Dedicated Problem Set — JEE/NEET Level
This is a curated set of 32 fully worked, exam-level problems covering every theme of Biomolecules — classifying carbohydrates, reducing vs non-reducing sugars, glucose/fructose structure, anomers and mutarotation, glycosidic linkages and the di-/polysaccharides, amino acids and the zwitterion/isoelectric point, the peptide bond and protein structure levels, denaturation, enzymes, vitamins and deficiency diseases, and the structure and base pairing of nucleic acids. Each solution shows the reasoning step by step. Work each one before reading the solution.
Carbohydrates — Classification & Structure
Example 1. Classify sucrose, maltose and starch and state which are reducing.
Solution: Sucrose and maltose are disaccharides; starch is a polysaccharide. Maltose is reducing; sucrose is non-reducing; starch is essentially non-reducing.
Example 2. Why does glucose not react with NaHSO or give the Schiff's test despite having a -CHO group?
Solution: In solution glucose exists almost entirely as the cyclic hemiacetal (pyranose), with only a trace of the free aldehyde, so the typical aldehyde reactions are not observed.
Example 3. Explain mutarotation in terms of the anomers of glucose.
Solution: Freshly dissolved pure alpha- or beta-D-glucose slowly interconverts through the open-chain form until an equilibrium mixture of both anomers is reached, so the optical rotation changes to a constant value — this is mutarotation.
Example 4. Why is sucrose called a non-reducing sugar while its hydrolysis product is reducing?
Solution: In sucrose both anomeric carbons are locked in the glycosidic linkage (non-reducing). On hydrolysis it gives free glucose and fructose, each with a free reducing group — so the products (invert sugar) are reducing.
Carbohydrates — Polysaccharides & Invert Sugar
Example 5. Distinguish amylose from amylopectin.
Solution: Amylose is a linear, water-soluble chain of alpha-glucose (about 15-20% of starch); amylopectin is a branched, water-insoluble chain of alpha-glucose (about 80-85%).
Example 6. Why can humans digest starch but not cellulose?
Solution: Starch has alpha-glycosidic linkages, which human enzymes (amylase) can hydrolyse; cellulose has beta-glycosidic linkages, for which humans lack the enzyme, so cellulose is indigestible.
Example 7. Sucrose is dextrorotatory but its hydrolysis is called inversion. Explain.
Solution: Hydrolysis gives glucose (+52.5°) and fructose (−92.4°); fructose's strong laevorotation dominates, so the mixture becomes laevorotatory — the sign of rotation inverts from (+) to (−).
Example 8. Name the storage polysaccharides of plants and animals and the structural one.
Solution: Plant storage = starch; animal storage = glycogen; structural = cellulose.
Amino Acids & Proteins
Example 9. Write the zwitterion form of alanine and explain its amphoteric nature.
Solution: Alanine's zwitterion is ⁺HN-CH(CH)-COO⁻. It is amphoteric because the -COO⁻ can accept a proton (acts as a base) and the -NH can donate a proton (acts as an acid).
Example 10. Define the isoelectric point.
Solution: The pH at which the amino acid exists mainly as the neutral zwitterion and does not migrate to either electrode in an electric field.
Example 11. Name the four levels of protein structure.
Solution: Primary (sequence), secondary (alpha-helix / beta-pleated sheet), tertiary (overall 3-D fold) and quaternary (arrangement of subunits).
Example 12. What is denaturation, and which structure survives it?
Solution: Denaturation is the loss of secondary/tertiary structure and biological activity (e.g. boiling egg white). The primary structure (peptide-bonded sequence) survives.
Proteins, Enzymes & Vitamins
Example 13. What is a peptide bond, and how is it formed?
Solution: A peptide bond is the amide linkage (-CO-NH-) formed between the -COOH of one amino acid and the -NH of the next, with the loss of water.
Example 14. Why are enzymes highly specific?
Solution: Each enzyme has an active site of a particular shape into which only the matching substrate fits (lock-and-key), so it catalyses one specific reaction.
Example 15. Name the deficiency diseases of vitamins A, C, D and B.
Solution: A → night blindness; C → scurvy; D → rickets/osteomalacia; B → beriberi.
Example 16. Which vitamins are fat-soluble, and what is a consequence of this?
Solution: A, D, E and K are fat-soluble; because they are stored in fat/liver, an excess can accumulate and become toxic.
Nucleic Acids
Example 17. Give the three components of a nucleotide.
Solution: A pentose sugar, a nitrogenous base and a phosphate group.
Example 18. Distinguish a nucleoside from a nucleotide.
Solution: Nucleoside = sugar + base; nucleotide = sugar + base + phosphate (a nucleoside plus a phosphate).
Example 19. State the complementary base-pairing rules and the number of hydrogen bonds in each pair.
Solution: A-T (2 hydrogen bonds) and G-C (3 hydrogen bonds) in DNA (A-U in RNA).
Example 20. Give three differences between DNA and RNA.
Solution: DNA: deoxyribose, base thymine, usually double-stranded. RNA: ribose, base uracil, usually single-stranded.
Tests, Distinctions & Functions
Example 21. How would you chemically distinguish glucose from sucrose?
Solution: Use Fehling's solution (or Tollens') and warm: glucose gives a red CuO (or silver mirror) — it is reducing; sucrose gives no reaction — it is non-reducing.
Example 22. How is starch distinguished from glucose?
Solution: Add iodine solution: starch gives a blue-black colour; glucose does not.
Example 23. Give the complementary strand for the DNA sequence T-A-G-C-A.
Solution: Pairing each base (T-A, A-T, G-C, C-G, A-T) gives A-T-C-G-T.
Example 24. What are the two main biological functions of DNA?
Solution: Replication (making exact copies of genetic information) and directing protein synthesis (transcription to mRNA, then translation).
Mixed JEE/NEET Challenge
Example 25. A disaccharide gives a positive Tollens' test and on hydrolysis yields only glucose. Identify it.
Solution: A reducing disaccharide of two glucose units = maltose.
Example 26. A carbohydrate gives a blue-black colour with iodine and yields only alpha-glucose on hydrolysis. Identify it.
Solution: Starch (a polymer of alpha-glucose that gives the iodine test).
Example 27. Why does fructose, a ketose, reduce Tollens' reagent?
Solution: In the basic medium of the test, fructose isomerises (via an enediol) to an aldose, which then reduces the reagent.
Example 28. A protein loses its biological activity on heating but on analysis still shows the same amino-acid sequence. Explain.
Solution: The protein has been denatured — its secondary/tertiary structure is destroyed (loss of activity), but the primary structure (sequence) is unchanged.
Example 29. In a DNA sample, 30% of the bases are adenine. What percentage is cytosine?
Solution: A = T = 30%, so A + T = 60%; thus G + C = 40%, and since G = C, cytosine = 20%. (Chargaff's rule via complementary pairing.)
Example 30. Why are amino acids high-melting, water-soluble solids unlike comparable amines or acids?
Solution: They exist as zwitterions (internal salts); the strong electrostatic forces make them high-melting and the charged form makes them water-soluble — behaving like ionic compounds.
Example 31. Name the linkage in (a) proteins, (b) carbohydrates (between sugar units) and (c) the DNA backbone.
Solution: (a) Peptide bond (amide); (b) glycosidic linkage; (c) phosphodiester (sugar-phosphate) linkage.
Example 32. Classify the bases adenine, guanine, cytosine, thymine and uracil as purines or pyrimidines.
Solution: Purines: adenine, guanine (double-ring). Pyrimidines: cytosine, thymine, uracil (single-ring).
JEE Main & Advanced Level Solved Examples
Biomolecules rewards structure-based reasoning and a few quantitative skills (optical activity, stereoisomer counts, base ratios). These problems target exactly those discriminators. Track which carbon is anomeric and whether a free -CHO/hemiacetal can form before you answer.
Example 33: Why glucose, fructose and mannose give the same osazone [JEE]
D-glucose, D-fructose and D-mannose all form an identical osazone with excess phenylhydrazine. What does this prove?
Solution: Osazone formation involves only C-1 and C-2 (three molecules of phenylhydrazine react there). Glucose and mannose are C-2 epimers (they differ only at C-2), and fructose is the corresponding 2-ketose, so once C-1 and C-2 are converted to the bis-phenylhydrazone the difference at C-2 disappears. Since C-3, C-4 and C-5 are identical in all three, they give the same osazone — proving these three sugars share the same configuration at C-3, C-4 and C-5.
Example 34: Epimers versus anomers [JEE]
Distinguish epimers from anomers, with one example of each.
Solution: Epimers are diastereomers that differ in configuration at one chiral carbon other than the anomeric carbon — e.g. D-glucose and D-mannose differ only at C-2 (C-2 epimers), while D-glucose and D-galactose differ only at C-4. Anomers are the pair that differ only at the anomeric carbon (C-1 in an aldose), formed on ring closure — e.g. alpha- and beta-D-glucose. So all anomers are epimers at the anomeric centre, but ordinary epimers differ at a non-anomeric carbon.
Example 35: Counting stereoisomers of a sugar [JEE]
How many optical isomers are possible for (a) an aldohexose and (b) a ketohexose?
Solution: The count is 2, where n is the number of chiral carbons.
- (a) Aldohexose (e.g. glucose) has chiral centres at C-2, C-3, C-4 and C-5 → n = 4 → 2 = 16 stereoisomers (8 D- and 8 L-forms).
- (b) Ketohexose (e.g. fructose) has the carbonyl at C-2, leaving chiral centres at C-3, C-4 and C-5 → n = 3 → 2 = 8 stereoisomers.
An aldopentose, by the same logic, has 3 chiral centres → 8 isomers.
JEE/Advanced — Optical Activity & Sugar Reactions
Example 36: Quantitative mutarotation [JEE Advanced]
Pure alpha-D-glucose has [alpha] = +112° and pure beta-D-glucose has [alpha] = +19°. At equilibrium the specific rotation is +52.7°. Find the percentage of each anomer at equilibrium.
Solution: Let the fraction of the alpha-anomer be x. Then 112x + 19(1 - x) = 52.7, so 93x = 33.7 and x = 0.36. Thus the equilibrium mixture is about 36% alpha- and 64% beta-D-glucose — the more stable beta form (equatorial -OH) predominates. This gradual change of rotation to the equilibrium value is mutarotation.
Example 37: Ascending and descending the sugar series [JEE Advanced]
Name and outline the reactions used to (a) lengthen an aldose chain by one carbon and (b) shorten it by one carbon.
Solution: (a) Kiliani-Fischer synthesis lengthens the chain: the aldose + HCN gives a cyanohydrin; hydrolysis to the aldonic acid, lactonisation, and controlled reduction give the next-higher aldose (as a pair of C-2 epimers). (b) Ruff degradation shortens the chain: the aldose is oxidised to its aldonic acid (e.g. by Br water), whose calcium salt is treated with HO/Fe to remove C-1 as CO, giving the next-lower aldose.
Example 38: Invert sugar and the sign of rotation [JEE]
Sucrose ([alpha] = +66.5°) is hydrolysed to glucose and fructose. Why is the product called invert sugar?
Solution: Hydrolysis gives an equimolar mixture of D-glucose ([alpha] = +52.7°) and D-fructose ([alpha] = -92°). The strongly laevorotatory fructose outweighs the dextrorotatory glucose, so the net rotation becomes negative (about -20°) — the sign has inverted from + to -. Because the optical rotation is inverted on hydrolysis, the 1:1 glucose-fructose mixture is called invert sugar.
Example 39: Why a methyl glycoside is non-reducing [JEE]
Glucose is a reducing sugar, but methyl alpha-D-glucoside is not. Explain.
Solution: Free glucose exists in equilibrium with a small amount of its open-chain aldehyde (through the hemiacetal at the anomeric C-1), so it can reduce Tollens' and Fehling's reagents and shows mutarotation. When the anomeric -OH is converted to -OCH (a glycoside, a full acetal), the ring can no longer open to the aldehyde form. With no free -CHO available, the methyl glucoside is non-reducing and shows no mutarotation. The anomeric carbon is the key to a sugar's reducing behaviour.
JEE/Advanced — Amino Acids & Peptides
Example 40: How many peptides? [JEE]
(a) How many tripeptides can be formed using three different amino acids, each exactly once? (b) How many distinct tripeptides are possible from the 20 standard amino acids if repetition is allowed?
Solution: A peptide is directional (it has distinct N- and C-terminal ends), so sequence order matters. (a) Arranging 3 different amino acids in order = 3! = 6 tripeptides. (b) With 20 choices at each of the 3 positions and repetition allowed = 20 x 20 x 20 = 20 = 8000 tripeptides. The directionality of the peptide bond is why order-based counting (not mere combinations) applies.
Example 41: Isoelectric point and electrophoresis [JEE]
What is the isoelectric point (pI) of an amino acid, and how does an amino acid behave in an electric field at a pH below its pI?
Solution: The isoelectric point is the pH at which the amino acid exists almost entirely as the zwitterion with no net charge, so in an electric field it does not migrate (and its solubility is minimum). At a pH below pI the medium is more acidic, the -COO is protonated, and the molecule carries a net positive charge, so it migrates toward the cathode (negative electrode). Above pI it is net negative and moves to the anode — the basis of separating amino acids by electrophoresis.
JEE/Advanced — Nucleic Acids
Example 42: Chargaff's rule [JEE]
In a double-stranded DNA sample, adenine is 30% of the bases. Find the percentages of thymine, guanine and cytosine.
Solution: By Chargaff's rule, in double-stranded DNA A pairs with T and G pairs with C, so %A = %T and %G = %C. Given %A = 30%, then %T = 30%. The remaining 40% is shared equally between G and C, so %G = %C = 20%. (Check: 30 + 30 + 20 + 20 = 100%.) The A=T and G=C equalities reflect complementary base pairing across the two strands.