Biomolecules Class 11 Biology NCERT Solutions

Step-by-step solution

1. 1Give the definition first, then the classification with the three heads, and then the examples under each, since the question asks for both the term and the instances.
2. 2Definition: macromolecules are the large molecules of the cell, and they are the compounds that have a molecular weight of more than one thousand dalton. NCERT states the figure as molecular weight above 1000 daltons, and this is the definition to quote, because the examiner expects the number, not only the adjective large. They are the acid soluble fraction and the acid insoluble fraction, and this division is a practical one, since the acid soluble fraction can be extracted by treatment with dilute acid and is made of the smaller molecules, whereas the acid insoluble fraction is the macromolecular fraction that requires stronger treatment to break down.
3. 3The examples fall into three groups, and this is the cleanest way to organise the answer. First, the polysaccharides, which are the carbohydrates, and the examples are starch, glycogen, cellulose, chitin and inulin. Second, the proteins, which are polypeptides, and the examples are the enzymes, the structural proteins such as the collagen of the connective tissue, the hormone insulin, and the antibodies, and NCERT also names keratin and myosin. Third, the nucleic acids, which are polynucleotides, and these are DNA and RNA. The lipids are a different case and must be mentioned separately, because they are not strictly macromolecules, and this is the point the question is testing: lipids have a molecular weight of less than 800 daltons, and so they are not true macromolecules, although they are insoluble and form part of the acid insoluble fraction along with the macromolecules.
4. 4The two points to close on, because both are examinable, are these. First, the lipids are not strictly macromolecules, having a molecular weight below 800 daltons, and they are described as micro-molecules, but they are still a part of the acid insoluble fraction, and this is because they are not truly polymers, since a triglyceride is formed by the esterification of glycerol with fatty acids and not by a long chain of repeating monomers. Second, the macromolecules of the cell are not all polymers, since the lipids are the exception, and apart from the lipids every macromolecule named above is built by the repetition of a small monomer unit, which is precisely what makes their synthesis by the cell a matter of repeating a unit rather than of assembling a novel structure.

Step-by-step solution

1. 1Define the three-dimensional structure of a single polypeptide, and make the contrast with the secondary structure explicit, because the whole point of the term tertiary is that it is the next level above the alpha helix and the beta pleated sheet and not simply any folding.
2. 2The definition: the tertiary structure of a protein is the three-dimensional shape of a single polypeptide chain, and it is the definite and stable shape that a protein assumes because of its primary, secondary and tertiary structure together, in the environment in which it is placed. It is a compact, folded and globular shape in the case of most functional proteins, and NCERT emphasises the word definite, because a given protein folds into one reproducible conformation rather than a random coil, and it is that reproducibility which makes the protein biologically active.
3. 3The reason this level of structure exists is secondary bonding, and this is the sentence to write. Secondary structure, that is the alpha helix and the beta pleated sheet, is stabilised only by hydrogen bonds between the backbone groups, and it is therefore a local and comparatively loose arrangement. The tertiary structure is held together by additional bonds involving the side chains, that is the R groups of the amino acids, and these include hydrophobic bonds, which are the chief stabilising force and are formed because the non-polar side chains are driven together away from water, disulphide bonds, which are covalent sulphur-to-sulphur bonds and are by far the strongest of the interactions, hydrogen bonds, ionic or electrostatic bonds and van der Waals forces, which are weak individual attractions but numerous in number.
4. 4Add the functional consequence, because it completes the answer. In the absence of these bonds the tertiary structure would not exist and the protein would be a floppy chain, so it is the side-chain interactions that produce the definite 3D shape, and it is that shape which allows the protein to function, since a protein is only functional when it is folded correctly, and the folding also depends on the medium, so an acidic or alkaline environment can denature the protein and destroy its shape and therefore its activity. NCERT makes the same point by saying that the tertiary structure is a function of the primary structure and of the environment, which is exactly the statement that protein shape and protein function are one and the same thing.

Step-by-step solution

1. 1This is a field-work and enquiry question, so the answer must actually deliver the ten structures and then the industry and the buyer, rather than only describing the method. Present it as a table, since the question asks for structures and the reader must be able to check each formula, and then write the industrial section as prose.
2. 2NCERT itself supplies the set of ten in the worked example of this exercise, and the ten to write are the amino acids glycine, alanine, valine, leucine and isoleucine, the sugars glucose, ribose and the sugar component of the nucleic acid nucleotides, the nucleotides adenosine monophosphate and guanosine monophosphate, the lipids a triglyceride and a phospholipid, and the vitamins vitamin A and vitamin C. Writing down any ten of these with their correct structures satisfies the question, and the structures are all simple to draw as displayed formulae, since the amino acids are a general formula with a variable R group, the sugars are straight or five-membered chains, the nucleotides are the three-part base, sugar and phosphate unit, the triglyceride is glycerol joined to three fatty acid chains, the phospholipid is glycerol joined to two fatty acids and a phosphate, and the vitamins have the standard ring or long-chain structures.
3. 3The structures to present, in a table with three columns of name, class and structure, are these. Glycine, amino acid, NH2-CH2-COOH, the simplest and the only one with no chiral carbon. Alanine, amino acid, NH2-CH(CH3)-COOH. Valine, amino acid, NH2-CH(CH(CH3)2)-COOH, a branched-chain amino acid. Leucine, amino acid, NH2-CH(CH2CH(CH3)2)-COOH. Isoleucine, amino acid, NH2-CH(CH(CH3))CH2CH3-COOH, with the amino and carboxyl groups on adjacent carbons. Glucose, monosaccharide, a six-carbon aldose with the chain CHO, (CHOH)4, CH2OH. Ribose, monosaccharide, a five-carbon aldose, CHO, (CHOH)3, CH2OH. Adenosine monophosphate, nucleotide, adenine joined to ribose joined to phosphate. Guanosine monophosphate, nucleotide, guanine joined to ribose joined to phosphate. A triglyceride, lipid, glycerol esterified with three fatty acids. A phospholipid, lipid, glycerol esterified with two fatty acids and one phosphate. Vitamin A, fat-soluble vitamin, a long isoprenoid chain with a β-ionone ring. Vitamin C, water-soluble vitamin, a six-membered lactone ring with an enediol group.
4. 4The industrial part. The amino acids are manufactured by isolation, and the classical example is the production of glutamic acid, which is isolated from a microbial broth and used as a flavouring agent under the name monosodium glutamate or MSG, and the buyers are the food-processing companies, that is the manufacturers of soups, snack foods, seasoning powders and instant noodles. Cysteine is similarly isolated and used in bread making, where it improves the dough. The sugars are the largest of the isolation industries, since sucrose is obtained from the sugar cane and the sugar beet and glucose is obtained from starch by acid or enzyme hydrolysis, and the buyers are the confectionery, soft-drink, bakery and pharmaceutical industries. The nucleotides and the nucleosides are isolated for use in the biochemical and pharmaceutical industries, as components of the culture media, of the reagents, and of the nucleotide drugs such as the antiviral, and the buyers are the biochemical laboratories and the pharmaceutical manufacturers.
5. 5Close with the observation the question is really after. The interesting point is that the human and other animal body cannot manufacture all of these and is dependent on dietary intake, and that the industries exist only because of that dependence, so the isolation industry and the medicine industry both rest on the fact that a small molecular weight biomolecule is made in only small quantity by the body or not at all, and is therefore obtained by isolation from plants, from bacteria or from tissue and sold to the buyers, who are the food and pharmaceutical industries. It is a neat illustration that a purely botanical or biochemical fact, that we cannot synthesise every biomolecule ourselves, is the commercial foundation of a whole set of industries.

Step-by-step solution

1. 1Split the answer into the two halves the question names, therapeutic agents and other applications, and within each half give the protein first and the use attached to it, because the protein is the item and the application is the point.
2. 2Therapeutic agents, the standard NCERT list. Trypsin and chymotrypsin are given for dissolving blood clots and for treating inflammation, and the enzyme bromelain, obtained from the pineapple, is used for treating bruises and inflammation, while papain, from the papaya, is used in the treatment of indigestion, and pepsin and trypsin are digestive enzymes of use in digestion itself. Cytochrome oxidase, with its function clarified as a redox enzyme and not as a cytochrome, is a protein used in the treatment of certain respiratory and metabolic disorders in the standard list, and the important thing is to get its name right. The hormones insulin, as a hormone, oxytocin, and the growth hormone or somatotropin are used in the treatment of diabetes, of disorders of childbirth and of growth-related disorders respectively, and the hormone of the parathyroid, thyroxine, is used against goitre and against hypothyroidism, while the steroid cortisone is a related therapeutic use and adrenaline and its derivatives are used in the treatment of bronchial asthma and of cardiac disorders. The antibodies and the antitoxins, the sera and the vaccines, are proteins used against infections, the antitoxin of the cobra and of the snake bites, the antitetanus serum and the antivenom, and the anti-Rh serum used in cases of erythroblastosis foetalis, all belong to this group. Albumin and plasma proteins are used to restore the blood volume in cases of severe loss of blood, and gelatin is used to repair cartilage and bone defects. Collagen, elastin and keratin are used in cosmetic and surgical preparations, and protein hydrolysates are used in the preparation of artificial diets.
3. 3Other applications, and this is where the question wants breadth. In cosmetics, collagen is added to face creams and to anti-wrinkle preparations because it gives elasticity and smoothness, elastin keeps the skin supple, keratin is used in hair creams and in polishes, and gelatin is used as a humectant. In nutrition, protein hydrolysates and casein are used in dietetic foods, and the milk proteins of the infant formula are proteins of this class. In agriculture, the casein of milk and the proteins of the seed are used in animal feed, and the microbial proteins from single-cell protein are a source for human food as well. In the textile and leather industries, casein, a milk protein, is used in the manufacture of rayon and in waterproofing, and in plastics and in the manufacture of buttons and adhesives, and in photography gelatin is used in the emulsion of the film. In the laboratory, albumin is used as a molecular weight marker in electrophoresis, and the enzymes are used in the detergent industry as the biological detergents already mentioned. Mention also the defensive proteins, the antibodies, and the transport proteins of the plasma, since these are applications too.
4. 4The two closing observations that make the answer look informed. First, note that the applications are of proteins, not of any single protein, and that a large fraction of the uses depend on the ability of a protein to act as an enzyme and to be highly specific, since it is the specificity of the enzyme that makes the digestive and the clot-dissolving applications possible at all. Second, note that the wide application list shows why the genetic engineering of the Recombinant DNA technology is commercially important, since human insulin, human growth hormone and the interferons are now produced in this way in bacteria, and the commercial demand is the reason for the effort.

Step-by-step solution

1. 1Name the components, then the linkage, then the shape and the property that follows from the shape, in that order. A triglyceride is a lipid, so no chirality and no isomerism is possible, and that is the single most examinable point about it.
2. 2Composition: a triglyceride is formed by the esterification of one molecule of glycerol with three molecules of fatty acids, and glycerol is a trihydric alcohol with the structure CH2OH-CHOH-CH2OH, that is it has three hydroxyl groups, while a fatty acid is a monocarboxylic acid with a long even-numbered hydrocarbon chain. The three components and their counts should be stated explicitly, one glycerol to three fatty acids, since the number three is what the name states and what the answer is tested on.
3. 3The linkage is the ester bond, and it is formed by a condensation or dehydration reaction between the hydroxyl group of the glycerol and the carboxyl group of the fatty acid, with the elimination of a molecule of water for each of the three linkages. So three ester bonds are formed and three molecules of water are lost, and this is the standard esterification of a fatty acid with an alcohol, and the resulting compound is an ester. This should be contrasted with the peptide and the glycosidic bonds, since this is the chapter in which the three are compared.
4. 4Shape and properties, and these carry the rest of the marks. The three fatty acid chains are long, and because the ester bonds are non-polar and the hydrocarbon chains dominate the molecule, the triglyceride is entirely non-polar and hydrophobic, and hence insoluble in water and soluble in organic solvents such as ether, chloroform and benzene. The chains are unbranched and because they are unbranched and identical in a simple triglyceride, the molecule is not chiral at any carbon, and the three chains can pack closely together, and this close packing is the physical basis of the grease being a semisolid or a solid. The fats are triglycerides that are largely solid at ordinary temperature and the oils are those that are largely liquid, and the difference is entirely in the length and the degree of saturation of the fatty acid chains, since a saturated or a shorter chain packs better and melts at a higher temperature, whereas an unsaturated chain with a double bond has a kink that prevents close packing and lowers the melting point. And because a triglyceride is an uncharged, non-polar, insoluble molecule with no groups able to form hydrogen bonds, it cannot form a membrane on its own, and this is why the phospholipid, which retains the glycerol-phosphate-head group and only two fatty acid chains, is the membrane molecule.

Step-by-step solution

1. 1Answer the question as an activity, that is describe what to do and what is found, and be honest that the models are the hard-sphere and the ball-and-stick types sold commercially. The answer should show that the exercise is about the geometry of the carbon skeleton and about the shapes the atoms can take, not about the chemistry of synthesis.
2. 2The materials: a commercially available ball-and-stick molecular model set, which contains coloured spheres representing the atoms of a given element in a fixed size for that element, and short rods, or sticks, representing the bonds, together with the instruction sheet that gives the valencies, so that hydrogen is valency one, oxygen is two, nitrogen is three and carbon is four. The board, the cellulosic or plastic spheres for the larger atoms and the smaller ones for hydrogen and oxygen, all come in the set, and the colour coding is fixed, so carbon is always one colour and hydrogen always another.
3. 3What to build, and the exercise should be done in the order of increasing difficulty, since each one shows a new geometric point. First the alanine of Q7, that is two carbons joined with an amino group, a carboxyl group and a methyl group and a hydrogen, which introduces the tetrahedral carbon and the four valencies. Then a saturated fatty acid, then a triglyceride, in which the three chains must all point away from the glycerol backbone, which is the first model that shows steric arrangement. Then a glucose chain to show the ring, and then a short fragment of a polypeptide to show the helical arrangement of the peptide bond, and finally a short piece of a double helix built as a ladder of paired bases, with the two strands joined by the hydrogen bonds, which is the model that makes the structure of DNA visible rather than merely described.
4. 4What the models show, and this is the conclusion to write. First, that the valencies of the atoms determine the shape, so carbon is tetrahedral because its four valencies point to the corners of a tetrahedron, and sulphur and phosphorus are pyramidal for the same reason, and this is why a straight chain of carbons zig-zags rather than lying flat. Second, that the models make visible the fact that the monomers of a macromolecule are joined in a specific sequence and a specific geometry, so that the same atoms in the same numbers can form different substances, since a starch and a cellulose are both glucose polymers differing only in the geometry of the glycosidic linkage, and a triglyceride and a phospholipid are both glycerol-fatty acid lipids differing only in the third group. Third, that the three-dimensional folding of a macromolecule determines its function, since only a correctly folded protein is active. And the limitation to state, because examiners like it, is that the ball-and-stick model is a rigid model and shows bonds and angles but not the movement of the electrons and not the real flexibility of the molecules, so it is a teaching aid and not a copy of reality.

Step-by-step solution

1. 1This is a drawing question, so the answer must say exactly which bonds to draw and which atoms to label, and should then give the structure in a form that can be copied onto the paper.
2. 2The identity of the molecule: alanine is the simplest amino acid after glycine, with a methyl group as its side chain, and it is therefore the compound to use for the demonstration of the general alpha-amino acid structure, in which the amino group and the carboxyl group are both attached to the same carbon, that is the alpha carbon. The word alpha matters, because the whole definition of an alpha-amino acid rests on it.
3. 3The general formula to draw and label first is the alpha-amino acid one, CH(NH2)(COOH) bonded to an R group, and it should be drawn with the central carbon in the middle, the carboxyl group to the right, the amino group above or below, the hydrogen on the same carbon, and the R group on the opposite side. The labels to write on the drawing are the central or alpha carbon, the amino group, the carboxyl group and the variable R or side chain, and the fourth bond to the hydrogen, so that the four substituents on the alpha carbon are all identified. This is the diagram that earns the first marks, because it shows the general structure and the labels.
4. 4Then the specific structure of alanine, which is the same diagram with the R group filled in as a methyl group, CH3, so the structure is CH3-CH(NH2)-COOH, or written with the bonds fully displayed, H3C-CH(NH2)-C(=O)-OH, and the carbon of the methyl group is the beta carbon while the carbon bearing the amino and the carboxyl groups is the alpha carbon. Alanine is a non-polar, aliphatic amino acid, since its R group is a hydrocarbon, and it is one of the twenty that are used by the cell to build proteins, and it is the amino acid that is the immediate precursor in the synthesis of the pyruvate and of the glucose in theglycolysis pathway, which is worth one closing sentence because it connects the structure to the chapter of metabolism.

Step-by-step solution

1. 1Answer the two halves in turn, and the first half is answered from the chemistry of the chapter, since a gum is a polysaccharide, while the second half is a comparison and needs a stated point of difference, which is that the gum is a carbohydrate and the adhesive is not.
2. 2Gums are polysaccharides. They are complex carbohydrates, that is polymers of monosaccharides, and they are formed largely by plants, being the dried exudates of the stems and of the trunks of trees, so a gum is a plant product and the tears or the resin-like exudate of an Acacia or a Sterculia is the usual source. The units are the pentoses and the hexoses, and the polymer may be a homopolysaccharide or a heteropolysaccharide, and in either case the bond is a glycosidic bond, which is the same linkage as in starch and cellulose. So the answer to the first half is that the gums are heteropolysaccharides, complex carbohydrates of plant origin, joined by glycosidic linkages, and they are non-reducing in the usual case and are used by the plant as a protection against injury and against desiccation, since the exudate hardens on exposure to air.
3. 3Now the comparison with Fevicol. Fevicol is a synthetic adhesive, and it is emphatically not a gum in the biological sense, because it is not a carbohydrate at all. It is a polyvinyl acetate emulsion, a synthetic polymer prepared from the petroleum-derived monomer vinyl acetate, and it is sold as an aqueous emulsion of white latex, so its molecular nature is a synthetic vinyl polymer rather than a polysaccharide. It is therefore a protein-free and a carbohydrate-free adhesive, and it is not edible and is not a food product, whereas a gum is a carbohydrate, is digestible to a degree and is a recognised food material.
4. 4The two points of difference to close with, so that the comparison is complete. The first is chemical: the gum is a natural polysaccharide, a carbohydrate of plant origin, joined by glycosidic bonds and having the general formula characteristic of a polymer of monosaccharides, whereas the Fevicol is a synthetic polyvinyl polymer, not a carbohydrate, and contains no monosaccharide units. The second is biological and practical: the gum is a product of the living plant, formed as an exudate and used by the plant itself, and it is a food for man, whereas Fevicol is made in a factory from a synthetic monomer and is used only as an adhesive and is not fit for food. A third, minor point is that both are used in practice as adhesives and thickeners, and that is the confusion the question is aiming at, so it is worth saying that the similarity is one of use and not of composition.

Step-by-step solution

1. 1This is a laboratory test question, so give the reagent, the procedure, the positive result and then the interpretation for each of the four, and close with the caution about the interference that makes the biology honest.
2. 2Proteins. The qualitative test is the Biuret test. The procedure is to take a small quantity of the sample in a test tube, add a few drops of freshly prepared dilute sodium hydroxide solution, that is 10 per cent NaOH, so that the sample is made alkaline, and then add two or three drops of a dilute copper sulphate solution, about one per cent, drop by drop with shaking, so that a blue or violet colour appears. The positive result is a violet or purple colour, and the negative result is no violet, only the blue of the excess copper sulphate remaining. The principle is that in an alkaline medium the copper ions form a coordination complex with the peptide bonds of the protein, and the violet is the colour of that complex, so the test detects the presence of peptide bonds, which means a substance with two or more peptide linkages, that is a polypeptide, will give it.
3. 3Fats and oils. The test is the grease spot or the paper test, and the soluble test is the emulsion test. For the grease spot, take a small quantity of the sample, place it on a piece of paper or a cotton cloth, and dry it, and if the paper becomes translucent and shows a permanent oily spot that does not disappear on drying and washing, the test is positive. The principle is that the fat molecules are non-polar and cannot be wetted by water, so they make a translucent spot on the cellulose of the paper. The emulsion test is the more convincing one, since fats and oils are insoluble in water, and if the sample is shaken with water in a test tube, a milky emulsion forms which settles on standing, and the presence of an emulsifier such as bile salts or soap is required for the emulsion to be stable.
4. 4Amino acids. The test is the Ninhydrin test, which is used for the free amino groups. A small quantity of the sample is boiled with a ninhydrin solution, and on cooling a deep blue or violet colour is produced, and this is the positive result, while the negative result is the yellow colour of the ninhydrin itself. The principle is that an amino acid, and any compound with a free alpha-amino group, reacts with ninhydrin to form a blue or purple product called Ruhemann's purple, so the test detects free amino groups, not peptide bonds, and the two tests are therefore complementary: the Biuret detects the peptide bond of a protein and the ninhydrin detects the free amino group of an amino acid.
5. 5The tests to be applied, with the results to expect, and the interpretation. Fruit juice, since it is a plant product, should be expected to give a positive ninhydrin test for free amino acids, and it should give a positive Biuret test if it contains soluble protein, though most fruit juices contain too little protein to give a strong Biuret unless they are protein-rich, and they should give a positive grease spot for the waxes, the oils and the lipids of the cuticle and the seed, so a filtered and clear juice will give a strong oil test and the pulp will give it more strongly. Saliva should give a positive Biuret test, because saliva contains the protein enzyme ptyalin, the salivary amylase, and it should also give a positive ninhydrin test for the free amino acids of the amino acids that are present in low concentration. Sweat should give a positive Biuret test, since the sweat contains the protein of the sweat, and a positive grease spot for the sebaceous material, and it should give a positive ninhydrin test for the free amino acids, and a positive test for the salts is also expected since the sweat carries sodium and chloride salts. Urine should give a positive ninhydrin test, since urine contains the free amino acids, and it should give a positive grease spot, since the lipids and the fatty casts occur, but it should normally give a negative Biuret test because the proteins of the plasma are too large to pass the glomerulus and so do not appear in a normal urine, and it is a positive Biuret test that is a pathological finding, being the basis of the clinical test for the kidney damage and the nephrotic condition, where the glomerular filter is damaged and protein leaks through.
6. 6The controls and the limitations to state, since these are what turn a result into a valid conclusion. Every test must be run with a known positive and a known negative, so a solution of egg white is the positive control for the Biuret, a solution of glycine the positive control for the ninhydrin and a drop of vegetable oil the positive control for the grease spot, and distilled water is the negative control for all three. Then note the limits: the Biuret needs an alkaline medium and a drop too much copper sulphate gives a blue precipitate of copper hydroxide that masks the violet, and the ninhydrin needs boiling and is interfered with by the ammonia of urine, and the grease spot is insensitive at low concentration, so a negative grease spot means the fat is below the limit of the test and not that it is absent, and the presence of a detergent or of an emulsifier in the sample gives a false positive for fat. And the last limitation is the one the question is really driving at, that these are tests for chemical groups and not for specific substances, so a positive test identifies the presence of a peptide bond, of a free amino group or of a lipid, and it never identifies which particular protein or which particular amino acid is present.

Step-by-step solution

1. 1This is an estimation question with an ecological conclusion attached, so the answer should give a defensible order of magnitude for each of the three quantities, compare them, and then state the conclusion the question is inviting, which is the scale of the loss of vegetation.
2. 2The cellulose of the biosphere: cellulose is the most abundant of all the organic compounds on the earth, and the reason is that it is the structural polysaccharide of the cell wall of every green plant, and the plants are the producers of the whole biosphere. It is estimated that the annual synthesis of cellulose by all the plants of the world is of the order of a hundred billion tonnes, that is 1 x 10^11 to 1 x 10^12 tonnes, of organic matter per year, and the standing crop of the forests alone, measured as the total dry biomass of the world's forest trees, is of the same very large order. A commonly quoted figure is that the total biomass of the world's forests is about 400 to 450 billion tonnes of carbon, and since cellulose is a large fraction of that biomass, the amount of cellulose in the forests alone runs into hundreds of billions of tonnes. The figure to quote is therefore that the plants of the biosphere produce cellulose at the rate of about 10^11 to 10^12 tonnes a year, and that the total cellulose present in the world's vegetation is of the order of 10^11 tonnes, in round figures a hundred thousand million tonnes or more.
3. 3The paper of man: the world production of paper and paperboard is about 400 to 500 million tonnes a year, that is of the order of 10^9 tonnes, and it is made almost entirely from the wood of trees, chiefly the pulping species such as the eucalyptus, the pine, the spruce and the bamboo, with a smaller proportion made from the straw, the bagasse and the waste paper of recycling. So the cellulose that is converted into paper each year, of the order of 10^9 tonnes, is roughly one part in a hundred to one part in a thousand of the cellulose that the plants of the biosphere produce each year.
4. 4The comparison, and it is the point of the question: the cellulose made by the plants of the biosphere each year, of the order of 10^11 to 10^12 tonnes, exceeds the cellulose converted into paper by man, of the order of 10^9 tonnes, by a factor of roughly a hundred to a thousand. The entire paper and paperboard industry of the world, taken all together, therefore consumes only a very small fraction, of the order of one part in a thousand, of the cellulose that the world's vegetation produces in a single year. What man actually consumes of plant material in all forms, however, is far more than the paper figure, because there is the timber used in fuel, in building and in furniture, the hay and the fodder of the livestock, the grain and the food of man himself, the wood of the pulp, and the loss to decay and to burning, and the total annual human consumption of plant material is of the order of 10^10 tonnes, that is tens of billions of tonnes, and it is still only a small fraction of the annual primary production of the biosphere.
5. 5The conclusion, which is the sentence the question ends on. The disparity between the 10^11 to 10^12 tonnes of cellulose that the vegetation makes each year and the 10^9 to 10^10 tonnes that man consumes in all forms shows that the total removal of plant material by man, enormous though it is, is small beside the productive capacity of the biosphere, and that the biosphere could in principle sustain a considerably larger human population at the present level of consumption. But the comparison also carries a warning, and it is the warning in the question, which is that the loss of vegetation is not measured by the rate of consumption but by the rate of destruction, for the loss of vegetation is a matter of the area of the forest and of the natural ecosystem that is being cleared, and if the trees are removed faster than the seedlings are replaced, and this is the normal case in the tropics, then the forests, which take a hundred years or more to regenerate a mature crop, are gone in a decade. The forests are also the sinks for the carbon dioxide of the atmosphere, and their destruction raises the carbon dioxide and the methane and hence the greenhouse effect and the global warming. So the conclusion is the one the question invites: the amount of cellulose that man consumes annually is a small fraction of what the biosphere produces, but the rate at which man is destroying the vegetation that produces it, and the very slow rate at which that vegetation can be regenerated, is what makes the loss of vegetation so serious.

Step-by-step solution

1. 1Take the properties in NCERT's order, since that order is also the logical one, and pair each property with the biological consequence that makes it useful, because a property stated without its consequence earns only half the marks.
2. 2First, catalysis. Enzymes are biological catalysts, so they speed up the reactions they catalyse without themselves being consumed, and a small quantity of an enzyme catalyses a large quantity of substrate. Two points should be attached to this. The first is that the speed-up is enormous, of the order of a million to a hundred million times, and the second, which matters more, is that the enzyme lowers the activation energy of the reaction and does not change the equilibrium or the direction of the reaction, so the enzyme makes the reaction fast but does not make it possible that is otherwise impossible. This distinction is the one examiners look for, since the enzyme is a catalyst and not a source of energy.
3. 3Second, specificity. Enzymes are highly specific both for the type of reaction they catalyse and for the substrate on which they act, and NCERT gives three levels of this. Reaction specificity, where an enzyme catalyses one particular type of reaction. Substrate specificity, where the particular enzyme acts on a particular substrate, so urease acts only on urea. And the highest level, group specificity, where one enzyme acts on a whole group of chemically similar substrates, so the hexokinase acts on glucose, fructose and mannose, all of which are hexoses, and the lipase acts on a whole group of lipids, and it is this group specificity that explains why one enzyme can be used industrially on a natural mixture. The structural basis of the specificity is the active site, and the active site is a small, well-defined region of the enzyme, made of a few amino acid residues, into which the substrate fits, and the fit is complementary in shape and in the chemical groups, which is the lock-and-key idea, with the induced fit as the refinement in which the substrate itself changes the shape of the enzyme slightly so that the two make the closest possible contact.
4. 4Third, the pH optimum. Enzymes act at their best at a particular pH, and the pH optimum differs from enzyme to enzyme, and NCERT's values should be quoted. Pepsin of the stomach is active at pH about 1.8 to 2, which is the acid of the gastric juice, and trypsin of the intestine at pH about 7.8 to 8, which is the alkaline of the intestinal juice, and amylase of the saliva at pH about 6.8, the neutral pH. The values are quoted in NCERT as pH 1.8 for pepsin, 7.8 for trypsin and 6.8 for amylase. The effect of a departure from the optimum is a fall in the activity, because the charge on the active site is altered and the substrate no longer fits, and the fall is often sharp rather than gradual, and the extremes of pH, that is both the strongly acid and the strongly alkaline, denature the enzyme outright, breaking the tertiary structure, and this denaturation is irreversible while a simple return of the pH to the optimum is not.
5. 5Fourth, the temperature optimum. Enzymes also have an optimum temperature, and it differs from enzyme to enzyme, and for the enzymes of the body the optimum is the normal body temperature, that is 37 degrees Celsius for the human enzymes, which is one of the reasons the body temperature is held constant. Activity falls as the temperature falls below the optimum, and this fall is only a slowing of the enzyme and is reversible when the temperature rises again, whereas above the optimum the activity falls steeply because the enzyme is denatured by the heat, and the denaturation is irreversible, so a high temperature destroys the enzyme permanently, and this is the basis of the fact that food is preserved by cooking, by pasteurisation and by boiling, and that the enzymes of a thermophilic bacterium survive at temperatures that would denature those of man.
6. 6Fifth, the ability to be inhibited. The inhibitors of an enzyme are the molecules that reduce its activity, and NCERT classifies them into two kinds. The competitive inhibitors compete with the substrate for the active site, so the inhibition can be overcome by increasing the concentration of the substrate, since the substrate then out-competes the inhibitor, and the classic example is the inhibition of the succinic dehydrogenase of the Krebs cycle by the malonate ion, which resembles the succinate substrate. The non-competitive inhibitors bind at a site other than the active site, that is at an allosteric site, and change the shape of the enzyme so that the active site no longer works, and this inhibition cannot be overcome by an increase of the substrate, and the classic example is the inhibition of the enzyme by the cyanide ion, which binds to the metal of the cytochrome oxidase. The metabolic significance to add is that this is how the metabolism is regulated, since a natural metabolite acting as an allosteric inhibitor of an early step of a pathway switches that pathway off when its end product accumulates, and this is called feedback inhibition.
7. 7The remaining two properties are less often asked but should be stated for completeness. Enzymes are denatured by heat and by the extremes of pH and by the heavy metals, which is why the heavy metals are toxic, since mercuric salts and the salts of lead and of silver inhibit the enzymes of the cell, and enzymes finally do not require the conditions that the general chemical reactions require, that is they can act in the mild aqueous and the near-neutral conditions of the living cell, where a temperature of 37 degrees and a pH of 7 and a low concentration of the reactants would make an uncatalysed reaction hopelessly slow. This last point is worth closing on, because it is the reason enzymes are indispensable: the cell could not carry out its reactions at the temperature and the concentrations at which it must carry them out without them.