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.

Introduction to Biochemistry - Quiz 2.6.4

Glucose is a reducing sugar, therefore we have a clear solution with methylene blue in reduced state.

With a 1, 4 glycosidic linkage, the sugar is still a reducing sugar, therefore we have a clear solution with methylene blue in reduced state.

With a 1,1 glycosidic linkage, the sugar is non-reducing, therefore we have a blue solution with methylene blue in oxidized state.

Introduction to Biochemistry - Quiz 2.6.3

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.

Introduction to Biochemistry - Quiz 2.6.2

If glucose has a reducing aldehyde group, it should re-color the Schiff reagent, therefore the answer is

Glucose does not have a reducing aldehyde group.

The carbons are numbered from top to bottom. The alcohol group attached to carbon 5 is going to attack the carbonxyl carbon C1 to form hemiacetal.

That way the ring would be a 6-member ring and I believe that has least strain.

The anomeric carbon is the carbonxyl carbon, therefore the answer is C.

Depending on whether the C5 hydroxyl group attacks the front or back of the anomeric carbon, it will yield different isomers.

Pyran is a 6 member ring and furan is a 5 member ring. The answer is the number of carbon and oxygen atom in the ring.

Introduction to Biochemistry - Quiz 2.6.1

They contain multiple hydroxyl groups and they contains an aldehyde or ketone group.

Ribose and Glucose are monosaccharides.

Ketose must contain a ketone group, a ketone group is a C double bond O NOT at the end of the carbon chain, C is the only molecule with that structure.

$ C_1 $ has two hydrogens - that cannot be a chiral center
$ C_2 $ is $ sp^2$ hybridized - that cannot be a chiral center
$ C_4 $ has two hydrogen - that cannot be a chiral center

$ C_3 $ is a chiral center - its four bonds links to four different groups.

Fischer convention use (D) and (L) designations, it is done by drawing the carbohydrate with the aldehyde/ketone group on the top and then see if the lowest chiral carbon has hydroxyl group on left or right under the Fischer's projection. Note that this has nothing to do with the sugar's optical property. It is just a convention.

The lowest chiral carbon has hydroxyl group of the left, therefore it is a L sugar. It is an aldehyde, therefore the answer is L-idose.

I think this is the answer - but edx gives me a wrong answer - it thinks the middle OH bond is on the same plane. I just do not agree - I believe that middle carbon is $ sp^3 $ hybridized, so there is no way for 3 bonds to be on the same plane.