Biomolecules

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

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.

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

• On hydrolysis, both oligopeptides and proteins yield α-amino acids, not β-amino acids.

• Oligopeptides give fewer α-amino acids, while proteins give a large number of α-amino acids.

• Hence, the mention of β-amino acids makes the statement incorrect.

• Denaturation of proteins by acids generally converts water-soluble globular proteins into insoluble coagulated forms.

• Fibrous proteins are already water-insoluble, so they are not converted from soluble to insoluble forms.

• Therefore, the statement is conceptually incorrect.

On hydrolysis of sucrose gives D-(+)-glucose and D-(–)-fructose while in Statement I D-(+)-fructose is given evince Statement–I is incorrect.
Statement–II : It is correct because sucrose on hydrolysis gives invert sugar.

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.

Ascorbic acid is another name for vitamin C.

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.

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

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

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.

- 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.

Maltose is composed of two glucose units . Sucrose has glycosidic linkage between $C_1$ of α-glucose and $C_2$ of β-fructose. . Lactose is composed of galactose and glucose . Amylose has glycosidic linkage between $C_1$ of α-glucose $C_4$ of α-glucose.

- 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.

Let numbers of units be: ribose = n₁, 2-deoxyribose = n₂, glucose = n₃.
Octasaccharide ⇒ n₁ + n₂ + n₃ = 8.

Total mass of monomers after complete hydrolysis:
150 n₁ + 134 n₂ + 180 n₃ = 1024 + 7×18 = 1150 g

Given 2-deoxyribose is 58.26% (w/w) of total monomers:
134 n₂ = 0.5826 × 1150 ≈ 670 ⇒ n₂ = 5

Then n₁ + n₃ = 3 and
150 n₁ + 180 n₃ = 1150 − 670 = 480

Put n₁ = 3 − n₃:
150(3 − n₃) + 180 n₃ = 480 ⇒ 450 + 30 n₃ = 480 ⇒ n₃ = 1, n₁ = 2.

Number of ribose units = 2.

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.

Gly ala val

Gly val ala

Val gly ala

Val ala gly

Ala val gly

Ala gly val

Total tri peptides = 6

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

Two nucleic acid chains are wound about each other and held together by H bonds between pair of
bases. Adenine from two hydrogen bonds with thymine and Guanine form three hydrogen bond with cytosine.
5' G–G–C–A–A–A–T–C–G–G–C–T–A–3'
In given DNA strand total seven guanine and cytosine bases which form total 21 H-bonds and six adenine and thymine base which will form total 12 H-bonds with other DNA strand.
Total no. of H bonds = 7 × 3 + 6 × 2 = 33

• Fehling's test is used to detect reducing sugars. Reducing sugars have a free anomeric carbon (aldehyde or ketone group) that can reduce the blue copper(II) ions ( ${\text{Cu}}^{2+}$ Cu2+ Sv6Kpe[]) in Fehling's solution to red copper(I) oxide ( ${\text{Cu}}_2\text{O}$ Cu2O Sv6Kpe[]) precipitate.

Sucrose is dextrorotatory.
On hydrolysis, it gives glucose and fructose. The mixture is called invert sugar and shows laevorotation because fructose has stronger optical activity.
Therefore, Statement I is true.

Statement II: It says laevorotation of glucose is more than dextrorotation of fructose.

This is incorrect because:

Glucose is dextrorotatory (+)
Fructose is laevorotatory (−)
Magnitude of fructose rotation is greater than glucose

So the mixture becomes laevorotatory due to fructose dominance, not glucose. Thus, Statement II is false.

- 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.

Vitamins A, D, E, and K are fat-soluble vitamins.

Sucrose is dextrorotatory.
On hydrolysis, it gives glucose and fructose. The mixture is called invert sugar and shows laevorotation because fructose has stronger optical activity.
Therefore, Statement I is true.

Statement II: It says laevorotation of glucose is more than dextrorotation of fructose.

This is incorrect because:

Glucose is dextrorotatory (+)
Fructose is laevorotatory (−)
Magnitude of fructose rotation is greater than glucose

So the mixture becomes laevorotatory due to fructose dominance, not glucose. Thus, Statement II is false.

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.

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.

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.

DNA & RNA are chiral molecules due to presence of chiral 2-deoxy-Ribose & Ribose sugar unit respectively.

On hydrolysis of sucrose gives D-(+)-glucose and D-(–)-fructose while in Statement I D-(+)-fructose is given evince Statement–I is incorrect.
Statement–II : It is correct because sucrose on hydrolysis gives invert sugar.

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.

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

Boiling an egg causes denaturation of its protein resulting in loss of its quaternary, tertiary and secondary structures, whereas primary structure remains intact.

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

A. Glucose / NaHCO₃ / Δ
Glucose has no -COOH group, so it cannot react with NaHCO₃ (only carboxylic acids release CO₂).
A $\to$ II (No reaction)
B. Glucose / ΗΝΟ₃
Hot nitric acid is a strong oxidising agent.
It oxidises both the aldehyde group and the primary alcohol group into dicarboxylic acid.
Product = Saccharic acid
B $\to$ IV (Saccharic acid)
C. Glucose / ΗΙ/Δ
Hot HI + red P → complete reduction of glucose (all C-O bonds reduced).
This gives n-hexane.

C $\to$ III (n-hexane)
D. Glucose / Bromine water
Br₂ water is a mild oxidising agent; it oxidises only the aldehyde group gluconic acid.
D$\to$ 1 (Gluconic acid)

A-II, B-III, C-I, D-IV

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

Gly ala val

Gly val ala

Val gly ala

Val ala gly

Ala val gly

Ala gly val

Total tri peptides = 6

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 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.

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

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.

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

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

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

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.

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.

Vitamins A, D, E, and K are fat-soluble vitamins.

Ascorbic acid is another name for vitamin C.

Let's analyze each statement carefully:

(A) Enzymes are biocatalysts →Correct

(B) Enzymes are non-specific and can catalyse different kinds of reactions → Incorrect
→ Enzymes are highly specific, not non-specific.

(C) Most enzymes are globular proteins → Correct

(D) Enzyme oxidase catalyses the hydrolysis of maltose into glucose → Incorrect
→ That reaction is catalyzed by maltase, not oxidase.

Incorrect statements: (B) and (D)

Correct option: B — (B) and (D)

Let's analyze each statement carefully:

(A) Enzymes are biocatalysts →Correct

(B) Enzymes are non-specific and can catalyse different kinds of reactions → Incorrect
→ Enzymes are highly specific, not non-specific.

(C) Most enzymes are globular proteins → Correct

(D) Enzyme oxidase catalyses the hydrolysis of maltose into glucose → Incorrect
→ That reaction is catalyzed by maltase, not oxidase.

Incorrect statements: (B) and (D)

Correct option: B — (B) and (D)

DNA & RNA are chiral molecules due to presence of chiral 2-deoxy-Ribose & Ribose sugar unit respectively.

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

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.

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