What are essential amino acids?

Essential amino acids cannot be synthesised by the human body and must be obtained from diet. There are 10 essential amino acids (9 in adults, 10 in children): Phe, Trp, Met, Thr, His, Ile, Leu, Lys, Val, Arg (mnemonic: PVT TIM HaLl). Complete proteins (eggs, milk, meat) contain all essential amino acids.

Samacheer Kalvi 12th Bio-Chemistry Chapter 3 Exercise Solutions

Exercise Solutions Available

Step-by-step answers for all textbook questions in this chapter

View Solutions →

Chapter 3 — Proteins and Amino Acids

Key Concepts Quick Recap

Term	Definition
Amino acid	Has –NH₂ (amino), –COOH (carboxyl), R group, and H on α-carbon
Peptide bond	Covalent bond between –COOH of one and –NH₂ of next amino acid (dehydration)
Isoelectric point (pI)	pH at which amino acid/protein has zero net charge; minimum solubility
α-helix	Secondary structure; H-bonds within same polypeptide chain
β-pleated sheet	Secondary structure; H-bonds between adjacent polypeptide chains
Denaturation	Loss of 3D structure (but not primary) due to heat, pH, chemicals
Coagulation	Irreversible denaturation (e.g. cooking egg white)

I. Choose the Best Answer (MCQs)

1. The bond that links amino acids in a protein chain is called: (a) Ester bond (b) Glycosidic bond (c) Peptide bond (d) Disulphide bond

Answer(c) Peptide bond Peptide bond: formed between –COOH of one amino acid and –NH₂ of the next with loss of water (condensation). This bond is planar and rigid (partial double bond character).

2. The α-helix is stabilised by: (a) Disulphide bonds (b) Hydrogen bonds within the same polypeptide (c) Ionic bonds (d) Van der Waals forces

Answer(b) Hydrogen bonds within the same polypeptide In α-helix: H-bonds between C=O of residue n and N-H of residue n+4 within the same chain. 3.6 residues per turn. Right-handed helix.

3. Denaturation of proteins destroys: (a) Primary structure (b) Secondary and tertiary structure (c) Amino acid composition (d) Peptide bonds

Answer(b) Secondary and tertiary structure Denaturation: disruption of H-bonds, ionic bonds, disulphide bonds → loss of 3D shape → loss of function. Primary structure (sequence of amino acids) is NOT destroyed. The protein can sometimes refold (renaturation). Example: heat denatures egg white.

4. The quaternary structure of haemoglobin consists of: (a) 1 polypeptide chain (b) 2 polypeptide chains (c) 4 polypeptide chains (2α + 2β) (d) 6 polypeptide chains

Answer(c) 4 polypeptide chains (2α + 2β) Haemoglobin: tetramer of 2α and 2β subunits, each with a haem group containing Fe²⁺. Quaternary structure is stabilised by non-covalent interactions between subunits.

5. An amino acid at its isoelectric point: (a) Has maximum solubility (b) Carries a positive charge (c) Carries no net charge (d) Is completely ionised

Answer(c) Carries no net charge At pI: +NH₃ group is balanced by –COO⁻ group → zwitterion (both charges, but net charge = 0). Minimum solubility and electrophoretic mobility = 0 at pI.

6. The essential amino acid that is also the precursor of serotonin and melatonin is: (a) Phenylalanine (b) Tyrosine (c) Tryptophan (d) Lysine

Answer(c) Tryptophan Tryptophan (essential) → serotonin (5-hydroxytryptamine) → melatonin. Also the precursor of nicotinic acid (Vitamin B3/Niacin). Deficiency leads to Pellagra.

7. Which test is used to detect proteins? (a) Benedict's test (b) Biuret test (c) Iodine test (d) Fehling's test

Answer(b) Biuret test Biuret test: proteins (2+ peptide bonds) + NaOH + CuSO₄ → violet/purple colour (Cu²⁺ complexes with peptide bonds). Positive for: proteins, peptides. Negative for: amino acids (only 1 peptide bond), carbohydrates.

II. Short Answer Questions

Q1. Describe the four levels of protein structure.

Answer

1. Primary structure: Linear sequence of amino acids joined by peptide bonds. Determined by gene (DNA). Example: insulin has 51 amino acids (2 chains: A=21, B=30).

2. Secondary structure: Local folding of polypeptide backbone:

α-helix: right-handed spiral; H-bonds within chain; 3.6 residues/turn. Example: myoglobin (75% α-helix).
β-pleated sheet: parallel or antiparallel strands; H-bonds between strands. Example: silk fibroin.

3. Tertiary structure: Overall 3D shape of entire polypeptide. Stabilised by: disulphide bonds (–S–S–, strongest), hydrophobic interactions, hydrogen bonds, ionic bonds. Example: myoglobin, ribonuclease.

4. Quaternary structure: Association of 2+ polypeptide subunits. Stabilised by non-covalent interactions. Example: haemoglobin (2α + 2β subunits), DNA polymerase.

Exam Checklist

Peptide bond: –COOH + –NH₂ → –CO–NH– + H₂O (condensation).
4 levels: primary (sequence) → secondary (α-helix/β-sheet) → tertiary (3D) → quaternary (subunits).
α-helix: H-bonds within chain; β-sheet: H-bonds between chains.
Denaturation: secondary/tertiary destroyed; primary intact.
Isoelectric point: net charge = 0; minimum solubility.
Biuret test: violet with proteins; Benedict's test: reducing sugars.
Essential amino acids (10): PVT TIM HaLL — Phe, Val, Trp, Thr, Ile, Met, His, Arg, Leu, Lys.

Frequently Asked Questions

1. Primary: sequence of amino acids linked by peptide bonds. 2. Secondary: local folding — α-helix (H-bonds within chain) and β-pleated sheet (H-bonds between chains). 3. Tertiary: 3D folding of entire polypeptide chain — stabilised by disulphide bonds, hydrophobic interactions, ionic bonds, H-bonds. 4. Quaternary: two or more polypeptide subunits (e.g. haemoglobin: 2α+2β subunits).