Question

Match the carbohydrates in Column I with the correct glycosidic linkages involved in Column II.

Column I	Column II
P. Amylose	1. \(\beta(1\rightarrow4)\)
Q. Sucrose	2. \(\alpha(1\rightarrow4)\)
R. Amylopectin	3. \(\alpha1\rightarrow\beta2\)
S. Cellulose	4. \(\alpha(1\rightarrow4), \alpha(1\rightarrow6)\)

• A. P-4; Q-3; R-2; S-1
• B. P-2; Q-1; R-4; S-3
• C. P-2; Q-3; R-4; S-1
• D. P-1; Q-3; R-4; S-2

Hint:

Amylose and cellulose are unbranched (differ only in alpha vs beta 1,4 links), amylopectin adds alpha 1,6 branches, and sucrose's link is alpha1 to beta2 between glucose and fructose.

Answer 1

The Correct Option is C

Step 1: Recall the linkage in each polysaccharide/disaccharide.
Amylose is the unbranched fraction of starch. It is a straight chain of glucose units joined only by \(\alpha(1\rightarrow4)\) glycosidic bonds.
Amylopectin is the branched fraction of starch. Its chain is built with \(\alpha(1\rightarrow4)\) bonds, and a new branch starts every 24 to 30 residues through an \(\alpha(1\rightarrow6)\) bond. So it needs both linkages together.
Cellulose is a straight chain of glucose units too, but the bond is \(\beta(1\rightarrow4)\), which is why it forms rigid, fibrous sheets instead of the helical coils of starch.
Sucrose is not a glucose polymer, it is a disaccharide of glucose and fructose. The bond forms between C1 of glucose (an \(\alpha\) anomeric carbon) and C2 of fructose (a \(\beta\) anomeric carbon), written as \(\alpha1\rightarrow\beta2\). Because both anomeric carbons are tied up in this bond, sucrose is a non-reducing sugar.

Step 2: Match each entry in Column I to Column II.
P. Amylose only has \(\alpha(1\rightarrow4)\) bonds, so P matches item 2.
Q. Sucrose has the \(\alpha1\rightarrow\beta2\) bond, so Q matches item 3.
R. Amylopectin has both \(\alpha(1\rightarrow4)\) and \(\alpha(1\rightarrow6)\) bonds, so R matches item 4.
S. Cellulose has the \(\beta(1\rightarrow4)\) bond, so S matches item 1.

Step 3: Check this against the given options.
The match P-2, Q-3, R-4, S-1 corresponds to option (C). Option (A) swaps P and R incorrectly (gives amylose the branched linkage). Option (B) assigns cellulose the \(\alpha1\rightarrow\beta2\) bond, which belongs to sucrose, and gives sucrose a plain \(\beta(1\rightarrow4)\) bond, which is wrong. Option (D) wrongly gives amylose the \(\beta(1\rightarrow4)\) bond of cellulose.

Final Answer:
P-2; Q-3; R-4; S-1, which is option (C). \[ \boxed{P\text{-}2;\ Q\text{-}3;\ R\text{-}4;\ S\text{-}1} \]