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Question #44996

Difficulty: 8/10 | 🟢 Free

The Question

An aqueous solution contains an equimolar mixture (1 mole each) of four disaccharides: Sucrose, Lactose, Maltose, and Cellobiose. Four identical aliquots of this solution (each containing 1 mole of all four disaccharides) are subjected to different hydrolytic conditions at 298 K298\text{ K} and optimal pH:

Aliquot I: Treated with Invertase (strictly specific to sucrose). Aliquot II: Treated with β\beta-galactosidase (strictly specific to β\beta-D-galactoside linkages). Aliquot III: Treated with Maltase (strictly specific to α\alpha-(1→\rightarrow4)-D-glucosidic linkages). Aliquot IV: Heated with dilute aqueous \ce{H2SO4} until complete chemical hydrolysis occurs.

Assume all enzymatic and chemical hydrolyses proceed to 100% completion. Let GiG_i be the total moles of free D-glucose molecules present in the final mixture of Aliquot ii. Let RiR_i be the total moles of reducing sugar molecules present in the final mixture of Aliquot ii.

Calculate the value of ∑i=14Gi+∑i=14Ri\sum_{i=1}^{4} G_i + \sum_{i=1}^{4} R_i.

Options

  • A.

    31

  • B.

    34

  • C.

    33

  • D.

    29

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Disaccharide Hydrolysis Products|Enzyme Specificity 'Lock and Key'|Glycosidic Linkage Geometry

Why This Question Is Tricky

Common Mistakes

This problem tests the 'Lock and Key' specificity of enzymes based on stereochemistry (α\alpha vs β\beta glycosidic linkages) and the strict definition of reducing sugars.

The primary trap is assuming Maltase will hydrolyze Cellobiose because both Maltose and Cellobiose are D-glucose dimers with 1,4-linkages.

However, Maltase is strictly specific to α\alpha-(1→\rightarrow4) linkages, while Cellobiose has a β\beta-(1→\rightarrow4) linkage, rendering it immune to Maltase.

Additionally, it tests the knowledge that Fructose (a ketose) is a reducing sugar due to its ability to isomerize via the Lobry de Bruyn-van Ekenstein transformation, and that Sucrose is a non-reducing sugar.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterBiomolecules - Class 12
ConceptDisaccharide Hydrolysis Products
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree