1. Chapter at a Glance
Biomolecules are the organic compounds that build and run living systems: carbohydrates, proteins, enzymes, vitamins, nucleic acids and hormones. The chapter is mostly factual, which makes it one of the easiest scoring chapters in JEE Main — one question comes almost every year. Focus on the structure of glucose, reducing and non-reducing sugars, amino acids and protein structure, vitamins and deficiency diseases, and the differences between DNA and RNA.
- Open-chain and cyclic structures of glucose; reactions that prove its structure.
- Reducing vs non-reducing sugars; glycosidic linkages in sucrose, maltose and lactose.
- Essential amino acids, zwitterions and levels of protein structure.
- Vitamins and deficiency diseases; DNA vs RNA bases.
2. Carbohydrates
Carbohydrates are polyhydroxy aldehydes or ketones, or compounds that give these on hydrolysis. They are classified as monosaccharides (glucose, fructose), oligosaccharides (sucrose, maltose, lactose — 2 to 10 units) and polysaccharides (starch, cellulose, glycogen).
Glucose (C₆H₁₂O₆), an aldohexose
- Heating with HI gives n-hexane → six carbons in a straight chain.
- Reacts with hydroxylamine (oxime) and HCN (cyanohydrin) → a carbonyl group is present.
- Bromine water oxidises it to gluconic acid → the carbonyl is an aldehyde.
- Acetylation gives a pentaacetate → five –OH groups.
- Nitric acid gives saccharic acid (both ends oxidised) → one end is a 1° alcohol.
- Glucose does not give Schiff's test or react with NaHSO₃, and its pentaacetate does not react with NH₂OH. These facts show that glucose mainly exists in a cyclic hemiacetal (pyranose) form, formed between C1 and the C5–OH.
- The cyclic form has two anomers, α and β, which differ only in the configuration at C1. Mutarotation: either anomer in water slowly changes to an equilibrium mixture.
- Open-chain glucose has 4 chiral carbons (C2–C5), so 2⁴ = 16 stereoisomeric aldohexoses exist. Natural glucose is D-(+)-glucose.
Fructose
A ketohexose (C=O at C2) and the sweetest natural sugar; it forms a five-membered (furanose) ring. It is laevorotatory: D-(−)-fructose.
Disaccharides and polysaccharides
| Sugar | Units | Linkage | Reducing? |
|---|---|---|---|
| Sucrose | α-D-glucose + β-D-fructose | C1 (glc) – C2 (fru) | No — both anomeric carbons used |
| Maltose | 2 α-D-glucose | α(1→4) | Yes |
| Lactose (milk sugar) | β-D-galactose + β-D-glucose | β(1→4) | Yes |
| Starch | Amylose (15–20%, linear α-1,4) + amylopectin (80–85%, branched with α-1,6) | α-glycosidic | — |
| Cellulose | β-D-glucose | β(1→4), linear | — |
| Glycogen | α-D-glucose, highly branched | α-1,4 and α-1,6 | Animal starch |
Invert sugar: hydrolysis of sucrose (dextrorotatory) gives an equimolar glucose–fructose mixture that is laevorotatory, because fructose's (−) rotation is larger than glucose's (+) rotation.
3. Amino Acids and Proteins
- α-Amino acids have –NH₂ and –COOH on the same carbon. All except glycine are optically active (glycine has two H on the α-carbon).
- Essential amino acids must come from the diet (valine, leucine, isoleucine, lysine, threonine, methionine, phenylalanine, tryptophan, histidine, arginine). Non-essential ones are made in the body.
- In water, amino acids exist as zwitterions (H₃N⁺–CHR–COO⁻), so they are high-melting, water-soluble solids. At the isoelectric point, the net charge is zero and they do not move in an electric field.
- Amino acids join by peptide bonds (–CO–NH–), losing water. Polypeptides with more than about 100 residues (mass above 10,000 u) are proteins.
| Level | What it describes | Held by |
|---|---|---|
| Primary | Sequence of amino acids | Peptide (covalent) bonds |
| Secondary | α-helix or β-pleated sheet | H-bonds between C=O and N–H |
| Tertiary | Overall 3D folding (fibrous or globular) | H-bonds, disulphide, ionic, van der Waals |
| Quaternary | Arrangement of two or more subunits (haemoglobin) | Non-covalent forces |
Denaturation (heat, pH change) destroys secondary and tertiary structure but not the primary structure — e.g. boiling an egg, curdling of milk. The protein loses its biological activity.
4. Enzymes, Vitamins and Hormones
- Enzymes are biological catalysts, mostly globular proteins. They are highly specific and lower the activation energy (e.g. maltase: maltose → glucose; invertase: sucrose → glucose + fructose).
- Vitamins: fat-soluble A, D, E, K (stored in liver and fat) and water-soluble B group and C (must be taken regularly; B₁₂ is the exception and is stored in the liver).
- Hormones: insulin (protein), thyroxine (amino-acid derivative), steroids such as oestrogen and testosterone.
| Vitamin | Deficiency disease |
|---|---|
| A | Night blindness, xerophthalmia |
| B₁ (thiamine) | Beriberi |
| B₂ (riboflavin) | Cheilosis (cracked lips), digestive disorders |
| B₆ (pyridoxine) | Convulsions |
| B₁₂ (cyanocobalamin) | Pernicious anaemia |
| C (ascorbic acid) | Scurvy (bleeding gums) |
| D | Rickets (children), osteomalacia (adults) |
| E | Fragile red blood cells, muscular weakness |
| K | Delayed blood clotting |
5. Nucleic Acids
- A nucleoside = base + sugar; a nucleotide = base + sugar + phosphate. Nucleotides join by phosphodiester links (3′ to 5′).
- DNA: sugar 2-deoxy-D-ribose; bases A, G, C, T; double helix (Watson and Crick). RNA: sugar D-ribose; bases A, G, C, U; usually single stranded (mRNA, tRNA, rRNA).
- Base pairing: A pairs with T (2 H-bonds), G pairs with C (3 H-bonds). So in DNA, %A = %T and %G = %C (Chargaff's rule).
- DNA carries genetic information and replicates itself; RNA helps in protein synthesis.
6. Master Table
| Item | Key fact |
|---|---|
| Glucose | Aldohexose, 4 chiral C (open chain), pyranose ring |
| Anomers | α and β differ at C1 |
| Non-reducing sugar | Sucrose |
| Optically inactive amino acid | Glycine |
| Peptide bond | –CO–NH– |
| Chargaff's rule | %A = %T, %G = %C |
| Fat-soluble vitamins | A, D, E, K |
7. Mind Map
- Glucose, fructose
- Sucrose, maltose, lactose
- Starch, cellulose
- Amino acids, zwitterion
- 1°–4° structure
- Denaturation
- Specific catalysts
- Deficiency diseases
- Hormones
- DNA vs RNA
- Base pairing
- Nucleotide structure
8. Tips and Tricks
- A sugar is reducing if it has a free anomeric carbon (hemiacetal/hemiketal) that can open to a carbonyl.
- "All Doctors Eat Kheer" → fat-soluble vitamins A, D, E, K.
- For Chargaff problems: %A + %G = 50%. If A = 20%, then T = 20%, G = C = 30%.
- RNA has U instead of T; DNA has deoxyribose (no –OH at C2′).
- Calling sucrose a reducing sugar.
- Thinking denaturation breaks peptide bonds.
- Confusing nucleoside (no phosphate) with nucleotide.
9. Important One-liners
- Cellulose is a linear polymer of β-D-glucose; humans cannot digest it.
- Amylopectin is branched and forms 80–85% of starch.
- Invert sugar is laevorotatory.
- Insulin is a protein hormone that controls blood glucose.
- Vitamin C is destroyed by cooking and is water soluble.
- The α-helix is stabilised by intramolecular hydrogen bonds.
10. PYQ-pattern Questions with Solutions
These questions are original and modelled on concepts frequently tested in JEE Main.
Which of the following is a non-reducing sugar?
- Glucose
- Maltose
- Lactose
- Sucrose
Show solution
In sucrose, C1 of glucose is joined to C2 of fructose — both anomeric carbons are locked in the glycosidic bond, so no free carbonyl can form.
Answer: (D)
The amino acid that is optically inactive is:
- Alanine
- Glycine
- Valine
- Serine
Show solution
Glycine (H₂N–CH₂–COOH) has two H atoms on the α-carbon, so it has no chiral centre.
Answer: (B)
α-D-Glucose and β-D-glucose differ in configuration at:
- C1
- C2
- C4
- C5
Show solution
They are anomers: the –OH on the anomeric carbon C1 is on opposite sides in the two forms. All other carbons are the same.
Answer: (A)
Which observation shows that glucose exists in a cyclic form?
- It gives gluconic acid with bromine water
- It forms a pentaacetate
- It does not react with NaHSO₃
- It gives n-hexane with HI
Show solution
A free aldehyde would add NaHSO₃. Glucose does not, because most molecules are in the cyclic hemiacetal form. The other three facts are explained by the open-chain structure.
Answer: (C)
Deficiency of vitamin B₁₂ causes:
- Scurvy
- Beriberi
- Pernicious anaemia
- Rickets
Show solution
Vitamin B₁₂ (cyanocobalamin) is needed to make red blood cells; lack of it causes pernicious anaemia. Scurvy — C; beriberi — B₁; rickets — D.
Answer: (C)
A double-stranded DNA sample contains 20% adenine. What is the percentage of cytosine?
Show solution
A = T = 20%, so A + T = 40%. G + C = 60% and G = C, so C = 30%.
Answer: 30
How many stereoisomers are possible for the open-chain form of an aldohexose such as glucose?
Show solution
Open-chain aldohexose: C2, C3, C4 and C5 are chiral (4 centres). Stereoisomers = 2⁴ = 16 (8 D and 8 L).
Answer: 16
When a protein is denatured by heating, which structure remains intact?
- Primary
- Secondary
- Tertiary
- Quaternary
Show solution
Heat breaks the weak H-bonds and other non-covalent forces holding the secondary, tertiary and quaternary structures. Covalent peptide bonds (primary structure) are not broken.
Answer: (A)
11. Quick Revision Summary
- Glucose: aldohexose, cyclic pyranose, α/β anomers at C1, mutarotation.
- Sucrose non-reducing; maltose, lactose reducing; invert sugar laevorotatory.
- Glycine achiral; zwitterions; peptide bond –CO–NH–.
- Protein levels 1°–4°; denaturation keeps the primary structure.
- Vitamins: A, D, E, K fat-soluble; B₁₂ → pernicious anaemia; C → scurvy.
- DNA: deoxyribose, A=T, G≡C; RNA: ribose, uracil.