Biomolecules

Statement 1 is true because fructose can give the Tollens’ test due to its ability to form an aldehyde group in an alkaline medium.

Statement 2 should be: When reacted with a base, fructose undergoes tautomerization to form an aldehyde group.

Glucose (being an α‐hydroxy aldehyde in equilibrium)

• does give Tollen’s test (A), react with HCN (C) and form an oxime with NH₂OH (D).

• It typically does not respond to Schiff’s reagent (B) or form a stable bisulphite adduct (E).
  Hence, B and E are the reagents with no characteristic reaction.

Only cysteine dimerizes readily via its –SH group to form a disulfide-linked dimer (cystine), while option (1) ignores amino acids with two chiral centers, option (2) neglects achiral glycine, and option (3) overlooks aspartic acid, which also has a side-chain –COOH.

Animal body stores carbohydrate as glycogen. It is also called animal starch because its structure is similar to amylopectin. It is present in brain, liver and muscles.

Phosphodiester linkage

Deficiency of riboflavin (vitamin $B_2$) can cause cheilosis and digestive disorder.

The C1-C4 glycosidic linkage is present in maltose, lactose, and amylopectin. However, sucrose has a C1-C2 glycosidic bond between glucose (α-1) and fructose (β-2). Thus, sucrose does not contain a C1-C4 linkage.

Correct answer: (D) Sucrose.

The C1-C4 glycosidic linkage is present in maltose, lactose, and amylopectin. However, sucrose has a C1-C2 glycosidic bond between glucose (α-1) and fructose (β-2). Thus, sucrose does not contain a C1-C4 linkage.

Correct answer: (D) Sucrose.

2, 4-Dinitrophenylhydrazine (DNPH) reacts specifically with free aldehydes and ketones to form hydrazones.

Since D-glucose pentaacetate does not have a free aldehyde group, it does not react with DNPH.

Therefore, Statement I is false.

Since glycine (gly) is achiral, alanine (Ala) is the C-terminal amino acid, therefore N terminal amino acid can be valine (val) or serine (ser). In the view of these condition, the following four sequences for the tetrapeptide is possible. (i) Val-gly-Ser-Ala (ii) Val-Ser-gly-Ala (iii) Ser-gly-Val-Ala (iv) Ser-Val-gly-Ala

- Starch, upon acidic hydrolysis, gives glucose molecules.
- Cane sugar (sucrose) is a disaccharide made up of glucose and fructose. Upon hydrolysis, it breaks down into equal amounts of glucose and fructose.
- Milk sugar (lactose) is a disaccharide made of glucose and galactose. Upon hydrolysis, it breaks down into glucose and galactose.
- Amylopectin is a polysaccharide composed of glucose units.
- Amylose, a type of starch, is a polysaccharide made up of glucose units, and upon hydrolysis, it yields only glucose.

The number of tetrapeptides possible involving each of these amino acids (glycine, alanine, valine, leucine); It has three (3) peptides linkage the number of permutations in which they can be arranged
= 4 × 3 × 2 × 1
= 24

$\mathrm{\alpha }$-helix and $\mathrm{\beta }$-pleated sheet structure belong to
secondary protein or 2° protein.

Ascorbic acid is another name for vitamin C.

Vitamin C (also known as ascorbic acid ) is a water-soluble vitamin found in citrus and other fruits, berries and vegetables.

The correct answer is (D) Secondary structure.

The α-helix and β-pleated sheet are secondary structures of proteins, stabilized by hydrogen bonds between the amide (−NH) and carbonyl (−C=O) groups of the peptide backbone. These structures provide proteins with stability and specific folding patterns, distinct from primary (sequence), tertiary (3D folding), and quaternary (subunit arrangement) structures.

The C1-C4 glycosidic linkage is present in maltose, lactose, and amylopectin. However, sucrose has a C1-C2 glycosidic bond between glucose (α-1) and fructose (β-2). Thus, sucrose does not contain a C1-C4 linkage.

Correct answer: (D) Sucrose.

No, glucose does not respond to the 2,4-dinitrophenyl hydrazine (2,4-DNP) test, as it does not contain any free aldehydic group in its cyclic form.

Commercially, Glucose is prepared by hydrolysis of starch by boiling it with dil .sulphuric acid.

$\left({\mathrm{C}}_6{\mathrm{H}}_{10}{\mathrm{O}}_5\right) +{\mathrm{nH}}_2\mathrm{O} \stackrel{\mathrm{dil} {\mathrm{H}}_2{\mathrm{SO}}_4/373\mathrm{K}}{\to } {\mathrm{nC}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$

The carbohydrate Ribose present in DNA is a pentose sugar, exists in the D-configuration, and functions as a reducing sugar when free. Ribose exists mainly in the furanose (5-membered ring) form.

Valine, Lysine, and Threonine are essential amino acids, whereas Proline and Tyrosine are non-essential.