Chapter 10: Biomolecules Short Questions | chemistry Class 12 CBSE

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Chapter 10: Biomolecules Short Questions | chemistry Class 12 CBSE | ByteNotes.in

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Question 1

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Define carbohydrates and give the chemical basis for why acetic acid is not classified as one despite fitting the general formula.

Carbohydrates are optically active polyhydroxy aldehydes or ketones, or compounds that yield such units on hydrolysis.
Acetic acid (CH3COOH) fits the general formula C2(H2O)2 but does not possess a polyhydroxy aldehyde or ketone structure.
Carbohydrates must contain specific functional groups (aldehyde or ketone along with multiple hydroxyl groups), which acetic acid lacks.

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Question 2

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Distinguish between aldose and ketose monosaccharides with one example each.

An aldose is a monosaccharide that contains an aldehyde (–CHO) functional group; glucose is an example of an aldohexose.
A ketose is a monosaccharide that contains a keto (>C=O) functional group; fructose is an example of a ketohexose.
Both types are further classified by the number of carbon atoms, e.g., triose, tetrose, pentose, hexose, etc.

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Question 3

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State the evidence from glucose reactions that confirms the presence of an aldehydic group in its open-chain structure.

Glucose reacts with hydroxylamine to form an oxime and adds hydrogen cyanide to give a cyanohydrin, confirming a carbonyl group.
On reaction with bromine water (a mild oxidising agent), glucose is oxidised to gluconic acid, a six-carbon carboxylic acid, indicating that the carbonyl is an aldehydic group.
Fehling's solution and Tollens' reagent are reduced by glucose, which is characteristic behaviour of aldehydes.

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Question 4

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What is the cyclic structure of glucose and why does it exist in two forms?

Glucose forms a six-membered cyclic hemiacetal (pyranose) ring when the –OH group at C-5 adds to the –CHO group.
The two cyclic forms, alpha and beta, differ only in the configuration of the –OH group at C-1 (the anomeric carbon), and are called anomers.
The alpha-form (m.p. 419 K) is obtained from concentrated solution at 303 K and the beta-form (m.p. 423 K) from hot saturated solution at 371 K; both exist in equilibrium with the open-chain structure.

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Question 5

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Why is sucrose a non-reducing sugar while maltose is a reducing sugar?

In sucrose, the glycosidic bond forms between C1 of alpha-D-glucose and C2 of beta-D-fructose, involving both reducing groups (aldehyde and ketone), so no free carbonyl group remains.
In maltose, the glycosidic bond is between C1 of one alpha-D-glucose and C4 of another; the free aldehyde group can be generated at C1 of the second glucose unit in solution.
Since maltose can produce a free aldehyde group in solution, it reduces Fehling's solution and Tollens' reagent, classifying it as a reducing sugar.

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Question 6

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Describe the structural difference between amylose and amylopectin.

Amylose is a long, unbranched chain polymer of alpha-D-glucose with 200–1000 units linked by C1–C4 glycosidic bonds; it is water-soluble and constitutes 15–20% of starch.
Amylopectin is a branched chain polymer of alpha-D-glucose where the main chain is formed by C1–C4 linkage and branching occurs at C1–C6 glycosidic linkage; it constitutes 80–85% of starch.
Amylopectin is insoluble in water due to its highly branched structure, in contrast to water-soluble amylose.

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Question 7

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How does glycogen differ from starch in structure and biological role?

Glycogen is the carbohydrate storage form in animals, while starch is the storage form in plants.
Glycogen has a structure similar to amylopectin but is more highly branched; it is found in liver, muscles, and brain.
When the body requires glucose, enzymes break glycogen down to glucose; starch serves the analogous role in plants.

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Question 8

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What is a zwitter ion? Explain its significance in the behaviour of amino acids.

A zwitter ion is a dipolar ionic form of an amino acid in which the carboxyl group (–COOH) has lost a proton and the amino group (–NH2) has accepted a proton, yielding a neutral species with both positive and negative charges.
In aqueous solution, amino acids predominantly exist as zwitter ions, which explains why they have high melting points and are water-soluble, behaving like salts rather than simple amines or acids.
The zwitter ionic form also explains the amphoteric behaviour of amino acids — they can react with both acids (via –NH3+) and bases (via –COO–).

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Question 9

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Distinguish between essential and non-essential amino acids with two examples of each.

Essential amino acids cannot be synthesised by the human body and must be supplied through diet; examples include valine and leucine.
Non-essential amino acids can be synthesised within the body and do not need to be obtained from food; examples include glycine and alanine.
There are approximately 10 essential amino acids among the 20 naturally occurring alpha-amino acids.

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Question 10

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What is a peptide bond? How is a dipeptide formed?

A peptide bond (–CO–NH–) is an amide linkage formed between the –COOH group of one amino acid and the –NH2 group of another with the elimination of a water molecule.
A dipeptide is formed when the carboxyl group of one amino acid condenses with the amino group of another amino acid, releasing water.
For example, glycine and alanine combine to form the dipeptide glycylalanine (Gly-Ala) with one peptide bond.

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Question 11

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Compare fibrous and globular proteins in terms of structure, solubility, and examples.

Fibrous proteins have polypeptide chains running parallel, held by hydrogen and disulphide bonds, forming fibre-like structures; they are generally insoluble in water.
Globular proteins have polypeptide chains coiled to give a spherical shape; they are usually soluble in water.
Keratin (hair, wool) and myosin (muscles) are fibrous proteins; insulin and albumins are globular proteins.

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Question 12

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What happens to a protein during denaturation? Which structural levels are affected?

Denaturation occurs when a native protein is subjected to changes in temperature or pH, disrupting hydrogen bonds and causing globules to unfold and helices to uncoil.
During denaturation, the secondary and tertiary structures of the protein are destroyed, resulting in loss of its biological activity.
The primary structure (sequence of amino acids) remains intact during denaturation; only the higher-order conformations are disrupted.

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Question 13

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What are enzymes? State two characteristic features of enzyme action.

Enzymes are biocatalysts, almost all of which are globular proteins, that speed up specific chemical reactions in living organisms under very mild conditions.
Enzymes are highly specific — they act on a particular substrate and catalyse a specific type of reaction; they are named using the suffix '-ase' after the substrate or reaction.
Enzymes reduce the activation energy of reactions; for example, sucrase reduces the activation energy for sucrose hydrolysis from 6.22 kJ/mol to 2.15 kJ/mol.

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Chapter 10: Biomolecules | Short Questions

Define carbohydrates and give the chemical basis for why acetic acid is not classified as one despite fitting the general formula.

Distinguish between aldose and ketose monosaccharides with one example each.

State the evidence from glucose reactions that confirms the presence of an aldehydic group in its open-chain structure.

What is the cyclic structure of glucose and why does it exist in two forms?

Why is sucrose a non-reducing sugar while maltose is a reducing sugar?

Describe the structural difference between amylose and amylopectin.

How does glycogen differ from starch in structure and biological role?

What is a zwitter ion? Explain its significance in the behaviour of amino acids.

Distinguish between essential and non-essential amino acids with two examples of each.

What is a peptide bond? How is a dipeptide formed?

Compare fibrous and globular proteins in terms of structure, solubility, and examples.

What happens to a protein during denaturation? Which structural levels are affected?

What are enzymes? State two characteristic features of enzyme action.

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