If a branched polysaccharide (e.g. amylopectin) has 7 branch points, how
many free -OH ends are available for reduction?
7
1
6
0
The correct answer is 1. In amylopectin, a branched polysaccharide, only one reducing end exists per molecule regardless of branch points, as branches create additional non-reducing ends while the original chain retains a single free anomeric -OH group capable of reduction.
Reducing Ends in Polysaccharides
Reducing ends feature a free anomeric carbon with a -OH group that can open to form an aldehyde, enabling reduction reactions like those with Fehling’s reagent. Linear polysaccharides like amylose have one reducing end; branching in amylopectin occurs via α-1,6 linkages, preserving just one reducing end at the chain’s start. Each branch point connects new chains but does not generate extra reducing ends, as the branching glucose lacks a free anomeric -OH.
Explanation of Options
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7: Incorrect. This assumes one reducing end per branch, but branches add non-reducing ends only.
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1: Correct. The molecule maintains a single reducing end irrespective of branches (e.g., 7 branch points yield 1).
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6: Incorrect. Might suggest reducing ends equal branches minus one, but topology confirms only one total.
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0: Incorrect. Applies to cyclic forms or fully non-reducing polymers, not amylopectin.
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Branching Structure Impact
Amylopectin’s structure includes a main chain with α-1,4-linked glucose and branches every 24-30 residues via α-1,6 bonds. With 7 branch points, 8 chain segments form (main + 7 branches), but all branch starts are non-reducing, leaving one free -OH at the core reducing end. This design aids enzymatic breakdown, where debranching enzymes access non-reducing ends first.
