A disaccharide has an $\alpha$-(1→4) glycosidic linkage. Which statement about this sugar is true?
A) Both monosaccharides in the disaccharide are linked at their anomeric carbon.
B) Both monosaccharides in the disaccharide are the $\alpha$ anomer.
C) No part of the disaccharide can undergo mutarotation.
D) The closed ring form of the sugar linked at its carbon-4 is in equilibrium with its open-chain form.
E) All of these statements are true.

Question

Quizplus · Accepted Answer

The glycosidic linkage between the two monosaccharides is between the anomeric carbon of the F1F1F1S1F1F1F1 monosaccharide and the carbon-4 of the F1F1F10 monosaccharide, indicating that the closed ring form of the sugar linked at its carbon-4 is in equilibrium with its open-chain form. Options A and B are incorrect as only one monosaccharide is the F1F1F1S1F1F1F1 anomer and is linked to the other monosaccharide at its carbon-4. Option C is incorrect because the F1F1F1S1F1F1F10 monosaccharide can undergo mutarotation around its anomeric carbon.

A Disaccharide Has An α\alphaα -(1→4) Glycosidic Linkage α\alphaα Anomer C) No Part of the Disaccharide Can Undergo Mutarotation

A Disaccharide Has An $\alpha$ -(1→4) Glycosidic Linkage $\alpha$ Anomer
C) No Part of the Disaccharide Can Undergo Mutarotation