Biomolecules

Riboflavin ( $\mathrm{B}_2$ ) deficiency leads to cheilosis (cracks at mouth corners), glossitis, and digestive issues. Vitamin A is related to vision; C for scurvy; Niacin for pellagra.

Given that, alanine is the C-terminal amino acid and N-terminal must be chiral. So, glycine is not used at the N-terminal because it is achiral. So, possible sequence of tetrapeptides are

Val-Gly-Ser-Ala

Val-Ser-Gly-Ala

Ser-Gly-Val-Ala

Ser-Val-Gly-Ala

Deficiency of riboflavin (vitamin $ B_2 $) can lead to conditions such as cheilosis, which is characterized by the cracking or scaling of the lips, as well as digestive disorders.

A tetrapeptide consists of four amino acids: alanine, serine, glycine, and valine. Given that alanine is the C-terminal amino acid and the N-terminal amino acid must be chiral, we cannot use glycine at the N-terminal position because it is achiral. This means the N-terminal amino acid can be either valine or serine, as both are chiral.

Considering these constraints, we have the following possible sequences for the tetrapeptide:

Val-Gly-Ser-Ala

Val-Ser-Gly-Ala

Ser-Gly-Val-Ala

Ser-Val-Gly-Ala

These sequences respect the condition of having a chiral N-terminal amino acid, with alanine fixed as the C-terminal amino acid.

Carbohydrates which differ in configuration at a carbon other than the anomeric carbon. Glucose and mannose are differ in configuration around $\mathrm{C}_2$ and hence these are $\mathrm{C}_2$ epimers.

To determine which compound is being described, we need to analyze the given chemical tests and their results:

1. Negative test with ninhydrin: Ninhydrin is a chemical reagent commonly used to detect amino acids and proteins. When it reacts with these compounds, it typically produces a deep purple color known as Ruhemann's purple. Therefore, a negative test with ninhydrin implies that the compound does not contain free amino groups, ruling out amino acids (Option A) and proteins (Option D).

2. Positive test with Benedict's solution: Benedict's solution is used to test for the presence of reducing sugars. If the compound is a reducing sugar, it will reduce the copper(II) ions in Benedict's solution to form a brick-red precipitate of copper(I) oxide when heated. Therefore, a positive test with Benedict's solution indicates the presence of a reducing sugar.

Given these observations, the compound must be a reducing sugar. Thus, the correct answer is:

Option B: a monosaccharide.

Hence, three molecules of phenyl hydrazine (C₆H₅NHNH₂) is used to yield osazone.

Maltase converts maltose into glucose by hydrolysis as follows

Maltose $\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{Maltase}^{Hydrolysis}} $ glucose + glucose

Statement (a) is incorrect. Because $\alpha$-D-fructose and $\beta$-D-fructose are anomers, diastereomers and geometrical isomers and not enantiomers.

Rest of the given statements are correct.