Question

A molecule of amylopectin contains 1500 glucose residues and is branched after every 30 residues. How many reducing ends are there?

• A. 0
• B. 1
• C. 2
• D. 5

Hint:

For any linear or branched polysaccharide chain: Reducing Ends = 1; Non-reducing Ends = Number of branches + 1. Calculation: 1500/30 = 50 branches, so 51 non-reducing ends but only 1 reducing end!

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A reducing end of a carbohydrate is the end of the polymer chain that has a free anomeric carbon (C-1) not involved in a glycosidic bond. This free anomeric carbon can exist in an open-chain form with a reactive aldehyde or ketone group, allowing it to reduce reagents like Fehling's or Benedict's solution.
Key Formula or Approach:
The structure of branched polysaccharides like amylopectin or glycogen is key. These molecules grow by adding new glucose units to the non-reducing ends. No matter how many branches are formed, they all originate from a single starting glucose unit.

Step 2: Detailed Explanation:
In amylopectin, glucose units are linked by \(\alpha\)(1\(\rightarrow\)4) glycosidic bonds in the linear portions and \(\alpha\)(1\(\rightarrow\)6) bonds at the branch points. When a branch is formed, the C-1 (anomeric carbon) of the first glucose in the new branch is bonded to the C-6 of a glucose unit in the main chain. Therefore, the anomeric carbons of all glucose residues in the branches, as well as all but one in the main chain, are occupied in glycosidic bonds. Only the very first glucose molecule at the "base" of the entire tree-like structure has a free C-1 anomeric carbon. This means that every single amylopectin molecule, regardless of its size (1500 residues) or the frequency of its branching (every 30 residues), has exactly one reducing end and many non-reducing ends (one at the tip of every branch).

Step 3: Final Answer:
A single molecule of amylopectin has exactly one reducing end at its structural origin.