Q.27 The product(s) resulting from the hydrolysis of maltose is/are
(A) a mixture of α-D-Glucose and β-D-Glucose
(B) a mixture of D-Glucose and L-Glucose
(C) α-D-Glucose only
(D) β-D-Glucose only
Maltose hydrolysis yields two molecules of D-glucose through cleavage of its α-1,4 glycosidic bond. This process occurs via enzymes like maltase or acid catalysis, fundamental in carbohydrate metabolism.
Correct Answer
The correct option is (A) a mixture of α-D-Glucose and β-D-Glucose. Upon hydrolysis, maltose breaks into two D-glucose units, which equilibrate in solution to form both α and β anomers.
Maltose Structure
Maltose consists of two α-D-glucose units linked by an α-1,4 glycosidic bond between C1 of one glucose and C4 of the other. Hydrolysis adds water across this bond, releasing free glucose molecules that can interconvert between anomeric forms.
Option Explanations
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(A) Correct: Free D-glucose exists mainly as a mixture of α-D (36%) and β-D (64%) forms in equilibrium via mutarotation. Maltose hydrolysis produces this mixture, not exclusively one anomer.
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(B) Incorrect: Glucose in nature is D-glucose; L-glucose is rare and not produced from maltose, a natural D-sugar disaccharide.
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(C) Incorrect: While maltose derives from α-D-glucose, hydrolysis yields free glucose that equilibrates to both α and β forms, not α only.
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(D) Incorrect: β-D-glucose forms part of the equilibrium but is not the sole product; α-D predominates in the ring structure before equilibration.
| Option | Key Reason Incorrect/Correct | Biological Relevance |
|---|---|---|
| (A) | Equilibrium mixture of anomers | Matches solution behavior |
| (B) | No L-glucose in nature | Irrelevant to biology |
| (C) | Ignores mutarotation | Oversimplifies products |
| (D) | Not exclusive product | Partial only |
