Stereochemistry from second to fifth carbon is R, S, S, R, which indicates D-galactose. The Haworth structure is a six-membered ring, so the molecule is in its pyranose form. The molecule has its anomeric hydroxyl group pointing down, so it's the alpha anomer.

Stereochemistry from second to fifth carbon is R, S, R, R, which indicates D-glucose. The Haworth structure is a five-membered ring, so the molecule is in its furanose form. The molecule has its anomeric hydroxyl group pointing down, so it's the alpha anomer.

A reducing sugar contains a hemiacetal/hemiketal group which means that in its open chain form it contains a ketone/aldehyde group. Sugars containing a free aldehyde group can be oxidized to a carboxylic acid, while sugars containing a free ketone group must be tautomerized to an aldehyde group through an ene-diol intermediate (shown below), and this can be oxidized to a carboxylic acid.

Reducing sugars are detectable with the formation of either a precipitate or a solution color change after addition of Tollens' Reagent, Benedict's solution, or Fehling's solution. 

Glucose in its open chain form (shown below), has a free aldehyde group (an aldose), and it contains six carbons (a hexose). Together, glucose is shown to be an aldohexose.

The anomeric carbon is formed from the original carbonyl carbon in the straight-chain form of the molecule being attacked by a hydroxyl group to form a hemiacetal. This is seen as a carbon that is bonded to two oxygen atoms. In the case of glucose, the carbon labelled A is a hemiacetal, and is considered to be the anomeric carbon.

In this question, we're given the structure of a sugar molecule, and we're asked to identify which answer choice represents the correct identification of this molecule.

To answer this question, there are two things we need to look at. For one, we need to determine whether it is an aldehyde sugar (aldose) or a ketone sugar (ketose). The first carbon atom in the molecule (shown at the very top in the image) is shown as . This means that the carbon contains a double bond to the oxygen. Furthermore, since it also contains a bond to a hydrogen, we can conclude that this is an aldehyde functional group. Consequently, we know that this must be an aldose sugar.

Next, we need to look at the number of carbon atoms in the molecule. In the image shown in the question, there are a total of seven carbon atoms. Thus, this sugar would be classified as a heptose sugar (seven carbons) rather than a hexose sugar (six carbons).

Putting these two pieces of information together, we know that the sugar is both an aldose and a heptose. Therefore, this sugar is an aldoheptose.

Lysine's R-group includes an amine (-NH₂), which can be ionized by picking up a hydrogen. This ability classifies lysine as basic.

Cysteine's R-group contains a sulfhydryl group (-SH), which can participate in the formation of a disulfide bridge in a protein's tertiary and/or quaternary structure. Cysteine is the only amino acid to contain a sulfur atom.

The only achiral amino acid is Glycene. Glycene's side chain is simply a hydrogen. Because a hydrogen already exists on the fundamental structure of an amino acid backbone, a side chain of a single hydrogen atom causes glycene to be achiral.

The nine essential amino acids are valine, leucine, isoleucine, histidine, phenylalanine, tryptophan, methionine, threonine, and lysine. These amino acids must be consumed in the diet, since they cannot be synthesized by adult humans.