Q.30 Choose the option with two reducing sugars.
(A) Lactose and Maltose (B) Trehalose and Sucrose
(C) Maltose and Trehalose (D) Lactose and Sucrose
Answer: (A) Lactose and Maltose
Reducing sugars possess a free anomeric carbon (hemiacetal group) that can open to form an aldehyde or ketone, enabling reduction of reagents like Benedict’s or Fehling’s solutions. Lactose (Gal-β1,4-Glc) and maltose (Glc-α1,4-Glc) are reducing disaccharides with one free anomeric carbon, while sucrose (Glc-α1,2-Fru) and trehalose (Glc-α1,1-Glc) are non-reducing due to full glycosidic bond involvement of both anomeric carbons.
Option Analysis
(A) Lactose and Maltose: Correct—both have one reducing end (free anomeric OH on glucose unit).
(B) Trehalose and Sucrose: Incorrect—both non-reducing; trehalose links both anomeric carbons α1↔1, sucrose α1↔2β.
(C) Maltose and Trehalose: Incorrect—maltose reduces, trehalose does not (symmetric α1,1 linkage).
(D) Lactose and Sucrose: Incorrect—lactose reduces, sucrose does not (fructose ketose linkage).
Introduction: Two Reducing Sugars Lactose Maltose
Reducing sugars identification distinguishes carbohydrates with free hemiacetal groups in biochemistry exams. This analysis confirms lactose and maltose as the only pair of reducing sugars among options, explaining glycosidic linkages and chemical tests for molecular biology students.
Reducing vs Non-Reducing Classification
| Sugar | Linkage | Reducing End? | Structure Detail |
|---|---|---|---|
| Lactose | β1,4 | Yes | Galactose-β1,4-Glucose (free C1) |
| Maltose | α1,4 | Yes | Glucose-α1,4-Glucose (free C1) |
| Sucrose | α1↔2β | No | Both anomeric carbons linked |
| Trehalose | α1↔1 | No | Symmetric, no free anomeric OH |
Chemical Test Confirmation
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Benedict’s Test: Reducing sugars reduce Cu²⁺→Cu₂O (red ppt); lactose/maltose positive, sucrose/trehalose negative.
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Open-chain form: Lactose/maltose equilibrate to aldehyde; others cannot.
Essential for carbohydrate metabolism and bioprocess engineering exam preparation.
