase, Cytochrome oxidase
c. Both in photosynthesis and respiration: ATPase

Q19. How does a tree trunk exchange gases with the environment although it lacks stomata?
Ans: In stems, the ‘living’ cells are organised in thin layers inside and beneath the bark. They also have openings called lenticels. The cells in the interior are dead and provide only mechanical support. Thus, most cells of a plant have at least a part of their surface in contact with air. This is also facilitated by the loose packing of parenchyma cells in leaves, stems and roots, which provide an interconnected network of air spaces.

Q20. Write two energy yielding reactions of glycolysis.
Ans: The conversion of BPGA to 3-phosphoglyceric acid (PGA), is an energy yielding process; this energy is trapped by the formation of ATP. Another ATP is synthesised during the conversion of PEP to pyruvic acid.

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Q21. Name the site(s) of pyruvate synthesis. Also, write the chemical reaction wherein pyruvic acid dehydrogenase acts as a catalyst.
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Q22.Mention the important series of events of aerobic respiration that occurs in the matrix of the mitochondrion as well as one that takes place in the inner membrane of the mitochondrion.
Ans: For aerobic respiration to take place within the mitochondria, the final product of glycolysis, pyruvate is transported from the cytoplasm into the mitochondria. The crucial events in aerobic respiration are:
The complete oxidation of pyruvate by the stepwise removal of all the hydrogen atoms, leaving three molecules of C02.

The passing on of the electrons removed as part of the hydrogen atoms to molecular 02 with simultaneous synthesis of ATP. What is interesting to note is that the first process takes place in the matrix of the mitochondria while the second process is located on the inner membrane of the mitochondria.

Q23.Respiratory pathway is believed to be a catabolic pathway. However, nature of TCA cycle is amphibolic. Explain.
Ans: During breakdown and synthesis of protein, respiratory intermediates form the link. Breaking down processes within the living organism is catabolism, and synthesis is anabolism. Because the respiratory pathway is involved in both anabolism and catabolism, it would hence be better to consider the respiratory pathway as an amphibolic pathway rather than as a catabolic one.

Long Answer Type Questions
Q1. In the following flow chart, replace the symbols a,b,c and d with appropriate terms. Briefly explain the process and give any two application of it.
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Q2. Given below is a diagram showing ATP synthesis during aerobic respiration, replace the symbols A, B,C,D and E by appropriate terms given in the box.
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Q3. Oxygen is critical for aerobic respiration. Explain its role with respect to ETS.
Ans: Although the aerobic process of respiration takes place only in the presence of oxygen, the role of oxygen is limited to the terminal stage of the process. Yet, the presence of oxygen is vital, since it drives the whole process by removing hydrogen from the system. Oxygen acts as the final hydrogen acceptor. Unlike photophosphorylation where it is the light energy that is utilised for the production of proton gradient required for phosphorylation, in respiration it is the energy of oxidation-reduction utilised for the same process. It is for this reason that the process is called oxidative phosphorylation.

Q4. 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.
Ans: The Respiratory Balance Sheet:
It is possible to make calculations of the net gain of ATP for every glucose molecule oxidised; but in reality this can remain only a theoretical exercise. These calculations can be made only on certain assumptions that:
There is a sequential, orderly pathway functioning, with one substrate forming the next and with glycolysis, TCA cycle and ETS pathway following one after another.
The NADH synthesised in glycolysis is transferred into the mitochondria and undergoes oxidative phosphorylation.
None of the intermediates in the pathway are utilised to synthesise any other compound.
Only glucose is being respired – no other alternative substrates are entering in the pathway at any of the intermediary stages. But this kind of assumptions are not really valid in a living system; all pathways work simultaneously and do not take place one after another; substrates enter the pathways and are withdrawn from it as and when necessary; ATP is utilised as and when needed; enzymatic rates are controlled by multiple means. Yet, it is useful to do this exercise to appreciate the beauty and efficiency of the living system in extraction and storing energy. Hence, there can be a net gain of 36 ATP molecules during aerobic respiration of one molecule of glucose.

Q5. Give an account of Glycolysis. Where does it occur? What are the end products? Trace the fate of these products in both aerobic and anaerobic respiration.
Ans: Glycolysis occurs in the cytoplasm of the cell and is present in all living organisms. In this process, glucose undergoes partial oxidation to form two molecules of pyruvic acid. In plants, this glucose is derived from sucrose, which is the end product of photosynthesis, or from storage carbohydrates. Sucrose is converted into glucose and fructose by the enzyme, invertase, and these two monosaccharides readily enter the glycolytic pathway. Glucose and fructose are phosphorylated to give rise to glucose-6-phosphate by the activity of the enzyme hexokinase. This phosphorylated form of glucose then isomerises to produce fructose-6-phosphate. Subsequent steps of metabolism of glucose and fructose are same. In glycolysis, a chain of ten reactions,
under the control of different enzymes, takes place to produce pyruvate from glucose. Pyruvic acid is then the key product of glycolysis. The metabolic fate of pyruvate depends on the cellular need. There are three major ways in which different cells handle pyruvic acid produced by glycolysis. These are lactic acid fermentation, alcoholic fermentation and aerobic respiration. Fermentation takes place under anaerobic conditions in many prokaryotes and unicellular eukaryotes. For the complete oxidation of glucose to C02 and H20, however, organisms adopt Krebs’ cycle which is also called as aerobic respiration. This requires 02 supply.

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NCERT Exemplar Class 11 Biology Chapter 12 Mineral Nutrition

NCERT Exemplar Class 11 Biology Chapter 12 Mineral Nutrition

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

NCERT Exemplar Class 11 Biology Chapter 12 Mineral Nutrition

Multiple Choice Questions

Q1. Which one of the following roles is not a characteristic of an essential element?
(a) Being a component of biomolecules
(b) Changing the chemistry of soil
(c) Being a structural component of energy-related chemical compounds
(d) – Activation or inhibition of enzymes
Ans: (b)
(i) Essential elements as components of biomolecules and hence structural elements of cells.
(ii) Essential elements that are components of energy-related chemical compounds in plants.
(iii) Essential elements that activate or inhibit enzymes.
(iv) Some essential elements can alter the osmotic potential of a cell.

Q2. Which one of the following statements can best explain the term critical concentration of an essential element?
(a) Essential element concentration below which plant growth is retarded
(b) Essential element concentration below which plant growth becomes enhanced
(c) Essential element concentration below which plant remains in the vegetative phase
(d) None of the above
Ans: (a) The concentration of the essential element below which plant growth is retarded is termed as critical concentration.

Q3. Deficiency symptoms of an element tend to appear first in young leaves. It indicates that the element is relatively immobile. Which one of the following elemental deficiency would show such symptoms?
(a) Sulphur (b) Magnesium (c) Nitrogen (d) Potassium
Ans: (a) The deficiency symptoms tend to appear first in the young tissues whenever the elements are relatively immobile and are not transported out of the mature organs, e.g., S and Ca.

Q4. Which one of the following symptoms is not due to manganese toxicity in plants?
(a) Calcium translocation in shoot apex is inhibited
(b) Deficiency in both Iron and Nitrogen is induced
(c) Appearance of brown spot surrounded by chlorotic veins
(d) None of the above
Ans: (b) Excess of manganese may, in fact, induce deficiencies of iron, magnesium and calcium.

Q5. Reaction carried out by N2 fixing microbes include
NCERT Exemplar Class 11 Biology Chapter 12 Mineral Nutrition Img 1
Which of the following statements about those equations is not true?
(a) Step (i) is carried out by Nitrosomonas or Nitrococcus
(b) Step (ii) is carried out by Nitrobacter
(c) Both steps (i) and (ii) can be called nitrification
(d) Bacteria carrying out these steps are usually photoautotrophs
Ans:(d) Bacteria carrying out these steps are usually chemoautotrophs.

Q6. With regard to the Biological Nitrogen Fixation by Rhizobium in association with soyabean, which one of the following statement/statements does not hold true?
(a) Nitrogenase may require oxygen to its functioning
(b) Nitrogenase is Mo-Fe protein
(c) Leghaemoglobin is a pink coloured pigment
(d) Nitrogenase helps to convert N2 gas into two molecules of ammonia
Ans:(a) Nitrogenase is highly sensitive to molecular oxygen (02), thus requires
anaerobic conditions. Nodules have adaptations that ensure that the enzyme is protected from 02. To protect nitrogehase, nodule contains an 02-scavenger celled leghaemoglobin.

Q7. Match the element with its associated functions/roles and choose the correct option among the given below.

A. Boron (i)

V

Splitting of H20 to liberate 02 during photosynthesis
B. Manganese (ii) Needed for synthesis of auxins
C. Molybdenum (iii) Component of nitrogenase
D. Zinc (iv) Pollen germination
E. Iron (v) Component of ferredoxin

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

A. Boron (iv) Pollen germination
B. Manganese (0 Splitting of H20 to liberate 02 during photosynthesis
C. Molybdenum (iii) Component of nitrogenase
D. Zinc (ii) Needed for synthesis of auxins
E. Iron (v) Component of ferredoxin

Q8. Plants can be grown in (Tick the incorrect option).
(a) Soil with essential nutrients
(b) Water with essential nutrients
(c) Either water or soil with essential nutrients
(d) Water or soil without essential nutrients
Ans: (d) Plants can be grown in soil with essential nutrients, water with essential nutrients and either water or soil with essential nutrients.

Very Short Answer Type Questions
Q1. Name a plant, which accumulates silicon.
Ans: Rice, sugarcane, etc.

Q2. Mycorrohiza is a mutualistic association. How do the organisms involved in this association gain from each other?
Ans: Mycorrhiza is a symbiotic association between a fungus and the roots of a vascular plant. Through mycorrhization, the plant obtains phosphate and other minerals, such as zinc and copper, from the soil. The fungus obtains nutrients, such as sugars, from the plant root.

Q3. Nitrogen fixation is shown by prokaryotes and not eukaryotes. Comment?
Ans: Very few living organisms can utilise the nitrogen in the form N2, available
abundantly in the air. Only certain prokaryotic species are capable of fixing nitrogen. The enzyme, nitrogenase which is capable of nitrogen reduction is present exclusively in prokaryotes. Such microbes are called N2-fixers.

Q4. Carnivorous plants like Nepenthes and Venus fly trap have nutritional adaptations. Which nutrient do they especially obtain and from where?
Ans: Carnivorous plants grow in nitrogen deficient soil but they utilise their nitrogen by killing the insect by some special structure.

Q5. Think of a plant which lacks chlorophyll. From where will it obtain nutrition? Give an example of such a type of plant.
Ans: Cuscuta, a parasitic plant that is commonly found growing on hedge plants, has lost its chlorophyll and leaves in the course of evolution. It derives its nutrition from the host plant which it parasitises.

Q6. Name an insectivorous angiosperm.
Ans: Nepenthes, Utricularia, Drosera, Dionea, etc.

Q7. A farmer adds Azotobacter culture to soil before sowing maize. Which mineral element is being replenished?
Ans: Nitrogen

Q8. What type of conditions are created by leghaemoglobin in the root nodule of a legume?
Ans: Anaerobic condition

Q9. What is common to Nepenthes, Utricularia and Drosera with regard to mode of nutrition?
Ans: All are carnivorous plant (angiosperms).

Q10. Plants with zinc deficiency show reduced biosynthesis of .
Ans: Auxin

Q11. Yellowish edges appear in leaves deficient in .
Ans: K (potassium)

Q12. Name the macronutrient which is a component of all organic compounds but is not obtained from soil.
Ans: Carbon

Q13. Name one non-symbiotic nitrogen fixing prokaryote.
Ans: (i) Free-living (non-symbiotic) and non photosynthetic aerobic N2-fixing microbes: Azotobacter and Beijernickia.
(ii) Free-living and anaerobic N2-fixing microbes: Rhodospirillum, Bacillus polymyxa and Clostridium.

Q14. Rice fields produce an important green house gas. Name it.
Ans: CH4 (methane) .

Q15. Complete the equation for reductive amination
NCERT Exemplar Class 11 Biology Chapter 12 Mineral Nutrition Img 2

Q16. Excess of Mn in soil leads to deficiency of Ca, Mg and Fe. Justify.
Ans: Manganese competes with iron and magnesium for uptake and with magnesium for binding with enzymes. Manganese also inhibits calcium translocation in shoot apex. Therefore, excess of manganese may, in fact, induce deficiencies of iron, magnesium and calcium. Thus, what appears as symptoms of manganese toxicity may actually be the deficiency symptoms of iron, magnesium and calcium.

Short Answer Type Questions
Q1. How is sulphur important for plants? Name the amino acids in which it is present.
Ans: Sulphur, besides being present in some amino acids essential for protein synthesis, is also a constituent of several coenzymes, vitamins and ferredoxin which are involved in some biochemical pathway.

Q2. How are organisms like Pseudomonas and Thiobacillus of great significance in nitrogen cycle?
Ans: Pseudomonas and Thiobacillus carry out denitrification process wherein the nitrate present in the soil is reduced to nitrogen thus contributing to the atmospheric nitrogen.

Q3. Carefully observe the following figure:
NCERT Exemplar Class 11 Biology Chapter 12 Mineral Nutrition Img 3
a. Name the technique shown in the figure and the scientist who demonstrated this technique for the first time.
b. Name at least 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?
Ans: a. Hydroponics, Julius von Sachs.
b. Tomato, seedless cucumber, lettuce.
c. Aerating tube ensures adequate aeration of the root for optimum growth of the plant. The funnel is used to release water and nutrients into the container with nutrient solution. This solution needs to be replaced every day or two for maximum growth.

Q4. Name the most crucial enzyme found in root nodules for N2 fixation. Does it require a special pink coloured pigment for its functioning? Elaborate.
Ans: Nitrogenase. Yes, it does require the presence of a pink-coloured pigment in the nodule called leghaemoglobin for its functioning. This pigment helps in scavenging oxygen as nitrogenase functions under anaerobic condition.

Q5. How are the terms ‘critical concentration’ and ‘deficient’ different from each other in terms of concentration of an essential element in plants? Can you find the values of ‘critical concentration’ and ‘deficient’ for minerals – Fe and Zn?
Ans: The concentration of the essential element below which plant growth is retarded is termed as critical concentration. The element is said to be deficient when present below the critical concentration. Yes. one can find the values of ‘critical concentration’ and ‘deficient’ for minerals – Fe and Zn through the hydroponics technique.

Q6. Carnivorous plants exhibit nutritional adaptation. Citing an example explain this fact.
Ans: Carnivorous plants have green leaves so they are autotrophic but they grow in nitrogen deficient soil. For nitrogen requirement they capture and digest the insects so they are partially heterotrophic nature.

Q7. A farrper adds/supplies 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.
Ans: This is due to the manganese toxicity. Many a times, excess of an element may inhibit the uptake of another element. For example, the prominent symptom of manganese toxicity is the appearance of brown spots surrounded by chlorotic veins. Manganese competes with iron and magnesium for uptake and with magnesium for binding with enzymes. Manganese also inhibits calcium translocation in shoot apex. Therefore, excess of manganese may, in fact, induce deficiencies of iron, magnesium and calcium.
• The farmer should not supplies Mn to his field.

Long Answer Type Questions
Q1. It is observed that deficiency of a particular element showed its symptoms initially in older leaves and then in younger leaves.
a. Does it indicate that the element is actively mobilised or relatively immobile?
b. Name two elements which are highly mobile and two which are relatively immobile.
c. How is the aspect of mobility of elements important to horticulture and agriculture?
Ans: a. It is actively mobilised.
b. Highly mobile—nitrogen, magnesium Relatively immobile—calcium, boron
c. Symptoms of deficiency of mobile elements are more pronounced in older leaves and symptoms of deficiency of relatively immobile element appear first in younger leaves. This information can be utilised by horticulturist and agriculturist to get a broad idea of the deficiency elements in plants.

Q2. 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 mycorrihiza 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.
Ans:  a. Yes, one can artificially induce the property of nitrogen fixation in a plant—leguminous or non-leguminous through genetic engineering which involves introduction of specific genes to the host plant that synthesises nitrogenase enzymes.
b. Symbiotic relationship
c. Yes, it is necessary for a microbe to be in close association with a plant to provide mineral nutrition as seen in leguminous plants. Species of rod-shaped Rhizobium has such relationship with the roots of several legumes such as alfalfa, sweet clover, sweet pea, lentils, garden pea, broad bean, clover beans, etc. The most common association on roots is as nodules. The nodule contains all the necessary biochemical components, such as the enzyme nitrogenase and leghaemoglobin. The enzyme nitrogenase is an Mo-Fe protein and catalyses the conversion of atmospheric nitrogen to ammonia.

Q3. What are essential elements for plants? Give the criteria of essentiality. How are minerals classified depending upon the amount in which they are needed by the plants?
Ans: Essential elements: carbon, hydrogen, oxygen, nitrogen, phosphorous, sulphur, potassium, calcium, magnesium, iron, manganese, copper, molybdenum, zinc, boron, chlorine and nickel.
Criteria for Essentiality:
The criteria for essentiality of an element are given below:
(a) The element must be absolutely necessary for supporting normal growth and reproduction. In the absence of the element, the plants do not complete their life cycle or set the seeds.
(b) The requirement of the element must be specific and not replaceable by another element. In other words, deficiency of any one element cannot be met by supplying some other element.
(c) The element must be directly involved in the metabolism of the plant. Based upon the above criteria only a few elements have been found to be absolutely essential for plant growth and metabolism. These elements are further divided into two broad categories based on their quantitative requirements.
(i) Macronutrients and
(ii) Micronutrients
Macronutrients are generally present in plant tissues in large amounts (in excess of 10 mmole kg~’ of dry matter). The macronutrients include carbon, hydrogen, oxygen, nitrogen, phosphorous, sulphur, potassium, calcium and magnesium. Of these, carbon, hydrogen and oxygen are mainly obtained from C02 and H20, while the others are absorbed from the soil as mineral nutrition.
Micronutrients or trace elements are needed in very small amounts (less than 10 mmole kg-1 of dry matter). These include iron, manganese, copper, molybdenum, zinc, boron, chlorine and nickel.

Q4. With the help of examples describe the classification of essential elements based on the function they perform.
Ans: Essential elements can also be grouped into four broad categories on the basis of their diverse functions. These categories are:
(i) Essential elements as components of biomolecules and hence structural elements of cells (e.g., carbon, hydrogen, oxygen and nitrogen).
(ii) Essential elements that are components of energy-related chemical compounds in plants (e.g., magnesium in chlorophyll and phosphorous in ATP).
(iii) Essential elements that activate or inhibit enzymes, for example Mg2+ is an activator for both ribulose bisphosphate carboxylase/oxygenase and phosphoenol pyruvate carboxylase, both of which are critical enzymes in photosynthetic carbon fixation; Zn +is an activator of alcohol dehydrogenase and Mo of nitrogenase during nitrogen metabolism.
(iv) Some essential elements can alter the osmotic potential of a cell. Potassium plays an important role in the opening and closing of stomata. Minerals also play role as solutes in determining the water potential of a cell.

Q5. We know that plants require nutrients. If we supply these in excess, will it be beneficial to the plants? If yes, how/ If no, why?
Ans: No, excess supply of nutrients is not beneficial for the plants. It is toxic to the plants. Any mineral ion concentration in tissues that reduces the dry weight of tissues by about 10% is considered toxic. Such critical concentrations vary widely among different micronutrients. The toxicity symptoms are difficult to identify. Toxicity levels for any element also vary for different plants. Many a times, excess of an element may inhibit the uptake of another element. For example, the prominent symptom of manganese toxicity is the appearance of brown spots surrounded by chlorotic veins. It is important to know that manganese competes with iron and magnesium for uptake and with magnesium for binding with enzymes. Manganese also inhibits calcium translocation in shoot apex. Therefore, excess of manganese may, in fact, induce deficiencies of iron, magnesium and calcium.

Q6. Trace the events starting from the coming in contact of Rhizobiwn to a leguminous root till nodule formation. Add a note on importance of leghaemoglobin.
Ans: Nodule Formation: Nodule formation involves a sequence of multiple interactions between Rhizobium and roots of the host plant. Principal stages in the nodule formation are summarised as follows: Rhizobia multiply and colonise the surroundings of roots and get attached to epidermal and root-hair cells. The root-hairs curl and the bacteria invade the root-hairs. An infection thread is produced carrying the bacteria into the cortex of the root, where they initiate the nodule formation in the cortex of the root. Then the bacteria are released from the thread into the cells which leads to the differentiation of specialised nitrogen fixing cells. The nodule thus formed, establishes a direct vascular connection with the host for exchange of nutrients.
Importance of leghemoglobin: The enzyme nitrogenase is highly sensitive to the molecular oxygen; it requires anaerobic conditions. The nodules have adaptations that ensure that the enzyme is protected from oxygen. To protect these enzymes, the nodule contains an oxygen scavenger called leghaemoglobin. It is interesting to note that these microbes live as aerobes under free-living conditions (where nitrogenase is not operational), but during nitrogen-fixing events, they become anaerobic (thus protecting the nitrogenase enzyme).

Q7. Give the biochemical events occurring in the root nodule of a pulse plant. What is the end product? What is its fate?
Ans: The nodule contains all the necessary biochemical components, such as the enzyme nitrogenase and leghaemoglobin. The enzyme nitrogenase is an Mo-Fe protein and catalyses the conversion of atmospheric nitrogen to ammonia, the first stable product of nitrogen fixation.
NCERT Exemplar Class 11 Biology Chapter 12 Mineral Nutrition Img 4
• Glutamic acid is the main amino acid from which transfer of NH2 (amino group) takes place and other amino acids are formed through transamination. Enzyme transaminase catalyses all such reactions.
For example, the two most important amides (asparagine and glutamine) found in plants are a structural part of proteins.
• Asparagine formed from aspartic acid and glutamine is formed from glutamic acid by addition of amino group to each. The hydroxyl part of the acid is replaced by another NH2 radicle.

Q8. Hydroponics have been shown to be a successful technique for growing of plants. Yet most of the crops are still grown on land. Why?
Ans: The technique of growing plants in anutrient solution is known as hydroponics. Since, then a number of improvised methods have been employed to try and determine the mineral nutrients essential for plants. The essence of all these methods involves the culture of plants in a soil-free, defined mineral solution. These methods require purified water and mineral nutrient salts. Hydroponics has been successfully employed as a technique for the commercial production of vegetables such as tomato, seedless cucumber and lettuce.
Yet most of the crops are still grown on land because it must be emphasised that the nutrient solutions must be adequately aerated to obtain the optimum growth. On land no such conditions are required.

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

NCERT Exemplar Class 11 Biology Chapter 15 Plant Growth and Development

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

NCERT Exemplar Class 11 Biology Chapter 15 Plant Growth and Development

Multiple Choice Questions

Q1. Ethylene is used for
(a) Retarding ripening of tomatoes
(b) Hastening of ripening of fruits
(c) Slowing down ripening of apples
(d) Both (b) and (c)
Ans: (b) Ethylene is used for hastening of ripening of fruits.

Q2. Coconut water contains
(a) ABA
(b) auxin
(c) cytokinin
(d) gibberellin
Ans: (c) Coconut water contains cytokinin.

Q3. The effect of apical dominance can be overcome by which of the following hormone?
(a) IAA
(b) Ethylene
(c) Gibberellin
(d) Cytokinin
Ans: (d) The effect of apical dominance can be overcome by cytokinin hormone.

Q4. Match the following:

A. IAA (0 Herring sperm DNA
B. ABA (ii) Bolting
C. Ethylene (iii) Stomatal closure
D. GA (iv) Weed-free lawns
E. Cytokinins (v) Ripening of fruits

Ans: (a)

A. IAA (iv) Weed-free lawns
B. ABA (iii) Stomatal closure
C. Ethylene (v) Ripening of fruits
D. GA (ii) Bolting
E. Cytokinins (i) Herring sperm DNA

 

Q5. Apples are generally wrapped in waxed paper to
(a) prevent sunlight for changing its colour
(b) prevent aerobic respiration by checking the entry of 02
(c) prevent ethylene formation due to injury
(d) make the apples look attractive
Ans: (b) Apples are generally wrapped in waxed paper to prevent aerobic respiration by checking the entry of 02.

Q6. Growth can be measured in various ways. Which of these can be used as parameters to measure growth?
(a) Increase in cell number
(b) Increase in cell size
(c) Increase in length and weight
(d) All the above
Ans: (d) Growth can be measured in various ways. Increase in cell number, increase in cell size and increase in length and weight are used as parameters to measure growth.

Q7. The term synergistic action of hormones refers to
(a) when two hormones act together but bring about opposite effects
(b) when two hormones act together and contribute to the same function
(c) when one hormone affects more than one function
(d) when many hormones bring about any one function
Ans: (b) The term synergistic action of hormones refers to when two hormones act together and contribute to the same function.

Q8. Plasticity in plant growth means that
(a) plant roots are extensible
(b) plant development is dependent on the environment
(c) stems can extend
(d) none of the above
Ans: (b) Plasticity in plant growth means that plant development is dependent on the environment.

Q9. To increase sugar production in sugarcanes, they are sprayed with
(a) IAA
(b) cytokinin
(c) gibberellin
(d) ethylene
Ans: (c) To increase sugar production in sugarcanes, they are sprayed with gibberellin.

Q10. ABA acts antagonistic to
(a) ethylene
(b) cytokinin
(c) gibberellic acid
(d) IAA
Ans: (c) ABA acts antagonistic to gibberellic acid.

Q11. Monocarpic plants are those which
(a) bear flowers with one ovary
(b) flower once and die
(c) bear only one flower
(d) all of the above
Ans: (b) Monocarpic plants are flower once and die.

Q12. The photoperiod in plants is perceived at
(a) meristem
(b) flower
(c) floral buds
(d) leaves
Ans: (d) The photoperiod in plants is perceived at leaves.

Very Short Answer type Questions
Q1. Fill in 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 to
obtain a good callus in plant tissue culture.
d. _________of a vegetative plants are the sites of photoperiodic perception.
Ans: a. Exponential/log phase of an S-curve.
b. Auxin/IAA
c. CytokininlKinetinl6 BAP/Zeatinletc.
d. Leaves

Q2. PJ.ant growth substances (PGS) have innumerable practical applications. Name the PGS you should use to
a. increase yield of sugar cane b. promote lateral shoot growth
c. cause sprouting of potato tuber d. inhibit seed germination
Ans: a. GA3/gibberellinlgibberellic acid
b. CytokininlzeatinlkinetinJKn
c. C2H4/Ethylene
d. ABA/Abscisic acid

Q3. A primary root grows from 5 cm to 19 cm in a week. Calculate the growth rate and relative growth rate over the period.
NCERT Exemplar Class 11 Biology Chapter 15 Plant Growth and Development Img 1

Q4. 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 priytohormone.
b. Which property of Gibberellin caused foolish seedling disease in rice?
Ans: a. GA3 is used to speed up the malting process in brewing industry. Gibberellins also promote bolting (internode elongation just prior to flowering) in beet, cabbages and many plants with rosette habit.
b. Gibberellin causes foolish seedling disease in rice because it has the property of internode elongation.

Q5. Gibberellins promote the formation of _________ flowers on genetically _________ plants in Cannabis whereas ethylene promotes formation of _______ flowers
on genetically ____ plants.
Ans: Gibberellins promote the formation of male flowers on genetically female plants in Cannabis whereas ethylene promotes formation of female flowers on genetically male plants.

Q6. Classify the following plants into Long-Day Plants (LDP), Short-Day Plants (SDP) and Day-Neutral Plants (DNP) Xanthium, Henbane (Hyoscyamus niger), Spinach, Rice, Strawberry, Bryophyllum, Sunflower, Tomato, Maize.
Ans: Xanthium: (SDP)
Henbane (Hyoscyamus niger): (LDP)
Spinach: (LDP)
Rice: (SDP)
Strawberry: (SDP)
Bryophyllum: LSDP (Long short day plants)
Sunflower: (DNP)
Tomato: (DNP)
Maize: (DNP)

Q7. A farmer grows cucumber plants in his field. He wants to increase the number of female flowers in them. Which can plant growth regulator be applied to achieve this?
Ans: Ethylene (C2H4)

Q8. Where are the following hormones synthesised in plants?
a. IAA
b. Gibberellins
c. Cytokinins
Ans: a. IAA: Shoot tips and apical bud
b. Gibberellins: Root tips and young leaves
c. Cytokinins: Meristematic zones like root tips

Q9. In botanical gardens and tea gardens, gardeners trim the plants regularly so that they remain bushy. Does this practice have any scientific explanation?
Ans: Mostly in higher plants, the growing apical bud inhibits the growth of the lateral (axillary) buds, a phenomenon called apical dominance. Removal of shoot tips (decapitation) usually results in the growth of lateral buds. Hence, in botanical gardens and tea gardens, gardeners trim the plants regularly so that they remain bushy.

Q10. Light plays an important role ‘in the life of all organisms. Name any three physiological processes in plants which are affected by light.
Ans: Photoperiodism, phototropism and photosynthesis.

Q11. In the figure of Sigmoid growth curve given below, label segments 1,2 and 3
NCERT Exemplar Class 11 Biology Chapter 15 Plant Growth and Development Img 2

Q12. Growth is one of the characteristics of all living organisms. Do unicellular organism also grow? If so, what are the parameters?
Ans: Increase in mass and increase in number of individuals are twin characteristics of growth. A multicellular organism grows by cell division. Unicellular organisms grow by cell division. One can easily observe this in in vitro cultures by simply counting the number of cells under the microscope.

Q13. The rice seedlings infected with fungus Gibberella fujikuroi is called foolish seedlings? What was the reason behind it?
Ans: The rice seedling infected with fungus Gibberella fujikuroi is called foolish seedlings because the fungus secreted a hormone gibberellin and causes excessive growth of rice plants. Plants become tall but unable to produce seeds so they are called foolish.

Short Answer Type Questions

Q1. Nicotiana tobacum, a short-day plant, when exposed to more than critical period of light fails to flower. Explain.
Ans: a. Some plants require a periodic exposure to alternate light and dark for its flowering response. This phenomenon is termed photoperiodism.
b. The requirement of light exposure is critical. The SDP plants, when exposed to light period in excess of critical period fail to flower,
c. Those plants which require exposure to light period at critical or more than critical period for its flowering response are called long-dayplant.
d. Nicotiana tabacum fails to flower if exposed to more than critical period of light because it is an SDP.

Q2. What are the structural characteristics of
a. Meristematic cells near root tip
b. The cells in the elongation zone of the root
Ans: a. The meristematic cells near root tip are characterised by
• rich protoplasm
• large conspicuous nucleus
• thin and cellulosic cell wall -primary in nature
• fewer vacuoles
• greater number of mitochondria
• numerous (abundant) plasmodesmata
b. The cells in the elongation zone of a root are characterized by
• increased vacuolation
• enlarged size/dimension
• deposition of new cellulosic cell walls

Q3. Does the growth pattern in plants differ from that in animals? Do all the parts of plant grow indefinitely? If not, name the regions of plant, which can grow indefinitely.
Ans: Yes, the growth pattern in plants differ from that in animals. Plant growth is unique because plants retain the capacity for unlimited growth throughout their life. This ability of the plants is due to the presence of meristems at certain locations in their body. The cells of such meristems have the capacity to divide and self-perpetuate. The product, however, soon loses the capacity to divide and such cells make up the plant body. This form of growth wherein new cells are always being added to the plant body by the activity of the meristem is called the open form of growth.

Q4. 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
Ans: a. Differentiation: The cells derived from root apical and shoot-apical meristems and cambium differentiate and mature to perform specific functions. This act leading to maturation is termed as differentiation. During differentiation, cells undergo few to major structural changes both in their cell walls and protoplasm. For example, to form a tracheary element, the cells would lose their protoplasm. They also develop a very strong, elastic, lignocellulosic secondary cell walls, to carry water to long distances even under extreme tension.
b. De-differentiation: The living differentiated cells that by now have lost the capacity to divide can regain the capacity of division under certain conditions. This phenomenon is termed as de-differentiation. For example, formation of meristems – interfascicular cambium and cork cambium from fully differentiated parenchyma cells.
c. Re-differentiation: While doing de-differentiation, such meristems/ tissues are able to divide and produce cells that once again lose the capacity to divide but mature to perform specific functions, i.e., get re-differentiated, e.g., secondary xylem and secondary cortex.

Q5. Auxins are growth hormones capable of promoting cell elongation. They have been used in horticulture to promote growth, flowering and rooting. Write a line to explain the meaning of the following terms related to auxins.
a. Auxin precursors
b. Anti-auxins
c. Synthetic auxins
Ans: a. Auxin precursors: The substances that produce the auxin are called auxin precursors. For example, tryptophan is the auxin precursor.
b. Anti-auxins: The substances which inhibit the synthesis or transport of auxin are called anti-auxins. For example, TIBA (Triiodobenzoic acid) is anti-auxin compound.
c. Synthetic auxins: The artificially synthesised chemicals having auxin-like property are called synthetic auxins. For example, NAA
– (Naphthalene acetic acid) and 2, 4-D (2, 4-Dichloro phenoxyacetic acid).

Q6. The role of ethylene and abscisic acid is both positive and negative. Justify the statement.
Ans: Positive roles of ethylene: Influences of ethylene on plants include horizontal growth of seedlings, swelling of the axis and apical hook formation in dicot seedlings. Ethylene breaks seed and bud dormancy, initiates germination in peanut seeds, sprouting of potato tubers.
• Negative roles of ethylene: Ethylene promotes senescence and abscission of plant organs especially of leaves and flowers.
• Positive roles of abscisic acid: ABA plays an important role in seed development and maturation.
• Negative roles of abscisic acid: It acts as a general plant growth inhibitor and an inhibitor of plant metabolism. ABA inhibits seed germination. ABA stimulates the closure of stomata in the epidermis and increases the tolerance of plants to various kinds of stresses.

Q7. While experimentation, why do you think it is difficult to assign any effect seen to any single hormone?
Ans: Many hormones have synergistic and antagonistic effect with each other. So, while experimentation, it is difficult to assign any effect seen to any single hormone.

Q8. What is the mechanism underlying the phenomenon by which the terminal/ apical bud suppresses the growth of lateral buds? Suggest measures to overcome this phenomenon.
Ans: The phenomenon by which the terminal/apical bud suppresses the growth of lateral buds is called apical dominance. Apical dominance is due to auxin hormone secreted by apical buds. This can be overcome by decapitation (removal of apical buds) or the application of cytokinin.

Q9. 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?
Ans: In plants, the hormones are formed by different tissues like shoot tips, root tips, meristematic tissues, leaves and apical buds, etc.
Hormones are translocated to the site of activity by vascular tissues (xylem and phloem)

Q10. Many discoveries in science have been accidental. This is true for plant hormones also. Can you justify this statement by giving an example? Also what term is used for such accidental findings?
Ans: The discovery of each of the five major groups of PGRs have been accidental. All this started with the observation of Charles Darwin and his son Francis Darwin when they observed that the coleoptiles of canary grass responded to unilateral illumination by growing towards the light source (phototropism). After a series of experiments, it was concluded that the tip of coleoptile was the site of transmittable influence that caused the bending of the entire coleoptile. Auxin was isolated by F.W. Went from tips of coleoptiles of oat seedlings. Such accidental findings or discoveries are known as serendipity.

Q11. To get carpet-like grass lawn are mowed regularly. Is there any scientific explanation for this?
Ans: To get a carpet-like grass lawns are mowed regularly because mowing causes decapitation which promotes the growth of lateral buds.

Q12. 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?
Ans: The meristematic cells are rich in protoplasm, possess large conspicuous nuclei. Their cell walls are primary in nature, thin and cellulosic with abundant plasmodesmatal connections. Cells incapable of dividing attain their maximal size in terms of wall thickening and protoplasmic modifications.

Q13. 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))
Ans: A rubber band stretches and reverts back to its original position, it is due to elasticity. Bubble gum stretches, but it would not return to its original position, this is due to plasticity.
• The meristematic cells are rich in protoplasm, possess large conspicuous nuclei. Their cell walls are primary in nature, thin, cellulosic and elastic with abundant plasmodesmatal connections.
• Plants follow different pathways in response to environment or phases of life to form different kinds of structures. This ability is called plasticity,
e. g., heterophylly in cotton, coriander and larkspur. In such plants, the leaves of the juvenile plant are different in shape from those in mature plants.

Q14. Label the diagram
a. This is which part of a dicotyledonous plant?
b. If we remove part 1 from the plant, what will happen?

NCERT Exemplar Class 11 Biology Chapter 15 Plant Growth and Development Img 3

Q15. 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?
Ans: Plant growth is unique because plants retain the capacity for unlimited growth throughout their life. This ability of the plants is due to the presence of meristems at certain locations in their body. The cells of such meristems have the capacity to divide and self-perpetuate. The product, however, soon loses the capacity to divide and such cells make up the plant body. This form of growth wherein new cells are always being added to the plant body by the
Plant Growtli and Development 167
activity of the meristem is called the open form of growth. Yes, this statement hold true for sponges also.

Q16. Define parthenocarpy. Name the plant hormone used to induce parthenocarpy.
Ans: Most fruits however develop only from the ovary and are called true fruits.
Although in most of the species, fruits are the results of fertilisation, there are a few species in which fruits develop without fertilisation. Such fruits are called parthenocarpic fruits. Banana is one such example. Parthenocarpy can be induced through the application of growth hormones (like gibberellin and auxin) and such fruits are seedless. Auxins induce parthenocarpy in tomatoes.

Q17. While eating watermelons, all of us wish it was seedless. As a plant physiologist can you suggest any method by which this can be achieved.
Ans: This can be achieved through parthenocarpy. Parthenocarpy can be induced through the application of growth hormones (like gibberellin and auxin) and such fruits are seedless.

Q18. A gardener finds some broad-leaved dicot weeds growing in his lawns. What can be done to get rid of the weeds efficiently?
Ans: The dicotyledonous plants grow by their apical shoot meristems while grasses (which make lawns) possess intercalary meristem. Certain auxins, such as synthetic 2, 4-Dichlorophenoxyacetic acid (2,4-D) when applied in excess can damage the shoot apical meristems but they do not cause any damage to the- intercalary meristems. Thus, when 2, 4-D is sprayed on lawns, only the dicots get killed and the lawns become free of weeds.

Q19. On germination a seed first produces shoots with leaves, flowers appear later,
a. Why do you think this happens?
b. How is this advantageous to the plant?
Ans: a. All organisms have to reach a certain stage of growth and maturity in their life, before they can reproduce sexually. That period of growth is called the juvenile phase. It is known as vegetative phase in plants. This phase is of variable durations in different organisms. The end of juvenile/ vegetative phase which marks the beginning of the reproductive phase can be seen easily in the higher plants when they come to flower.
b. This enables the plant to have sufficient time to reach maturity.

Q20. Fill in the blanks:
a. Maximum growth is observed in phase.
b. Apical dominance is due to .
c. initiate rooting. .
d. Pigment involved in Photoperception in flowering plants is .
Ans: a. Maximum growth is observed in log/exponential phase.
b. Apical dominance is due to auxin.
c. Auxins initiate rooting.
d. Pigment involved in Photoperception in flowering plants is
phytochrome.

Long Answer Type Questions ‘
Q1. 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?
Ans: There are plants for which flowering is either quantitatively or qualitatively dependent on exposure to low temperature. This phenomenon is termed vernalisation. It prevents precocious reproductive development late in the growing season, and enables the plant to have sufficient time to reach maturity. Vernalisation refers specially to the promotion of flowering by a period of low temperature. Some important food plants, wheat, barley, rye have two kinds of varieties: winter and spring varieties. The ‘spring’ variety are normally planted in the spring and come to flower and produce grain before the end of the growing season. Winter varieties, however, if planted in spring would normally fail to flower or produce mature grain within a span of a flowering season. Hence, they are planted in autumn. They germinate, and over winter come out as small seedlings, resume growth in the spring, and are harvested usually around mid-summer.

Q2. It is known that some varieties of wheat are sown in autumn but are harvested around next mid-summer.
a. ■ What could be the probable reason for this?
b. What term is used for this promotion of flowering under low temperature?
c. Which plant hormone can replace the cold treatment?
Ans: a. Winter varieties, if planted in spring would normally fail to flower or produce mature grain within a span of a flowering season. Hence, they are planted in autumn. They germinate, and over winter come out as small seedlings, resume growth in the spring, and are harvested usually around mid-summer.
b. Vernalisation
c. Gibberellin

Q3. 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
Ans: a. Gaseous in nature: Ethylene (C2H4)
b. Responsible for phototropism: Auxin
c. Induces femaleness in flowers of cucumber: Ethylene (C2H4)
d. Used for killing weeds (dicots): Auxin
e. Induces flowering in long day plants: Gibberellin

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NCERT Exemplar Class 11 Biology Chapter 4 Animal Kingdom

NCERT Exemplar Class 11 Biology Chapter 4 Animal Kingdom are part of NCERT Exemplar Class 11 Biology. Here we have given NCERT Exemplar Class 11 Biology Chapter 4 Animal Kingdom.

NCERT Exemplar Class 11 Biology Chapter 4 Animal Kingdom

Multiple Choice Questions

Q1. In some animal groups, the body is found divided into compartments with serial repetition of at least some organs. This characteristic feature is called
(a) Segmentation (b) Metamerism
(c) Metagenesis (d) Metamorphosis
Ans: (b) Metamerism: In some animals, the body is externally and internally divided into segments with a serial repetition of at least some organs. For example, in earthworm, the body shows this pattern called metameric segmentation and the phenomenon is known as metamerism or true segmentation. Metamerism is found in 3 animal phylums—Annelida, Arthropoda and Chordata.

Q2. Given below are the types of cells present in some animals. Which of the following cells can differentiate to perform different functions?
(a) Choanocytes (b) Interstitial cells
(c) Gastrodermal cells (d) Nematocytes
Ans: (b) Interstitial cells can differentiate to perform different functions. Choanocytes is the characteristics cells of porifera.
Gastrodermal cells and Nematocytes are found in Hydra.

Q3. Which one of the following sets of animals share a four chambered heart?
(a) Amphibian, Reptiles, Birds (b) Crocodiles, Birds, Mammals
(c) Crocodiles, Lizards, Turtles (d) Lizards, Mammals, Birds
Ans: (b) Crocodiles, Birds, Mammals are the set of animals that share a four chambered heart
• Amphibian—two chambered heart
• Reptiles and lizards—three chambered heart except crocodiles

Q4. Which of the following pairs of animals has non-glandular skin?
(a) Snake and Frog (b) Chameleon and Turtle
(c) Frog and Pigeon (d) Crocodile and Tiger
Ans: (b) Chameleon and Turtle has non-glandular skin.

Q5. Birds and mammals share one of the following characteristics as a common feature.
(a) Pigmented skin (b) Pneumatic bones
(c) Viviparity (d) Warm blooded nature
Ans: (d) Birds and mammals both are warm blooded.

Q6. Which one of the following sets of animals belong to a single taxonomic group?
(a) Cuttlefish, Jellyfish, Silverfish, Dogfish, Starfish
(b) Bat, Pigeon, Butterfly
(c) Monkey, Chimpanzee, Man
(d) Silkworm, Tapeworm, Earthworm
Ans: (c) Monkey, Chimpanzee and Man belong to a single taxonomic group mammalia (class).

Q7. Which one of the following statements is incorrect? ,
(a) Mesoglea is present in between ectoderm and endoderm in Obelia
(b) Asterias exhibits radial symmetry
(c) Fasciola is a pseudocoelomate animal
(d) Taenia is a triploblastic animal
Ans: (c) Fasciola is an acoelomate animal.

Q8. Which one of the following statements is incorrect?
(a) In cockroaches and prawns excretion of waste material occurs through Malpighian tubules.
(b) In ctenophore, locomotion is mediated by comb plates.
(c) In Fasciola flame cells take part in excretion.
(d) Earthworms are hermaphrodites and yet cross fertilization take place among them.
Ans: (a)
• In cockroaches excretion of waste material occurs through Malpighian tubules.
• In prawns excretion of waste material occurs through green glands or antennal glands.

Q9. Which one of the following is oviparous?
(a) Platypus (b) Flying fox (Bat)
(c) Elephant (d) Whale
Ans: (a) Prototherians have evolved from reptiles. Mammary gland lacks nipples or teats.Corpus callosum is absent. Prototherians are oviparous/egg laying mammals. E.g.: Omithorynchus (Duck-billed platypus) is monotreme mammal.

Q10. Which one of the following is not a poisonous snake?
(a) Cobra (b) Viper (c) Python (d) Krait
Ans: (c) Python (Ajgar) is a non-poisonous snake. Poisonous snakes: Naja (Cobra), Bangarus coeruleus (Krait), Vipera (Viper), Hydrophis (Sea snake)
Najahaunaha: King Cobra. King Cobra is only snake which builds its nest. Nest forming snake is terrestrial and poisonous.

Q11. Match the following list of animals with their level of organization.

Division of Labour Animal
A. Organ level (i) Pheretima
B. Cellular aggregate level (ii) Fasciola
C. Tissue level (iii) Spongilla
D. Organ system level (iv) Obelia

Choose the correct match showing division of labour with animal example.

  • B—(i), C—(ii), D—(iii), A—(iv)
  • B—(i), D—(ii), C—(iii), A—(iv)
  • D—(i), A—(ii), B—(iii), C—(iv)
  • A—(i), D—(ii), C—(iii), B—(iv)

Ans. (c)

A. Organ level (ii) Fasciola
B. Cellular aggregate level (iii) Spongilla
C. Tissue level (iv) Obelia
D. Organ system level (i) Pheretima

Q12. Body cavity is the cavity present between body wall and gut wall. In some animals the body cavity is not lined by mesoderm. Such animals are called (a) Acoelomate (b) Pseudocoelomate .
(c) Coelomate (d) Haemocoelomate
Ans: (b)
(i) Acoelomate: The animals in which the body cavity is absent are
called acoelomates, e.g.: Porifers, Coelentrates, Ctenophores and Platyhelminthes. .
(ii) Pseudocoelomate: In some animals, the body cavity is not lined by mesoderm, instead,the mesoderm is present as scattered pouches in between the ectoderm and endoderm. Here body cavity is directly connected to archenteron. Such a body cavity is called pseudocoelom and the animals possessing them are called pseudocoelomates. E.g.: Aschelminthes (Ascaris). Pseudocoelom is derived from blastocoel.
(iii) Coelomate: The body cavity, which is lined by mesoderm (on both sides) is called coelom. Animals possessing coelom are called coelomates, or coelom is cavity between alimentary canal and body wall enclosed by mesoderm on both sides. E.g.: Annelids, Arthropods, Molluscs, Echinoderms, Hemichordates and Chordates.
Depending upon its origin, true coelom or eucoelom is of two types:
(a) Schizocoelous: The coelom is formed by splitting of mesoderm. E.g.: Annelida, Arthropoda and Mollusca.
Note: The cavity filled with blood is called haemocoel. It is found in Arthropods (cockroach) and Molluscs (Pila).
(b) Enterocoelom: The coelom develops as an outgrowth of the enteron or embryonic gut. E.g.: Deuterostomia (Echinodermata and Chordata). Echinodermata is an enterocoelomate invertebrate.

Q13. Match the column A with column B and choose the correct option.

Column A Column B
A. Porifera (i) Canal system
B. Aschelminthes (ii) Water-vascular system
C. Annelida (iii) Muscular Pharynx
D. Arthropoda (iv) Jointed appendages
E. Echinodermata (v) Metamers
  • A—(ii), B—(iii), C—(v), D—(iv), E—(i)
  • A—(ii), B—(v), C—(iii), D—(iv), E—(i)
  • A—(i), B—(iii), C—(v), D—(iv), E—(ii)
  • A—(i), B—(v), C—(iii), D—(iv), E—(ii)

Ans: (c)

A. Porifera (i) Canal system                             .
B. Aschelminthes (iii) Muscular Pharynx
C. Annelida (v) Metamers
D. Arthropoda (iv) Jointed appendages
E. Echinodermata (ii) Water-vascular system

Very Short Answer Type Questions

Q1. Identify the phylum in which adults exhibit radial symmetry and larva exhibit bilateral symmetry.
Ans: In phylum echinodermata, adults show radial symmetry whereas larva show bilateral symmetry.

Q2. What is the importance of pneumatic’ bOnes and air sacs in Aves?
Ans: Pneumatic bones in Aves keep the body light and thus help in flight. Air sacs help in respiration and buoyancy.

Q3. What is metagenesis? Mention an example which exhibits this phenomenon.
Ans: Alteration of generation is known as metagenesis. Obelia exhibits this phenomenon.

Q4. What is the role of feathers?
Ans: Feathers keep the body light and thus help in flight.

Q5. Which group of chordates possess sucking and circular mouth without jaws?
Ans: Cyclostomes have a sucking and circular mouth without jaws.

Q6. Give one example each for an animal possessing placoid scales and that with cycloid scales.
Ans: Placoid scales—Pristis, Trygon, Torpedo. Cycloid scales—Labeo Catla, Clarias

Q7.Mention two modifications in reptiles required for terrestrial mode of life.
Ans: 1. Their body is covered by dry and comified skin, epidermal scales or scutes.
2. Cleidoic eggs

Q8. Mention one example each for animals with chitinous exoskeleton and those covered by a calcareous shell.
Ans: The body of arthropods (like cockroach) is covered by chitinous exoskeleton and molluscan (like Pila) body is covered by a calcareous shell.

Q9. What is the role of radula in molluscs?
Ans: Radula is a file-like rasping organ help in feeding.

Q10. Name the animal, which exhibits the phenomenon of bioluminescence. Mention the phylum to which it belongs.
Ans: Bioluminescence (the property of a living organism to emit light) is well- marked in ctenophores.
Examples: Pleurobrachia, Ctenoplana, Beroe, Coeloplana and Velamen.

Q11. Write one example each of the following in the space provided.
a. Cold blooded animal
b. Warm blooded animal
c. Animal possessing dry and comified skin
d. Dioecious animal
Ans: Examples:
a. Cold blooded animal: amphibians (Frog)
b. Warm blooded animal: mammals (Human)
c. Animal possessing dry and comified skin: reptiles (Lizard)
d. Dioecious animal: aschelminthes (Ascaris)

Q12. Differentiate between a diploblastic and a triploblastic animal.
Ans: (i) Diploblastic: Animals in which the cells are arranged in two embryonic layers, an external ectoderm and an internal endoderm, are called diploblastic animals, e.g. Porifers, coelenterates and ctenophores.
(ii) Triploblastic: Those animals in which the developing embryo has a third germinal layer, mesoderm, in between the ectoderm and endoderm, are called triploblastic animals. E.g.: Platyhelminthes to Chordates.

Q13. Give an example for the following:
a. Roundworm
b. Fish possessing poison sting
c. A limbless reptile/amphibian .
d. An oviparous mammal
Ans: Examples:
a. Roundworm: Ascaris
b. Fish possessing poison sting: Trygon
c. A limbless reptile/ amphibian: Snakelfcthyophis
d. An oviparous mammal: Ornithorynchus (Duck-billed platypus)

Q14. Provide appropriate technical term in the space provided.
a. Blood-filled cavity in arthropods _________
b. Free-floating form of cnidaria _________
c. Stinging organ of jelly fishes _________
d. Lateral appendages in aquatic annelids _________
Ans: Provide appropriate technical term in the space provided.
a. Blood-filled cavity in arthropods: Haemocoel
b. Free-floating form of cnidaria: Medusa
c. Stinging organ of jelly fishes: Nematocyst
d. Lateral appendages in aquatic annelids: Parapodia
Q15. Match the following:

Animals Locomotory organ
a. Octopus              – (i) Limbs
b. Crocodile (ii) Comb plates
c. Catla (iii) Tentacles
d. Ctenoplana (iv) Fins

Ans:

Animals Locomotory organ
a.’ Octopus (iii) Tentacles
b. Crocodile (i) Limbs
c. Catla (iv) Fins
d. Ctenoplana (ii) Comb plates

Short Answer Type Questions
Q1. Differentiate between:
a. Open circulatory system and closed circulatory system
b. Oviparous and viviparous characteristic
c. Direct development and Indirect development
Ans: a. Open Circulatory System and Closed Circulatory System:

Open Circulatory System Closed Circulatory System
The blood is pumped out of the heart into sinuses and the cells and tissues are directly bathed in it. The blood is circulated within a network of vessels.

b. Oviparous and Viviparous:

Oviparous Viviparous
Animals which lay eggs are called oviparous. Animals which give birth to their young ones are called viviparous.

C. Direct Development and Indirect Development:

Direct Development Indirect Development
Animals which do not have a larval stage in their development are said to exhibit direct development. Animals which have a larval stage, which do not resemble the adult in their development are said to exhibit indirect development

Q2. Sort out the animals on the basis of their symmetry (radial or bilateral) coelenterates, ctenophores, annelids, arthropods, and echinoderms.
Ans: Coelenterates: Radial Ctenophores: Radial Annelids: Bilateral Arthropods: Bilateral Echinoderms: Radial

Q3. There has been an increase in the number of chambers in heart during evolution of vertebrates. Give the names of the class of vertebrates having two, three or four-chambered heart.
Ans: Two chambered heart: chondrichthyes and osteichthyes
Three chambered heart: Amphibia and Reptilia
Four chambered heart: Aves and mammalia

Q4. Fill up the blank spaces appropriately.

Phylum/

Class

Excretory

Organ

Circulatory

Organ

Respiratory

Organ

Arthropoda A B . Lungs/Gills/ Tracheal System
C Nephridia Closed Skin/Parapodia
D Metanephridia Open E
Amphibia F Closed Lung

Ans.
A = Malpighian Tubule/ coxal glands/
antemary glands/ green glands
B = Open
C = Annelida
D = Mollusca
E = Feather like gills
F = Kidney

Q5. Match the following:

a. Amphibia (i) Air bladder
b. Mammals (ii) Cartilagenous notochord
c. Chondrichthyes (hi) Mammary glands
d. Osteichthyes (iv) Pneumatic bones
e. Cyclostomata (v) Dual habitat
f. Aves (vi) Sucking and circular mouth without jaws

 Ans:

a. Amphibia (v) Dual habitat
b. Mammals (iii) Mammary glands
c. Chondrichthyes (ii) Cartilagenous notochord
d. Osteichthyes (i) Air bladder
e. Cyclostomata (vi) Sucking and circular mouth without jaws
f. Aves (iv) Pneumatic bones

Q6. Endoparasites are found inside the host body. Mention the special structure, possessed by these and which enables them to survive in those conditions.
Ans: The life cycles of endoparasities are more complex because of their extreme specialisation. Their morphological and anatomical features are greatly simplified while emphasizing their reproductive potential.
In accordance with their life style parasites evolve special adaptations such as:
1. Loss of unnecessary sense organs.
2. Loss of digestive system.
3. High reproductive capacity.
4. Presence of adhesive organs or suckers to cling on to the host.

Q7. Match the following and write correct choice in space provided.

Animal Characteristics
a. Pila (i) Jointed appendages
b. Cockroach (ii) Perching
c. Asterias . (iii) Water vascular system
d. Torpedo (iv) Electric organ
e. Parrot (v) Presence of shell
f. Dog fish (vi) Placoid scales

Ans.

Animal . Characteristics
a. Pila (v) Presence of shell
b. Cockroach (i) Jointed appendages
c. Asterias (iii) Water vascular system
d. Torpedo (iv) Electric organ
e. Parrot (ii) Perching
f. Dog fish (vi) Placoid scales

Q8. Differentiate between:
a. Open and closed circulatory system .
b. Oviparity and viviparity
c. Direct and indirect development
d. Aceolomate and pseudocoelomate
e. Notochord and nerve cord
f. Polyp and medusa
Ans: a. The circulatory system may be of two types:
1. Open type in which the blood is pumped out of the heart and the cells and tissues are directly bathed in it. E.g.: Arthropoda, Mollusca and Hemichordata.
2. Closed type in which the blood is circulated through a series of vessels of varying diameters (arteries, veins and capillaries). E.g.: Annelida and Chordata.
b. Oviparous animals give birth to an egg while viviparous animals are those that give birth to the live young ones.
c. Direct development: It is a type of development in which an embryo develops into a mature individual without involving a larval stage. Indirect development: It is a type of development that involves a sexually-immature larval stage.
d. Acoelomate: The animals in which the body cavity is absent are •called acoelomates, e.g.: Porifers, Coelentrates, Ctenophores and Platyhelminthes. ,
Pseudocoelomate: In some animals, the body cavity is not lined by mesoderm, instead, the mesoderm is present as scattered pouches in between the ectodenp and endoderm. E.g., aschelminthes
e. Notochord is a mesodermally derived rod-like-structure formed on the
dorsal side during embryonic development in some animals and it is the part of skeletal system. .
Nerve cord is the part of nervous system.
f. The polyp is a sessile and cylindrical fonn like Hydra, Adamsia, etc., whereas, the medusa is umbrella-shaped and free-swimming like Aurelia or jelly fish.

Q9. Give the characteristic features of the following citing one example of each:
a. Chondrichthyes and osteichthyes
b. Urochordata and cephalochordata
Ans: a. Chondrichthyes-. They are marine animals with streamlined body and have cartilaginous endoskeleton. Mouth is located ventrally. Notochord is persistent throughout life. Gill slits are separate and without operculum (gill cover). The skin is tough, containing minute placoid scales. Teeth are modified placoid scales which are backwardly directed. Their jaws are very powerful. These animals are predaceous. Due to the absence of air bladder, they have to swim constantly to avoid sinking. Scoliodon (Dog fish), Pristis (Saw fish), Carcharodon (Great white shark), Trygon (Sting ray).
Osteichthyes: It includes both marine and fresh water fishes with bony endoskeleton. Their body is streamlined. Mouth is mostly terminal. They have four pairs of gills which are covered by an operculum on each side. Skin is covered with cycloid/ctenoid scales. Air bladder is present which regulates buoyancy.
Examples: Marine—Exocoetus (Flying fish), Hippocampus (Sea horse); Freshwater—Labeo (Rohu), Catla (Katla), Clarias (Magur); Aquarium—Betta (Fighting fish), Pterophyllum (Angel fish),
b. Subphyla Urochordata and Cephalochordata are often referred to as protochordates and are exclusively marine. In Urochordata, notochord is present only in larval tail, while in Cephalochordata, it extends from head to tail region and is persistent throughout their life.
Examples: Urochordata—Ascidia, Salpa, Doliolum; Cephalochordata— Branchiostoma (Amphioxus or Lancelet).

Q10. Mention two similarities between:
Aves and mammals
A frog and crocodile ‘
A turtle and pila
Ans: Similarities between aves and mammalian

(i) Both are homeothermic animals (warm blooded)
(ii) Both have four chambered heart
b. Similarities between frog and crocodile
(i) Both are poikilothermic animals (cold blooded)
(ii) Both are oviparous animals
c. Similarities between turtle and Pila
(i) Both are poikilothermic animals (cold blooded)
(ii) Both are oviparous animals

Q11. Name
a. A limbless animal ‘
b. A cold blooded animal
c. A warm blooded animal
d. An animal possessing dry and comified skin
e. An animal having canal system and spicules
f. An animal with cnidoblasts

Ans: a. A limbless animal: Icthyophis
b. A cold blooded animal: Trygon (sting ray)
c. A warm blooded animal: Macaca (monkey)
d. An animal possessing dry and comifiedskin: Naja (Cobra)
e. An animal having canal system and spicules: Euspongia (bath sponge)
f. An animal with cnidoblasts: Hydra.
Q12. Give an example for each of the following:
a. A viviparous animal
b. A fish possessing a poison sting
c. A fish possessing an electric organ
d. An organ, which regulates buoyancy
e. Animal, which exhibits alternation of generation
f. Oviparous animal with mammary gland.
Ans: a. A viviparous animal: Panthera leo (lion)
b. A fish possessing a poison sting: Trygon (sting ray)
c. A fish possessing an electric organ: Torpedo
d. An organ, which regulates buoyancy: Air bladder
e. Animal, which exhibits alternation of generation: Obelia (Sea-fur)
f. Oviparous animal with mammary gland: Echidna (Platypus).

Q13. Excretory organs of different animals are given below. Choose correctly and write in the space provided.

Animal Excetory Organ/Unit
a. Balanoglossus (i) Metanephridia
b. Leech (ii) Nephridia
c. Locust (iii) Flame cells
d. Liver fluke (iv) Absent
e. Sea urchin (v) Malpighian tubule
f. Pila (vi) Proboscis gland

Ans.

a. Balanoglossus (vi) Proboscis gland
b. Leech (ii) Nephridia
c. Locust (v) Malpighian tubule
d. Liver fluke (iii) Flame cells
e. Sea urchin (iv) Absent
f. Pila (0 Metanephridia

Long Answer Type Questions
Q1. Give three major differences between chordates and non-chordates and draw a schematic sketch of a chordate showing those features.
Ans: Animals belonging to phylum Chordata are fundamentally characterised by the presence of a notochord, a dorsal hollow nerve cord and paired
pharyngeal gill slits. These are bilaterally symmetrical, triploblastic, coelomate with organ-system level of organisation. They possess a post anal tail and a closed circulatory system.

S. No. Chordates Non-chordates
1. Notochord present. Notochord absent.
2. Central nervous system is dorsal, hollow and single. Central nervous system is ventral, solid and double.
3. Pharynx perforated by gill slits. Gill slits are absent.
4. Heart is ventral. Heart is dorsal (if present).
5. A post-anal part (tail) is present. Post-anal tail is absent.

NCERT Exemplar Class 11 Biology Chapter 4 Animal Kingdom Img 1

Q2. What is the relationship between germinal layers and the formation of body cavity in case of coelomate, acoelomates and pseudocoetomates?
Ans: Presence or absence of a cavity between the body wall and the gut wall is very important in classification. The body cavity, which is lined by mesoderm is called coelom. Animals possessing coelom are called coelomates, e.g., annelids, molluscs, arthropods, echinoderms, hemichordates and chordates. In some animals, the body cavity is not lined by mesoderm, instead, the mesoderm is present as scattered pouches in between the ectoderm and endoderm. Such a body cavity is called pseudocoelom and the animals possessing them are called pseudocoelomates, e.g., aschelminthes. The animals in which the body cavity is absent are called acoelomates, e.g., platyhelminthes

NCERT Exemplar Class 11 Biology Chapter 4 Animal Kingdom Img 2

Q3. Comment upon the habitats and external features of animals belonging to class, amphibia and reptilia.
Ans: Class—Amphibia As the name indicates (Gr., Amphi: dual, bios, life), amphibians can live in aquatic as well as terrestrial habitats. Most of them have two pairs of limbs. Body is divisible into head and trunk. Tail may be present in some. The amphibian skin is moist (without scales). The eyes have eyelids. A tympanum represents the ear.
Examples: Bufo (Toad), Rana (Frog), Hyla (Tree frog), Salamandra (Salamander), Ichthyophis (Limbless amphibia).
Class-Reptilia :The class name refers to their creeping or crawling mode of locomotion (Latin, repere or reptum, to creep or crawl). They are mostly terrestrial animals and their body is covered by dry and comified skin, epidermal scales or scutes. They do not have external ear openings. Tympanum represents ear. Limbs, when present, are two pairs.
Examples: Chelone (Turtle), Testudo (Tortoise), Chameleon (Tree lizard), Calotes (Garden lizard), Crocodilus (Crocodile), Alligator (Alligator). Hemidactylus (Wall lizard), Poisonous snakes – Naja (Cobra), Bangarus (Krait), Vipera (Viper).

Q4. Mammals are most adapted among the vertebrates. Elaborate.
Ans: Mammals are most adapted among the vertebrates as they are found in a variety of habitats – polar ice caps, deserts, mountains, forests, grasslands and*dark caves. Some of them have adapted to fly or live in water.
They have two pairs of limbs, adapted for walking, running, climbing,. burrowing, swimming or flying. The skin of mammals is unique in possessing hair. External ears or pinnae are present. Different types of teeth are present in the jaw. Heart is four chambered.
They are ‘homoiot’hermous. Respiration is by lungs. Sexes are separate and fertilisation is internal. They are viviparous with few exceotions and development is direct.

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NCERT Exemplar Class 11 Biology Chapter 20 Locomotion and Movement

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

NCERT Exemplar Class 11 Biology Chapter 20 Locomotion and Movement

Multiple Choice Questions

Q1. Match the following and mark the correct option.

Column I Column II
A. Fast muscle fibres (i) Myoglobin
B. Slow muscle fibres (ii) Lactic acid
C. Actin filament (iii) Contractile unit
D. Sarcomere (iv) I-band

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

Ans. (c)

A. Fast muscle fibres (ii) Lactic acid
B. Slow muscle fibres (i) Myoglobin
C. Actin filament (iv) I-band
D. Sarcomere (iii) Contractile unit

Q2. Ribs are attached to
(a) Scapula (b) Sternum (c) Clavicle (d) Ilium
Ans: (b) Ribs are attached to sternum.

Q3. What is the type of movable joint present between the atlas and axis?
(a) Pivot (b) Saddle (c) Hinge (d) Gliding
Ans: (a) Pivot joint: Between atlas and axis called atlanto-axial joint.

Q4. ATPase of the muscle is located in
(a) Actinin (b) Troponin (c) Myosin (d) Actin
Ans: (c) ATPase of the muscle is located in head of myosin.

Q5. Intervertebral disc is found in the vertebral column of
(a) Birds (b) Reptiles (c) Mammals (d) Amphibians
Ans: (c) Intervertebral disc is found in the vertebral column of mammals.

Q6. Which one of the following is showing the correct sequential order of vertebrae in the vertebral column of human beings? ‘
(a) Cervical — lumbar — thoracic — sacral — coccygeal
(b) Cervical — thoracic — sacral — lumbar — coccygeal
(c) Cervical — sacral — thoracic — lumbar — coccygeal
(d) Cervical — thoracic — lumbar — sacral — coccygeal
Ans: (d) Cervical—thoracic—lumbar—sacral—coccygeal is the correct sequential order of vertebrae in the vertebral column of human beings.

Q7. Which one of the following options is incorrect?
(a) Hinge joint—between humerus and pectoral girdle
(b) Pivot joint—between atlas, axis and occipital condyle
(c) Gliding joint—between the carpals
(d) Saddle joint—between carpel and metacarpals of thumb
Ans: (a) Hinge joint—Knee joint and elbow joint

Q8. Knee joint and elbow joints are examples of
(a) Saddle joint (b) Ball and socket joint
(c) Pivot joint (d) Hinge joint
Ans: (d) Knee joint and elbow joints are examples of hinge joint.

Q9. Macrophages and leucocytes exhibit
(a) Ciliary movement
(b) Flagellar movement
(c) Amoeboid movement
(d) Gliding movement
Ans: (c) Amoeboid movements: Some specialised cells in our body like macrophages and leucocytes in blood exhibit amoeboid movement. It is effected by pseudopodia formed by the streaming of protoplasm (as in Amoeba). Cytoskeletal elements like microfilaments are also involved in amoeboid movement.

Q10. Which one of the following is not a disorder of bone?
(a) Arthritis
(b) Osteoporosis
(c) Rickets
(d) Atherosclerosis
Ans: (d) Atherosclerosis is a disorder of circulatory system.

Q11. Which one of the following statement is incorrect?
(a) Heart muscles are striated and involuntary
(b) The muscles of hands and legs are striated and voluntary
(c) The muscles located in the inner walls of alimentary canal are striated and involuntary
(d) Muscles located in the reproductive tracts are unstriated and involuntary
Ans:(c) The muscles located in the inner walls of alimentary canal are non- striated and involuntary.

Q12. Which one of the following statements is-true?
(a) Head of humerus bone articulates with acetabulum of pectoral girdle
(b) Head of humerus bone articulates with glenoid cavity of pectoral girdle
(c) Head of humerus bone articulates with a cavity called acetabulum of pelvic girdle
(d) Head of humerus bone articulates with a glenoid cavity of pelvic girdle
Ans: (b) Below the acromion is a depression called the glenoid cavity which articulates with the head of the humerus to form the shoulder joint.

Q13. Muscles with characteristic striations and involuntary are
(a) Muscles in the wall of alimentary canal
(b) Muscles of the heart
(c) Muscles assisting locomotion
(d) Muscles of the eyelids
Ans: (b) Muscles with characteristic striations and involuntary are muscles of the heart (Cardiac muscles).

Q14. Match the followings and mark the correct option.

Column I .                  Column II
A. Sternum (i) Synovial fluid
B. Glenoid cavity (ii) Vertebrae
C. Freely movable joint (iii) Pectoral girdle
D. Cartilaginous joint (iv) Flat bones

Options:

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

Ans. (b)

A. Sternum (iv) Flat bones
B. Glenoid cavity (iii) Synovial fluid
C. Freely movable joint (i) Pectoral girdle
D. Cartilaginous joint (ii) Yertebrae

 

Very Short Answer Type Questions
Q1. Name the cells/tissues in human body which
a. exhibit amoeboid movement
b. exhibit ciliary movement
Ans: a. Macrophages and leucocytes
b. Trachea, fallopian tube and bronchiole

Q2. Locomotion requires a perfect‘coordinated activity of muscular _____________, _______ , systems.
Ans: Skeletal and Neural

Q3. Sarcolemma, sarcoplasm and sarcoplasmic reticulum refer to a particular type of cell in our body. Which is this cell and to what parts of that cell do these names refer to?
Ans: Each muscle fibre or muscle cell is lined by the plasma membrane called sarcolemma enclosing the sarcoplasm. A muscle fibre is a syncytium as the sarcoplasm (cytoplasm) contains many nuclei. The endoplasmic reticulum, i. e., sarcoplasmic reticulum of the muscle fibres is the store house of calcium ions.

Q4 .Label the different components of actin filament in the diagram given
NCERT Exemplar Class 11 Biology Chapter 20 Locomotion and Movement Img 1
Q5. The. three tiny bones present in middle ear are called ear ossicles. Write them in correct sequence beginning from eardrum.
Ans: Malleus, incus and stapes.

Q6. What is the difference between the matrix of bones and cartilage?
Ans: Bones have a hard and non-pliable ground substance rich in calcium salts and collagen fibres which give bone its strength.
The inter-cellular material of cartilage is solid and pliable which resists compression. Cell of cartilage are called chondrocytes which are enclosed in a small cavities (lacunae) within the matrix secreted by them.

Q7. Which tissue is afflicted by Myasthenia gravis? What is the underlying cause?
Ans: Myasthenia gravis: Auto-immune disorder affecting neuromuscular junction leading to fatigue, weakening and paralysis of skeletal muscle.

Q8. How do our bone joints function without grinding noise and pain?
Ans: Our bone joints function without grinding noise and pain due to the presence of synovial fluid between bones.

Q9. Give the location of a ball and socket joint in a human body.
Ans: Ball and socket joint: Between humerus and pectoral girdle (shoulder joint). Between femur and acetabulum of pelvic girdle (hip joint). Total 4 ball and socket joints present in human body -2 shoulder joint and 2 hip joint.

Q10. Our forearm is made of three different bones. Comment.
Ans: The bones of the forearm are humerus, radius and ulna.

Q1. With respect to rib cage, explain the following:
a. Bicephalic ribs
b. True ribs
c. Floating ribs
Ans: a. Bicephalic ribs: Each rib is a thin flat bone connected dorsally to the vertebral column and ventrally to the sternum. It has two articulation surfaces on its dorsal end and is hence called bicephalic.
b. True ribs: First seven pairs of ribs are called true ribs. Dorsally, they are attached to the thoracic vertebrae and ventrally connected to the sternum with the help of hyaline cartilage.
c. Floating ribs: Last 2 pairs (11th and 12th) of ribs are not connected ventrally and are therefore, called floating ribs.

Q2. In old age, people often suffer from stiff and inflamed joints. What is this condition called? What are the possible reasons for these symptoms?
Ans: In old age, people suffer from stiff and inflamed joints, it is due to rheumatoid arthritis (autoimmune disorder)
Causes: (i) Inflammation of synovial membrane
(ii) Genetic factors (50% cases) .
(iii) Smoking
(iv) Vitamin-D deficiency

Q3. Exchange of calcium between bone and extracellular fluid takes place under the influence of certain hormones.

  1. What will happen if more of Ca + is in extracellular fluid?
  2. What will happen if very less amount of Ca++ is in the extracellular fluid?

Ans: a. If more of Ca++ is in extracellular fluid then it will be accumulated on the bones under the influence of thyrocalcitonin (TCT).
b. If very less amount of Ca++ is in the extracellular fluid then parathyroid hormone (PTFI) acts on bones and stimulates the process of bone resorption (dissolution/demineralisation). PTH also stimulates reabsorption of Ca2+ by the renal tubules and increases Ca2+ absorption from the digested food.

Q4. Name at least two hormones which result in fluctuation of Ca++

Ans: Thyrocalcitonin (TCT) and Parathyroid Hormone (PTH).

Q5. Rahul exercises regularly by visiting a gymnasium. Of late he is gaining weight. What could be the reason? 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 bdcause he is accumulating water in the body.
Ans: Rahul has gained weight due to increased muscle and less of fat.

Q6. 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 exercise?
b. How did her breathing rate change?
Ans: a. Repeated activation of the muscles can lead to the accumulation of lactic acid due to anaerobic breakdown of glycogen in them, causing fatigue.
b. During strenuous exercise demand of oxygen also increases so breathing rate has been changed.

Q7. Write a few lines about Gout.
Ans: When metabolic waste-uric acid crystals are accumulated in bones, then it results into inflammation of bone and joints thereby causing pain. This disorder of skeletal system is called gout.

Q8. What is the source of energy for muscle contraction?
Ans: ATP (Adenosine Triphosphate)

Q9. What are the points for articulation of pelvic and pectoral girdles?
Ans: The components of pelvic girdle are ilium, ischium and pubis. It articulates with, femur through acetabulum. The components of pectoral girdle are scapula and clavicle. It is the glenoid cavity of pectoral girdle in which head . of humerus articulates.

Long Answer Type Questions

Q1. Calcium ion concentration in blood affects muscle contraction. Does it lead to tetany in certain cases? How will you correlate fluctuation in blood calcium with tetany?
Ans: Muscle contraction is initiated by a signal sent by the central nervous system (CNS) via a motor neuron. A neural signal reaching this junction releases a neurotransmitter (acetyl choline) which generates an action potential in the sarcolemma. This spreads through the muscle fibre and causes the release of calcium ions into the sarcoplasm. Increase in Ca++ level leads to the binding of calcium with a subunit of troponin on actin filaments and thereby remove the masking of active sites for myosin. Utilising the energy from ATP hydrolysis, the myosin head now binds to the exposed active sites on actin to form a cross-bridge. This pulls the attached actin filaments towards the centre of ‘A’ band. The ‘Z’ line attached to these actins are also pulled inwards thereby causing a shortening of the sarcomere, i.e., contraction. The process continues till the Ca++ ions are pumped back to the sarcoplasmic cisternae resulting in the masking of actin filaments.

Tetany: Rapid spasms (wild contractions) in muscle due to low Ca in body fluid.

Q2. An elderly woman slipped in th£ bathroom and had severe pain in her lower back. After X-ray examination doctors told her it is due to a slipped disc. What does that mean? How does it affect our health?
Ans: Displacement of intervertebral disc from’ their normal position is called slipped disc.
Effects:
i. Neck or lower back pain
ii. Muscular weakness
iii. Paralysis
iv. Sciatica

Q3. Explain sliding filament theory of muscle contraction with neat sketches.
Ans. Mechanism of muscle contraction: Mechanism of muscle contraction is best explained by the sliding filament theory which states that contraction of a muscle fibre takes place by the sliding of the thin filaments over the thick filaments. Muscle contraction is initiated by a signal sent by the Central Nervous System (CNS) via a motor neuron. A motor neuron alongwith the muscle fibres connected to it constitute a motor unit. The junction between a motor neuron and the sarcolemma of the muscle fibre is called the neuromuscular junction or motor-end plate. A neural signal reaching this junction releases a neurotransmitter (acetyl choline) which generates an action potential in the sarcolemma. This spreads through the muscle fibre and causes the release of calcium ions into the sarcoplasm. Increase in Ca++ level leads to the binding of calcium with a subunit of troponin on actin filaments and thereby remove the masking of active sites for myosin.Utilising the energy from ATP hydrolysis, the myosin head now binds to the exposed active sites on actin to form a cross-bridge. This pulls the attached

NCERT Exemplar Class 11 Biology Chapter 20 Locomotion and Movement Img 2

actin filaments towards the centre of ‘A’ band. The ‘Z’ line attached to these actins are also pulled inwards thereby causing a shortening of the sarcomere, i.e., contraction. It is clear from the above steps, that during shortening of the muscle, i.e., contraction, the ‘I’ bands get reduced, whereas the ‘A’ bands retain the length. The myosin, releasing the ADP and P, goes back to its relaxed state. A new ATP binds and the cross-bridge is broken. The ATP is again hydrolysed by the myosin head and the cycle of cross-bridge formation and breakage is repeated causing further sliding. The process continues till the Ca++ ions are pumped back to the sarcoplasmic cistemae resulting in the masking of actin filaments. This causes the return of ‘Z’ lines back to their original position, i.e., relaxation.

Q5. Discuss the role of Ca2+ ions in muscle contraction. Draw neat sketches to illustrate your answer.
Ans: Muscle contraction is initiated by a neural signal, which after reaching neuromuscular junction or motor end plate releases a neurotransmitter, as a result an action potential in the sarcolemma is generated. Action potential spreads through muscle fibre and causes the release of calcium ions into the sarcoplasm. Increase in Ca2+ level leads to the binding of calcium with a subunit of troponin on actin filaments and thereby removes the masking of active sites for myosin. Utilising the energy from ATP hydrolysis, the myosin head now binds to the exposed active site on actin to form a cross-bridge. This pulls the attached actin filaments towards the centre of ‘A’ band. The ‘Z’ line attached to these actins are also pulled inwards thereby causing shortening of the sarcomere, i.e., contraction.
NCERT Exemplar Class 11 Biology Chapter 20 Locomotion and Movement Img 3
A new ATP binds to myosin head and the cross-bridge is broken. The ATP is again hydrolysed by the myosin head and the cycle of cross-bridge formation and breakage is repeated causing further sliding. The process continues till the Ca++ ions are pumped back to the sarcoplasmic cistemae resulting in masking of actin filaments and breakage of all cross-bridges. This cause the return of ‘Z’ lines along with filaments back to their original position, i.e., relaxation.

Q6. Differentiate between pectoral and pelvic girdle.
Ans: Pectoral and pelvic girdle help in the articulation of upper and lower limbs respectively. Each girdle is made of two equal halves. Each half of a pectoral girdle consists of clavicle and scapula. Scapula is a large triangular flat bone. There is glenoid cavity at the joint of scapula, clavicle and acromian process, which articulates with the head of humerus to form the shoulder joint. Each half of pelvic girdle is formed by three bones—ilium, ischium and pubis. At the point of their fusion; there is a cavity called acetabulum to which the head of femur articulates.

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

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.

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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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.

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.

NCERT Exemplar Class 11 Biology Chapter 21 Neural Control and Coordination Img 7

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NCERT Exemplar Class 11 Biology Chapter 7 Structural Organisation in Animals

NCERT Exemplar Class 11 Biology Chapter 7 Structural Organisation in Animals are part of NCERT Exemplar Class 11 Biology. Here we have given NCERT Exemplar Class 11 Biology Chapter 7 Structural Organisation in Animals.

NCERT Exemplar Class 11 Biology Chapter 7 Structural Organisation in Animals

Multiple Choice Questions

Q1. Which one of the following types of cell is involved in making of the inner walls of large blood vessels?
(a) Cuboidal epithelium
(b) Columnar epithelium
(c) Squamous epithelium
(d) Stratified epithelium
Ans: (c) The squamous epithelium (pavement epithelium) is made up of a single thin layer of flattened cells with irregular boundaries. They are found in the walls of blood vessels and air sacs of lungs and are involved in a function like forming a diffusion boundary.

Q2. To which one of the following categories does adipose tissue belong?
(a) Epithelial
(b) Connective
(c) Muscular
(d) Neural
Ans: (b) Loose connective tissue has cells and fibres loosely arranged in a semi¬fluid ground substance, for example, areolar tissue present beneath the skin. Adipose tissue is a type of loose connective tissue located mainly beneath the skin.

Q3. Which of the following is not a connective tissue?
(a) Bone
(b) Cartilage
(c) Blood
(d) Muscles
Ans: (d) Cartilage, bones and blood are various types of specialised connective tissues.

Q4. The clitellum is a distinct part in the body of earthworm, it is found in
(a) Segments 13-14—15
(b) Segments 14-15-16
(c) Segments 12-13-14
(d) Segments 15-16-17
Ans: (b) The ditellum, a prominent circular band of glandular nature is found from 14th-16th segments. It secretes mucus and albumin, which helps in the formation of cocoon.

Q5. Setae help in locomotion in earthworm but are not uniformly present in all the segments. They are present in
(a) 1st segment
(b) Last segment
(c) Clitellar segment
(d) 20th-22nd segment
Ans: (d) Setae help in locomotion in earthworm, it is present in all the segments except 1st segment, last segment and clitellar segment.

Q6. Which one of the following statements is true for cockroach?
(a) The number of ovarioles in each ovary are ten.
(b) The larval stage is called caterpillar.
(c) Anal styles are absent in females.
(d) They are ureotelic.
Ans: (c) In cockroach, anal styles are absent in females.

Q7. Match the following and choose the correct option.

A. Adipose tissue – (i) Nose
B. Stratified epithelium (ii) Blood
C. Hyaline cartilage (iii) Skin
D. Fluid connective tissue (iv) Fat storage


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

Ans. (b)

A. Adipose tissue (iv) Fat storage
B. Stratified epithelium (Hi) Skin
C. Hyaline cartilage (i) Nose
D. Fluid connective tissue (ii) Blood

Q8. Match the following and choose the correct answer.

A. Hermaphrodite (0 Produces blood cells and haemoglobin
B. Direct development (ii) Testis and ovary in the same animal
G. Chemoreceptor (iii) Larval form absent
D. Blood gland in earthworm (iv) Sense of chemical substances

 

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

A. Hermaphrodite (ii) Testis and ovary in the same animal
B. Direct development (iii) Larval form absent
C. Chemoreceptor (iv) Sense of chemical substances
D. Blood gland in earthworm (0 Produces blood cells and haemoglobin

 

Q9. Match the following with reference to cockroach and choose the correct option.

A. Phallomere (i) Chain of developing ova
B. Gonopore (ii) Bundles of sperm
C. Spermatophore (iii) Opening of the ejaculatory duct
D. Ovarioles (iv) The external genitalia

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

 10. Match the followings and choose the correct answer.

 

A. Touch (i) Nasal epithelium
B. Smell (ii) Foramen magnum
C. Cranial nerves (iii) Sensory papillae
D. Medulla oblongata (iv) Peripheral nervous system

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

Ans: (d)

A. Touch (iii) Sensory papillae
B. Smell (i) Nasal epithelium
C. Cranial nerves (iv) Peripheral nervous system
D. Medulla oblongata (ii) Foramen magnum

Very Short Answer Type Questions

Q1. State the number of segments in earthworm which are covered by a prominent dark band or clitellum.
Ans: 14th—16th segments.

Q2. Where are sclerites present in Cockroach?
Ans: In each segment of the body of cockroach, exoskeleton has hardened plates called sclerites.

Q3. How many times do nymphs moult to reach the adult form of cockroach?
Ans: The nymph grows by moulting about 13 times to reach the adult form.

Q4. Identify the sex of a frog in which sound producing vocal sacs are present.
Ans: Male frog

Q5. Name the process by which a tadpole develops into an adult frog.
Ans: By metamorphosis a tadpole develops into an adult frog.

Q6. What is the scientific term given to earthworm’s body segments?
Ans: Metamers

Q7. A muscle fibre tapers at both ends and does not show striations. Name the muscle fibre.
Ans: Smooth muscle fibre

Q8. Name the different cell junctions found in tissues.
Ans: a. Tight Junctions
b. Adhering Junctions
c. Gap Junctions

Q9. Give two identifying features of an adult male frog.
Ans: a. Vocal Sacs
b. Thumb pads/copulatory pads in thumb

Q10. Which mouth part of cockroach is comparable to our tongue?
Ans: Hypopharynx

Q11. The digestive system of a frog is made of the following parts. Arrange them in an order beginning from mouth.
Mouth, oesophagus, buccal cavity, stomach, intestine, cloaca, rectum, cloacal aperture
Ans: Mouth, buccal cavity, oesophagus, stomach, intestine, rectum, cloaca and cloacal aperture.

Q12. What is the difference between cutaneous and pulmonary respiration?
Ans: Frogs respire on land and in the water by two different methods. In water,
skin acts as aquatic respiratory organ (cutaneous respiration). Dissolved oxygen in the water is exchanged through the skin by diffusion. On land, the buccal cavity, skin and lungs act as the respiratory organs. The respiration by lungs is called pulmonary respiration.

Q13. Special venous connection between liver and intestine, and between kidney and intestine is found in frog, what are they called?
Ans: Hepatic portal system and renal portal system respectively.

Short Answer Type Questions

Q1. Give the location of hepatic caeca in a Cockroach. What is their function?
Ans: A ring of 6-8 blind tubules called hepatic caeca are present at the junction of foregut and midgut. Hepatic caeca secretes digestive juice.

Q2. Frogs are beneficial for mankind, justify the statement.
Ans: Frogs are beneficial for mankind as they can eat some crop pests and protect crop. Frogs maintain ecological balance as they are an important component of food chain and food web in the ecosystem. They are edible in some countries.

Q3. The body of sponges does not possess tissue level of organisation though it is made of thousands of cells. Comment.
Ans: Though all members of Animalia are multicellular, all of them do not exhibit the same pattern of organisation of cells. For example, in sponges, the cells are arranged as loose cell aggregates, i.e. they exhibit cellular level of organisation. Some division of labour (activities) occur among the cells. Hence, the body of sponges does not possess tissue level of organisation.

Q4. Structural organisation in animals attains different levels as cell—organ— organ system. What is missing in this chain? Mention the significance of such an organisation.
Ans: Tissue is missing in this chain. In multicellular animals, a group of similar cells along with intercellular substances perform a specific function. Such an organisation is called tissue. Tissues are organised in specific proportion and pattern to form an organ like stomach, lung, heart and kidney. When two or more organs perform a common function by their physical and/or chemical interaction, they together form organ system, e.g., digestive system, respiratory system, etc. Cells, tissues, organs and organ systems split up the work in a way that exhibits division of labour and contribute to the survival of the body as a whole.

Q5. Stratified epithelial cells have limited role in secretion. Justify their role in our skin.
Ans: Stratified epithelium is made of more than one layer of cells and thus has a limited role in secretion and absorption. Their main function is to provide protection against chemical and mechanical stresses. They cover the dry surface of the skin.

Q6. How does a gap junction facilitate intercellular communication?
Ans: Gap junctions facilitate the cells to communicate with each other by connecting the cytoplasm of adjoining cells, for rapid transfer of ions, small molecules and sometimes big molecules.

Q7. Why are blood, bone and cartilage called connective tissue?
Ans: Blood, bone and cartilage are called connective tissue because of their special function of linking and supporting other tissues/organs of the body.

Q8. Why are neurons called excitable cells? Mention special features of the membrane of the neuron?
Ans: Neurons are called excitable cells because when a neuron is suitably stimulated, an electrical disturbance is generated which swiftly travels along its plasma membrane. Arrival of the disturbance at the neuron’s endings, or output zone, triggers events that may cause stimulation or inhibition of adjacent neurons and other cells.

Q9. Why earthworm is called the friend of farmer?
Ans: Earthworms are known as ‘friends of farmers’ because they make burrows in the soil and make it porous which helps in respiration and penetration of the developing plant roots. The process of increasing fertility of soil by the earthworms is called vermicomposting.

Q10. How do you distinguish between dorsal and ventral surface of the body of earthworm?
Ans: The dorsal surface of the body of earthworm is marked by a dark median mid dorsal line (dorsal blood vessel) along the longitudinal axis of the body. The ventral surface is distinguished by the presence of genital openings (pores).

Q11. Correct the wrong statements among the following.
a. In earthworm, a single male genital pore is present.
b. Setae help in locomotion of earthworm.
c. Muscular layer in the body wall of earthworm is made up of only circular muscles.
d. Typhlosole is the part of intestine of earthworm.
Ans: a. In earthworm, a pair of male genital pores is present.
b. Correct statement
c. Muscular layer in the body wall of earthworm is made up of circular muscles and longitudinal muscles.
d. Correct statement

Q12. Why nephridia in earthworm that are basically similar in structure classified into three types? Mention the names of each.
Ans: The excretory organs occur as segmentally arranged coiled tubules called nephridia (sing.: nephridium). They are of three types: (i) septal nephridia, present on both the sides of intersegmental septa of segment 15 to the last that open into intestine, (ii) integumentary nephridia, attached to lining of the body wall of segment 3 to the last that open on the body surface and (iii) pharyngeal nephridia, present as three paired tufts in the 4th, 5th and 6th segments. These different types of nephridia are basically similar in structure.
Nephridia regulate the volume and composition of the body fluids. A nephridium starts out as a funnel that collects excess fluid from coelomic chamber. The funnel connects with a tubular part of the nephridium which delivers the wastes through a pore to the surface in the body wall into the digestive tube.

Q13. Common name of some animals are given in Column A, write their scientific name in Column B.

Column A Column B
a. Tiger
b. Peacock
c. Housefly

Ans: 

a. Tiger Panthera tigris
b. Peacock Pavo cristatus
c. Housefly Musca domestica

Q14. Complete the following statement:
a. In Cockroach grinding of food particle is performed by
b. Malpighian tubules help in removal of
c. Hind gut of Cockroach js differentiated into
d. In Cockroach blood vessels open into spaces called
Ans: a. . In Cockroach grinding of food particle is performed by gizzard.
b. Malpighian tubules help in removal of nitrogenous waste products.
c. Hindgut of Cockroach is differentiated into ileum, colon and rectum.
d. In Cockroach blood vessels open into spaces called sinuses.

Q15. Mention special features of eye in Cockroach.
Ans: The compound eyes are situated at the dorsal surface of the head. Each eye consists of about 2000 hexagonal ommatidia (sing.: ommatidium). With the help of several ommatidia, a cockroach can receive several images of an object. This kind of vision is known as mosaic vision with more sensitivity but less resolution, being common during night (hence called nocturnal vision).

Q16. Frog is a poikilotherm, exhibits camouflage and undergoes aestivation and hibernation, how are all these beneficial to it?
Ans: Frog have the ability to change colour which hides them from their enemies (camouflage). This protective coloration is called mimicry.
Frogs are not seen during peak summer and winter. During this period they take shelter in deep burrows to protect them from extreme heat and cold. This is called as summer sleep (aestivation) and winter sleep (hibernation).

Q17. Write the functions in brief in column B, appropriate to the structures given in column A.

Column A Column B
a. Nictitating membrane i.
b. Tympanum ii.
c. Copulatory pad iii.

Ans:

a. Nictitating membrane i. Protects them while in water.
b. Tympanum ii. Receives sound signals
c. Copulatory pad iii. Hold the female during copulation

Q18. Write the appropriate type of tissues in column B according to the functions mentioned in column A.

Column A Column B
a. Secretion and absorption i.
b. Protective covering ii.
c. Linking and supporting framework iii.

Ans:

Column A Column B
a. Secretion and absorption i. Cuboidal epithelium
b. Protective covering ii. Compound epithelium
c. Linking and supporting framework iii. Connective tissue

Q19. Using appropriate examples, differentiate between false and true body segmentation.
Ans: (i) Metamerism: In some animals, the body is externally and internally divided into segments with a serial repetition of at least some organs. For example, in earthworm, the body shows this pattern called metameric segmentation and the phenomenon is known as metamerism or true segmentation.
Metamerism is found in 3 animal phylums—Annelida, Arthropoda and chordata.
(ii) Pseudometamerism or false segmentation: In tapeworm, the body is externally divisible into parts called proglottis. They develop from the neck but are not embryonic in origin, this segmentation is called pseudometamerism.

Q20. What is special about tissue present in the heart?
Ans: Cardiac muscle tissue is a contractile tissue present only in the heart. Cell junctions fuse the plasma membranes of cardiac muscle cells and make them stick together. Communication junctions (intercalated discs) at some fusion points allow the cells to contract as a unit, i.e. when one cell receives a signal to contract, its neighbours are also stimulated to contract.

Long Answer Type Questions

Q1. Classify and describe epithelial tissue on the basis of structural modifications of cells.
Ans: There are two types of epithelial tissues namely simple epithelium and compound epithelium. Simple epithelium is composed of a single layer of cells and functions as a lining for body cavities, ducts, and tubes. The compound epithelium consists of two or more cell layers and has protective function as it does in our skin. On the basis of structural modification of the cells, simple epithelium is further divided into three types. These are (i) Squamous, (ii) Cuboidal and (iii) Columnar
• The squamous epithelium is made of a single thin layer of flattened cells with irregular boundaries. They are found in the walls of blood vessels and air sacs of lungs and are involved in functions like forming a diffusion boundary. The cuboidal epithelium is composed of a single layer of cube-like cells. This is commonly found in ducts of glands and tubular parts of nephrons in kidneys and its main functions are secretion and absorption. The epithelium of proximal convoluted tubule (PCT) of nephron in the kidney has microvilli. The columnar epithelium is composed of a single layer of tall and slender cells. Their nuclei are located at the base. Free surface may have microvilli. They are found in the lining of stomach and intestine and help in secretion and absorption.
• Compound epithelium is made of more than one layer (multi-layered) of cells and thus has a limited role in secretion and absorption. Their main function is to provide protection against chemical and mechanical stresses. They cover the dry surface of the skin, the moist surface of buccal cavity, pharynx, inner lining of ducts of salivary glands and of pancreatic ducts.

Q2. Write down the common features of the connective tissue. On the basis of structure and function, differentiate between bones and cartilages. .
Ans: Connective tissues are most abundant and widely distributed in the body of complex animals. They are named connective tissues because of their special function of linking and supporting other tissues/organs of the body. They range from soft connective tissues to specialised types, which include cartilage, bone, adipose, and blood. In all connective tissues except blood, the cells secrete fibres of structural proteins called collagen or elastin. The fibres provide strength, elasticity and flexibility to the tissue. These cells also
secrete modified polysaccharides, which accumulate between cells and fibres and act as matrix (ground substance).
• The intercellular material of cartilage is solid and pliable and resists compression. Cells of this tissue (chondrocytes) are enclosed in small cavities within the matrix secreted by them. Most of the cartilages in vertebrate embryos are replaced by bones in adults. Cartilage is present in the tip of nose, outer ear joints, between adjacent bones of the vertebral column, limbs and hands in adults.
• Bones have a hard and non-pliable ground substance rich in calcium salts and collagen fibres which give bone its strength. It is the main tissue that provides structural frame to the body. Bones support and protect softer tissues and organs. The bone cells (osteocytes) are present in the spaces called lacunae. Limb bones, such as the long bones of the legs, serve weight-bearing functions. They also interact with skeletal muscles attached to them to bring about movements. The bone marrow in some bones is the site of production of blood cells.

Q3. Comment upon the gametic exchange in earthworm during mating.
Ans: A mutual exchange of sperm occurs between two worms during mating. One worm has to find another worm and they mate juxtaposing opposite gonadal openings exchanging packets of sperms called spermatophores. Mature sperm and egg cells, and nutritive fluid are deposited in cocoons produced by the gland cells of clitellum. Fertilisation and development occur within the cocoons which are deposited in soil. The ova (eggs) are fertilised by the sperm cells within the cocoon which then slips off the worm and is deposited in or on the soil. The cocoon holds the worm embryos. After about 3 weeks, each cocoon produces two to twenty baby worms with an average of four. Development of earthworms is direct, i.e. there is no larva formed.

Q4. Explain the digestive system of Cockroach with the help of a labelled sketch.
Ans: The digestive system consists of alimentary canal and digestive glands. The alimentary canal of cockroach is divided into foregut, midgut and hindgut. The mouth opens into a short tubular pharynx, leading to a narrow tubular passage, the oesophagus, which opens into a sac like crop used for storing food. The crop is followed by a gizzard or proventriculus. Gizzard consists of six chitinous plates called teeth which helps in grinding food. The entire foregut is lined by cuticle. A ring of six to eight blind tubular structures called hepatic or gastric caecae is present at the junction of foregut and midgut which secrete digestive juice. At the junction of midgut and hindgut is present another ring of yellow colored thin filamentous malpighian tubules which help in removal of excretory products from haemolymph. The hindgut is broader than midgut and is differentiated into ileum, colon and rectum. The rectum opens out through anus.
NCERT Exemplar Class 11 Biology Chapter 7 Structural Organisation in Animals Img 1

Q5. Draw a neat and well labelled diagram of male reproductive system of a frog.
NCERT Exemplar Class 11 Biology Chapter 7 Structural Organisation in Animals Img 2

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NCERT Exemplar Class 11 Biology Chapter 22 Chemical Coordination and Integration

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

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 18 Body Fluids and Circulation

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

NCERT Exemplar Class 11 Biology Chapter 18 Body Fluids and Circulation

Multiple Choice Questions

Q1. Which of the following cells does not exhibit phagocytic activity?
(a) Monocytes (b) Neutrophil (c) Basophil (d) Macrophage
Ans: (c) Basophils secrete histamine, serotonin, heparin, etc., and are involved in inflammatory reactions.

Q2. One of the common symptoms observed in people infected with Dengue fever is
(a) significant decrease in RBC count
(b) significant decrease in WBC count
(c) significant decrease in platelets count
(d) significant increase in platelets count
Ans: (c) One of the common symptoms observed in people infected with Dengue fever is significant decrease in platelets count.

Q3. Which among the followings is correct during each cardiac cycle?
(a) The volume of blood pumped out by the Rt and Lt ventricles is same
(b) The volume of blood pumped out by the Rt and Lt ventricles is different
(c) The volume of blood received by each atrium is different
(d) The volume of blood received by the aorta and pulmonary artery is different
Ans: (a) The volume of blood pumped out by the Rt and Lt ventricles is same.

Q4. The cardiac activity could be moderated by the autonomous neural system. Tick the correct answer
(a) The parasympathetic system stimulates heart rate and stroke volume
(b) The sympathetic system stimulates heart rate and stroke volume
(c) The parasympathetic system decreases the heart rate but increase stroke volume
(d) The sympathetic system decreases the heart rate but increase stroke volume
Ans: (b) A special neural centre in the medulla oblongata can moderate the cardiac function through autonomic nervous system (ANS). Neural signals through the sympathetic nerves (part of ANS) can increase the rate of heart beat, the strength of ventricular contraction and thereby the cardiac output. On the other hand, parasympathetic neural signals (another component of ANS) decrease the rate of heart beat, speed of conduction of action potential and thereby the cardiac output.

Q5. Mark the pair of substances among the following which is essential for coagulation of blood.
(a) Heparin and calcium ions (b) Calcium ions and platelet factors
(c) Oxalates and citrates (d) Platelet factors and heparin
Ans: (b) Calcium ions and platelet factors are essential for coagulation of blood.

Q6. ECG depicts the depolarisation and repolarisation processes during the cardiac cycle. In the ECG of a normal healthy individual one of the following waves is not represented.
(a) Depolarisation of atria (b) Repolarisation of atria
(c) Depolarisation of ventricles (d) Repolarisation of ventricles
Ans: (b) ECG depicts the depolarisation and repolarisation processes during the cardiac cycle. In the ECG of a normal healthy individual repolarisation of atria is not represented.

Q7. Which one of the following types of cells lack nucleus in humans
(a) RBC (b) Neutrophils (c) Eosinophils (d) Erythrocytes
Ans: (a and d) RBCs or erythrocytes lack nucleus in humans.

Q8. Which one of the following blood cells is involved in antibody production?
(a) B-lymphocytes (b) T-Lymphocytes
(c) RBC (d) Neutrophils
Ans: (a) B-lymphocytes cells are involved in antibody production.

Q9. The cardiac impulse is initiated and conducted further up to ventricle. The correct sequence of conduction of impulse is
NCERT Exemplar Class 11 Biology Chapter 18 Body Fluids and Circulation Img 1

Q10. Agranulocytes responsible for immune response of the body are
(a) Basophils (b) Neutrophils (c) Eosinophils (d) Lymphocytes
Ans: (d) Lymphocytes (20-25%) are of two major types—‘B’ and ‘T’ forms. Both B and T lymphocytes are responsible for immune responses of the body.

Q11. The second heart sound (dub) is associated with the closure of
(a) Tricuspid valve (b) Semilunar valves
(c) Bicuspid valve (d) Tricuspid and bicuspid valves
Ans: (b) The first heart sound (lub) is associated with the closure of the tricuspid and bicuspid valves whereas the second heart sound (dub) is associated with the closure of the semilunar valves.

Q12. Which of the following correctly explains a phase/event in cardiac cycle in a standard electrocardiogram?
(a) QRS complex indicates atrial contraction
(b) QRS complex indicates ventricular contraction
(c) Time between S and T represents atrial systole
(d) P-wave indicates beginning of ventricular contraction
Ans: (b)
• P-wave—Atrial depolarisation (atrial systole/contraction)
• QRScomplex—Ventriculardepolarisation(ventricularsystole/contraction)
• T-wave—Ventricular repolarisation (ventricular relaxation)

Q13. Which of the following statements is incorrect?
(a) A person of ‘O’ blood group has anti ‘A’ and anti ‘B’ antibodies in his blood plasma
(b) A person of ‘B’ blood group cannot donate blood to a person of ‘A’ blood group
(c) Blood group is designated on the basis of the presence of antibodies in the blood plasma
(d) A person of AB blood group is universal recipient
Ans: (c) Blood group is designated on the basis of the antigen is present on the RBCs surface

Q14. What would be the cardiac output of a person having 72 heart beats per minute and a stroke volume of 50 mL?
(a) 360 mL (b) 3600 mL (c) 7200 mL (d) 5000 mL
Ans: (b) Cardiac output = Stroke volume x Heart rate = 50 x 72 = 3600 mL/min

Q15. Match the terms given under Column ‘A’ with their functions given under Column ‘B’ and select the answer from the options given below:

Column A Column B
A. Lymphatic system (i) Carries oxygenated blood
B. Pulmonary vein (ii) Immune Response
C. Thrombocytes (iii) To drain back the tissue fluid to the circulatory system
D. Lymphocytes (iv) Coagulation of blood

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

Ans: (b)

A. Lymphatic System (iii) To drain back the tissue fluid to the circulatory system
B. Pulmonary vein (i) Carries oxygenated blood
C. Thrombocytes (iv) Coagulation of blood
D. Lymphocytes (ii) Immune Response

 

Q16. Read the following statements and choose the correct option.
Statement 1: Atria receive blood from all parts of the body which subsequently flows to ventricles.
Statement 2: Action potential generated at sino-atrial node passes from atria to ventricles.
(a) Action mentioned in Statement 1 is dependent on action mentioned in Statement 2.
(b) Action mentioned in Statement 2 is dependent on action mentioned in Statement 1.
(c) Actions mentioned in Statements 1 and 2 are independent of each other.
(d) Actions mentioned in Statements 1 and 2 are synchronous.
Ans: (b) Statement 1: Atria receive blood from all pans of the body which subsequently flows to ventricles.
Statement 2: Action potential generated at sino-atrial node passes from atria to ventricles.
Action mentioned in Statement 2 is dependent on action mentioned in Statement 1.
Very Short Answer Type Questions
Q1. Name the blood component which is viscous and straw coloured fluid.
Ans: Plasma .

Q2. 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.

Ans: a. Plasma without clotting factors is called serum.
b. Neutrophils and monocytes are phagocytic cells.
c. Eosinophils are associated with allergic reactions.
d. Calcium ions play a significant role in clotting.
e. One can determine the heart beat rate by counting the number of QRS complex in an ECG.

Q3. Given below is the diagrammatic representation of a standard ECG. Label its different peaks
NCERT Exemplar Class 11 Biology Chapter 18 Body Fluids and Circulation Img 2

Q4. Name the vascular connection that exists between the digestive tract and liver.
Ans: Hepatic portal system

Q5. 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
Ans: a. Acute chest pain due to failure of 02 supply to heart muscles—Angina
b. Increased systolic pressure—Hypertension/high blood pressure

Q6. Which is coronary artery diseases caused due to narrowing of the lumen of arteries?
Ans: Atherosclerosis

Q7. Define the following terms and give their-locations?
a. Purkinje fibre
b.Bundle of His
Ans: a. Purkinje fibre—Right and left bundles give rise to minute fibres throughout the ventricular musculature of the respective sides and are called purkinje fibres. .
b. Bundle of His—Purkinje fibres alongwith right and left bundles are known as bundle of His and present in ventricles.

Q8. State the functions of the following in blood:
a. Fibrinogen b. Globulin
c. Neutrophils d. Lymphocytes
Ans: a. Fibrinogen—Fibrinogens are needed for clotting or coagulation of blood. ■
b. Globulin—Globulins primarily are involved in immunity, i.e., defense mechanisms of the body.
c. Neutrophils—Phagocytosis
d. Lymphocytes—Immunity

Q9. What physiological circumstances lead to’erythroblastosis foetalis?
Ans: A special case of Rh incompatibility (mismatching) has been observed between the Rh-negative blood of a pregnant mother with Rh-positive blood of the foetus.

Q10. Explain the consequences of a situation in which blood does not coagulate.
Ans: This situation leads to excessive loss of blood from body due to injury which
can be fatal.
Q11. What is the significance of time gap in the passage of action potential from sino-atrial node to the ventricle?
Ans: This time gap is significant for ventricular systole.

Q12. How will you interpret an electrocardiogram (ECG) in which time taken in QRS complex is higher.
Ans: Period of ventricular systole increases

Short Answer Type Questions

Q1. The walls of ventricles are much thicker than atria. Explain.
Ans: The walls of ventricles are much thicker than atria because they pump blood more strongly than the atria.

Q2. Differentiate between:
a.  Blood and Lymph
b. Basophils and Eosinophils
c. Tricuspid and Bicuspid valves
Ans: a. Blood and Lymph: Blood is a connective tissue consisting of a fluid matrix, plasma and formed elements (RBCs, WBCs and Platelets). Blood flows in blood vascular system comprising heart, arteries and veins.
Lymph is a colourless fluid containing specialised lymphocytes (imparting immunity to the body), but devoid of RBCs. Lymph flows in the lymphatic system and it absorbs fats.

a. Blood    ‘ Lymph
1. It contains plasma, RBCs, WBCs and platelets 1. It contains plasma and lymphocytes
2. It is red in colour 2. It is colourless
3. Haemoglobin is present 3. Haemoglobin is absent
4. It transports nutrients and gases from heart to tissues and vice-versa 4. It transports infection fighting white blood cells from tissues to lymph nodes
b. Basophils Eosinophils
1. They constitute about 0.5-1% of WBCs 1. They constitute about 2-3% of WBCs
2. They secrete heparin,

histamine and serotonin

2. They resist infection
3. They are involved in inflammatory reaction 3. These are associated with allergic reactions
c. Tricuspid valve Bicuspid valve
1. It has three cusps. 1. It have two cusps.
2. It is present between right atrium and right ventricle 2. It is present between left atrium and left ventricle.

 

Q3. Briefly describe the following:
a. Anaemia
b. Angina Pectoris
c. Atherosclerosis
d. Hypertension
e. Heart failure
f. Erythroblastosis foetalis
Ans: a. Anaemia: Decrease in oxygen carrying capacity of blood either due to reduced RBCs production or low haemoglobin content is called anaemia.
b. Angina Pectoris: A symptom of acute chest pain appears when no enough oxygen is reaching the heart muscle. Angina can occur in men and women of any age but it is more common among the middle-aged ” and elderly. It occurs due to conditions that affect the blood flow.
c. Atherosclerosis: Sometimes deposition of calcium, fat, cholesterol and fibrous tissues occurs in the blood vessel (e.g., coronary artery) supplying blood to the heart muscles. This condition makes the lumen of arteries narrower affecting blood supply to heart; which leads to Coronary Artery Disease (CAD) also referred to as atherosclerosis.
d. Hypertension: If repeated checks of blood pressure of an individual is 140/90 (140 over 90) or higher, it shows hypertension. High blood pressure leads to heart diseases and also affects vital organs like brain and kidney.
e. Heart failure: Heart failure means the state of heart when it is not pumping blood effectively enough to meet the needs of the body. It is sometimes called congestive heart failure because congestion of the lungs is one of the main symptoms of this disease.
f. Erythroblastosis foetalis: A special case of Rh incompatibility (mismatching) has been observed between the Rh-negative blood of a pregnant mother with Rh-positive blood of the foetus! Rh antigens of the foetus do not get exposed to the Rh-negative blood of the mother in the first pregnancy as the two bloods are well separated by the placenta. However, during the delivery of the first child, there is a possibility of exposure of the maternal blood to small amounts of the Rh-positive blood from the foetus. In such cases, the mother starts preparing antibodies against Rh antigen in her blood. In case of her subsequent pregnancies, the Rh antibodies from the mother (Rh-negative) can leak
into the blood of the foetus (Rh-positive) and destroy the foetal RBCs. This could be fatal to the foetus or could cause severe anaemia and jaundice to the baby. This condition is called erythroblastosis foetalis. This can be avoided by administering anti-Rh antibodies to the mother immediately after the delivery of the first child.

Q4. Explain the advantage of the complete partition of ventricle among birds and mammals and hence leading to double circulation.
Ans: Complete partition of ventricle among birds and mammals is advantageous because there is no mixing of oxygenated and deoxygenated blood in the ventricle, so tissues of the body receive more oxygenated blood.

Q5. What is the significance of hepatic portal system in the circulatory system?
Ans: The hepatic portal vein carries blood from intestine to liver before it is
delivered to systemic circulation. This is significant because excess of nutrients like glucose is converted into glycogen in liver and stored there.

Q6. Explain the functional significance of lymphatic system?
Ans: As the blood passes through the capillaries in tissues, some water along with many small water soluble substances move out into the spaces between the cells of tissues leaving the larger proteins and most of the formed elements in the blood vessels. This fluid released out is called the interstitial fluid or tissue fluid. It has the same mineral distribution as that in plasma. Exchange of nutrients, gases, etc., between the blood and the cells always occur through this, fluid. An elaborate network of vessels called the lymphatic system collects this fluid and drains it back to the major veins. The fluid present in the lymphatic system is called the lymph. Lymph is a colourless fluid containing specialised lymphocytes which are responsible for the immune responses of the body. Lymph is also an important carrier for nutrients, hormones, etc. Fats are absorbed through lymph in the lacteals present in the intestinal villi.

Q7. Write the features that distinguish between the two
a. Plasma and Serum
b. Open and Closed circulatory system
c. Sino-atrial node and Atrio-ventricular node

a. Plasma Serum
i. Blood without              formed

elements is called plasma

1. Plasma without clotting factor is called serum
2. Plasma has clotting factors 2. Serum does not have clotting factors
3. Plasma involved in blood coagulation 3. Serum does not involve in blood coagulation
b. Open circulatory system Closed circulatory system
1. Blood pumped by heart passes through large vessels into open spaces or body cavities called sinuses 1. Blood pumped by the heart is always circulated through a closed network of blood vessels
2. Less advantageous 2. More advantageous
3. Flow of fluid cannot be more precisely regulated 3. Flow of fluid can be more precisely regulated
4. It is present in arthropods molluscs, and hemichordates 4. It is found in annelids and chordates
c. Sino-atrial node Atrio-ventricular node
1. SA node is present in the right upper comer of the right atrium 1. AV node is present in the lower left comer of the right atrium
2. It initiates and maintains the rhythmic contractile activity of the heart 2. It passes the electrical impulses from SA node to AV bundle
3. It is also called pace-maker 3. It is also called pace-setter

 

Q9. Answer the following:
a. Name the major site where RBCs are formed.
b. Which part of heart is responsible for initiating and maintaining its rhythmic activity?
c. What is specific in the heart of crocodiles among reptilians?
Ans: a. Bone marrow
b. Sino-Atrial Node (SA Node)
c. Reptiles have 3-chambered heart but crocodiles have 4-chambered heart.

Long Answer Type Questions
Q1. Explain Rh-incompatibility in humans.
Ans: A special case of Rh-incompatibility (mismatching) has been observed between the Rh-negative blood of a pregnant mother with Rh-positive blood of the foetus. Rh antigens of the foetus do not get exposed to the Rh-negative blood of the mother in the first pregnancy as the two bloods are well separated by the placenta. However, during the delivery of the first child, there is a possibility of exposure of the maternal blood to small amounts of the Rh – positive blood from the foetus. In such cases, the mother starts preparing antibodies against Rh antigen in her blood. In case of her subsequent pregnancies, the Rh antibodies from the mother (Rh-negative) can leak into the blood of the foetus (Rh-positive) and destroy the foetal RBCs. This could be fatal to the foetus or could cause severe anaemia and jaundice to the baby. This condition is called erythroblastosis foetalis. This can be avoided by administering anti-Rh antibodies to the mother immediately after the delivery of the first child.

Q2. Describe the events in cardiac cycle. Explain “double circulation”.
Ans:
• Cardiac cycle: To begin with, all the four chambers of heart are in a relaxed state, i.e., they are in joint diastole. As the tricuspid and bicuspid valves are open, blood from the pulmonary veins and vena cava flows into the left and the right ventricle respectively through the left and right atria. The semilunar valves are closed at this stage. The SAN now generates an action potential which stimulates both the atria to undergo a simultaneous contraction—the atrial systole. This increases the flow of blood into the ventricles by about 30%. The action potential is conducted to the ventricular side by the AVN and AV bundle from where the bundle of His transmits it through the entire ventricular musculature. This causes the ventricular muscles to contract (ventricular systole), the atria undergoes relaxation (diastole), coinciding with the ventricular systole. Ventricular systole increases the ventricular pressure causing the closure of tricuspid and bicuspid valves due to attempted backflow of blood into the atria. As the ventricular pressure increases further, the semilunar valves guarding the pulmonary artery (right side) and the aorta (left side) are forced open, •allowing the blood in the ventricles to flow through these vessels into the circulatory pathways.
The ventricles now relax (ventricular diastole) and the ventricular pressure falls causing the closure of semilunar valves which prevents the backflow of blood into the ventricles. As the ventricular pressure declines further, the tricuspid and bicuspid valves are pushed open by the pressure in the atria exerted by the blood which was being emptied into them by the veins. The blood now once again moves freely to the ventricles. The ventricles and atria are now again in a relaxed (joint diastole) state, as earlier. Soon the SAN generates a new action potential and the events described above are repeated in that sequence and the process continues.
• Double circulation: The blood pumped by the right ventricle enters the pulmonary artery, whereas the left ventricle pumps blood into the aorta. The deoxygenated blood pumped into the pulmonary artery is passed on to the lungs from where the oxygenated blood is carried by the pulmonary veins into the left atrium. This pathway constitutes the pulmonary circulation. The oxygenated blood entering the aorta is carried by a network of arteries, arterioles and capillaries to the tissues from where the deoxygenated blood is collected by a system of venules, veins and vena cava and emptied into the right atrium. This is the systemic circulation. The systemic circulation provides nutrients, 02 and other essential substances to the tissues and takes C02 and other harmful substances away for elimination.

Q3. Explain different types of blood groups and donor compatibility by making a table.
Ans: ABO blood grouping is based* on the presence or absence of two surface antigens on the RBCs namely A and B. Similarly, the plasma of different individuals contain two natural antibodies anti-A and anti-B. Blood group ‘A’ carries antigen-A and antibodies-B. The donor’s group for blood group A are A and O. Blood group B carries antigen-B and antibodies-A. The donor’s group for blood group B are B and O. Blood group AB carries antigens A and B but no corresponding antibodies so, the compatible donor’s group for blood group AB are A, B, AB and O hence, blood group ‘AB’ is also called as “universal acceptor’’. Blood group ‘O’ carries no antigens but carries antibodies both A and B hence its compatible donor’s group is only ‘O’ but it is a compatible donor group for all the blood groups. A, B, AB and O hence, blood group ‘O’ is called as ‘universal donor’.

Blood groups and donor compatibility

Blood Group . Antigen on RBCs Antibodies in Plasma Donor’s

Compatibility

A A Anti-B A, O
B B Anti-A B, 0
AB A. B Nil AB, A, B, 0
0 ‘ Nil Anti – A, B 0

Q4. Write a short note on the following:
a. Hypertension
b. Coronary Artery Disease
Ans: a. Hypertension: If repeated checks of blood pressure of an individual is 140/90 (140 over 90) or higher, it shows hypertension. High blood pressure leads to heart diseases and also affects vital organs like brain and kidney.
b. Coronary Artery Disease: Coronary Artery Disease, often referred to as atherosclerosis, affects the vessels that supply blood to the heart muscle. It is caused by deposits of calcium, fat, cholesterol and fibrous tissues, which makes the lumen of arteries narrower.

Q5. In the diagrammatic presentation of heart given below, mark and label, SAN, AVN, AV bundles, bundle of His and Purkinje fibres.

NCERT Exemplar Class 11 Biology Chapter 18 Body Fluids and Circulation Img 3

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