Q. 24

Differentiate bet

Answer :



Amylose


(i) Straight chain polymer


(ii) Test with iodine shows blue stains


(iii) 20% water soluble


Amylopectin


(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




OR


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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