(A) (a) (i) Denaturation of proteins
Definition : Denaturation is the process in which a protein loses its native three-dimensional structure, and with it its biological activity, when subjected to a change in physical or chemical conditions such as heating or a change in pH.
What is destroyed : the secondary and tertiary structures, as the hydrogen bonds and other weak forces holding the folded shape are broken.
What survives : the primary structure remains intact, since the strong covalent peptide bonds are not broken. The globular protein simply uncoils into a random, fibrous form.
Everyday examples : the coagulation of egg white on boiling; the curdling of milk.
(A) (a) (ii) Oligosaccharides
Definition : Carbohydrates that on hydrolysis give two to ten monosaccharide units are called oligosaccharides.
They are classified further by the number of units produced :
- Disaccharides — 2 units (sucrose, maltose, lactose)
- Trisaccharides — 3 units (raffinose)
- Tetrasaccharides — 4 units (stachyose)
(Carbohydrates giving more than ten units are polysaccharides, e.g. starch, cellulose and glycogen.)
(A) (b) Amylose versus Amylopectin
Both are components of starch and both are polymers of $\alpha$-D-glucose, but they differ in structure :
| Amylose | Amylopectin |
|---|
| Linear, unbranched chain | Highly branched structure |
| Glucose units joined only by $\alpha$-1,4 glycosidic linkages | $\alpha$-1,4 linkages along the chains, with $\alpha$-1,6 linkages at the branch points |
| Makes up about 15-20% of starch | Makes up about 80-85% of starch |
| Water soluble | Water insoluble |
OR
(B) (a) Hydrolysis of a thymine-containing DNA nucleotide
A nucleotide has three components, and complete hydrolysis releases all three :
$$\text{Nucleotide} + H_2O \rightarrow \text{Thymine} + \beta\text{-D-2-deoxyribose} + \text{Phosphoric acid}$$
The three products are :
- Thymine — the nitrogenous base
- $\beta$-D-2-deoxyribose — the pentose sugar (note the deoxy, since this is DNA)
- Phosphoric acid, $H_3PO_4$
Partial hydrolysis would stop at the nucleoside (thymidine) plus phosphoric acid.
(B) (b) Evidence for five $-OH$ groups on different carbons
Evidence : acetylation with acetic anhydride
$$C_6H_{12}O_6 + 5(CH_3CO)_2O \rightarrow \underset{\text{glucose pentaacetate}}{C_6H_7O(OCOCH_3)_5} + 5CH_3COOH$$
Two separate conclusions follow :
How many $-OH$ groups : exactly five acetyl groups are taken up, so glucose must contain five hydroxyl groups.
Why they must be on different carbons : two $-OH$ groups on the same carbon would form a gem-diol, which is unstable and immediately loses water to give a carbonyl group. Since all five survive and are acetylated, each must sit on a separate carbon atom.
(B) (c) Glucose with HCN
The aldehyde group undergoes nucleophilic addition of HCN to give a cyanohydrin :
$$\underset{\text{glucose}}{CHO-(CHOH)_4-CH_2OH} + HCN \rightarrow \underset{\text{glucose cyanohydrin}}{CH(OH)(CN)-(CHOH)_4-CH_2OH}$$
Mechanism : the $CN^-$ ion attacks the electron-deficient carbonyl carbon, and the resulting alkoxide picks up a proton.
What it proves : glucose contains a carbonyl group, since cyanohydrin formation is characteristic of aldehydes and ketones.
