What this chapter is about
Biomolecules are the organic molecules that form the structural and functional basis of all living organisms. This chapter introduces the four major classes of biomolecules: carbohydrates, proteins, nucleic acids and vitamins. A Class 12 student meets this topic because it connects organic chemistry concepts (functional groups, stereochemistry, condensation reactions) to biological systems, showing how chemistry explains life processes at the molecular level.
After studying this chapter, you should be able to classify carbohydrates and write their open-chain and cyclic structures, explain how amino acids link to form proteins, describe the primary to quaternary structure of proteins, distinguish between DNA and RNA, and state the functions of different vitamins. You will also understand enzyme action and the role of these molecules in metabolism.
The chapter builds on your knowledge of aldehydes, ketones, carboxylic acids and amines from earlier organic chemistry, applying those ideas to molecules essential for life.
Key ideas
- Carbohydrates are polyhydroxy aldehydes or ketones, or compounds that yield them on hydrolysis; they are classified as monosaccharides, disaccharides and polysaccharides based on the number of sugar units.
- Monosaccharides exist in open-chain and cyclic (Haworth) forms; glucose forms a six-membered pyranose ring, while fructose forms a five-membered furanose ring. The α and β anomers differ in the orientation of the –OH group at C-1.
- Amino acids contain both an amino group (–NH₂) and a carboxyl group (–COOH); in aqueous solution they exist as zwitterions (H₃N⁺–CHR–COO⁻). They are linked by peptide bonds (–CO–NH–) to form proteins.
- Protein structure has four levels: primary (sequence of amino acids), secondary (α-helix or β-pleated sheet held by hydrogen bonds), tertiary (overall 3-D folding stabilised by disulphide bridges, ionic and hydrophobic interactions), and quaternary (association of multiple polypeptide chains).
- Enzymes are biological catalysts, mostly proteins, that are highly specific and lower the activation energy of biochemical reactions. They have an active site where substrate binding occurs.
- Nucleic acids (DNA and RNA) are polymers of nucleotides; each nucleotide consists of a nitrogenous base, a pentose sugar and a phosphate group. DNA uses deoxyribose and bases A, G, C, T; RNA uses ribose and bases A, G, C, U.
- Vitamins are organic compounds required in small amounts for normal metabolism; they are classified as water-soluble (B-complex, C) and fat-soluble (A, D, E, K). Deficiency diseases result when intake is insufficient.
Formulas and facts to remember
- Glucose molecular formula: C₆H₁₂O₆ — an aldohexose with an aldehyde group at C-1.
- Fructose molecular formula: C₆H₁₂O₆ — a ketohexose with a ketone group at C-2.
- Peptide bond formation: amino acid 1 + amino acid 2 → dipeptide + H₂O (condensation reaction between –COOH of one and –NH₂ of another).
- Isoelectric point: the pH at which an amino acid exists as a zwitterion with no net charge.
- DNA double helix: two antiparallel strands held by hydrogen bonds; A pairs with T (two H-bonds), G pairs with C (three H-bonds).
- Reducing sugars: carbohydrates with a free aldehyde or ketone group that can reduce Fehling's solution or Tollens' reagent (e.g., glucose, maltose).
- Non-reducing sugars: sucrose is non-reducing because both anomeric carbons are involved in the glycosidic bond.
- Vitamin deficiency examples: Vitamin A deficiency causes night blindness; Vitamin C deficiency causes scurvy; Vitamin D deficiency causes rickets.
Worked examples
Example 1: Identifying a reducing sugar
Problem: An unknown carbohydrate gives a positive Tollens' test (silver mirror) and on hydrolysis yields only glucose. Identify whether it is a monosaccharide, disaccharide or polysaccharide, and name a possible compound.
Solution: Step 1: A positive Tollens' test means the carbohydrate has a free or potentially free aldehyde group — it is a reducing sugar. Step 2: Hydrolysis yields only glucose, so the compound is made up of glucose units. Step 3: If hydrolysis is needed to release glucose, the compound has more than one sugar unit; thus it is not a monosaccharide. Step 4: A disaccharide that is reducing and gives only glucose on hydrolysis is maltose (two glucose units linked α-1,4 with a free anomeric carbon on the second unit).
Example 2: Calculating peptide bonds in a polypeptide
Problem: A polypeptide chain contains 150 amino acid residues. How many peptide bonds are present in this chain?
Solution: Step 1: Each peptide bond links two amino acids by removing one molecule of water. Step 2: If there are n amino acids, the number of peptide bonds = n − 1. Step 3: Number of peptide bonds = 150 − 1 = 149 peptide bonds.
Example 3: Base pairing in DNA
Problem: A sample of double-stranded DNA has 22 % adenine. What are the percentages of the other three bases?
Solution: Step 1: In double-stranded DNA, adenine pairs with thymine, so % A = % T. Therefore, % T = 22 %. Step 2: Guanine pairs with cytosine, so % G = % C. Step 3: Total = % A + % T + % G + % C = 100 %. 22 + 22 + % G + % C = 100 → % G + % C = 56 %. Step 4: Since % G = % C, each = 56 / 2 = 28 %.
Answer: T = 22 %, G = 28 %, C = 28 %.
Common mistakes
- Confusing aldoses and ketoses → Remember: aldoses have –CHO at terminal carbon; ketoses have C=O at C-2.
- Writing the wrong number of H-bonds in base pairs → A–T has 2 hydrogen bonds; G–C has 3 hydrogen bonds.
- Calling sucrose a reducing sugar → Sucrose is non-reducing because both anomeric carbons are engaged in the glycosidic linkage.
- Mixing up α and β anomers → In α-anomer, the –OH at anomeric carbon is on the opposite side of CH₂OH in the Haworth projection; in β-anomer, they are on the same side.
- Forgetting that enzymes are substrate-specific → Each enzyme acts on a particular substrate due to the shape of its active site (lock-and-key model).
Quick revision
- Carbohydrates: polyhydroxy aldehydes (aldoses) or ketones (ketoses); classified by number of sugar units.
- Amino acids join via peptide bonds (–CO–NH–); proteins have primary, secondary, tertiary and quaternary structures.
- DNA: deoxyribose + A, G, C, T; RNA: ribose + A, G, C, U; base pairing A–T (2 H-bonds), G–C (3 H-bonds).
- Vitamins: water-soluble (B, C) and fat-soluble (A, D, E, K); deficiency causes specific diseases.
- Enzymes lower activation energy and are highly specific to substrates.