Match List-I with List-II. List - I (Saccharides) List - II (Glycosidic-linkages found) (A) Sucrose (I) α 1 - 4 (B) Maltose (II) α 1 - 4 and α 1 - 6 (C) Lactose (III) α 1 - β 2 (D) Amylopectin (IV) β 1 - 4 Choose the correct answer from the options given below :

(A)-(III), (B)-(I), (C)-(IV), (D)-(II)

(A)-(I), (B)-(II), (C)-(III), (D)-(IV)

(A)-(II), (B)-(IV), (C)-(III), (D)-(I)

(A)-(IV), (B)-(II), (C)-(I), (D)-(III)

Biomolecules PYQ — JEE Main 2025 · 28 Jan · Shift 2 · Q6 with Solution

Match List-I with List-II.

List - I (Saccharides)	List - II (Glycosidic-linkages found)
(A) Sucrose	(I) α 1 $-$ 4
(B) Maltose	(II) α 1 $-$ 4 and α 1 $-$ 6
(C) Lactose	(III) α 1 $-$ β 2
(D) Amylopectin	(IV) β 1 $-$ 4

Choose the correct answer from the options given below :

View written solutionFree

Correct answer: A

Identify the glycosidic linkage in each saccharide:

(A) Sucrose
- Sucrose is formed from $\alpha$ -D-glucose and $\beta$ -D-fructose.
- The linkage is between C-1 of glucose and C-2 of fructose.
- Hence, linkage = $\alpha 1-\beta 2$
- So, (A) $\to$ (III)
(B) Maltose
- Maltose consists of two glucose units.
- The glycosidic bond is $\alpha(1\to 4)$ .
- So, (B) $\to$ (I)
(C) Lactose
- Lactose consists of galactose and glucose.
- The glycosidic bond is $\beta(1\to 4)$ .
- So, (C) $\to$ (IV)
(D) Amylopectin
- Amylopectin is a branched polymer of glucose.
- It contains $\alpha(1\to 4)$ linkages in the main chain and $\alpha(1\to 6)$ at branch points.
- So, (D) $\to$ (II)
Final matching:
$(A)-(III),\quad (B)-(I),\quad (C)-(IV),\quad (D)-(II)$
Compare with options:
- This corresponds to Option A.
Comparison with stored correct answer:
- Stored correct answer = A
- Derived answer = A
- Therefore, they agree.

Match List-I with List-II.

List - I (Saccharides)	List - II (Glycosidic-linkages found)
(A) Sucrose	(I) α 1 $-$ 4
(B) Maltose	(II) α 1 $-$ 4 and α 1 $-$ 6
(C) Lactose	(III) α 1 $-$ β 2
(D) Amylopectin	(IV) β 1 $-$ 4

Choose the correct answer from the options given below :

View written solutionFree

Correct answer: A

Identify the glycosidic linkage in each saccharide:

(A) Sucrose
- Sucrose is formed from $\alpha$ -D-glucose and $\beta$ -D-fructose.
- The linkage is between C-1 of glucose and C-2 of fructose.
- Hence, linkage = $\alpha 1-\beta 2$
- So, (A) $\to$ (III)
(B) Maltose
- Maltose consists of two glucose units.
- The glycosidic bond is $\alpha(1\to 4)$ .
- So, (B) $\to$ (I)
(C) Lactose
- Lactose consists of galactose and glucose.
- The glycosidic bond is $\beta(1\to 4)$ .
- So, (C) $\to$ (IV)
(D) Amylopectin
- Amylopectin is a branched polymer of glucose.
- It contains $\alpha(1\to 4)$ linkages in the main chain and $\alpha(1\to 6)$ at branch points.
- So, (D) $\to$ (II)
Final matching:
$(A)-(III),\quad (B)-(I),\quad (C)-(IV),\quad (D)-(II)$
Compare with options:
- This corresponds to Option A.
Comparison with stored correct answer:
- Stored correct answer = A
- Derived answer = A
- Therefore, they agree.