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Question #45005
The Question
A non-reducing trisaccharide () is subjected to the following enzymatic and chemical analyses at 298 K and pH 7.4:
-
Complete exhaustive methylation of followed by acid hydrolysis yields equimolar amounts of:
- 2,3,4,6-tetra--methyl-D-galactose
- 2,3,4,6-tetra--methyl-D-glucose
- 2,3,6-tri--methyl-D-glucose
-
Hydrolysis of with lactase (-galactosidase) yields D-galactose and a disaccharide . does not reduce Tollens' reagent and is completely resistant to maltase (-glucosidase).
-
Hydrolysis of with emulsin (-glucosidase) yields D-glucose and a disaccharide . reduces Tollens' reagent and undergoes mutarotation.
Based on these observations, identify the correct IUPAC structure of the trisaccharide .
Options
- A.
-\beta-D-galactopyranosyl-(1\rightarrow4)--\beta-D-glucopyranosyl-(1\leftrightarrow1)-\beta-D-glucopyranoside
- B.
-\beta-D-galactopyranosyl-(1\rightarrow4)--\alpha-D-glucopyranosyl-(1\leftrightarrow1)-\beta-D-glucopyranoside
- C.
-\alpha-D-galactopyranosyl-(1\rightarrow4)--\beta-D-glucopyranosyl-(1\leftrightarrow1)-\alpha-D-glucopyranoside
- D.
-\beta-D-glucopyranosyl-(1\rightarrow4)--\beta-D-galactopyranosyl-(1\leftrightarrow1)-\beta-D-glucopyranoside
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
Enzyme Specificity 'Lock and Key'|Glycosidic Linkage Geometry|Disaccharide Hydrolysis Products
Why This Question Is Tricky
Common Mistakes
The question tests the student's ability to integrate methylation analysis with the strict stereospecificity of glycosidase enzymes.
The 'Hidden Variable' is the anomeric configuration of the (11) linkage.
Students often overlook that a non-reducing disaccharide like can have , , or configurations.
The resistance to maltase (an -glucosidase) acts as the critical negative evidence, proving the absence of an -linkage and confirming the geometry.
Furthermore, it tests the understanding that enzymatic cleavage of a non-reducing sugar can expose a latent anomeric carbon, generating a reducing sugar () that undergoes mutarotation.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Biomolecules - Class 12 |
| Concept | Enzyme Specificity 'Lock and Key' |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |