Question

i) Write the chemical names of water soluble vitamins and the diseases caused by their deficiency.
ii) Explain the cyclic structure of glucose. [2+3]
OR
i) Explain the primary, secondary, tertiary and quaternary structures of protein.
ii) Write a note on denaturation of protein. [4+1]

Hint:

Water soluble vitamins are the B-complex and vitamin C (e.g. C deficiency causes scurvy). Glucose forms a six-membered pyranose ring via a C-5 OH to C-1 aldehyde hemiacetal, giving the anomeric carbon. For proteins, recall primary sequence, secondary helix/sheet, tertiary 3-D fold, quaternary sub-units, and that denaturation destroys secondary and tertiary structure but not the peptide bonds.

Answer 1

Option 1

i) Water soluble vitamins and deficiency diseases:
Water soluble vitamins are the B-group vitamins and vitamin C. They dissolve in water, are not stored in the body and are excreted in urine, so they must be supplied regularly in the diet.
Step 1: Vitamin B₁ (Thiamine) → deficiency causes Beri-beri.
Step 2: Vitamin B₂ (Riboflavin) → deficiency causes cheilosis (cracking at the corners of the mouth) and digestive disorders.
Step 3: Vitamin B₃ (Nicotinic acid / Niacin) → deficiency causes Pellagra (dermatitis, diarrhoea).
Step 4: Vitamin B₆ (Pyridoxine) → deficiency causes convulsions and anaemia.
Step 5: Vitamin B₁₂ (Cyanocobalamin) → deficiency causes pernicious anaemia.
Step 6: Vitamin C (Ascorbic acid) → deficiency causes Scurvy (bleeding gums).

ii) Cyclic structure of glucose:
Step 1: In the open-chain form, glucose is a straight chain with an aldehyde group (–CHO) at C-1 and hydroxyl (–OH) groups at C-2 to C-6.
Step 2: The –OH group on C-5 attacks the carbonyl carbon (C-1) of the aldehyde. This intramolecular addition forms a cyclic hemiacetal, producing a six-membered ring (pyranose ring) containing one oxygen atom.
Step 3: On ring closure, C-1 becomes a new chiral (asymmetric) centre and is called the anomeric carbon. It gives two forms: \( \alpha \)-D-glucose (the C-1 –OH is on the opposite side to the C-6 –CH₂OH in the Haworth projection, i.e. down) and \( \beta \)-D-glucose (C-1 –OH up). These are called anomers.
Step 4: This cyclic form explains why glucose does not give the Schiff's test and does not react with NaHSO₃, and why glucose shows mutarotation (a slow change in optical rotation as \( \alpha \) and \( \beta \) forms interconvert in solution through the open chain).

Option 2

i) Structures of protein:
Primary structure: The exact sequence (order) in which the amino acids are linked to one another in a polypeptide chain through peptide bonds. Any change in this order changes the protein.
Secondary structure: The way the polypeptide chain folds or coils locally, stabilised by hydrogen bonds between the C=O and N–H groups of the backbone. Two common shapes are the \( \alpha \)-helix (coil) and the \( \beta \)-pleated sheet.
Tertiary structure: The overall three-dimensional folding of the whole polypeptide chain, stabilised by hydrogen bonds, disulphide (–S–S–) bridges, ionic (electrostatic) attractions and hydrophobic (van der Waals) forces. It gives fibrous (e.g. keratin) or globular (e.g. insulin) shapes.
Quaternary structure: The arrangement in which two or more polypeptide sub-units come together to form one functional protein, e.g. haemoglobin has four sub-units.

ii) Denaturation of protein:
When a protein is exposed to a physical change (heat, UV radiation) or a chemical change (strong acid or alkali, salts of heavy metals), the hydrogen bonds are broken, the globule unfolds and the helix uncoils. The secondary and tertiary structures are destroyed but the primary structure (the peptide bonds) stays intact. The protein loses its biological activity. Common examples: coagulation of egg white on boiling and curdling of milk.