During the formation of glycosidic linkage, the hemiacetal group of one cyclized monosaccharide reacts with the hydroxyl group of a second monosaccharide.
These groups reacts and form acetal, that is the glycosidic linkage.
A sugar is reducing only if it has an exposed hemiacetal/hemiketal group. In A and D, the rightmost carbon expose the hemiacetal group. In C, the rightmost carbon is not a hemiketal group because it doesn't have an hydroxyl group attached to it.
The answer is E - the rightmost carbon has a hemiacetal group that allows it to react with other hydroxyl groups.
The anomeric carbon has number 1 for aldose - this happens only on B.
Remember the anomeric carbon is carbon 2 for ketose, the glycosidic linkage bonds the two anomeric carbon together, therefore it is a 1,2 glycosidic linkage.
The answer is D and E
D is a 1,1 glycosidic linkage, two anomeric carbons of two aldoses are bonded together
E is a 1,2 glycosidic linkage, again, the anomeric carbon of an aldose and the anomeric carbon of a ketose is bonded together.
Glycogen synthase catalyze the formation of the chain, therefore it is 1,4 glycosidic linkage.
Glycogen synthase can join a UDP-Glucose to the non-reducing end of glycogen to elongate the chain.
The reducing end of a glycogen chain has the hemiacetal group.
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