JEE Biomolecules Questions

Nylon 6,6 is prepared by condensation of hexamethylenediamine and adipic acid.
Only primary amines give the carbylamine test, therefore the monomer $$X$$ must be the diamine:

$$X \;=\; NH_2\!-\!(CH_2)_6\!-\!NH_2 \;$$ (hexamethylenediamine)

Each molecule of $$X$$ contains two nitrogen atoms.

Number of moles of $$X$$ taken  =  10

Moles of nitrogen atoms present
$$= 10 \times 2 = 20 \text{ mol of N-atoms}$$

In the Dumas method all the nitrogen is converted to molecular nitrogen: $$2\,\text{N-atoms} \rightarrow N_2$$.
Hence, moles of $$N_2$$ formed
$$= \frac{20}{2} = 10 \text{ mol}$$

Mass of $$N_2$$ evolved
$$= 10 \text{ mol} \times 28 \text{ g mol}^{-1} = 280 \text{ g}$$

Therefore, the amount of nitrogen gas obtained is 280 g.

Let the octasaccharide contain $$r$$ ribose units, $$d$$ 2-deoxyribose units and $$g$$ glucose units.

Because it is an octasaccharide, the total number of monosaccharide residues is
$$r + d + g = 8 \qquad -(1)$$

During hydrolysis all the glycosidic bonds break. In a linear chain of 8 units there are $$8-1 = 7$$ bonds, each bond having been formed by the loss of one water molecule (molar mass $$18\text{ g mol}^{-1}$$).
Hence,

total mass of isolated monosaccharides, $$M = (\text{molar mass of octasaccharide}) + 7 \times 18$$
$$M = 1024 + 126 = 1150\text{ g mol}^{-1} \qquad -(2)$$

The masses contributed by the individual monosaccharides are

$$M = 150r + 134d + 180g \qquad -(3)$$

The question states that 2-deoxyribose accounts for 58.26 % of the total mass of the hydrolysis products, i.e.

$$\frac{134d}{M} = 0.5826 \qquad -(4)$$

Substituting $$M = 1150$$ from (2) into (4):
$$134d = 0.5826 \times 1150 = 669.99 \approx 670$$
$$d = \frac{670}{134} = 5 \qquad -(5)$$

Insert $$d = 5$$ into (1):
$$r + g = 8 - 5 = 3 \qquad -(6)$$

Insert $$d = 5$$ into (3) and use $$M = 1150$$:
$$150r + 134(5) + 180g = 1150$$
$$150r + 670 + 180g = 1150$$
$$150r + 180g = 480$$
Divide by 30:
$$5r + 6g = 16 \qquad -(7)$$

Solve simultaneous equations (6) and (7).
From (6): $$r = 3 - g$$
Substitute in (7):
$$5(3 - g) + 6g = 16$$
$$15 - 5g + 6g = 16$$
$$15 + g = 16$$
$$g = 1$$
Therefore $$r = 3 - 1 = 2$$.

Thus, the number of ribose units present in one molecule of the octasaccharide is 2.

P is glucose. Glucose possesses a free $$\text{-CHO}$$ (aldehydic) group in the open-chain form. Hence it behaves as a reducing sugar and gives Tollens’/Fehling’s/Benedict’s test, producing a silver mirror or red $$Cu_2O$$ precipitate. This description corresponds to observation 2.

Q is cellulose. Cellulose is a straight-chain (linear) homopolysaccharide made up solely of $$\beta$$-D-glucopyranose units joined by $$\beta\,(1\!\rightarrow\!4)$$ glycosidic bonds. “Linear polymer of $$\beta$$-D-glucose” matches observation 5.

R is glycine. Among all naturally occurring $$\alpha$$-amino acids, glycine is the only one whose side chain is just a hydrogen atom. Therefore its $${\rm C_\alpha}$$ is not chiral and the molecule is optically inactive. This is exactly observation 1.

S is starch. Starch (because of its amylose component) forms an intense blue-black complex when treated with iodine solution. Hence it fits observation 3.

Thus the correct matching is
$$P\to2,\;Q\to5,\;R\to1,\;S\to3$$

Option B is therefore the right choice.

We analyze both statements about glucose derivatives and starch.

Statement I: D-glucose pentaacetate reacts with 2,4-dinitrophenylhydrazine (2,4-DNP).

2,4-DNP reacts with the carbonyl group ($$C=O$$) of aldehydes and ketones. In D-glucose pentaacetate, all five hydroxyl groups (including the one at C-1) are acetylated. The C-1 hydroxyl is converted to an acetate ester, meaning the hemiacetal/open-chain aldehyde form is blocked. Since there is no free aldehyde group available, D-glucose pentaacetate does NOT react with 2,4-DNP. Statement I is FALSE.

Statement II: Starch, on heating with concentrated sulfuric acid at 100 degrees C and 2-3 atm pressure, produces glucose.

Starch undergoes acid-catalyzed hydrolysis. When heated with dilute or concentrated sulfuric acid under the given conditions, the glycosidic bonds in starch break down to yield glucose (the monomer unit). Statement II is TRUE.

The answer is Option A) Statement I is false but Statement II is true.

Let us analyze both statements:

Statement I: Fructose does not contain an aldehydic group but still reduces Tollen's reagent.

Fructose is a ketose sugar (contains a keto group, not an aldehyde group). However, in basic medium (Tollen's reagent provides a basic environment), fructose undergoes tautomerism and can reduce Tollen's reagent. This statement is true.

Statement II: In the presence of base, fructose undergoes rearrangement to give glucose.

In basic medium, fructose undergoes Lobry de Bruyn-van Ekenstein transformation, converting to glucose (and mannose) through an enediol intermediate. This statement is true.

Both statements are true, and Statement II correctly explains Statement I (the base in Tollen's reagent converts fructose to glucose, which then reduces the reagent).

The correct answer is Option 1: Both Statement I and Statement II are true.

Proteins and peptides are condensation polymers of $$\alpha$$-amino acids, that is, every constituent amino acid possesses its $$-NH_2$$ group on the carbon atom next to the carboxyl group (the $$\alpha$$-carbon).

Case 1: Checking Statement I
Oligopeptides contain only a few peptide (amide) linkages, so on complete hydrolysis they furnish only a small number of constituent amino acids. Full-sized proteins are very long chains, so their hydrolysis yields a large number of amino-acid units. However, in every case the amino acids obtained are $$\alpha$$-amino acids, never $$\beta$$-amino acids (where the $$-NH_2$$ group would be on the second carbon away from $$-COOH$$).
Hence the phrase “large number of $$\beta$$-amino acids” makes Statement I wrong.

Case 2: Checking Statement II
Acids, bases, heat, or heavy-metal ions can denature proteins. During denaturation a water-soluble globular protein loses its native three-dimensional shape and often precipitates as an insoluble fibrous mass. Fibrous proteins are themselves normally water insoluble. Therefore acid treatment converts the water-soluble form of a globular protein into an insoluble fibrous form, not “the water-soluble form of fibrous proteins”. Hence Statement II is also wrong.

Both statements are incorrect, so the correct choice is Option C.

The sugar unit present in DNA is $$\mathbf{2\text{-}deoxyribose}$$.

Step 1: Count of carbon atoms.
2-Deoxyribose contains five carbon atoms, so it is a pentose. ⇒ Statement A is correct.

Step 2: Ring size in nucleic acids.
Inside DNA the sugar occurs as a five-membered $$\mathbf{furanose}$$ ring, not a six-membered pyranose ring. ⇒ Statement B is incorrect.

Step 3: Absolute (D/L) configuration.
Naturally occurring ribose and deoxyribose possess the configuration related to $$\mathbf{D\text{-}glyceraldehyde}$$. ⇒ Statement C is correct.

Step 4: Reducing property when free.
A sugar is reducing if it has a free hemiacetal (aldehyde) group. Free 2-deoxyribose has such a group, so it reduces Tollens’ and Fehling’s reagents. ⇒ Statement D is correct.

Step 5: Anomeric configuration in DNA.
The base is attached to the anomeric carbon through a $$\mathbf{\beta}$$-N-glycosidic bond. Hence the sugar in DNA is the $$\beta$$-anomer, not the $$\alpha$$-anomer. ⇒ Statement E is incorrect.

Collecting the correct statements: A, C and D.

Therefore the correct option is Option B.

Vitamins are usually grouped on the basis of their solubility:

• Fat-soluble vitamins = $$A, D, E, K$$
• Water-soluble vitamins = All $$B$$-group vitamins and vitamin $$C$$

Comparing this fact with the vitamins listed in the question:

$$\begin{array}{ll} \text{A.}&\; \text{Vitamin } B_1 \;(\text{water-soluble})\\[2pt] \text{B.}&\; \text{Vitamin } C \;(\text{water-soluble})\\[2pt] \text{C.}&\; \text{Vitamin } E \;(\textbf{fat-soluble})\\[2pt] \text{D.}&\; \text{Vitamin } B_{12} \;(\text{water-soluble})\\[2pt] \text{E.}&\; \text{Vitamin } K \;(\textbf{fat-soluble}) \end{array}$$

Therefore, among the five choices given, only Vitamin E (choice C) and Vitamin K (choice E) are fat-soluble.

The option that contains C & E only is Option D.

Hence, the correct answer is Option D (C & E Only).

To determine the correct dipeptide X, we first identify the amino acids Y and Z obtained on hydrolysis.

The first clue states that amino acid Y produces lactic acid on treatment with aqueous $$\mathrm{HNO_2}$$.

Nitrous acid converts the primary amino group $$\left(-NH_2\right)$$ of an aliphatic amino acid into a hydroxyl group $$\left(-OH\right)$$.

Since the product formed is lactic acid,

$$\mathrm{CH_3-CH(OH)-COOH},$$

the original amino acid must have been

$$\mathrm{CH_3-CH(NH_2)-COOH},$$

which is alanine.

Hence,

• $$Y =$$ Alanine.

The second clue states that amino acid Z forms the given cyclic compound on heating.

When $$\alpha$$-amino acids are heated, two molecules condense to form a six-membered cyclic diamide known as a diketopiperazine.

From the structure of the cyclic product, the carbon atoms between the nitrogen and carbonyl groups are simple $$CH_2$$ units with no alkyl substituents.

This is possible only if the amino acid is glycine,

$$\mathrm{H_2N-CH_2-COOH}.$$

Therefore,

• $$Z =$$ Glycine.

The required dipeptide X is obtained by condensation of alanine and glycine.

Hence,

• $$Y =$$ Alanine
• $$Z =$$ Glycine
• $$X =$$ Alanine-Glycine

Therefore, the correct answer is Alanine-Glycine.

The α-helix and β-pleated sheet structures of protein are associated with the secondary structure of proteins.

Primary structure: sequence of amino acids

Secondary structure: local folding patterns (α-helix, β-pleated sheet) stabilized by hydrogen bonds between peptide bonds

Tertiary structure: overall 3D shape

Quaternary structure: arrangement of multiple polypeptide subunits

The correct answer is Option 3: secondary structure.

Essential amino acids are those that cannot be synthesized by the human body and must be obtained from the diet.

The essential amino acids are: Valine, Leucine, Isoleucine, Lysine, Methionine, Phenylalanine, Threonine, Tryptophan, and Histidine.

Checking each option:

(A) Valine - Essential ✓

(B) Proline - Non-essential ✗

(C) Lysine - Essential ✓

(D) Threonine - Essential ✓

(E) Tyrosine - Non-essential (conditionally essential) ✗

The correct answer is Option 2: (A), (C) and (D) only.

We need to identify which acid is a vitamin.

Analyzing each option:

(A) Adipic acid - This is a dicarboxylic acid ($$HOOC(CH_2)_4COOH$$) used in nylon production. It is NOT a vitamin.

(B) Ascorbic acid - This is Vitamin C, an essential nutrient required for tissue repair and enzymatic production.

(C) Saccharic acid - This is glucaric acid, a sugar acid. It is NOT a vitamin.

(D) Aspartic acid - This is an amino acid. It is NOT a vitamin.

The correct answer is Option 2: Ascorbic acid (Vitamin C).

We test each statement one by one, recalling the standard structures and properties of the 20 common (protein-forming) amino acids.

Case A: “All naturally occurring amino acids except glycine contain one chiral centre.”
Glycine has the structure $$NH_2\!-\!CH_2\!-\!COOH$$; since the $$\alpha$$-carbon bears two identical $$H$$ atoms, it is achiral.
However, amino acids such as isoleucine and threonine possess two chiral centres each (one at the $$\alpha$$-carbon and another in the side chain). Thus the phrase “one chiral centre” is wrong. Hence statement A is incorrect.

Case B: “All naturally occurring amino acids are optically active.”
Because glycine is achiral, it is not optically active. Therefore this statement is also incorrect.

Case C: “Glutamic acid is the only amino acid that contains a -COOH group at the side chain.”
Glutamic acid has the side chain $$-CH_2-CH_2-COOH$$, but aspartic acid also contains an extra carboxyl group in its side chain $$-CH_2-COOH$$. Thus glutamic acid is not unique in this respect, so statement C is incorrect.

Case D: “Amino acid, cysteine easily undergo dimerization due to the presence of free SH group.”
Cysteine contains the side chain $$-CH_2-SH$$. Two cysteine molecules readily oxidise to form a disulfide linkage:
$$2 \,HS-CH_2-CH(NH_2)COOH \;\longrightarrow\; HOOCCH(NH_2)-CH_2-S-S-CH_2-CH(NH_2)COOH + 2H^+ + 2e^-$$
The dimer formed is called cystine. Hence statement D is correct.

Therefore the correct choice is Option D (Option 4).

An α-amino acid contains two main acid-base centres:
• the carboxylic group $${-COOH}$$ with $$pK_a \approx 2$$ (behaves as an acid)
• the amino group $${-NH_2}$$ with $$pK_a \approx 9.5$$ (behaves as a base)

General protonation rule: the group is protonated when $$\text{pH} \lt pK_a$$ and de-protonated when $$\text{pH} \gt pK_a$$.

Case 1:

$$\text{pH}=2$$ is lower than the amino group’s $$pK_a$$ but equal to the carboxyl group’s $$pK_a$$.
• Carboxyl group: at (or just below) its $$pK_a$$ it stays protonated as $$-COOH$$.
• Amino group: since $$2 \lt 9.5$$ it is fully protonated to $$-NH_3^+$$.

Therefore structure $$A$$ is $$H_3^+N-CH\bigl(CH(CH_3)_2\bigr)-COOH$$, written shortly as $$H_3^+N-CH-COOH$$.

Case 2:

$$\text{pH}=10$$ is higher than both $$pK_a$$ values.
• Carboxyl group: $$10 \gt 2$$ ⇒ completely de-protonated to $$-COO^-$$.
• Amino group: $$10 \gt 9.5$$ ⇒ largely de-protonated to the neutral $$-NH_2$$ form.

Hence structure $$B$$ is $$H_2N-CH\bigl(CH(CH_3)_2\bigr)-COO^-$$, written briefly as $$H_2N-CH-COO^-$$.

Comparing with the given options:

Option A A = $$H_3^+N-CH-COOH$$, B = $$H_2N-CH-COO^-$$

This matches the derived forms.

So, the correct choice is Option A.

The dipeptide Gly-Ala is written as
$$\text{NH}_2{-}CH_2{-}CO{-}NH{-}CH(CH_3){-}COOH$$

Key observations about this structure:

1. The N-terminal (left-hand) residue is Gly (glycine); its $$\text{NH}_2$$ group is free.
2. The peptide (amide) bond is between the carboxyl group of Gly and the amino group of Ala.
3. The carboxyl group of Ala remains free at the C-terminal end.

To prepare this dipeptide in one step with elimination of $$HCl$$, we need the following.

• The carboxyl function of Gly must be activated as an acyl chloride so it can supply the $$\text{CO}$$ part of the peptide bond and furnish the $$\text{Cl}$$ that leaves as $$HCl$$.
• The amino group of Ala must be available as a nucleophile (unprotected $$NH_2$$) so it can attack the activated acyl chloride and form the $$CONH$$ linkage.

Let us inspect each option in this light.

Case A: $$NH_2{-}CH_2{-}COCl$$ (Gly, acyl chloride) and $$NH_2{-}CH(CH_3){-}COOH$$ (Ala, free amine)
• Gly provides the acyl chloride; Ala provides the free $$NH_2$$.
• Nucleophilic acyl substitution gives
$$NH_2{-}CH_2{-}COCl + NH_2{-}CH(CH_3){-}COOH \;\longrightarrow\; NH_2{-}CH_2{-}CO{-}NH{-}CH(CH_3){-}COOH + HCl$$
This product is exactly Gly-Ala.

Case B: $$NH_2{-}CH_2{-}COCl$$ (Gly, acyl chloride) and $$NH_3^+{-}CH(CH_3){-}COCl$$ (Ala, acyl chloride)
• Both molecules are acyl chlorides; there is no free $$NH_2$$ to act as nucleophile.
• Peptide bond formation cannot occur directly, so this pair is unsuitable.

Case C: $$NH_2{-}CH_2{-}COOH$$ (Gly, free amine) and $$NH_2{-}CH(CH_3){-}COCl$$ (Ala, acyl chloride)
• Here the acyl part comes from Ala and the amine from Gly, giving
$$NH_2{-}CH(CH_3){-}CO{-}NH{-}CH_2{-}COOH$$
This is the dipeptide Ala-Gly, not Gly-Ala.

Case D: Both reactants are simple amino acids with carboxyl groups (no acyl chloride).
• Their direct condensation would eliminate $$H_2O$$, not $$HCl$$, and the reaction is not favoured under the stated condition.

Only Case A fulfils all requirements and yields the desired Gly-Ala dipeptide with elimination of $$HCl$$.

Hence, the correct choice is Option A.

We are asked to find how many different tetrapeptide sequences can be formed when the peptide is made of exactly four different amino acids: glycine (Gly), alanine (Ala), valine (Val) and leucine (Leu), each appearing exactly once.

Key idea: A peptide sequence is an ordered arrangement of the constituent amino acids. Hence, the problem reduces to counting the number of ways to arrange four distinct objects in a line.

For $$n$$ distinct objects, the number of possible linear arrangements (permutations) is given by the factorial formula$$n!$$.

Here $$n = 4$$, so

$$\text{Number of sequences} = 4! = 4 \times 3 \times 2 \times 1 = 24$$

Therefore, $$24$$ different tetrapeptide sequences are possible.

Hence, the correct choice is Option D.

The general acid hydrolysis reaction of a glycosidic bond is given by $$ \text{R1-O-R2} + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{R1-OH} + \text{R2-OH} $$.

Below we examine the hydrolysis products of each option.

Option (A): Starch, composed of amylose and amylopectin, yields only glucose upon acidic hydrolysis and is therefore incorrect.

Option (B): Cane sugar (sucrose) hydrolyzes to equal amounts of glucose and fructose.
$$ \text{sucrose} + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{glucose} + \text{fructose} $$ This option is correct $$-(1)$$.

Option (C): Milk sugar (lactose) hydrolyzes to glucose and galactose.
$$ \text{lactose} + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{glucose} + \text{galactose} $$ This option is correct $$-(2)$$.

Option (D): Amylopectin yields only glucose units upon hydrolysis and does not produce fructose, making it incorrect.

Option (E): Amylose hydrolyzes to glucose units only and is therefore correct.

Collecting the correct conversions gives Options (B), (C), and (E).

Final Answer: Option B

The given strand runs from the 5′-end to the 3′-end as
$$5' - G - G - C - A - A - A - T - C - G - G - C - T - A - 3'$$

To count the hydrogen bonds we need two facts:
1. $$G$$ pairs with $$C$$ through $$3$$ hydrogen bonds.
2. $$A$$ pairs with $$T$$ through $$2$$ hydrogen bonds.

Step 1 Count each type of base in the given strand:

  • $$G$$ appears $$4$$ times (positions 1, 2, 9, 10).
  • $$C$$ appears $$3$$ times (positions 3, 8, 11).
  • $$A$$ appears $$4$$ times (positions 4, 5, 6, 13).
  • $$T$$ appears $$2$$ times (positions 7, 12).

Step 2 Find the number of base pairs of each kind:

  • Total $$G\text{-}C$$ pairs $$= 4 + 3 = 7$$.
  • Total $$A\text{-}T$$ pairs $$= 4 + 2 = 6$$.

Step 3 Multiply by the corresponding hydrogen bonds:

  • Hydrogen bonds from $$G\text{-}C$$ pairs $$= 7 \times 3 = 21$$.
  • Hydrogen bonds from $$A\text{-}T$$ pairs $$= 6 \times 2 = 12$$.

Step 4 Add the two contributions:
$$\text{Total hydrogen bonds} = 21 + 12 = 33$$

Hence, the DNA double helix formed with this strand will have $$33$$ hydrogen bonds.

Aspartame is named L-aspartyl-L-phenylalanine-1-methyl ester. The systematic name already tells us how every group is connected, so we decode it step by step.

1. “L-aspartyl” means the N-terminal amino acid is L-aspartic acid. Aspartic acid contains
  • an α-amino group $$NH_2$$ attached to $$CH$$ (the α-carbon),
  • an α-carboxyl group $$COOH$$, and
  • a side-chain carboxyl group $$CH_2COOH$$.

2. “L-phenylalanine” is the second amino acid. Its α-carbon carries
  • an α-amino group (but this is consumed in peptide bonding),
  • a benzyl side chain $$CH_2\!\!\!- \!\!\!\bigcirc$$, and
  • an α-carboxyl group $$COOH$$.

3. “Methyl ester” tells us that the carboxyl group of phenylalanine is esterified with methanol, converting $$COOH$$ into $$COOCH_3$$.

Putting these facts together:

• The peptide bond must be between the α-carboxyl of aspartic acid and the α-amino of phenylalanine (standard N→C direction in peptides).
• The side-chain carboxyl of aspartic acid remains free as $$COOH$$.
• The phenylalanine α-carboxyl is not a free acid; it is present as $$COOCH_3$$.

Only one of the given structures shows simultaneously
  ✔ a peptide linkage $$-CO-NH-$$ between the correct α-centres,
  ✔ a free second $$COOH$$ on the aspartyl side chain, and
  ✔ the $$COOCH_3$$ ester on the phenylalanine end,
with both chiral centres having the L-configuration. That structure is shown in Option B.

Hence, the correct structure of aspartame is given in Option B.

Answer: Option B which is: the L-aspartyl-L-phenylalanine-1-methyl ester structure

The given strand of the double-stranded DNA is
$$5' \; A \; G \; T \; C \; A \; C \; G \; T \; A \; A \; G \; T \; C \; 3'$$

Step 1 : Count the number of each nucleotide in this strand.

A (adenine) = 4 (at positions 1, 5, 9, 10)
T (thymine) = 3 (at positions 3, 8, 12)
G (guanine) = 3 (at positions 2, 7, 11)
C (cytosine) = 3 (at positions 4, 6, 13)

Step 2 : Identify the number of base pairs.

An A on one strand pairs with T on the complementary strand, and vice-versa; this gives an A-T base pair wherever the original strand has A or T. Similarly, G pairs with C, giving a G-C base pair wherever the original strand has G or C.

Number of A-T base pairs $$= 4 + 3 = 7$$
Number of G-C base pairs $$= 3 + 3 = 6$$

Step 3 : Convert “energy per H-bond” to “energy per base pair”.

Each A-T pair contains 2 H-bonds, energy per bond $$= 1.0\; \text{kcal}\, \text{mol}^{-1}$$ Energy per A-T pair $$= 2 \times 1.0 = 2.0\; \text{kcal}\, \text{mol}^{-1}$$

Each G-C pair contains 3 H-bonds, energy per bond $$= 1.5\; \text{kcal}\, \text{mol}^{-1}$$ Energy per G-C pair $$= 3 \times 1.5 = 4.5\; \text{kcal}\, \text{mol}^{-1}$$

Step 4 : Calculate the total energy required to break all H-bonds in the double strand.

Total energy $$= (7 \times 2.0) + (6 \times 4.5)$$
$$= 14.0 + 27.0$$
$$= 41.0\; \text{kcal}\, \text{mol}^{-1}$$

Hence, the energy required to separate the two DNA strands is 41 kcal mol−1.

We need to identify the type of amino acids obtained by hydrolysis of proteins.

Understand the structure of proteins.

Proteins are polymers of amino acids linked by peptide bonds ($$-CO-NH-$$). The peptide bond is formed between the $$\alpha$$-carboxyl group ($$-COOH$$) of one amino acid and the $$\alpha$$-amino group ($$-NH_2$$) of another.

Identify the type of amino acids.

In $$\alpha$$-amino acids, the amino group ($$-NH_2$$) is attached to the $$\alpha$$-carbon (the carbon atom directly adjacent to the carboxyl group). The general structure is:

$$H_2N - \underset{|}{C}H(R) - COOH$$

where $$R$$ is the side chain that varies between different amino acids.

Why $$\alpha$$-amino acids specifically?

All naturally occurring amino acids that make up proteins are $$\alpha$$-amino acids (with the exception of proline, which is technically an $$\alpha$$-imino acid). While $$\beta$$, $$\gamma$$, and $$\delta$$-amino acids exist (where the amino group is on the 2nd, 3rd, or 4th carbon from the carboxyl group respectively), they are not found in natural proteins.

The correct answer is Option (2): $$\alpha$$.

Coagulation of egg on heating is a common example discussed in the context of protein chemistry.

Key concept: Proteins have four levels of structure — primary (peptide bonds), secondary (hydrogen bonds forming alpha-helix, beta-sheets), tertiary (3D folding), and quaternary. When a protein is heated, the secondary, tertiary, and quaternary structures are disrupted, but the primary structure (peptide bonds) remains intact. This process is called denaturation.

When an egg is heated:

1. The proteins (mainly albumin) in the egg white undergo denaturation.

2. The hydrogen bonds, disulfide bonds, and hydrophobic interactions maintaining the native protein structure are broken.

3. The protein unfolds and aggregates, causing the egg white to become opaque and solid (coagulation).

4. The peptide bonds in the primary structure are NOT broken during this process.

The biological activity of the protein is lost upon denaturation.

The correct answer is Option (2): Denaturation of protein occurs.

We need to find the complementary strand for the DNA sequence ATGCTTCA. In DNA, the complementary base pairs are $$A \leftrightarrow T$$ (Adenine pairs with Thymine) and $$G \leftrightarrow C$$ (Guanine pairs with Cytosine).

For the given strand oriented 5' to 3' as A - T - G - C - T - T - C - A, each base pairs respectively to T - A - C - G - A - A - G - T. Reading this complementary strand from 3' to 5' yields TACGAAGT.

The correct answer is Option B) TACGAAGT.

Fehling's reagent is used to test for reducing sugars. A reducing sugar has a free aldehyde or ketone group (or can form one in solution) that can reduce $$Cu^{2+}$$ ions in Fehling's reagent to $$Cu_2O$$ (reddish-brown precipitate).

Analysing each option:

Glucose: Glucose is an aldose with a free aldehyde group. It is a reducing sugar and gives a positive Fehling's test (reddish-brown precipitate).

Maltose: Maltose is a disaccharide with a free anomeric carbon. It is a reducing sugar and gives a positive Fehling's test.

Lactose: Lactose is a disaccharide with a free anomeric carbon on the glucose unit. It is a reducing sugar and gives a positive Fehling's test.

Sucrose: Sucrose is a disaccharide in which the anomeric carbons of both glucose and fructose are involved in the glycosidic bond. Therefore, sucrose has no free aldehyde or ketone group. It is a non-reducing sugar and does not give the reddish-brown precipitate with Fehling's reagent.

The correct answer is Option 1: Sucrose.

Examining each statement about glucose:

(A) "Glucose is soluble in water because of having aldehyde functional group" — This is INCORRECT. Glucose is soluble in water primarily due to the multiple hydroxyl (-OH) groups that form hydrogen bonds with water, not due to the aldehyde group.

(B) Glucose in aqueous solution exists in multiple isomeric forms (α and β anomers and open chain) — TRUE.

(C) Glucose is one of the monomer units in sucrose (glucose + fructose) — TRUE.

(D) Glucose is an aldohexose — TRUE.

The correct answer is Option 1.

Two nucleotides in a nucleic acid (DNA or RNA) are joined together by a phosphodiester linkage.

This linkage is formed between the 3'-hydroxyl group of one nucleotide's sugar and the 5'-phosphate group of the next nucleotide, creating a phosphodiester bond.

The answer is phosphodiester linkage, which corresponds to Option (1).

We need to identify which level of protein structure remains intact after coagulation (denaturation) of egg white on boiling. Since proteins have four levels of structure, it is helpful to recall their definitions: (i) Primary structure is the sequence of amino acids in the polypeptide chain, held together by covalent peptide bonds ($$-CO-NH-$$). (ii) Secondary structure refers to local folding patterns such as $$\alpha$$-helices and $$\beta$$-sheets, which are stabilized by hydrogen bonds between backbone atoms. (iii) Tertiary structure is the overall three-dimensional shape of the protein, stabilized by hydrophobic interactions, hydrogen bonds, ionic bonds, and disulfide bridges. (iv) Quaternary structure arises from the arrangement of multiple polypeptide subunits, stabilized by non-covalent interactions.

When egg white is boiled, the protein albumin undergoes denaturation, during which the secondary, tertiary, and quaternary structures are disrupted because the non-covalent interactions—hydrogen bonds, hydrophobic interactions, and ionic bonds—that maintain these structures are broken by heat.

However, the primary structure remains intact because it depends on strong covalent peptide bonds, and breaking these bonds requires hydrolysis (for example, with strong acids or specific enzymes) rather than just the temperature of boiling water (100°C).

Therefore, the correct answer is Option (1): Primary.

Fat-soluble vitamins (stored): A, D, E, K = 4. Water-soluble vitamin B₁₂ can also be stored in liver. So stored vitamins from the list: A, D, E, K, B₁₂ = 5.

The answer is $$\boxed{5}$$.

Find the number of tripeptides formed by 3 different amino acids, each used once.

A tripeptide is a chain of 3 amino acids linked by peptide bonds. The sequence matters because each position in the chain is chemically distinct (N-terminal, middle, C-terminal).

We have 3 distinct amino acids (say A, B, C) and need to arrange all 3 in a sequence of 3 positions. This is a permutation problem:

$$ P(3,3) = 3! = 3 \times 2 \times 1 = 6 $$

The 6 tripeptides are: ABC, ACB, BAC, BCA, CAB, CBA.

The answer is 6.

Checking statements about nucleic acids:

A. RNA is regarded as the reserve of genetic information — FALSE (DNA is the genetic reserve).

B. DNA molecule self-duplicates during cell division — TRUE.

C. DNA synthesizes proteins in the cell — FALSE (RNA/ribosomes synthesize proteins).

D. The message for synthesis of particular proteins is present in DNA — TRUE.

E. Identical DNA strands are transferred to daughter cells — TRUE.

Correct statements: B, D, E = 3.

Therefore, the answer is $$\boxed{3}$$.

Bromine water is a very mild oxidising agent. It oxidises only those carbohydrates that possess a free aldehydic $$-CHO$$ group (or its equivalent $$\alpha$$-hydroxy-ketone) present in the open-chain (hemiacetal/hemiketal) form. Such carbohydrates are called reducing sugars.

If a disaccharide does not react with bromine water, both of its anomeric carbon centres must be tied up in the glycosidic linkage, so that no monomer can open out to the free $$-CHO$$ form. Hence X is a non-reducing disaccharide.

Among the common disaccharides asked in JEE, the only non-reducing member is sucrose (glucose-α-1→2-β-fructose). Lactose and maltose are reducing; cellobiose is also reducing.

To cross-check, let us look at the optical activity on hydrolysis.
Sucrose $$\xrightarrow[\text{H}_2O]{\text{acid}}$$ glucose + fructose

The specific rotations are:
  • D-glucose : $$+52.5^\circ$$
  • D-fructose : $$-92.0^\circ$$

Stoichiometrically 1 mol sucrose gives 1 mol each of glucose and fructose. The net rotation after complete hydrolysis is therefore

$$\alpha_{\text{mix}}=\frac{(+52.5)+(-92.0)}{2}= -19.75^\circ$$

The mixture is laevorotatory. The change of sign from dextrorotatory sucrose $$(+66.5^\circ)$$ to a laevorotatory hydrolysate is called inversion; the product mixture is known as invert sugar.

Thus all given observations are satisfied only by sucrose.

Option A which is: sucrose

Matching enzymatic reactions with their enzymes:

A. Sucrose → Glucose and Fructose: This hydrolysis of sucrose (inversion of sugar) is catalysed by the enzyme Invertase → Match with III

B. Glucose → Ethyl alcohol and CO₂: This fermentation reaction is catalysed by the enzyme Zymase → Match with I

C. Starch → Maltose: This hydrolysis of starch is catalysed by the enzyme Diastase (amylase) → Match with IV

D. Proteins → Amino acids: This hydrolysis of proteins is catalysed by the enzyme Pepsin → Match with II

The correct matching is: A-III, B-I, C-IV, D-II.

Assertion A: Glycine with Cl₂ and red P undergoes Hell-Volhard-Zelinsky reaction, giving α-chloro amino acid which has a chiral carbon. True.

Reason R: A molecule with 2 chiral carbons is always optically active — False (meso compounds have 2 chiral carbons but are optically inactive).

A is true but R is false.

We need to identify which statements are true about antagonists in pharmacology.

An antagonist is a drug that binds to a receptor but does not activate it. Instead, it blocks the receptor and prevents the natural agonist (messenger) from binding and producing its effect.

Let us evaluate each statement:

Statement A: "They bind to the receptor site."

This is TRUE. Antagonists bind to the receptor site on the cell surface. This binding is essential for their blocking action.

Statement B: "Get transferred inside the cell for their action."

This is FALSE. Antagonists act at the receptor level on the cell surface. They do not need to enter the cell to exert their effect. They simply block the receptor from the outside.

Statement C: "Inhibit the natural communication of the body."

This is TRUE. By blocking the receptor, antagonists prevent the natural messenger (agonist) from binding and transmitting its signal. This effectively inhibits the natural communication pathway of the body.

Statement D: "Mimic the natural messenger."

This is FALSE. It is the agonists that mimic the natural messenger and activate the receptor. Antagonists do the opposite -- they block the receptor without activating it.

Therefore, statements A and C are true about antagonists.

The correct answer is Option C: A and C.

We need to identify the product formed when caprolactam is heated at high temperature in the presence of water.

Caprolactam is a cyclic amide (lactam) with 6 carbon atoms. When heated with water at high temperature, it undergoes ring-opening polymerization:

The water initiates the hydrolysis of the amide bond in caprolactam, opening the ring. The opened ring monomers then undergo condensation polymerization to form a long-chain polyamide.

$$n \text{ Caprolactam} \xrightarrow{\Delta, \text{H}_2\text{O}} \text{Nylon 6}$$

The product is Nylon 6 (polycaprolactam), which is a polyamide with the repeating unit $$-[\text{NH}(\text{CH}_2)_5\text{CO}]_n-$$.

Note: Nylon 6,6 is made from hexamethylenediamine and adipic acid, not from caprolactam.

The correct answer is Option 4: Nylon 6.

Assertion A: BHA (Butylated hydroxyl anisole) is an antioxidant food additive that prevents oxidation of fats and oils, thereby increasing shelf life of butter. This is correct.

Reason R: BHA is more reactive towards oxygen than the food components (it preferentially gets oxidized). This is correct and IS the correct explanation of why it works as an antioxidant.

Both A and R are correct and R is the correct explanation of A.

The correct answer is Option 1.

There are two open-chain aldotetroses: Erythrose and Threose.

• When reduced with $$\text{NaBH}_4$$, the $$\text{-CHO}$$ group is converted into a primary alcohol ($$\text{-CH}_2\text{OH}$$):
  • Erythrose reduces to erythritol, which has a plane of symmetry and is an optically inactive meso (achiral) compound.
  • Threose reduces to threitol, which has no plane of symmetry and remains an optically active chiral compound.

Since the reaction of 'X' with $$\text{NaBH}_4$$ yields a Chiral compound ('B'), X must be Threose.

for the L-isomer of X.

• In carbohydrate chemistry (Fischer projections), "L" means the $$\text{-OH}$$ group on the bottom-most chiral carbon (Carbon-3 in this case) points to the Left.
• Since it's a Threose, the two middle $$\text{-OH}$$ groups must point to opposite sides. If Carbon-3 points to the left, Carbon-2 must point to the right.

Correct Option B

We need to identify which compound is an antacid.

(A) Ranitidine: It is an H$$_2$$-receptor antagonist that reduces stomach acid production. It is used as an antacid to treat conditions like peptic ulcers and GERD. ✓

(B) Meprobamate: It is a tranquilizer (anxiolytic) used for treating anxiety and muscle spasms. Not an antacid.

(C) Terfenadine: It is an antihistamine used for treating allergies. Not an antacid.

(D) Brompheniramine: It is an antihistamine used for treating cold and allergy symptoms. Not an antacid.

The correct answer is Option 1: Ranitidine.

We need to identify the correct statement about human consumption of food and air.

An average human consumes approximately 1.5-2 kg of food per day, 2-3 L of water per day, and about 15,000-20,000 L of air per day (at ~15 breaths/min, ~0.5 L per breath = 10,800 L/day).

By mass, air consumption $$\approx 15,000 \times 1.2 \text{ g/L} \approx 18$$ kg per day, while food consumption $$\approx 1.5$$ kg per day (excluding water).

Thus, $$\dfrac{\text{Air consumed}}{\text{Food consumed}} \approx \dfrac{18}{1.5} \approx 12-15$$ times, indicating that an average human consumes nearly 15 times more air than food.

The correct answer is Option B: An average human being consumes nearly 15 times more air than food.

Statement I: "Morphine is a narcotic analgesic. It helps in relieving pain without producing sleep."

The first part is correct — morphine IS a narcotic analgesic. However, the second part is incorrect. Morphine, being a narcotic analgesic, relieves pain AND causes drowsiness/sleep. Non-narcotic analgesics (like aspirin) relieve pain without causing sleep, but morphine does induce sleep.

Statement I is false.

Statement II: "Morphine and its derivatives are obtained from opium poppy."

This is true. Morphine is the principal alkaloid obtained from the opium poppy (Papaver somniferum).

The correct answer is Statement I is false but Statement II is true.

Match the polymers with their correct classifications.

A. Elastomeric polymer — These are polymers with elastic properties (rubber-like). Neoprene (IV) is a well-known elastomer (synthetic rubber).

B. Fibre polymer — These have strong intermolecular forces and high tensile strength. Polyester (III) is a common fibre.

C. Thermosetting polymer — These form cross-linked structures and cannot be remoulded. Urea formaldehyde resin (I) is a thermosetting polymer.

D. Thermoplastic polymer — These can be repeatedly softened and moulded. Polystyrene (II) is a thermoplastic polymer.

A → IV (Neoprene), B → III (Polyester), C → I (Urea formaldehyde resin), D → II (Polystyrene). The correct answer is Option C: A-IV, B-III, C-I, D-II.

Let us match each item in List I with the correct item in List II:

(A) Tranquilizers - Tranquilizers are Antidepressant drugs (III). They are used to treat anxiety, tension, and mental diseases by reducing stress and inducing calmness.

(B) Aspirin - Aspirin acts as an Anti blood clotting agent (I). Besides being an analgesic and antipyretic, aspirin inhibits platelet aggregation and is used to prevent heart attacks and strokes.

(C) Antibiotic - The example given is Salvarsan (II). Salvarsan (arsphenamine) was one of the first effective antimicrobial drugs, used to treat syphilis and trypanosomiasis.

(D) Antiseptic - The example given is Soframicine (IV). Soframicine (framycetin) is an antibiotic used topically as an antiseptic for treating skin infections and wounds.

Therefore, the correct matching is: (A)-III, (B)-I, (C)-II, (D)-IV.

We need to identify the incorrect statement about antibiotics.

Option A: "An antibiotic must be a product of metabolism."

This is correct. Antibiotics are naturally produced by microorganisms as metabolic products.

Option B: "An antibiotic is a synthetic substance produced as a structural analogue of naturally occurring antibiotic."

This describes semi-synthetic antibiotics and is an acceptable characterization. While the original definition of antibiotics refers to natural products, modern usage includes synthetic analogues.

Option C: "An antibiotic should promote the growth or survival of microorganisms."

This is incorrect. By definition, antibiotics inhibit the growth of or destroy microorganisms. They do not promote growth — that would be the opposite of their function.

Option D: "An antibiotic should be effective in low concentrations."

This is correct. A key characteristic of antibiotics is that they are effective even at very low concentrations.

The incorrect statement is Option C.

Equanil is the brand name for meprobamate, which is a tranquilizer drug.

Tranquilizers are a class of drugs used to treat mental diseases. Equanil is specifically used to control depression and hypertension.

It acts on the central nervous system and is classified as a sedative-hypnotic drug. It helps reduce anxiety, tension, and elevated blood pressure associated with stress.

Therefore, a patient prescribed Equanil would likely have symptoms of depression and hypertension, which is Option D.

Buna-S is the most common synthetic rubber. The name tells everything about its composition:
• “Bu” comes from $$1,3$$-butadiene
• “Na” indicates that sodium metal was the first catalyst used for its polymerisation
• “S” stands for styrene

Hence Buna-S is obtained by copolymerising $$1,3$$-butadiene\,(approx. $$75\%$$) with styrene\,(approx. $$25\%$$).

The structural repeat unit of the copolymer is therefore made by alternating segments derived from the two monomers:

$$\Bigl[\,CH_2\;{\text{-}}\;CH{=}\!CH\;{\text{-}}\;CH_2\,\Bigr]_x \;\;\;\Bigl[\,CH_2\;{\text{-}}\;CH(C_6H_5)\,\Bigr]_y$$

Any option that shows these two fragments—one with conjugated double bond units from butadiene and the other with a phenyl-substituted $$CH$$ unit from styrene—joined into the same main chain is the correct representation.

Among the four drawings given in the question, Option B (Option 2) is the only one that displays a polymer chain containing both
$$\;-\!CH_2-CH=CH-CH_2-\;$$ and $$\;-\!CH_2-CH(C_6H_5)-\;$$ links. Therefore,

Answer: Option B (Option 2).

Matching monomers to polymers:

A. Tetrafluoroethene (CF₂=CF₂): Polymerizes to form polytetrafluoroethylene → Teflon (III)

B. Acrylonitrile (CH₂=CHCN): Polymerizes to form polyacrylonitrile → Orlon (I)

C. Caprolactam: Ring-opening polymerization gives → Nylon-6 (IV)

D. Isoprene (2-methyl-1,3-butadiene): Polymerizes to form cis-1,4-polyisoprene → Natural rubber (II)

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

This matches option 1.

We need to match each polymer in List I with its correct use in List II.

A. Glyptal: Glyptal is a polyester resin formed by the condensation of ethylene glycol and phthalic acid. It is widely used in paints and lacquers due to its excellent adhesive and film-forming properties. This matches III.

B. Neoprene: Neoprene (polychloroprene) is a synthetic rubber known for its chemical resistance, flexibility, and durability. It is commonly used to make gaskets and seals. This matches IV.

C. Acrilan: Acrilan is an acrylic fibre made from polyacrylonitrile. It is used as a substitute for natural wool, i.e., synthetic wool, in clothing and textiles. This matches II.

D. LDP (Low Density Polyethylene): LDPE is a thermoplastic with a flexible, soft structure. It is used to manufacture flexible pipes, squeeze bottles, and packaging films. This matches I.

Final Matching:

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

The correct answer is Option B: A-III, B-IV, C-II, D-I.

Matching polymers with their types:

A. Nylon-2-Nylon-6: This is a biodegradable polymer (II) - a copolymer of glycine and aminocaproic acid.

B. Buna-N: This is a synthetic rubber (III) - a copolymer of butadiene and acrylonitrile.

C. Urea-formaldehyde resin: This is a thermosetting polymer (I) - once set, cannot be remolded.

D. Dacron: This is a polyester (IV) - polyethylene terephthalate (PET).

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

The correct answer is Option 2.

Therefore, Option D is Correct

Orlon is a synthetic fibre made from polyacrylonitrile (PAN). It is prepared by the addition polymerization of acrylonitrile (CH$$_2$$=CH-CN):

$$n\;\text{CH}_2=\text{CH-CN} \xrightarrow{\text{polymerization}} (-\text{CH}_2-\text{CH(CN)}-)_n$$

Orlon is used as a wool substitute in making sweaters, blankets, and other textile products.

So, the answer is Polyacrylonitrile.

Comparing the sweetness values of artificial sweeteners relative to cane sugar:

- Aspartame: approximately 200 times sweeter than sugar

- Saccharin: approximately 550 times sweeter than sugar

- Sucralose: approximately 600 times sweeter than sugar

- Alitame: approximately 2000 times sweeter than sugar

Among the given options, Alitame has the highest sweetness value in comparison to cane sugar.

We need to evaluate the Assertion-Reason pair about Seliwanoff's test.

Assertion (A): Ketoses give Seliwanoff's test faster than Aldoses.

This is TRUE. In Seliwanoff's test (using resorcinol and HCl), ketoses react faster and give a cherry-red color more quickly than aldoses. This is because ketoses are more readily dehydrated than aldoses.

Reason (R): Ketoses undergo $$\beta$$-elimination followed by formation of furfural.

This is FALSE. Ketoses (like fructose, a hexose) undergo dehydration to form hydroxymethylfurfural (HMF), not furfural. Furfural is formed from pentoses. The hydroxymethylfurfural then reacts with resorcinol to give the cherry-red color.

Since A is true but R is false:

The correct answer is Option 3: (A) is true but (R) is false.

Statement I: Ethane at 333-343 K, 6-7 atm with Ziegler-Natta catalyst (AlEt₃ + TiCl₄) gives LDP — False. These conditions actually give HDPE (High Density Polyethylene), not LDPE. LDPE requires high pressure (1000-2000 atm) with peroxide initiator.

Statement II: Caprolactam at 533-543 K in H₂O polymerizes by ring-opening to give Nylon 6 — True.

Statement I is false but Statement II is true.

We analyze each statement:

Statement I: Noradrenaline is a neurotransmitter.

This is CORRECT. Noradrenaline (norepinephrine) is an important neurotransmitter in the sympathetic nervous system and the brain.

Statement II: Low level of noradrenaline is not the cause of depression in humans.

This is INCORRECT. Low levels of noradrenaline are indeed associated with depression. Noradrenaline deficiency is one of the known causes of depression, and many antidepressant drugs work by increasing noradrenaline levels.

Statement I is correct but Statement II is incorrect.

The correct answer is Option B: Statement 1 is correct but statement 2 is incorrect.

A 6-membered ring containing one oxygen atom is a pyranose ring, whereas a 5-membered ring containing one oxygen atom is a furanose ring.

For D-sugars:

• If the anomeric $$-OH$$ group is on the opposite side of the ring from the terminal $$-CH_2OH$$ group, the sugar is the $$\alpha$$-anomer.

• If the anomeric $$-OH$$ group is on the same side as the terminal $$-CH_2OH$$ group, the sugar is the $$\beta$$-anomer.

For the first structure:

• The ring is a 6-membered pyranose ring.

• The arrangement of substituents corresponds to glucose.

• The anomeric $$-OH$$ group is opposite to the $$-CH_2OH$$ group.

Hence, it is $$\alpha\text{-D-(+) Glucopyranose}$$ and matches with (iii).

For the second structure:

• The ring is a 6-membered pyranose ring.

• The configuration corresponds to glucose.

• The anomeric $$-OH$$ group and $$-CH_2OH$$ group are on the same side.

Hence, it is $$\beta\text{-D-(+) Glucopyranose}$$ and matches with (iv).

For the third structure:

• The ring is a 5-membered furanose ring.

• The presence of two $$-CH_2OH$$ groups identifies it as fructose.

• The anomeric $$-OH$$ group is opposite to the terminal $$-CH_2OH$$ group.

Hence, it is $$\alpha\text{-D-(-) Fructofuranose}$$ and matches with (i).

For the fourth structure:

• The ring is a 5-membered furanose ring.

• The configuration corresponds to fructose.

• The anomeric $$-OH$$ group and terminal $$-CH_2OH$$ group are on the same side.

Hence, it is $$\beta\text{-D-(-) Fructofuranose}$$ and matches with (ii).

Therefore,

$$1 \rightarrow (iii), \quad 2 \rightarrow (iv), \quad 3 \rightarrow (i), \quad 4 \rightarrow (ii)$$

Hence, the correct matching is

$$\boxed{1-(iii),\;2-(iv),\;3-(i),\;4-(ii)}$$

Matching vitamins with their deficiency diseases:

A. Vitamin A → Xerophthalmia (III): Deficiency of Vitamin A causes dryness of the eyes (xerophthalmia) and night blindness.

B. Thiamine (Vitamin B₁) → Beri-Beri (I): Deficiency of thiamine causes beri-beri, characterised by weakness, nerve damage, and cardiovascular problems.

C. Ascorbic acid (Vitamin C) → Scurvy (IV): Deficiency of Vitamin C causes scurvy, characterised by bleeding gums, weakness, and poor wound healing.

D. Riboflavin (Vitamin B₂) → Cheilosis (II): Deficiency of riboflavin causes cheilosis (cracking of skin at corners of the mouth and lips).

The correct matching is: A-III, B-I, C-IV, D-II.

The standard one-letter codes for amino acids are:

A. Arginine → R (IV)

B. Aspartic acid → D (I)

C. Asparagine → N (II)

D. Alanine → A (III)

The correct matching is: (A)-IV, (B)-I, (C)-II, (D)-III.

We need to match the antimicrobials with their correct names.

(A) Narrow Spectrum Antibiotic → (III) Penicillin-G

Penicillin-G is effective against a limited range of bacteria, primarily Gram-positive bacteria. Hence, it is a narrow spectrum antibiotic.

(B) Antiseptic → (I) Furacin

Furacin (nitrofurazone) is used as an antiseptic for skin infections. Antiseptics are applied to living tissues to prevent infection.

(C) Disinfectant → (II) Sulphur dioxide

Sulphur dioxide is used as a disinfectant. Disinfectants are applied to non-living objects and surfaces to destroy microorganisms.

(D) Broad Spectrum Antibiotic → (IV) Chloramphenicol

Chloramphenicol is effective against a wide range of both Gram-positive and Gram-negative bacteria, making it a broad spectrum antibiotic.

The correct matching is: (A)-III, (B)-I, (C)-II, (D)-IV.

The correct answer is Option 1.

Sulphur-containing amino acids have sulphur atoms in their side chains.

(a) Isoleucine: Contains only C, H, O, N in its side chain (branched hydrocarbon). No sulphur. ✗

(b) Cysteine: Contains a thiol (-SH) group in its side chain (HSCH₂-). Contains sulphur. ✓

(c) Lysine: Contains an amino group (-NH₂) in its side chain. No sulphur. ✗

(d) Methionine: Contains a thioether (-S-CH₃) group in its side chain. Contains sulphur. ✓

(e) Glutamic acid: Contains a carboxyl group (-COOH) in its side chain. No sulphur. ✗

The sulphur-containing amino acids are b and d (cysteine and methionine).

The secondary and tertiary structures of proteins are stabilized by:

- Hydrogen bonding (key in secondary structure: alpha-helix and beta-sheet)

- Disulphide linkage (-S-S-) (covalent bond between cysteine residues)

- Van der Waals forces (weak intermolecular interactions)

- Ionic bonds (salt bridges)

- Hydrophobic interactions

The -O-O- linkage (peroxide linkage) does NOT play a role in stabilizing protein structures. It is not a natural bond found in protein stabilization.

The correct answer is Option 3: -O-O- linkage.

The polymer X has the following characteristics:

- Consists of linear molecules

- Closely packed

- Prepared using trimethylaluminium and titanium tetrachloride (Ziegler-Natta catalyst) under low pressure

These are the defining features of High Density Polythene (HDPE):

- HDPE is prepared by coordination polymerisation of ethylene using Ziegler-Natta catalyst (combination of triethylaluminium/trimethylaluminium and TiCl$$_4$$) at low pressure (6-7 atm) and moderate temperature (333-343 K).

- The resulting polymer has a linear structure with very little branching, allowing the chains to pack closely together, resulting in high density (0.94-0.97 g/cm$$^3$$).

In contrast, Low Density Polythene (LDPE) is prepared by free radical polymerisation at high pressure (1000-2000 atm) and has a highly branched structure.

Therefore, polymer X is High Density Polythene.

Arginine is an amino acid. Let us check each statement:

Option 1: High melting point - TRUE (amino acids exist as zwitterions and have high melting points)

Option 2: Multiple pKa values - TRUE (arginine has 3 pKa values: α-COOH, α-NH₃⁺, and guanidinium group)

Option 3: High solubility in benzene - FALSE (amino acids are ionic/zwitterionic and are insoluble in nonpolar solvents like benzene)

Option 4: Crystalline solid - TRUE (amino acids are crystalline solids)

The statement that is NOT true is option 3: High solubility in benzene.

In most keto-enol tautomeric systems, the keto form is more stable than the enol form because the $$C=O$$ bond is generally more stable than a $$C=C$$ bond accompanied by an $$-OH$$ group.

However, vitamin C (L-ascorbic acid) is an exception because its enediol form is highly stabilized by resonance.

In the given structure, the $$C=C$$ bond is conjugated with the carbonyl group of the lactone ring. This forms an $$\alpha,\beta$$-unsaturated lactone system in which the $$\pi$$ electrons are delocalized over multiple atoms.

This extended conjugation lowers the energy of the molecule and significantly increases its stability.

The two hydroxyl groups attached to the double bond constitute an enediol system. The lone pairs present on the oxygen atoms can participate in resonance with the conjugated ring system, providing additional stabilization.

Any tautomeric form that disrupts this conjugation would lose a substantial amount of resonance stabilization and would therefore be less stable.

Thus, the enediol structure containing the conjugated $$\alpha,\beta$$-unsaturated lactone system is the most stable form of vitamin C.

Hence, the correct answer is the enediol form shown in Option A.

To find the number of cyclic tripeptides formed from amino acids A and B, we first note that a linear tripeptide consists of three residues. Therefore, there are AAA, AAB, ABA, ABB, BAA, BAB, BBA, BBB — 8 total ($$2^3$$).

Since cyclic peptides are invariant under rotation, sequences that are cyclic rotations of one another represent the same molecule. Consequently, the sequence AAA is invariant under any rotation, yielding one distinct peptide, and similarly BBB yields one. Next, the three sequences with two A's and one B (AAB, ABA, BAA) are cyclically equivalent, as are the three with one A and two B's (ABB, BAB, BBA), with each group contributing one distinct cyclic peptide. From the above, the total number of distinct cyclic tripeptides is $$1 + 1 + 1 + 1 = 4$$, giving the answer as Option D: 4.

The molecular weight of uracil is 112g

The weight of Nitrogen is 28g.

28/112 = 0.25 x 100 = 25%

We need to identify the tranquilizers from the given options.

A. Chlorodiazepoxide: This is a benzodiazepine derivative used as a tranquilizer. It was the first benzodiazepine to be discovered (marketed as Librium).

B. Veronal: Veronal (barbital/barbitone) is a barbiturate used as a hypnotic/sedative, not classified as a tranquilizer. It is a sleep-inducing drug (hypnotic).

C. Valium: Valium (diazepam) is a benzodiazepine and is a well-known tranquilizer.

D. Salvarsan: Salvarsan is an organoarsenic compound used as an antimicrobial/antisyphilitic drug. It is not a tranquilizer.

Chlorodiazepoxide and Valium are tranquilizers. That gives us 2 tranquilizers.

The correct answer is 2.

Alanylglycylphenylalanylisoleucine is Ala-Gly-Phe-Ile (a tetrapeptide).

sp² hybridized carbons:

- 3 peptide bond carbonyl carbons (C=O in each peptide bond)

- 1 terminal carboxyl carbon (-COOH)

- 6 aromatic carbons in the phenylalanine benzene ring

Total = 3 + 1 + 6 = 10

The one-letter codes for amino acids are:

(A) Glutamic acid → E (matches III)

(B) Glutamine → Q (matches I)

(C) Tyrosine → Y (matches IV)

(D) Tryptophan → W (matches II)

The correct matching is: A-III, B-I, C-IV, D-II.

The answer is 10.

Benedict's solution tests for reducing sugars (those with free aldehyde or ketone groups that can reduce Cu²⁺).

Reducing sugars (give orange-red precipitate): Glucose, maltose, ribose, 2-deoxyribose, lactose

Non-reducing (do NOT give precipitate): Sucrose (glycosidic bond between reducing ends of both glucose and fructose) and amylose (polysaccharide, does not give positive Benedict's test)

Number of compounds not producing orange-red precipitate = 2

A tripeptide consists of 3 amino acid residues linked together.

With 2 amino acids (valine V and proline P), each position in the tripeptide can be either V or P.

Total number of tripeptides = $$2^3 = 8$$

The possible tripeptides are: VVV, VVP, VPV, VPP, PVV, PVP, PPV, PPP.

The answer is $$8$$.

The starting compound is 1,3,5-tris(4-nitrophenyl)benzene.

Step-1 Sn/HCl reduces each $$\!NO_2$$ group to $$\!NH_2$$, giving 1,3,5-tris(4-aminophenyl)benzene.

Step-2 NaNO$$\_2$$/HCl at 0 °C converts each $$\!NH_2$$ into a diazonium group (-$$N_2^+Cl^-$$).  The product obtained is denoted as $$P$$.

Step-3 Pure water at room temperature hydrolyses every diazonium group to an $$OH$$ group (Sandmeyer hydrolysis), giving $$Q =$$ 1,3,5-tris(4-hydroxyphenyl)benzene.

Step-4 Bromination of $$Q$$ in aqueous medium produces $$R$$.
Phenol rings undergo electrophilic substitution chiefly at the two ortho positions (positions 2 and 6 with respect to the $$OH$$) because the para position (4) is already occupied by the C-C bond that connects the side ring to the central benzene.

Mass-difference data supplied in the problem: $$\text{molar-mass}(R)\;-\;\text{molar-mass}(Q)=474\;\text{g mol}^{-1}$$

Each bromination step replaces an H atom (1 g mol$$^{-1}$$) by a Br atom (80 g mol$$^{-1}$$).
Mass increase per Br introduced = $$80-1=79\;\text{g mol}^{-1}$$.

Number of Br atoms introduced = $$\frac{474}{79}=6$$.

There are three phenolic rings in $$Q$$, so every ring receives $$\frac{6}{3}=2$$ bromine atoms, exactly the two ortho positions expected. Hence $$R$$ is 1,3,5-tris(2,6-dibromo-4-hydroxyphenyl)benzene.

Counting hetero-atoms in one molecule of $$R$$ (all atoms other than C and H):
• Oxygen: 3 (one $$O$$ in each $$OH$$ group)
• Bromine: 6 (two per phenolic ring × three rings)
Total hetero-atoms = $$3+6=9$$.

Therefore, the number of hetero-atoms present in one molecule of $$R$$ is 9.

The starting material is 1,3,5-tris(4-nitrophenyl)benzene.

Step 1 : Reduction
Sn/HCl converts each $$\;p\text{-}NO_2\;$$ group to $$\;p\text{-}NH_2\;$$.
Product: 1,3,5-tris(4-aminophenyl)benzene.

Step 2 : Diazotisation
Excess $$NaNO_2/HCl$$ at $$0^{\circ}\text{C}$$ changes every $$-NH_2$$ to a diazonium chloride group $$-N_2^+Cl^-$$.
Compound P: 1,3,5-tris(4-diazonium-chloride-phenyl)benzene.

Step 3 : Hydrolysis
Warming P with water replaces each $$-N_2^+Cl^-$$ by $$-OH$$, giving Q = 1,3,5-tris(4-hydroxyphenyl)benzene.

Step 4 : Bromination
In aqueous medium the phenolic ring gives electrophilic substitution at both ortho positions.
Each phenol ring of Q picks up two bromine atoms $$\;(2\times79)\;$$ and loses two hydrogens $$\;(2\times1)$$.
Mass increase per ring $$= 2\,(79-1)=158$$.
For three rings, total increase $$=3\times158=474\;\text{g mol}^{-1}$$, exactly the value given. Thus R contains six bromine atoms.

Step 5 : Azo coupling
Compound P reacts with excess phenol in basic medium. Each diazonium chloride group couples with a phenoxide ion at its para position:

$$\text{Ar}-N_2^+Cl^- + C_6H_5O^- \rightarrow \text{Ar}-N=N-C_6H_4OH + Cl^- + H^+$$

Mass change for one diazonium site:
• Phenol adds $$C_6H_5OH$$: $$6\times12 + 6\times1 + 16 = 94$$.
• A proton is lost during coupling: $$94-1 = 93$$.
• Chloride ion $$Cl^-$$ of mass $$35.5$$ is lost.
Net gain per site $$=93-35.5 = 57.5\;\text{g mol}^{-1}$$.
Given total gain $$172.5\;\text{g mol}^{-1}$$, the reaction involves three identical sites $$\;(172.5/57.5 = 3)$$, confirming that every diazonium group has coupled.

Compound S is therefore
1,3,5-tris[4-(p-hydroxyphenylazo)phenyl]benzene.

Step 6 : Counting atoms in one molecule of S

Carbons:
• Central benzene ring: $$6$$ C.
• Three inner phenyl rings: $$3\times6 = 18$$ C.
• Three outer phenol rings: $$3\times6 = 18$$ C.
Total carbon atoms $$= 6 + 18 + 18 = 42$$.

Heteroatoms (all atoms other than C and H):
• Azo nitrogens: each $$-N=N-$$ has $$2$$ N; for three links $$2\times3 = 6$$ N.
• Phenolic oxygens: one $$O$$ on each outer ring $$\Rightarrow 3$$ O.
Total heteroatoms $$= 6 + 3 = 9$$.

Step 7 : Required total
Carbon atoms $$+$$ heteroatoms $$= 42 + 9 = 51$$.

Hence, the total number of carbon atoms and heteroatoms present in one molecule of S is 51.

Explanation: Compound $$A$$ has the molecular formula $$C_4H_8O_4$$.

A positive Tollens’ test confirms the presence of an aldehyde group $$-CHO$$.

Formation of a triacetate derivative shows the presence of three hydroxyl groups $$-OH$$.

Thus, Compound $$A$$ is an aldotetrose with the structure:

$$CHO-CH(OH)-CH(OH)-CH_2OH$$

Since the compound is an $$L$$-isomer, the $$-OH$$ group on the bottom-most chiral carbon must be on the left side in the Fischer projection.

This eliminates the $$D$$-isomers.

On oxidation with nitric acid $$HNO_3$$, both the aldehyde group and primary alcohol group are converted into carboxylic acids.

The resulting compound is optically inactive, meaning it must be a meso compound possessing an internal plane of symmetry.

For Option A, the oxidized product has the $$-OH$$ groups on opposite sides, so no plane of symmetry exists and the product remains optically active.

For Option B, the oxidized product has both $$-OH$$ groups on the same side, producing a meso compound with an internal plane of symmetry.

Thus, the nitric acid oxidation product from Option B is optically inactive.

Oxidation with bromine water converts only the aldehyde group into a carboxylic acid while leaving the terminal $$CH_2OH$$ group unchanged.

Because the two ends of the molecule are now different, the product cannot possess symmetry and remains optically active.

Therefore, Option B satisfies all the given conditions.

We need to evaluate the Assertion-Reason statement about Dacron.

The assertion states that "Dacron is an example of polyester polymer." Dacron is the trade name for polyethylene terephthalate (PET), which is indeed a polyester formed by condensation polymerization involving ester linkages; hence the assertion is correct.

The reason states that "Dacron is made up of ethylene glycol and terephthalic acid monomers." Dacron (PET) is formed by the condensation reaction between ethylene glycol (HOCH₂CH₂OH), a diol, and terephthalic acid (benzene-1,4-dicarboxylic acid), a dicarboxylic acid:

$$n\text{HOCH}_2\text{CH}_2\text{OH} + n\text{HOOC-C}_6\text{H}_4\text{-COOH} \rightarrow \text{[-OCH}_2\text{CH}_2\text{OOC-C}_6\text{H}_4\text{CO-]}_n + 2n\text{H}_2\text{O}$$

The reason is correct and explains why Dacron is a polyester, since the formation of ester bonds from a diol and a dicarboxylic acid defines it as a polyester.

The answer is Option A: Both A and R are correct and R is the correct explanation of A.

We need to identify the correct statement about Buna-N.

Understand what Buna-N is.

Buna-N is a synthetic rubber. Let us break down the name:

- Bu stands for butadiene (1,3-butadiene)

- Na stands for sodium (Na), which is used as a catalyst/initiator in polymerization

- N stands for nitrile (acrylonitrile, $$CH_2=CH-CN$$)

Evaluate each option.

Option A: "It is a linear polymer of 1,3-butadiene." — This is incorrect. A linear polymer of only 1,3-butadiene would be Buna rubber, not Buna-N. Buna-N is a copolymer.

Option B: "It is obtained by copolymerization of 1,3-butadiene and styrene." — This is incorrect. The copolymer of 1,3-butadiene and styrene is called Buna-S (where S stands for styrene), not Buna-N.

Option C: "It is obtained by copolymerization of 1,3-butadiene and acrylonitrile." — This is correct. Buna-N is obtained by the copolymerization of 1,3-butadiene ($$CH_2=CH-CH=CH_2$$) and acrylonitrile ($$CH_2=CH-CN$$).

Option D: "The suffix N in Buna-N stands for its natural occurrence." — This is incorrect. The N stands for nitrile (acrylonitrile), not natural. Buna-N is a synthetic rubber.

The correct answer is Option C.

We need to identify which of the given options is an example of a polyester.

Option A: Butadiene-styrene copolymer

This is Styrene-Butadiene Rubber (SBR), which is a copolymer formed by addition polymerization of 1,3-butadiene and styrene. It is a synthetic rubber, not a polyester.

Option B: Melamine polymer

Melamine polymer (melamine-formaldehyde resin) is formed by condensation of melamine with formaldehyde. It is a thermosetting polymer with cross-linked structure, not a polyester.

Option C: Neoprene

Neoprene is polychloroprene, formed by addition polymerization of chloroprene (2-chloro-1,3-butadiene). It is a synthetic rubber, not a polyester.

Option D: Poly-$$\beta$$-hydroxybutyrate-co-$$\beta$$-hydroxyvalerate (PHBV)

PHBV is a copolymer of 3-hydroxybutyric acid and 3-hydroxyvaleric acid (3-hydroxypentanoic acid). These monomers contain both $$-OH$$ and $$-COOH$$ groups. They undergo condensation polymerization to form ester linkages ($$-COO-$$), making PHBV a polyester. It is also a biodegradable polymer used in packaging and medical applications.

The correct answer is Option D: Poly-$$\beta$$-hydroxybutyrate-co-$$\beta$$-hydroxyvalerate.

We need to evaluate the Assertion and Reason about natural rubber. The Assertion states: "Natural rubber is a linear polymer of isoprene called cis-polyisoprene with elastic properties." This is true because natural rubber is indeed cis-1,4-polyisoprene, a linear polymer of isoprene (2-methyl-1,3-butadiene) that exhibits elastic properties.

The Reason states: "The cis-polyisoprene molecules consist of various chains held together by strong polar interactions with coiled structure." This is false since natural rubber is a non-polar hydrocarbon and its chains are held together by weak van der Waals forces rather than strong polar interactions. Although the cis configuration causes the chains to adopt a coiled structure, the intermolecular forces are not "strong polar interactions."

Since the Assertion is true and the Reason is false, the correct choice is: Assertion is true but Reason is false. Hence, the correct answer is Option C.

We need to match the polymers in List-I with their uses in List-II.

Since Nylon 6,6 is a polyamide known for its high tensile strength and rigidity, commonly used to make bristles of brushes, textiles, and ropes, it corresponds to (III) Bristles of brushes, giving A - Nylon 6,6 → (III) Bristles of brushes.

Similarly, Low density polythene (LDPE) has a branched structure, making it soft and flexible, and is used for making squeeze bottles, toys, and packaging materials, so B - Low density polythene → (IV) Toys.

High density polythene (HDPE) has a linear structure, making it harder and more rigid than LDPE, and is used for making buckets, pipes, and bottles, which indicates that C - High density polythene → (I) Buckets.

Finally, Teflon (polytetrafluoroethylene, PTFE) is chemically inert and has a very low coefficient of friction, and is used as a coating in non-stick utensils, hence D - Teflon → (II) Non-stick utensils.

Therefore, the final matching is A - III, B - IV, C - I, D - II, and the correct option is Option B.

We need to match polymers with their commercial names.

The condensation polymer of phenol and formaldehyde is commonly known as Novolac (or Bakelite, depending on the conditions). Novolac is the linear polymer formed with excess phenol under acidic conditions, and therefore phenol-formaldehyde resin corresponds to (II) Novolac.

The synthetic rubber formed by copolymerization of 1,3-butadiene and styrene is known as Buna-S (Bu = butadiene, Na = sodium catalyst, S = styrene). It is also called SBR (Styrene Butadiene Rubber), so the copolymer of 1,3-butadiene and styrene corresponds to (III) Buna-S.

The polyester formed from ethylene glycol and phthalic acid (a dicarboxylic acid) is known as Glyptal. It is used in paints and lacquers, hence the polyester of glycol and phthalic acid corresponds to (I) Glyptal.

The polyester formed from ethylene glycol and terephthalic acid is known as Dacron (also called PET - polyethylene terephthalate). It is used to make polyester fibres, so the polyester of glycol and terephthalic acid corresponds to (IV) Dacron.

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

The correct answer is Option B: A-II, B-III, C-I, D-IV.

We need to match enzymes in List I with their functions in List II.

Invertase catalyzes the hydrolysis of sucrose (cane sugar) into glucose and fructose:
$$C_{12}H_{22}O_{11} + H_2O \xrightarrow{\text{Invertase}} C_6H_{12}O_6 \text{(glucose)} + C_6H_{12}O_6 \text{(fructose)}$$
So A matches with IV (Cane sugar into glucose).

Zymase is an enzyme complex found in yeast that catalyzes the fermentation of glucose into ethanol and carbon dioxide:
$$C_6H_{12}O_6 \xrightarrow{\text{Zymase}} 2C_2H_5OH + 2CO_2$$
So B matches with III (Glucose into ethanol).

Diastase (also known as amylase) catalyzes the hydrolysis of starch into maltose:
$$2(C_6H_{10}O_5)_n + nH_2O \xrightarrow{\text{Diastase}} nC_{12}H_{22}O_{11} \text{(maltose)}$$
So C matches with I (Starch into maltose).

Maltase catalyzes the hydrolysis of maltose into glucose:
$$C_{12}H_{22}O_{11} + H_2O \xrightarrow{\text{Maltase}} 2C_6H_{12}O_6 \text{(glucose)}$$
So D matches with II (Maltose into glucose).

Therefore, the final matching is A - IV, B - III, C - I, D - II.

Option C

We need to match each type of drug with its function.

Recall the definitions.

(A) Antipyretic: The prefix "anti" means against, and "pyretic" relates to "pyrexia" (fever). So antipyretics are drugs that reduce fever. Examples: Aspirin, Paracetamol.

$$\Rightarrow$$ Matches with (III) Reduces fever

(B) Analgesic: The word comes from Greek "an" (without) and "algos" (pain). So analgesics are drugs that reduce pain. Examples: Aspirin, Morphine.

$$\Rightarrow$$ Matches with (I) Reduces pain

(C) Tranquilizer: Tranquilizers are drugs that calm the mind and reduce stress, anxiety, and tension. Examples: Valium, Chlordiazepoxide.

$$\Rightarrow$$ Matches with (II) Reduces stress

(D) Antacid: The prefix "ant" means against, and "acid" refers to stomach acid. So antacids are drugs that reduce acidity in the stomach. Examples: Ranitidine, Aluminium hydroxide.

$$\Rightarrow$$ Matches with (IV) Reduces acidity (stomach)

Write the final matching.

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

The correct answer is Option A.

We need to match each polymer with its monomer.

A. Neoprene

Neoprene is a synthetic rubber made by polymerization of chloroprene (2-chloro-1,3-butadiene).

$$\therefore$$ A $$\rightarrow$$ II (Chloroprene)

B. Teflon

Teflon (polytetrafluoroethylene, PTFE) is formed by polymerization of tetrafluoroethene ($$CF_2 = CF_2$$).

$$\therefore$$ B $$\rightarrow$$ III (Tetrafluoroethene)

C. Acrilan

Acrilan is an acrylic fiber made by polymerization of acrylonitrile ($$CH_2 = CH - CN$$).

$$\therefore$$ C $$\rightarrow$$ I (Acrylonitrile)

D. Natural rubber

Natural rubber is a polymer of isoprene (2-methyl-1,3-butadiene).

$$\therefore$$ D $$\rightarrow$$ IV (Isoprene)

The correct matching is: A-II, B-III, C-I, D-IV

Hence, the correct answer is Option A.

Polymer $$\mathrm{A}$$ represents:

$$\mathrm{Neoprene}$$

Neoprene is a synthetic rubber.

Rubbers possess weak intermolecular forces, allowing polymer chains to stretch easily.

Hence, neoprene is an:

$$\mathrm{Elastomer}$$

Thus:

$$\mathrm{A \rightarrow III}$$

Polymer $$\mathrm{B}$$ represents:

$$\mathrm{Nylon\ 6,6}$$

Nylon contains strong intermolecular hydrogen bonding between polymer chains.

This gives high tensile strength and close packing of chains.

Hence, nylon behaves as a:

$$\mathrm{Fiber}$$

Thus:

$$\mathrm{B \rightarrow II}$$

Polymer $$\mathrm{C}$$ represents:

$$\mathrm{Polyvinyl\ Chloride\ (PVC)}$$

PVC softens on heating and hardens on cooling without undergoing chemical change.

This is the characteristic property of:

$$\mathrm{Thermoplastic\ Polymers}$$

Thus:

$$\mathrm{C \rightarrow IV}$$

Polymer $$\mathrm{D}$$ represents:

$$\mathrm{Phenol\text{-}Formaldehyde\ Resin}$$

On heating, extensive cross-linking occurs, producing a rigid three-dimensional network.

Once formed, it cannot be remelted or reshaped.

Hence, it is a:

$$\mathrm{Thermosetting\ Polymer}$$

Thus:

$$\mathrm{D \rightarrow I}$$

Therefore, the correct matching is:

$$\mathrm{A\text{-}III,\ B\text{-}II,\ C\text{-}IV,\ D\text{-}I}$$

Let us match each polymer with its correct application.

A. Bakelite → II (Electrical switches)

Bakelite is a thermosetting polymer (phenol-formaldehyde resin) with high heat resistance and electrical insulation properties. It is widely used for making electrical switches, plugs, and handles of utensils.

B. Glyptal → III (Paints and Lacquers)

Glyptal is a polyester resin made from ethylene glycol and phthalic acid. It is primarily used in making paints, lacquers, and surface coatings due to its good adhesion and film-forming properties.

C. PVC → IV (Water pipes)

PVC (polyvinyl chloride) is a versatile thermoplastic polymer. It is extensively used for making water pipes, raincoats, and vinyl flooring due to its durability, chemical resistance, and low cost.

D. Polystyrene → I (Radio and television cabinets)

Polystyrene is a rigid thermoplastic polymer used for making radio and television cabinets, toys, and packaging materials due to its good insulating properties and ease of moulding.

The correct matching is: A-(II), B-(III), C-(IV), D-(I)

Hence, the correct answer is Option B.

Formation of Melamine-Formaldehyde Polymer

The reaction depicts the synthesis of a widely used thermosetting plastic, commonly known as melamine resin. This process occurs in two distinct stages: a condensation reaction followed by polymerization.

Step 1: Formation of the Resin Intermediate

• Reactants: The process begins with Melamine (a heterocyclic aromatic compound containing three primary amine groups) and Formaldehyde ($HCHO$).
• Reaction: An initial addition reaction takes place where one of the amino groups ($-NH_2$) of melamine reacts with the carbonyl carbon of formaldehyde.
• Product: This forms a methylol derivative known as the Resin intermediate. Structurally, one of the $-NH_2$ groups has been converted into an $-NHCH_2OH$ group.

Step 2: Polymerisation

• Process: The intermediate molecules undergo condensation polymerization.
• Mechanism: Under appropriate conditions (usually heating), the hydroxyl group ($-OH$) from the methylol unit of one intermediate molecule reacts with an active hydrogen from an unreacted amino group ($-NH_2$) of an adjacent molecule. This results in the elimination of a water molecule ($H_2O$).
• Linkage: This condensation forms a strong methylene bridge ($-CH_2-$) between the triazine rings, creating a robust polymer chain.
• Final Product: Because melamine possesses three reactive amine sites, it can form linkages in multiple directions. This leads to the Melamine polymer, a highly cross-linked, three-dimensional network. This heavy cross-linking gives the final material its characteristic hardness, heat resistance, and durability (commonly used in unbreakable plastic crockery).
  
  Thus, the correct option is B.

Terylene (also known as Dacron or PET — Polyethylene terephthalate) is a polyester polymer.

It is formed by the condensation polymerization of:

• Ethane-1,2-diol (ethylene glycol, $$HOCH_2CH_2OH$$)
• Benzene-1,4-dicarboxylic acid (terephthalic acid, $$HOOCC_6H_4COOH$$)

The reaction proceeds as:

$$nHOCH_2CH_2OH + nHOOC-C_6H_4-COOH \to [-OCH_2CH_2OOC-C_6H_4-CO-]_n + 2nH_2O$$

The ester linkage ($$-COO-$$) is formed between the $$-OH$$ groups of the diol and the $$-COOH$$ groups of the dicarboxylic acid, with the elimination of water.

Key point: The dicarboxylic acid must be the para isomer (benzene-1,4-dicarboxylic acid), not the ortho (1,2) or meta (1,3) isomer, to form the linear polymer chain characteristic of terylene.

Therefore, the correct answer is Option C.

We need to identify what the mixture of chloroxylenol and terpineol is an example of. The mixture is commercially known as Dettol.

Dettol is an antiseptic used on living tissues (skin) to prevent infection from cuts, wounds, etc.

Note the difference:
- Antiseptic: Applied on living tissue (skin, wounds) to prevent infection.
- Disinfectant: Applied on non-living surfaces (floors, instruments) to kill microorganisms.

Dettol (chloroxylenol + terpineol) is primarily used as an antiseptic, so the answer is Option A: Antiseptic.

Antibiotics effective mainly against $$\mathrm{Gram\text{-}positive}$$ or $$\mathrm{Gram\text{-}negative}$$ bacteria are called $$\mathrm{Narrow\ Spectrum\ Antibiotics}$$.

We need to identify which polymer is NOT a condensation polymer.

Condensation polymers are formed by the condensation reaction between monomers, with the elimination of small molecules like water ($$H_2O$$) or hydrochloric acid ($$HCl$$).

Addition polymers are formed by the addition of monomers containing double bonds, without the loss of any small molecules.

Option A: Nylon-6,6 - Formed by the condensation of hexamethylenediamine and adipic acid, with elimination of water. This is a condensation polymer.

Option B: Nylon-6 - Formed by the ring-opening polymerization of caprolactam, which involves internal amide bond formation. This is classified as a condensation polymer.

Option C: Dacron (Polyethylene terephthalate, PET) - Formed by the condensation of ethylene glycol and terephthalic acid, with elimination of water. This is a condensation polymer.

Option D: Buna-S - This is a copolymer of 1,3-butadiene and styrene, formed by addition polymerization (free radical polymerization). No small molecule is eliminated during the process. This is an addition polymer, NOT a condensation polymer.

The correct answer is Option D: Buna-S.

We need to identify the type of drugs that bind to receptors, inhibit their natural function, and block a message.

Drugs interact with receptors in different ways, including agonists, antagonists, allosteric modulators, and lead compounds. Agonists bind to receptors and mimic the natural messenger, activating the receptor and producing a biological response. Antagonists bind to receptors and block the natural messenger from binding, thereby inhibiting the receptor's natural function without activating it. Allosteric modulators bind at a site different from the active site, modifying the receptor's response. Lead compounds are chemical compounds that show potential therapeutic activity and serve as starting points for drug development.

The question describes drugs that "bind to receptors, inhibiting its natural function and blocking a message." This is the exact definition of antagonists.

Hence, the correct answer is Option A.

Vitamin K (List-I, item A) is required for the synthesis of prothrombin that helps in blood clotting.
Deficiency of vitamin K therefore leads to uncontrolled bleeding or $$\text{haemorrhage}$$ (List-II, item III).
Hence $$A \rightarrow III$$.

Vitamin A (List-I, item B) maintains the health of the cornea and is necessary for the formation of the visual pigment rhodopsin.
Its prolonged deficiency produces dryness of the conjunctiva and cornea, a disorder called $$\text{xerophthalmia}$$ (List-II, item IV).
Hence $$B \rightarrow IV$$.

Vitamin B$$_1$$ (thiamine, List-I, item C) plays a key role in the oxidative decarboxylation of pyruvic acid during carbohydrate metabolism.
Lack of thiamine results in impaired energy production leading to the disease $$\text{beri beri}$$ (List-II, item I).
Hence $$C \rightarrow I$$.

Vitamin E (tocopherol, List-I, item D) functions as an antioxidant and protects cell membranes, especially in reproductive tissues.
In males, its deficiency can cause degeneration of germinal epithelium giving rise to $$\text{sterility}$$ (List-II, item II).
Hence $$D \rightarrow II$$.

Collecting all the matches:
$$A-III,\; B-IV,\; C-I,\; D-II$$.

The option that shows this sequence is Option C (the third option).

Cationic detergent $$\mathrm{CH_3(CH_2)_{15}-N^+(CH_3)_3Br^-}$$ is used in hair conditioners.

Thus, $$\mathrm{A \rightarrow IV}$$

Anionic detergent $$\mathrm{CH_3(CH_2)_{11}-C_6H_4-SO_3^-Na^+}$$ is commonly used in toothpastes.

Thus, $$\mathrm{B \rightarrow II}$$

Soap formulation containing sodium stearate $$\mathrm{(C_{17}H_{35}COO^-Na^+)}$$ along with builders such as $$\mathrm{Na_2CO_3}$$ and rosinate is used as laundry soap.

Thus, $$\mathrm{C \rightarrow III}$$

Non-ionic detergent $$\mathrm{CH_3(CH_2)_{16}COO(CH_2CH_2O)_nCH_2CH_2OH}$$ is used in dishwashing liquids and powders.

Thus, $$\mathrm{D \rightarrow I}$$

Final Match:

$$\mathrm{A\text{-}(IV),\ B\text{-}(II),\ C\text{-}(III),\ D\text{-}(I)}$$

We need to match each enzyme with its correct source.

(i) Zymase — Found in Yeast (B). Zymase is the enzyme complex in yeast that catalyzes the fermentation of sugars into ethanol and carbon dioxide.

(ii) Urease — Found in Soyabean (D). Urease was first crystallized by James Sumner from jack beans (soyabean family). It catalyzes the hydrolysis of urea.

(iii) Diastase — Found in Malt (C). Diastase is the enzyme found in germinated barley (malt) that converts starch into maltose.

(iv) Pepsin — Found in Stomach (A). Pepsin is a digestive enzyme secreted by the gastric glands in the stomach for protein digestion.

(i)-B; (ii)-D; (iii)-C; (iv)-A

The correct answer is Option A.

Enzymes typically possess an active site where the substrate binds, but certain drugs interact with a different location on the enzyme.

When a drug attaches away from the active site, it induces a conformational change that disrupts the active site's function. This phenomenon is known as allosteric inhibition, and the binding location is the allosteric site.

To determine the correct term, we evaluate the given options:

- Non-active site: This is not a standard scientific term.

- Allosteric site: This is the correct term for a regulatory site on an enzyme that is distinct from the active site.

- Competitive site: This would refer to the active site itself (competitive inhibition).

- Therapeutic site: This is not a standard biochemistry term.

Therefore, Option B, the allosteric site, is the correct answer.

Let us evaluate each statement about enzyme inhibitor drugs:

(A) "There are Competitive and Non-competitive inhibitor drugs."

This is correct. Enzyme inhibitors are broadly classified into competitive and non-competitive inhibitors.

(B) "These can bind at the active sites and allosteric sites."

This is correct. Competitive inhibitors bind at the active site of the enzyme, while non-competitive inhibitors bind at the allosteric site (a site different from the active site). So collectively, enzyme inhibitor drugs can bind at both types of sites.

(C) "Competitive Drugs are allosteric site blocking drugs."

This is incorrect. Competitive inhibitors block the active site, not the allosteric site. They compete with the substrate for binding at the active site.

(D) "Non-competitive Drugs are active site blocking drugs."

This is incorrect. Non-competitive inhibitors bind at the allosteric site, not the active site. They change the shape of the enzyme so that the substrate cannot fit into the active site.

Statements (C) and (D) have the descriptions interchanged — each incorrectly describes the other type of inhibitor.

Therefore, only statements A and B are correct, and the answer is Option C: A, B only.

We need to identify the type of soap/detergent formed when stearic acid reacts with polyethylene glycol.

Detergents are classified into three types based on the nature of their hydrophilic head: (i) anionic detergents carry a negative charge, for example sodium alkyl sulphates and sodium alkylbenzene sulphonates; (ii) cationic detergents carry a positive charge, such as quaternary ammonium salts like cetyltrimethylammonium bromide; and (iii) non-ionic detergents have a hydrophilic part with no charge and are formed by the reaction of a long-chain fatty acid with polyethylene glycol.

Stearic acid $$CH_3(CH_2)_{16}COOH$$ is a long-chain fatty acid, and when it reacts with polyethylene glycol $$HO(CH_2CH_2O)_nH$$ it forms an ester:

This product has a long hydrocarbon chain serving as the hydrophobic tail from stearic acid, and a polyethylene glycol chain serving as the hydrophilic head which is polar due to multiple ether linkages but carries no ionic charge.

Since the hydrophilic part does not carry any charge (neither positive nor negative), this is a non-ionic detergent.

The product formed is a non-ionic detergent; it is not a soap, which requires a metal salt of a fatty acid, nor is it anionic or cationic.

Hence, the correct answer is Option D: Non-ionic detergent.

Tagamet (Cimetidine) is a well-known histamine $$\text{H}_{2}$$-receptor antagonist used extensively as an antacid to treat heartburn and peptic ulcers:

The drug $$\mathrm{Cimetidine\ (Tagamet)}$$ was designed to prevent interaction of histamine with the receptors present in the stomach wall.

Cimetidine contains a modified imidazole ring similar to histamine.

Specifically, it contains a $$\mathrm{5\text{-}methylimidazole}$$ ring, which helps the molecule bind to the $$\mathrm{H_2}$$ receptor.

The side chain contains a sulfur atom, forming a $$\mathrm{Thioether}$$ linkage.

This longer chain helps the drug interact with additional binding sites on the receptor.

The most important structural modification is replacement of the strongly basic primary amine group of histamine with a $$\mathrm{Cyanoguanidine}$$ group.

Unlike the primary amine, the cyanoguanidine group remains largely neutral at physiological pH.

By this, Cimetidine binds strongly to the receptor without activating it and blocks the action of histamine.

Thus, D is the right option.