Biomolecules Class 12: Carbohydrates, Proteins, Nucleic Acids and Board Answers

What Biomolecules is about

Weightage: This chapter is Unit X, Biomolecules, which carries 7 of the 70 theory marks.Source: CBSE Chemistry (043) syllabus 2026-27, course structure, Unit X Biomolecules

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 7 marks per paper. See the questions

Key concepts

Classification of carbohydrates

NCERT section 10.1 Carbohydrates (10.1.1 Classification of Carbohydrates; 10.1.2 Monosaccharides)

Carbohydrates are optically active polyhydroxy aldehydes or ketones, or compounds that produce such units on hydrolysis. Many have the formula $C_x(H_{2}O{)}_y$, which once suggested 'hydrates of carbon', but not all compounds with this formula are carbohydrates (acetic acid, $C_2{H}_4{O}_2$, is not) and some carbohydrates do not fit it (rhamnose, $C_6{H}_{12}{O}_5$).

By hydrolysis behaviour:

• Monosaccharides: cannot be hydrolysed into simpler polyhydroxy aldehydes or ketones: glucose, fructose, ribose.
• Oligosaccharides: give 2 to 10 monosaccharide units: disaccharides (sucrose, maltose, lactose), trisaccharides and so on.
• Polysaccharides: give a large number of monosaccharide units: starch, cellulose, glycogen, gums. They are not sweet and are called non-sugars.

Reducing and non-reducing sugars: carbohydrates that reduce Fehling's solution and Tollens' reagent are reducing sugars. All monosaccharides (aldoses and ketoses) are reducing sugars. In a disaccharide, if the reducing groups of both monosaccharides are bonded together (as in sucrose), it is non-reducing; if one remains free (maltose, lactose), it is reducing.

Monosaccharides are further named by the number of carbons and the type of carbonyl: an aldohexose (glucose) has six carbons and an aldehyde group; a ketohexose (fructose) has six carbons and a ketone group.

Worked example

Classify as monosaccharide, disaccharide or polysaccharide, and as reducing or non-reducing: (a) fructose (b) sucrose (c) maltose (d) starch. (2 marks)

Try it first, then show the solutionSolution

1. (a) Fructose: monosaccharide (ketohexose), reducing. (½)
2. (b) Sucrose: disaccharide; the reducing groups of glucose (C1) and fructose (C2) are joined, so it is non-reducing. (½)
3. (c) Maltose: disaccharide; C1 of the second glucose is free, so it is reducing. (½)
4. (d) Starch: polysaccharide; non-reducing (not a sugar). (½)

Answer: (a) mono, reducing (b) di, non-reducing (c) di, reducing (d) poly, non-reducing

Glucose: preparation and open-chain structure

NCERT section 10.1.2.1 Glucose (Preparation; Structure)

Preparation:

• From sucrose: boiling with dilute HCl or H₂SO₄ in alcoholic solution gives glucose and fructose in equal amounts: $C_{12}{H}_{22}{O}_{11}+H_{2}O\stackrel{H^{+}}{\to }{C}_6{H}_{12}{O}_6+C_6{H}_{12}{O}_6$.
• From starch (commercial): boiling starch with dilute H₂SO₄ at 393 K under 2–3 atm pressure.

Glucose is an aldohexose, also called dextrose. Its open-chain structure was worked out from these facts:
1. Molecular formula $C_6{H}_{12}{O}_6$.
2. Prolonged heating with HI gives n-hexane: all six carbons are in a straight chain.
3. It forms an oxime with hydroxylamine and adds HCN to give a cyanohydrin: a carbonyl group is present.
4. Bromine water (a mild oxidant) converts it to gluconic acid, a six-carbon acid: the carbonyl is an aldehyde group.
5. Acetylation with acetic anhydride gives glucose pentaacetate: there are five –OH groups, on different carbons (the compound is stable).
6. Nitric acid oxidises glucose (and gluconic acid) to saccharic acid, a dicarboxylic acid: there is a primary alcohol (–CH₂OH) group.

So glucose is $OHC-(CHOH{)}_4-C{H}_{2}OH$. Fischer worked out the arrangement of the –OH groups. Its configuration is D, because the –OH on the lowest asymmetric carbon (C5) is on the right, as in D-(+)-glyceraldehyde; D-(+)-glucose is dextrorotatory. D and L refer to configuration, not to the direction of rotation.

Worked example

What do the following reactions of glucose tell us about its structure? (a) With HI on prolonged heating, it gives n-hexane. (b) With bromine water, it gives gluconic acid. (c) With acetic anhydride, it gives a pentaacetate. (3 marks)

Try it first, then show the solutionSolution

1. (a) The six carbon atoms are linked in a straight chain. (1)
2. (b) A mild oxidant converts it to a six-carbon carboxylic acid, so the carbonyl group is an aldehyde (–CHO) group. (1)
3. (c) Glucose has five –OH groups, each on a different carbon atom. (1)

Answer: (a) Straight chain of six carbons (b) aldehyde group (c) five –OH groups on different carbons.

Cyclic structure of glucose and fructose

NCERT section 10.1.2.1 Glucose (Cyclic structure of glucose); 10.1.2.2 Fructose

The open-chain structure cannot explain some facts:
1. Despite having an aldehyde group, glucose does not give Schiff's test and does not form the hydrogensulphite addition product with NaHSO₃.
2. Glucose pentaacetate does not react with hydroxylamine, so there is no free –CHO.
3. Glucose exists in two crystalline forms, α and β: α-form (m.p. 419 K) from a concentrated solution at 303 K, and β-form (m.p. 423 K) from a hot saturated solution at 371 K.

Explanation: the –OH at C5 adds to the –CHO group to form a cyclic hemiacetal, a six-membered ring. C1 becomes a new asymmetric carbon, so two forms exist: α and β, differing only in the configuration of the –OH at C1. C1 is the anomeric carbon and the two forms are anomers. The six-membered ring (five carbons and one oxygen) is called a pyranose ring, by analogy with pyran, and is best drawn as a Haworth structure.

Fructose is a ketohexose with the keto group at C2; it is D-(−)-fructose (D-series, laevorotatory). It forms a five-membered ring when the –OH at C5 adds to the C=O at C2: a furanose ring (by analogy with furan), again with α and β anomers.

Worked example

Give two reasons why the open-chain structure of glucose is incomplete, and explain what anomers are. (3 marks)

Try it first, then show the solutionSolution

1. Glucose does not give Schiff's test or react with NaHSO₃, although it has a –CHO group. (1)
2. Glucose pentaacetate does not react with NH₂OH, showing that no free –CHO is present (or: glucose exists in two forms, α and β). (1)
3. Anomers are the α and β cyclic forms that differ only in the configuration of the –OH group at C1, the anomeric carbon. (1)

Answer: No Schiff's test/no NaHSO₃ product, and pentaacetate does not react with NH₂OH; anomers differ only at C1.

Disaccharides and polysaccharides

NCERT section 10.1.3 Disaccharides; 10.1.4 Polysaccharides; 10.1.5 Importance of Carbohydrates

Two monosaccharides join through an oxide link formed by loss of a water molecule: a glycosidic linkage.

Disaccharides:

• Sucrose: α-D-glucose (C1) linked to β-D-fructose (C2). Both reducing groups are used, so sucrose is non-reducing. Sucrose is dextrorotatory, but its hydrolysis gives dextrorotatory glucose (+52.5°) and laevorotatory fructose (−92.4°); fructose's laevorotation is larger, so the mixture is laevorotatory. The sign of rotation changes from (+) to (−), and the product is called invert sugar.
• Maltose: two α-D-glucose units, C1 of one linked to C4 of the other. The free aldehyde at C1 of the second unit can be produced in solution, so maltose is reducing.
• Lactose (milk sugar): β-D-galactose and β-D-glucose, C1 of galactose to C4 of glucose; reducing.

Polysaccharides:

• Starch: the main storage carbohydrate of plants and a major food for humans (cereals, roots, tubers). It is a polymer of α-glucose with two components:
• Amylose: water-soluble, 15–20% of starch, an unbranched chain of 200–1000 α-D-(+)-glucose units joined by C1–C4 glycosidic links.
• Amylopectin: water-insoluble, 80–85% of starch, a branched chain: C1–C4 links in the chain and C1–C6 links at the branches.
• Cellulose: the main component of plant cell walls; an unbranched chain of β-D-glucose units joined by C1–C4 glycosidic links.
• Glycogen: the storage carbohydrate of animals (animal starch), more highly branched than amylopectin; found in liver, muscles and brain, and also in yeast and fungi. When the body needs glucose, enzymes break glycogen down.

Worked example

Why is sucrose called a non-reducing sugar, and why is the product of its hydrolysis called invert sugar? (2 marks)

Try it first, then show the solutionSolution

1. In sucrose, C1 of α-D-glucose is bonded to C2 of β-D-fructose; both reducing groups are tied up in the glycosidic linkage, so it does not reduce Fehling's or Tollens' reagent. (1)
2. Hydrolysis gives glucose (+52.5°) and fructose (−92.4°); the mixture is laevorotatory, so the sign of rotation is inverted from (+) to (−). (1)

Answer: Both reducing groups are in the glycosidic bond; the rotation changes from dextro to laevo on hydrolysis.

Amino acids

NCERT section 10.2.1 Amino Acids; 10.2.2 Classification of Proteins (amino acids)

Amino acids contain an amino group (–NH₂) and a carboxyl group (–COOH). Proteins are made from α-amino acids, in which both groups are on the same carbon: $R-CH(N{H}_2)-COOH$. They are named by their common names (glycine, alanine) with three-letter and one-letter codes (Gly, G).

Classification:

• By the number of amino and carboxyl groups: neutral (equal numbers), acidic (more –COOH, e.g. aspartic acid) and basic (more –NH₂, e.g. lysine).
• By diet: non-essential amino acids can be made by the body; essential amino acids cannot and must come from food. NCERT marks ten as essential (for example valine, leucine, isoleucine, lysine, methionine, phenylalanine, threonine, tryptophan, arginine, histidine).

Properties: amino acids are usually colourless, crystalline solids, water-soluble and high-melting, behaving like salts rather than simple amines or acids. In aqueous solution the carboxyl group loses a proton and the amino group gains one, giving a dipolar ion, the zwitterion: $H_3{N}^{+}-CH(R)-CO{O}^{-}$. It is amphoteric: it reacts with both acids and bases.

Except glycine ($H_{2}NC{H}_{2}COOH$), all naturally occurring α-amino acids are optically active, because the α-carbon is asymmetric; most have the L-configuration.

Worked example

(a) What is a zwitterion? Write the zwitterion form of glycine. (b) Why are amino acids high-melting and water-soluble? (c) Why is glycine optically inactive? (3 marks)

Try it first, then show the solutionSolution

1. (a) A dipolar ion with both positive and negative charges, formed when –COOH loses H⁺ to –NH₂: glycine gives $H_3{N}^{+}C{H}_{2}CO{O}^{-}$. (1)
2. (b) Because of their dipolar (salt-like) nature, strong electrostatic attractions hold them in the crystal, and they interact strongly with water. (1)
3. (c) Its α-carbon carries two H atoms, so it is not asymmetric (not chiral). (1)

Answer: (a) H₃N⁺CH₂COO⁻ (b) salt-like zwitterions (c) no asymmetric carbon.

Proteins: peptide bond, structure and denaturation

NCERT section 10.2.3 Structure of Proteins; 10.2.4 Denaturation of Proteins

Proteins are polymers of α-amino acids joined by peptide bonds: an amide link, –CO–NH–, formed between the –COOH of one amino acid and the –NH₂ of the next with loss of water. Two amino acids give a dipeptide (glycine + alanine → glycylalanine), three a tripeptide, and so on; with more than ten amino acids the product is a polypeptide, and a polypeptide of more than about a hundred amino acids (molar mass above about 10,000 u) is called a protein. Insulin contains 51 amino acids.

Types:

• Fibrous proteins: chains run parallel, held by hydrogen and disulphide bonds, giving fibre-like structures; generally insoluble in water: keratin (hair, wool, silk) and myosin (muscles).
• Globular proteins: chains coil into a spherical shape; usually water-soluble: insulin and albumins.

Levels of structure:

• Primary: the sequence of amino acids in each chain. Any change in it creates a different protein.
• Secondary: the shape in which a long chain exists, due to regular folding of the backbone and hydrogen bonds between >C=O and –NH– groups of peptide bonds: the α-helix (a right-handed screw with hydrogen bonds between the –NH of each amino acid and the >C=O of a residue further along the turn) and the β-pleated sheet (chains stretched out side by side, like the folds of drapery).
• Tertiary: further folding of the chain, giving fibrous or globular shapes; held by hydrogen bonds, disulphide links, van der Waals forces and electrostatic attractions.
• Quaternary: the spatial arrangement of two or more polypeptide chains (subunits) with respect to each other, as in haemoglobin.

Denaturation: physical changes (heat) or chemical changes (pH) disturb the hydrogen bonds; globules unfold and helices uncoil, and the protein loses its biological activity. Secondary and tertiary structures are destroyed, but the primary structure stays intact. Examples: coagulation of egg white on boiling, and curdling of milk by lactic acid produced by bacteria.

Worked example

(a) What is a peptide bond? Show its formation between glycine and alanine. (b) What happens to the structure of a protein when an egg is boiled? (3 marks)

Try it first, then show the solutionSolution

1. (a) An amide link (–CO–NH–) formed when the –COOH of one amino acid reacts with the –NH₂ of another, with loss of water. (1)
2. $H_{2}NC{H}_{2}COOH+H_{2}NCH(C{H}_3)COOH\to H_{2}NC{H}_{2}CONHCH(C{H}_3)COOH+H_{2}O$ (glycylalanine). (1)
3. (b) Heat breaks hydrogen bonds; secondary and tertiary structures are lost (denaturation) and the albumin coagulates, while the primary sequence is unchanged. (1)

Answer: (a) –CO–NH– amide link; glycine + alanine → glycylalanine + H₂O. (b) Denaturation: secondary and tertiary structures lost, primary kept.

Enzymes and vitamins

NCERT section 10.3 Enzymes; 10.4 Vitamins

Enzymes are biological catalysts; almost all are globular proteins. They are very specific for a particular reaction and substrate: maltase catalyses only the hydrolysis of maltose to glucose. Many are named after the substrate or reaction, with the ending -ase (oxidoreductases catalyse oxidation of one substrate with reduction of another). They are needed in small amounts and work by lowering activation energy: acid hydrolysis of sucrose needs 6.22 kJ mol⁻¹, but hydrolysis by the enzyme sucrase needs only 2.15 kJ mol⁻¹.

Vitamins are organic compounds needed in small amounts in the diet for specific biological functions; most cannot be made by our body (gut bacteria make some). The name came from 'vital amine', but most vitamins have no amino group. Excess vitamins are also harmful.

Classification:

• Fat-soluble: A, D, E and K. Stored in liver and adipose (fat-storing) tissue.
• Water-soluble: B group and C. Excreted readily in urine and not stored (except B₁₂), so they must be supplied regularly in the diet.

Sources and deficiency diseases (NCERT Table 10.3):

• A: fish liver oil, carrots, butter, milk; xerophthalmia (hardening of the cornea) and night blindness
• B₁ (thiamine): yeast, milk, green vegetables, cereals; beri beri
• B₂ (riboflavin): milk, egg white, liver, kidney; cheilosis (fissuring at corners of mouth and lips), digestive disorders, burning sensation of skin
• B₆ (pyridoxine): yeast, milk, egg yolk, cereals, grams; convulsions
• B₁₂: meat, fish, egg, curd; pernicious anaemia
• C (ascorbic acid): citrus fruits, amla, green leafy vegetables; scurvy (bleeding gums)
• D: sunlight, fish, egg yolk; rickets (children) and osteomalacia (adults)
• E: vegetable oils (wheat germ, sunflower); increased fragility of RBCs and muscular weakness
• K: green leafy vegetables; increased blood clotting time

Worked example

(a) Why must vitamin C be supplied regularly in the diet? (b) Name the deficiency diseases of vitamins B₁₂ and K. (c) Why are fat-soluble vitamins not needed daily? (3 marks)

Try it first, then show the solutionSolution

1. (a) Vitamin C is water-soluble, so it is excreted in urine and cannot be stored in the body. (1)
2. (b) B₁₂: pernicious anaemia; K: increased blood clotting time. (1)
3. (c) Fat-soluble vitamins (A, D, E, K) are stored in the liver and adipose tissue, so the body has a reserve. (1)

Answer: (a) Water-soluble, excreted, not stored (b) pernicious anaemia; increased blood clotting time (c) stored in liver and fat tissue.

Nucleic acids

NCERT section 10.5 Nucleic Acids (10.5.1 Chemical Composition; 10.5.2 Structure; 10.5.3 Biological Functions)

Nucleic acids are long-chain polymers of nucleotides (polynucleotides) found in chromosomes. They are of two types: DNA (deoxyribonucleic acid) and RNA (ribonucleic acid).

Composition: complete hydrolysis gives a pentose sugar, phosphoric acid and nitrogen-containing heterocyclic bases.

• Sugar: β-D-2-deoxyribose in DNA; β-D-ribose in RNA.
• Bases: DNA has adenine (A), guanine (G), cytosine (C) and thymine (T); RNA has A, G, C and uracil (U) in place of thymine.

Building blocks:

• Nucleoside = base + sugar (base attached at C1′ of the sugar).
• Nucleotide = nucleoside + phosphoric acid at the 5′ position of the sugar.
• Nucleotides are joined by phosphodiester linkages between the 5′ and 3′ carbons of the sugars.

Structure: the sequence of nucleotides is the primary structure. DNA is a double helix (Watson and Crick): two strands coiled around each other, held by hydrogen bonds between specific base pairs: A with T and C with G, so the strands are complementary. RNA molecules are single-stranded (though they may fold into a double helix in parts). RNA is of three types: messenger RNA (m-RNA), ribosomal RNA (r-RNA) and transfer RNA (t-RNA).

Functions: DNA is the chemical basis of heredity and holds the genetic information; it can replicate during cell division, so identical copies go to daughter cells. Nucleic acids also direct protein synthesis: proteins are made by the various RNA molecules, but the message for each protein is in DNA.

Worked example

Give three differences between DNA and RNA. (3 marks)

Try it first, then show the solutionSolution

1. Sugar: DNA has β-D-2-deoxyribose; RNA has β-D-ribose. (1)
2. Bases: DNA has thymine; RNA has uracil instead (both have A, G and C). (1)
3. Structure: DNA is a double helix; RNA is single-stranded (also: DNA replicates and carries heredity; RNA mainly carries out protein synthesis). (1)

Answer: Deoxyribose vs ribose; thymine vs uracil; double helix vs single strand.

Hormones

NCERT section 10.6 Hormones

Hormones are molecules that act as intercellular messengers. They are made by endocrine glands and released directly into the blood, which carries them to their site of action.

Chemical types:

• Steroids: estrogens and androgens.
• Polypeptides: insulin and endorphins.
• Amino acid derivatives: epinephrine and norepinephrine.

Functions:

• Insulin keeps blood glucose within a narrow range; it is released when blood glucose rises rapidly. Glucagon tends to increase blood glucose. Together they regulate it.
• Epinephrine and norepinephrine mediate responses to external stimuli.
• Thyroxine, made in the thyroid gland, is an iodinated derivative of the amino acid tyrosine. Low thyroxine causes hypothyroidism (lethargy and obesity); high thyroxine causes hyperthyroidism. Low iodine in the diet can cause hypothyroidism and enlargement of the thyroid gland, which is controlled by adding sodium iodide to table salt (iodised salt).
• Steroid hormones come from the adrenal cortex and gonads. Glucocorticoids control carbohydrate metabolism, modulate inflammation and are involved in stress responses; mineralocorticoids control excretion of water and salt by the kidney. Failure of the adrenal cortex can cause Addison's disease (hypoglycaemia, weakness, greater susceptibility to stress).
• Testosterone (main male sex hormone) and estradiol (main female sex hormone) are responsible for secondary sex characteristics; estradiol also takes part in the control of the menstrual cycle, and progesterone prepares the uterus for implantation of the fertilised egg.

Worked example

(a) Name one hormone each that is a steroid, a polypeptide and an amino acid derivative. (b) How do insulin and glucagon together regulate blood glucose? (2 marks)

Try it first, then show the solutionSolution

1. (a) Steroid: estrogen or testosterone; polypeptide: insulin or endorphin; amino acid derivative: epinephrine, norepinephrine or thyroxine. (1)
2. (b) Insulin is released when blood glucose rises and lowers it; glucagon raises blood glucose; their opposite actions keep it in a narrow range. (1)

Answer: (a) e.g. testosterone, insulin, epinephrine (b) insulin lowers and glucagon raises blood glucose.

Chemistry in real life: Biomolecules

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Glucose: Structure and Reactions
• Proteins: Amino Acids, Peptide Bond and Structure
• Vitamins and Deficiency Diseases

Biomolecules formula sheet

Hydrolysis of sucrose

$C_{12}{H}_{22}{O}_{11}+H_{2}O\stackrel{H^{+}}{\to }{C}_6{H}_{12}{O}_6+C_6{H}_{12}{O}_6$

Glucose (+52.5°) + fructose (−92.4°): invert sugar (laevorotatory).

Open-chain glucose

$OHC-(CHOH{)}_4-C{H}_{2}OH$

Aldohexose; D-(+)-glucose.

Key reactions of glucose

$\text{HI: n-hexane};{\text{NH}}_2\text{OH: oxime};{\text{Br}}_2\text{ water: gluconic acid};{\text{HNO}}_3\text{: saccharic acid}$

Acetic anhydride gives glucose pentaacetate (five –OH groups).

Formation of a dipeptide

$H_{2}NC{H}_{2}COOH+H_{2}NCH(C{H}_3)COOH\to H_{2}NC{H}_{2}CONHCH(C{H}_3)COOH+H_{2}O$

Glycine + alanine → glycylalanine; the –CO–NH– is the peptide bond.

Zwitterion

$H_{2}N-CH(R)-COOH\rightleftharpoons H_3{N}^{+}-CH(R)-CO{O}^{-}$

Amphoteric; explains high melting points and solubility.

Glycosidic links to remember

$\text{Sucrose: }\alpha C1-\beta C2;\text{Maltose: }\alpha C1-C4;\text{Lactose: }\beta C1-C4$

Sucrose is non-reducing; maltose and lactose are reducing.

Starch and cellulose

$\text{Amylose: }\alpha C1-C4(15\text{-}20\%);\text{Amylopectin: }\alpha C1-C4+C1-C6(80\text{-}85\%);\text{Cellulose: }\beta C1-C4$

Glycogen is like amylopectin but more branched.

Nucleoside and nucleotide

$\text{nucleoside}=\text{base}+\text{sugar};\text{nucleotide}=\text{nucleoside}+\text{phosphate}$

Nucleotides join by 5′–3′ phosphodiester links.

Base pairing in DNA

$A=T,G\equiv C$

RNA has U in place of T.

Enzymes lower activation energy

$E_a(\text{acid})=6.22{\text{kJ mol}}^{-1}\to E_a(\text{sucrase})=2.15{\text{kJ mol}}^{-1}$

Hydrolysis of sucrose (NCERT values).

Biomolecules: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the board asks Biomolecules

The question patterns to expect, and the method that works for each.

Tricks to remember Biomolecules

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Glucose detective clues

  HI → Hexane (straight chain); Hydroxylamine/HCN → C=O; Bromine water → aldehyde; Acetic anhydride → five OH; Nitric acid → CH₂OH. Remember 'Straight, Carbonyl, Aldehyde, Five, Primary'.

  Use it when: Questions on the open-chain structure.

• Sucrose locks both doors

  Sucrose joins glucose C1 to fructose C2, the two reducing 'doors', so it is non-reducing. Maltose and lactose leave one door open.

  Use it when: Reducing and non-reducing sugars.

• Amylose Alone, Amylopectin Branches

  AmyLose = Linear (C1–C4), soluble, 15–20%. AmyloPectin = Plenty of branches (C1–C6), insoluble, 80–85%.

  Use it when: Starch components.

• All Dogs Eat Kibble

  A, D, E, K are the fat-soluble vitamins; B and C are water-soluble.

  Use it when: Vitamin classification.

• AT two, GC three

  A pairs with T, G pairs with C; in RNA, U replaces T. (A–T: two hydrogen bonds, G–C: three, as usually drawn.)

  Use it when: Base pairing questions.

Biomolecules: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Biomolecules

The slips examiners see most often, and what to write instead.

• ✗Mistake: Linking D/L with the sign of rotation

  Wrong working

  D-fructose must be dextrorotatory.

  Right working

  D-fructose is laevorotatory, D-(−)-fructose; D refers to configuration, (+)/(−) to the direction of rotation.

  ✓Do this: Write both labels separately: D or L for configuration, (+) or (−) for rotation.

  Usually costs: 1 mark.

• ✗Mistake: Saying denaturation destroys the primary structure

  ✓Do this: Denaturation destroys secondary and tertiary structures; the amino acid sequence (primary structure) stays intact.

  Usually costs: 1 mark.

• ✗Mistake: Writing that cellulose is made of α-glucose

  ✓Do this: Cellulose is β-D-glucose units with C1–C4 links; starch is α-glucose.

  Usually costs: ½ to 1 mark.

• ✗Mistake: Mixing up nucleoside and nucleotide

  Wrong working

  Nucleoside = base + sugar + phosphate

  Right working

  Nucleoside = base + sugar; nucleotide = base + sugar + phosphate

  ✓Do this: Nucleotide has the 'T' for phosphaTe.

  Usually costs: 1 mark.

• ✗Mistake: Wrong vitamin–disease pairs

  ✓Do this: Learn the NCERT table: A night blindness and xerophthalmia; B₁ beri beri; B₂ cheilosis; B₆ convulsions; B₁₂ pernicious anaemia; C scurvy; D rickets and osteomalacia; E RBC fragility; K slow clotting.

  Usually costs: ½ mark per wrong pair.

NCERT exercise guide: Biomolecules

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Biomolecules: previous year questions

• Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 7 marks per paper; 33 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Biomolecules

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

Every Biomolecules question, by year

Practice questions on Biomolecules

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   Which vitamin deficiency causes rickets in children?
   (a) A (b) C (c) D (d) K

   Hint

   The sunshine vitamin.

   Answer

   (c) Vitamin D

2. Q2 · 1 mark · Assertion-reasonBoard level

   Assertion (A): Denaturation of a protein destroys its biological activity.
   Reason (R): During denaturation the primary structure of the protein is destroyed.
   (a) Both A and R are true and R is the correct explanation of A.
   (b) Both A and R are true but R is not the correct explanation of A.
   (c) A is true but R is false.
   (d) A is false but R is true.

   Hint

   Which structures are lost?

   Answer

   (c) Secondary and tertiary structures are lost; the primary structure stays intact.

3. Q3 · 2 marks · Very short answerBoard level

   What is a glycosidic linkage? Name the linkage in maltose. (2 marks)

   Hint

   Oxide link with loss of water.

   Answer

   The oxide linkage formed between two monosaccharide units by loss of a water molecule (1). In maltose, C1 of one α-D-glucose is linked to C4 of the next (α C1–C4) (1).

4. Q4 · 2 marks · Very short answerBoard level

   Name the vitamins whose deficiency causes (a) night blindness, (b) beri beri, (c) increased blood clotting time. (2 marks)

   Hint

   Use the NCERT table.

   Answer

   (a) Vitamin A (b) Vitamin B₁ (thiamine) (c) Vitamin K (½ + ½ + 1).

5. Q5 · 3 marks · Short answerHarder board level

   Write the reactions of glucose with (a) HI, (b) bromine water and (c) concentrated HNO₃, naming the products and stating what each shows. (3 marks)

   Hint

   Chain, aldehyde, primary alcohol.

   Answer

   (a) n-Hexane: straight chain of six carbons (1). (b) Gluconic acid: the carbonyl is an aldehyde (1). (c) Saccharic acid (a dicarboxylic acid): a primary –CH₂OH group is present (1).

6. Q6 · 3 marks · Short answerHarder board level

   Distinguish between amylose and amylopectin, and between starch and glycogen. (3 marks)

   Hint

   Branching, solubility, percentage, organism.

   Answer

   Amylose: water-soluble, 15–20% of starch, unbranched α C1–C4 chain (1). Amylopectin: insoluble, 80–85%, branched with C1–C6 links at branches (1). Starch is the plant storage carbohydrate; glycogen is the animal storage carbohydrate, more highly branched, stored in liver, muscles and brain (1).

7. Q7 · 2 marks · Very short answerHarder board level

   Why do amino acids behave like salts rather than simple amines or acids? Why is glycine the only common amino acid that is optically inactive? (2 marks)

   Hint

   Zwitterion; asymmetric carbon.

   Answer

   They exist as dipolar zwitterions (H₃N⁺–CHR–COO⁻), which are ionic, giving high melting points and water solubility (1). Glycine's α-carbon has two H atoms, so it is not chiral (1).

8. Q8 · 5 marks · Long answerHarder board level

   (a) Name the sugar and bases present in RNA. (b) What are the three types of RNA? (c) What is the role of DNA in heredity and in protein synthesis? (d) Name the linkage that joins nucleotides. (5 marks)

   Hint

   Ribose; m, r, t; replication; phosphodiester.

   Answer

   (a) β-D-ribose; adenine, guanine, cytosine, uracil (1). (b) m-RNA, r-RNA, t-RNA (1). (c) DNA carries the genetic information and replicates during cell division so identical copies pass to daughter cells (1); the message for each protein is in DNA, while the proteins are made by RNA molecules (1). (d) Phosphodiester linkage between 5′ and 3′ carbons of the sugars (1).

9. Q9 · 3 marks · Short answerHOTS / competency

   Suppose a nutritionist tells a teenager that a single large dose of vitamin A each month is fine, but vitamin C must be eaten every day. Use the classification of vitamins to explain this advice, and give one danger of taking vitamin pills without advice. (3 marks)

   Hint

   Where are fat-soluble vitamins stored?

   Answer

   Vitamin A is fat-soluble and is stored in the liver and adipose tissue, so the body can draw on its reserve (1). Vitamin C is water-soluble, is excreted in urine and is not stored, so it must be taken regularly (1). Excess vitamins are also harmful, so pills should be taken only on a doctor's advice (1).

10. Q10 · 2 marks · Very short answerHOTS / competency

    The enzyme sucrase lowers the activation energy of sucrose hydrolysis from 6.22 to 2.15 kJ mol⁻¹. (a) What is the general role of an enzyme? (b) Why is an enzyme called specific? Give an NCERT example. (2 marks)

    Hint

    Catalysts; one substrate.

    Answer

    (a) Enzymes are biological catalysts that speed up reactions by lowering activation energy (1). (b) Each enzyme acts on a particular substrate or reaction, e.g. maltase hydrolyses only maltose to glucose (1).

11. Q11 · 2 marks · Very short answerHOTS / competency

    A tripeptide is made from glycine, alanine and glycine (Gly-Ala-Gly). (a) How many peptide bonds does it have? (b) Is it optically active? Explain. (2 marks)

    Hint

    Look for an asymmetric carbon.

    Answer

    (a) Two (1). (b) Yes: the alanine unit's α-carbon carries H, CH₃, –NH– and –CO–, four different groups, so it is chiral (1).

12. Q12 · 1 mark · MCQHarder board level

    Which of the following is a fibrous protein?
    (a) Insulin (b) Albumin (c) Keratin (d) Haemoglobin

    Hint

    Hair and wool.

    Answer

    (c) Keratin

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Define carbohydrates.Show answer

   Optically active polyhydroxy aldehydes or ketones, or compounds that produce such units on hydrolysis.

2. 2.What product does glucose give with bromine water?Show answer

   Gluconic acid.

3. 3.What is the anomeric carbon in glucose?Show answer

   C1, whose –OH configuration differs between the α and β forms.

4. 4.What are the hydrolysis products of lactose?Show answer

   β-D-galactose and β-D-glucose.

5. 5.Which component of starch is water-soluble?Show answer

   Amylose (15–20% of starch).

6. 6.What type of protein is keratin, and what type is insulin?Show answer

   Keratin is fibrous; insulin is globular.

7. 7.What holds the α-helix together?Show answer

   Hydrogen bonds between >C=O and –NH– groups of the peptide bonds.

8. 8.Which vitamin deficiency causes pernicious anaemia?Show answer

   Vitamin B₁₂.

9. 9.Which base is in RNA but not DNA?Show answer

   Uracil.

10. 10.Thyroxine is a derivative of which amino acid?Show answer

    Tyrosine (an iodinated derivative).

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you explain what each of HI, NH₂OH, Br₂ water, acetic anhydride and HNO₃ shows about glucose?
• ☐Can you give three facts the open-chain structure of glucose cannot explain?
• ☐Can you draw α-D-glucopyranose as a Haworth structure?
• ☐Can you explain why sucrose is non-reducing and why its hydrolysis gives invert sugar?
• ☐Can you state the differences between amylose, amylopectin, cellulose and glycogen?
• ☐Can you describe the four levels of protein structure and denaturation?
• ☐Can you name the deficiency disease of each vitamin in the NCERT table?
• ☐Can you give three differences between DNA and RNA and the functions of insulin and thyroxine?

Biomolecules: frequently asked questions