Chapter 9: Biomolecules Long Questions | biology Class 11 CBSE

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Chapter 9: Biomolecules Long Questions | biology Class 11 CBSE | ByteNotes.in

Long Answer

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Question 1

Long Form

Describe the four levels of protein structure, providing the type of bonds or interactions that stabilise each level and a relevant example where applicable.

The primary structure is the linear sequence of amino acids in a polypeptide chain linked by peptide bonds; the left end is called the N-terminal amino acid and the right end is the C-terminal amino acid. This sequence determines all higher-order structures.
The secondary structure arises from folding of the polypeptide thread into regular repeating patterns, the most common being the right-handed alpha-helix (resembling a revolving staircase) and the beta-pleated sheet; hydrogen bonds between backbone atoms stabilise these structures.
The tertiary structure is the overall three-dimensional folding of the polypeptide chain upon itself, forming a compact globular shape like a hollow woolen ball; it is stabilised by hydrogen bonds and disulphide bonds between amino acid side chains.
At the tertiary level, the chain criss-crosses to create crevices and pockets; one such pocket is the active site of an enzyme — for example, the active site of carbonic anhydrase — making tertiary structure essential for biological activity.
The quaternary structure applies to proteins composed of more than one polypeptide subunit and describes the spatial arrangement of these subunits relative to each other; adult human haemoglobin, comprising two alpha and two beta subunits, exemplifies this level.
Each level of structure is dependent on the preceding one: a change in primary structure (mutation) can disrupt all higher-order structures and abolish biological function, illustrating the hierarchical nature of protein architecture.

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Question 2

Long Form

Compare the properties, structure, and biological roles of the three major polysaccharides — cellulose, starch, and glycogen — discussed in the chapter.

All three polysaccharides are homopolymers of glucose but differ in their linkages, branching, and functions; cellulose and starch are found in plants while glycogen is the animal storage carbohydrate.
Cellulose is a linear polymer forming the structural component of plant cell walls; paper and cotton fibre are cellulosic; it does not form helical structures and hence cannot hold iodine, giving no colour with I2.
Starch is a plant energy-storage polysaccharide that forms helical secondary structures capable of trapping iodine molecules in the helical portion, producing a characteristic blue colour — the basis of the starch-iodine test.
Glycogen is the branched polysaccharide found in animal tissues; in a glycogen chain, the right end is called the reducing end and the left end the non-reducing end; its branched structure allows rapid mobilisation of glucose.
Chitin, another polysaccharide mentioned in the chapter, is a complex homopolymer built from amino sugars (e.g., N-acetyl galactosamine) and forms the exoskeleton of arthropods, illustrating structural diversity in polysaccharides.
The diversity of polysaccharide function — from structural roles (cellulose, chitin) to energy storage (starch, glycogen) — reflects the versatility achieved by polymerising the same or modified monosaccharide units in different ways.

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Long Answer

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Question 3

Long Form

Analyse the mechanism of enzyme action from substrate binding to product release, explaining the role of the transition state and activation energy.

Enzymes are proteins with a specific active site — a crevice or pocket in the tertiary structure — into which the substrate (S) fits with high specificity; the substrate must diffuse to the active site, forming an obligatory enzyme-substrate (ES) complex.
The binding of the substrate induces the enzyme to alter its shape, fitting more tightly around the substrate; this conformational change positions the substrate precisely for bond breaking and formation.
During the ES complex state, the substrate is converted into a high-energy transition state structure — an unstable intermediate between S and P — through which the reaction must pass; all intermediate structural states between S and P are unstable.
Activation energy is the energy difference between the substrate's ground state and the transition state; enzymes lower this activation energy barrier, making it far easier for S to reach the transition state at physiological temperatures.
After bond breaking and making, the enzyme-product (EP) complex forms; the product is then released from the active site, regenerating the free enzyme, which is now available to catalyse another catalytic cycle (E + S → ES → EP → E + P).
The enormous catalytic power of enzymes is illustrated by carbonic anhydrase: without enzyme, only about 200 molecules of H2CO3 form per hour, but with the enzyme, about 600,000 molecules form per second — an acceleration of approximately 10 million times.

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Question 4

Long Form

Explain how the elemental and chemical composition of living tissues is determined experimentally, and what the results reveal about the chemistry of life.

To identify organic compounds, living tissue is ground in trichloroacetic acid (Cl3CCOOH) using a mortar and pestle to obtain a thick slurry that is then strained through cheesecloth, yielding an acid-soluble pool (filtrate) and an acid-insoluble fraction (retentate).
The acid-soluble pool contains thousands of small organic compounds (MW 18–800 Da) such as amino acids, sugars, nucleotides, and fatty acids, as well as inorganic ions and compounds like sulphate and phosphate.
The acid-insoluble fraction contains the four macromolecular classes: proteins, nucleic acids, polysaccharides, and lipids; together these two fractions represent the complete chemical composition of living tissues.
To determine elemental composition, tissue is dried (removing water), and then fully burnt; all carbon compounds are oxidised to CO2 and water vapour, leaving an ash that contains inorganic elements such as calcium, magnesium, sodium, and potassium.
Comparison of living tissue with earth's crust reveals that while the same elements are present in both, living organisms show a much higher relative abundance of carbon (18.5% vs 0.03%), hydrogen (9.5% vs 0.14%), and nitrogen (3.3% vs trace amounts).
This difference in relative elemental abundance — not the presence of unique elements — distinguishes living from non-living matter and confirms that life is a special organisation of common elements, with water being the most abundant component at 70–90% of cellular mass.

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Question 5

Long Form

Discuss the classification of enzymes into six classes based on the reactions they catalyse, providing the reaction type and an example for each class.

Enzymes are classified into six classes based on the type of reactions they catalyse; each class is divided into 4–13 subclasses and is assigned a four-digit classification number for precise identification.
Oxidoreductases (or dehydrogenases) catalyse oxidation-reduction reactions between two substrates S and S': S(reduced) + S'(oxidised) → S(oxidised) + S'(reduced); these enzymes are involved in cellular respiration and other redox pathways.
Transferases catalyse the transfer of a specific group G (other than hydrogen) from one substrate S to another substrate S': S-G + S' → S + S'-G; aminotransferases that transfer amino groups are important examples.
Hydrolases catalyse hydrolysis of ester, ether, peptide, glycosidic, C-C, C-halide, or P-N bonds by addition of water; digestive enzymes like trypsin (peptide bond hydrolysis) and lipases (ester bond hydrolysis) belong to this class.
Lyases catalyse the removal of groups from substrates by mechanisms other than hydrolysis, leaving double bonds in the product; isomerases catalyse interconversion of optical, geometric, or positional isomers within a single substrate molecule.
Ligases catalyse the joining together of two compounds by forming C-O, C-S, C-N, or P-O bonds, typically coupled with ATP hydrolysis; DNA ligase (which joins DNA fragments) is an important example from molecular biology.

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Question 6

Long Form

Distinguish between micromolecules and biomacromolecules in living systems. Why are lipids an exception to the general rule for biomacromolecules?

Micromolecules (or small biomolecules) are compounds found in the acid-soluble pool with molecular weights ranging from 18 to approximately 800 daltons; they include amino acids, monosaccharides, fatty acids, nucleotides, vitamins, and various inorganic ions.
Biomacromolecules are compounds found in the acid-insoluble fraction with molecular weights typically above 10,000 daltons; the true biomacromolecules are proteins, nucleic acids, and polysaccharides, which are all polymeric (made of repeating building-block monomers).
Lipids are an exception: their individual molecular weights do not exceed 800 Da, placing them in the small-molecule range, yet they are found in the acid-insoluble (macromolecular) fraction during tissue analysis.
This anomaly occurs because lipids are not present only as free molecules in cells; they are organised into membrane structures (cell membranes, organelle membranes) where they form lipid bilayers and other assemblies.
When tissue is ground for analysis, these membranes are disrupted and break into small vesicles that are not water-soluble; these vesicles sediment with the acid-insoluble fraction even though the individual lipid molecules are small.
Therefore, lipids are not strictly macromolecules — the chapter explicitly states this — but appear in the macromolecular fraction due to their membrane-associated organisation, distinguishing them from true polymeric macromolecules like proteins and nucleic acids.

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Question 7

Long Form

Analyse the factors that affect enzyme activity and explain the significance of optimum temperature and pH in enzyme function.

Enzyme activity is affected by temperature, pH, substrate concentration, and the binding of specific chemicals (inhibitors or activators) that alter the tertiary structure of the protein and therefore the shape of the active site.
Each enzyme shows maximum activity at a specific optimum temperature; activity declines both below and above this optimum. Low temperature places the enzyme in a temporarily inactive but reversible state, while high temperatures (above ~40°C) irreversibly denature the protein.
Thermophilic organisms living in extreme heat environments (e.g., hot vents, sulphur springs) produce thermostable enzymes that retain catalytic power at 80°–90°C, showing that the optimum temperature varies with the organism's habitat.
Similarly, each enzyme has an optimum pH at which its activity is highest; deviations alter the ionisation state of the amino acid residues at the active site, distorting the three-dimensional structure and reducing or abolishing activity.
Substrate concentration also governs reaction velocity: as substrate concentration increases, reaction velocity rises until all enzyme molecules are saturated, reaching Vmax; beyond this, adding more substrate has no effect because no free enzyme molecules remain.
Enzyme activity is also regulated by inhibitors: competitive inhibitors (structurally similar to the substrate) compete for the active site, while non-competitive inhibitors bind elsewhere on the enzyme, altering its shape; inhibitors thus provide a mechanism for metabolic regulation and are exploited in drug design against bacterial pathogens.

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Question 8

Long Form

Describe the structure and properties of amino acids, explaining their classification based on the nature of the R group and their ionisable nature.

Amino acids are organic compounds in which both an amino group (–NH2) and a carboxyl group (–COOH) are attached to the same carbon atom, called the alpha-carbon; they are therefore called alpha-amino acids and can be viewed as substituted methanes with four substituents: H, –COOH, –NH2, and an R group.
The R group (variable group) determines the identity and properties of each amino acid; of the many possible amino acids, only 20 types are proteinaceous (found in proteins); R groups include hydrogen (glycine), methyl (alanine), and hydroxymethyl (serine).
Based on the number of amino and carboxyl groups, amino acids are classified as acidic (e.g., glutamic acid, with an extra –COOH), basic (e.g., lysine, with an extra –NH2), or neutral (e.g., valine, with balanced groups); aromatic amino acids (tyrosine, phenylalanine, tryptophan) contain aromatic ring systems.
A key property of amino acids is the ionisable nature of both the –NH2 and –COOH groups; in solutions of different pH, these groups gain or lose protons, changing the overall charge and structure of the molecule.
At an intermediate pH (isoelectric point), the amino acid exists as a zwitterion (form B): the amino group is protonated (–NH3+) and the carboxyl group is deprotonated (–COO–), making the molecule electrically neutral yet dipolar.
Amino acids can be essential (must be supplied through diet because the body cannot synthesise them) or non-essential (synthesised by the body); this distinction is nutritionally important as dietary proteins are the source of essential amino acids.

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Question 9

Long Form

Explain the structure of nucleic acids from the level of nitrogenous bases to the complete polynucleotide, clearly distinguishing between DNA and RNA.

Nucleic acids are polynucleotides and the building block of each is a nucleotide, which has three chemically distinct components: a heterocyclic nitrogenous base, a monosaccharide pentose sugar, and a phosphoric acid (phosphate) group.
The nitrogenous bases in nucleic acids are adenine, guanine, uracil, cytosine, and thymine; adenine and guanine are purines (double-ring structure), while cytosine, uracil, and thymine are pyrimidines (single-ring structure).
When a nitrogenous base is attached to a sugar, the product is a nucleoside (e.g., adenosine = adenine + ribose; uridine = uracil + ribose; thymidine = thymine + deoxyribose); when a phosphate group is esterified to the sugar of a nucleoside, the product is a nucleotide (e.g., adenylic acid, thymidylic acid, uridylic acid).
The sugar in a polynucleotide is either ribose (a pentose monosaccharide, C5H10O5) or 2-deoxyribose (ribose with one fewer oxygen at the 2' position); a nucleic acid containing deoxyribose is deoxyribonucleic acid (DNA), while one containing ribose is ribonucleic acid (RNA).
Together with polysaccharides and polypeptides (proteins), nucleic acids constitute the true macromolecular fraction of any living tissue; they are found in the acid-insoluble fraction and have molecular weights in the range of tens of thousands of daltons or more.
DNA and RNA function as genetic material, carrying hereditary information; nucleic acids are passed from the parental generation to progeny, making them central to inheritance, gene expression, and the continuity of life.

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Question 10

Long Form

Examine the role of cofactors in enzyme function, distinguishing between prosthetic groups, co-enzymes, and metal ions with specific examples from the chapter.

Many enzymes require non-protein constituents called cofactors to be catalytically active; the protein portion of such an enzyme is called the apoenzyme, and the apoenzyme becomes active only when associated with its cofactor.
Prosthetic groups are organic cofactors that are tightly and permanently bound to the apoenzyme; they are distinguished from other cofactors by this tight association and often form part of the active site itself.
Haem is an example of a prosthetic group: it is found in peroxidase and catalase, enzymes that catalyse the breakdown of hydrogen peroxide to water and oxygen; haem is integral to the active site and directly participates in the catalytic mechanism.
Co-enzymes are also organic compounds, but they associate with the apoenzyme only transiently during the catalytic cycle rather than being permanently bound; they serve as cofactors in many different enzyme-catalysed reactions, giving them versatility.
The essential chemical components of many co-enzymes are vitamins; for example, nicotinamide adenine dinucleotide (NAD) and NADP both contain the vitamin niacin as an essential component and function as electron carriers in oxidation-reduction reactions.
Metal ions constitute a third type of cofactor; they form coordination bonds with amino acid side chains at the active site and simultaneously coordinate with the substrate, facilitating catalysis; zinc, for example, is the metal ion cofactor for carboxypeptidase, a proteolytic enzyme; removal of any cofactor destroys catalytic activity, confirming its essential role.

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Chapter 9: Biomolecules | Long Questions

Describe the four levels of protein structure, providing the type of bonds or interactions that stabilise each level and a relevant example where applicable.

Compare the properties, structure, and biological roles of the three major polysaccharides — cellulose, starch, and glycogen — discussed in the chapter.

Analyse the mechanism of enzyme action from substrate binding to product release, explaining the role of the transition state and activation energy.

Explain how the elemental and chemical composition of living tissues is determined experimentally, and what the results reveal about the chemistry of life.

Discuss the classification of enzymes into six classes based on the reactions they catalyse, providing the reaction type and an example for each class.

Distinguish between micromolecules and biomacromolecules in living systems. Why are lipids an exception to the general rule for biomacromolecules?

Analyse the factors that affect enzyme activity and explain the significance of optimum temperature and pH in enzyme function.

Describe the structure and properties of amino acids, explaining their classification based on the nature of the R group and their ionisable nature.

Explain the structure of nucleic acids from the level of nitrogenous bases to the complete polynucleotide, clearly distinguishing between DNA and RNA.

Examine the role of cofactors in enzyme function, distinguishing between prosthetic groups, co-enzymes, and metal ions with specific examples from the chapter.

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