All naturally occurring polysaccharides have one terminal residue, which contains a free anomeric carbon. Why do these polysaccharides not give a positive chemical test for a reducing sugar? The chemical test for reducing sugars is based on aldehyde oxidation, which will not work for polysaccharides. The structure of the polysaccharide does not allow for the lactone product of oxidation to form. The reducing ends are diluted out by the many sugar monomers and thus are difficult to detect. A different reaction occurs, the product of which cannot be detected.

Question

Carolyn Meadows

Biology

21 August, 13:45

All naturally occurring polysaccharides have one terminal residue, which contains a free anomeric carbon. Why do these polysaccharides not give a positive chemical test for a reducing sugar? The chemical test for reducing sugars is based on aldehyde oxidation, which will not work for polysaccharides. The structure of the polysaccharide does not allow for the lactone product of oxidation to form. The reducing ends are diluted out by the many sugar monomers and thus are difficult to detect. A different reaction occurs, the product of which cannot be detected.

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Answers (1)

Know the Answer?

Parker Costa · Answer 1

Option (A).

Explanation:

Polysaccharides may be defined as the carbohydrates that can be hydrolyzed into large number of the single units. The polysaccharides are always non reducing in nature.

Although all polysaccharides have the anomeric carbon but they do not have the free aldehyde and ketone group. Their functional group is is involved in the formation of the glycosidic bond. For the sugar to show the reducing test, they must have aldehyde or kteone group and it is absent in case of polysaccharides.

Thus, the correct answer is option (A).