Introduction to Biochemistry - End of session 2.6 assessment

Only D is a carbohydrate, it is a ketose.

The top carbon is not a chiral center - it is not $ sp^3 $ hybridized. The bottom carbon is also not a chiral center because it has two hydrogen bonded to it. All other carbons are chiral centers. Therefore there are 4 chiral centers.

D sugars are characterized by having OH on the right of the bottom-most chiral carbon, that is the case for D ane E

The drawing of the molecule is available above - note that by the suffix ulose, we know it is a ketose, the only possible ketose with three carbon is to have a double bond O in the middle carbon, the rest follows.

The carbonyl carbon is always carbon 2 in a ketose. Therefore the answer is C2.

The anomeric carbon is also the carbonyl carbon, which is A.

It has 6 carbon with a 5 member ring, therefore it is fructofuranose.

It is a 1,1 glycosidic linkage. Both carbon involved in the linkage is the anomeric carbon in an aldose.

1, 6 glycosidic linkage, this is what make glycogen branches.

This doner molecule contributes a glucose unit to the non-reducing end of glycogen.

Glycogen has more non-reducing ends than reducing ends.
The majority of glucose monomers found in glycogen are joined by 1,4 glycosidic linkages, while branch points are joined by 1,6 glycosidic linkages.
Glycogen chains are extended by glycogen-synthase, and glycogen synthesis is initiated by glycogenin.