Q. 24

Differentiate bet

Answer :


(i) Straight chain polymer

(ii) Test with iodine shows blue stains

(iii) 20% water soluble


(i) Branched chain polymer

(ii) Iodine test shows re-brown stains.

(iii) 80-85% water solubility

(iv) has 1-4 glycosidic linkage

(iv) Has α -1,4 and α-1,6-glycosidic linkage.

Peptide linkage

(i) The proteins are linked by peptide bond.

(ii) It consists of ─CONH2─ bond between two molecules.

(iii) Hydrolysis of peptide bond gives two amino acids

Glycosidic linkage

(i) Monosaccharides are linked by glycosidic bond

(ii) The molecules are bonded by (─O─) link.

(iii) Hydrolysis gives two Monosaccharides.

Fibrous proteins

(i) Has linear poly-peptide chain arrangement

(ii) Insoluble in water, e.g. myosin

Globular proteins

(i) Coil arrangement of peptide chains.

(ii) Soluble n water, e.g. albumins, insulin


D-glucose is obtained from hydrolysis or decomposition of sucrose by enzyme invertase. It has a straight chain, five alcohol groups and aldehyde as a carbonyl group as seen in the structure given below:

(i) the reaction of reduction of alcohol groups which results into hexane justifies the straight chain in the structure

(ii) Formation of penta-acetate when treatment with acetic anhydride suggests the presence of 5 alcohol groups.

(iii) Mild oxidation of glucose leads to formation of gluconic acid which can be detected using neutralisation in a basic medium.

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