Chemistry - Biomolecules Question with Solution | TestHub
LIST I | LIST II |
|---|---|
(P) Sucrose | (1) -(D) glucose unit is present |
(Q) Maltose | (2) Form brick red solid with fehling solution |
(R) Lactose( Milk Sugar) | (3) No anomeric hydroxyl group is present |
(S) Amylose | (4) Glycosidic bond is formed between reducing ends of sugar present in molecule. |
| (5) Even number of chiral carbons are present in each polymeric unit |
Which of the following is correct
Options:
Answer:
Solution:
P (Sucrose): (1) -D-glucose unit, (3) No anomeric OH, (4) Glycosidic bond between reducing ends. (5) 8 chiral carbons (even). So P is (1,3,4,5). Option A has (1,3,4).
Q (Maltose): (1) -D-glucose unit, (2) Reducing sugar, (5) 10 chiral carbons (even). So Q is (1,2,5). Option A has (1,2,5).
R (Lactose): (2) Reducing sugar, (5) 10 chiral carbons (even). So R is (2,5). Option A has (2,5).
S (Amylose): (1) -D-glucose unit. (2) Weakly reducing. So S is (1,2). Option A has (1,2).