JEE Chemistry - MCQ Practice Questions

1/[A] = 2 + 0.4(5) = 2 + 2 = 4; [A] = $\frac{1}{4}$ = 0.25 M

For second-order: t₁/₂ = 1/(k[A]₀); 100 = 1/(k × 0.5); k = 0.04 M⁻¹s⁻¹

Understanding:

We need to identify the correct statement about the $\alpha$-helix structure in proteins.

Step 1: Recall structural levels of proteins

Proteins have primary, secondary, tertiary, and quaternary structures. The $\alpha$-helix is a secondary structure.

Step 2: Identify the stabilizing force in $\alpha$-helix

The $\alpha$-helix is stabilized by intramolecular hydrogen bonds. These hydrogen bonds form between the $\text{N-H}$ group of one amino acid residue and the $\text{C=O}$ group of an amino acid four residues ahead in the sequence (not side chains).

Step 3: Eliminate incorrect options

Answer:

The $\alpha$-helix is a secondary structure stabilized by hydrogen bonds between the $\text{N-H}$ of one peptide bond and the $\text{C=O}$ of the peptide bond four residues earlier.

Quick Tip:

Always distinguish: hydrogen bonds → secondary structure ($\alpha$-helix, $\beta$-sheet); disulfide bridges → tertiary structure.

Understanding:

We need to identify the anomeric carbon and the position of the $\text{-OH}$ group in $\alpha$-$\text{D}$-glucose.

Step 1: Define the anomeric carbon

The anomeric carbon is the carbon that becomes a new chiral centre during ring closure (hemiacetal/hemiketal formation). In glucose (a pyranose), this is ${\text{C}}_1$ — the former aldehyde carbon.

Step 2: Distinguish $\alpha$ and $\beta$ forms

In the Haworth projection of $\text{D}$-glucose:

Step 3: Apply to the question

In $\alpha$-$\text{D}$-glucose, the $\text{-OH}$ at ${\text{C}}_1$ (the anomeric carbon) is in the axial position in the chair conformation.

Answer:

The anomeric carbon in $\alpha$-$\text{D}$-glucose is ${\text{C}}_1$ with $\text{-OH}$ in the axial position.

Quick Tip:

"$\alpha$ is axial" — a quick mnemonic to remember that in $\alpha$-$\text{D}$-glucose the $\text{-OH}$ at ${\text{C}}_1$ is axial.

Understanding:

We need to find a disaccharide that:

(i) hydrolyzes to give two glucose molecules, and

(ii) is a reducing sugar.

Step 1: Analyze each option

Step 2: Identify the correct answer

Only Maltose satisfies both conditions: it gives two glucose molecules on hydrolysis and is a reducing sugar.

Answer:

Maltose gives two glucose molecules on hydrolysis and is a reducing sugar due to its free anomeric ${\text{C}}_1$.

Quick Tip:

A disaccharide is a reducing sugar only if at least one anomeric carbon (${\text{C}}_1$ in aldoses) is free (not involved in the glycosidic bond).

Understanding:

We need to identify which biomolecule does NOT reduce Fehling's solution (i.e., is a non-reducing sugar).

Step 1: Principle of Fehling's test

Fehling's solution contains ${\text{Cu}}^{2+}$ ions (as a tartrate complex). Reducing sugars (those with a free aldehyde or ketone group, or a free anomeric carbon) reduce ${\text{Cu}}^{2+}$ to ${\text{Cu}}_2\text{O}$ (brick-red precipitate).

Step 2: Classify each option

Step 3: Conclusion

Sucrose does not reduce Fehling's solution because both anomeric carbons are involved in the $\alpha ,\beta$-1,2-glycosidic linkage.

Answer:

Sucrose is the only non-reducing sugar among the options and will NOT give a positive Fehling's test.

Quick Tip:

Sucrose is the classic example of a non-reducing disaccharide. It does not mutarotate, does not reduce Tollens' or Fehling's reagent.

Understanding:

We need to match each vitamin correctly with the disease caused by its deficiency.

Step 1: Recall vitamin-deficiency disease pairs

Step 2: Check each option

Answer:

Vitamin $\text{C}$ (Ascorbic acid) deficiency causes Scurvy.

Quick Tip:

A useful mnemonic: "${\text{B}}_1$ Beriberi, ${\text{B}}_3$ Pellagra, $\text{C}$ Scurvy, $\text{D}$ Rickets, $\text{A}$ Night-blindness" — memorize in this order.

Understanding:

We need to identify which of the given vitamins is fat-soluble.

Step 1: Classify vitamins by solubility

Vitamins are classified into two groups:

Step 2: Analyze each option

Answer:

Vitamin $\text{K}$ is a fat-soluble vitamin.

Quick Tip:

Remember "ADEK" — the four fat-soluble vitamins. All other common vitamins (B-complex and C) are water-soluble.

Understanding:

We need to calculate the isoelectric point (pI) of glycine.

Formula:

For a simple amino acid with one amino and one carboxyl group, the isoelectric point is:

Step 1: Substitute the values

Step 2: Interpret the result

At $\text{pH}=5.97$, glycine carries equal positive and negative charges (exists as ${\text{H}}_3{\text{N}}^{+}{\text{-CH}}_2{\text{-COO}}^{-}$, the zwitterion), so its net charge is zero.

Answer:

The isoelectric point of glycine is $5.97$.

Quick Tip:

For acidic amino acids (e.g., aspartic acid), use the two acidic $\text{p}{K}_a$ values; for basic amino acids (e.g., lysine), use the two basic $\text{p}{K}_a$ values to calculate pI.

Understanding:

We need to identify the INCORRECT statement about the DNA double helix structure.

Step 1: Verify each statement

Step 2: Analyze option D

The backbone of DNA is made of alternating deoxyribose sugar and phosphate groups. These are connected by $3^{\prime },5^{\prime }$-phosphodiester bonds (not $\text{N}$-glycosidic bonds). The $\text{N}$-glycosidic bond connects the nitrogenous base to the sugar (within a nucleotide), not the backbone linkage between nucleotides.

Step 3: Conclusion

Option D is incorrect because the backbone is held together by phosphodiester bonds, not $\text{N}$-glycosidic bonds.

Answer:

The statement that the backbone of DNA is held by $\text{N}$-glycosidic bonds is incorrect; the backbone uses phosphodiester bonds.

Quick Tip:

N-glycosidic bonds attach bases to sugars (within nucleosides); phosphodiester bonds link nucleotides together in the backbone.

Understanding:

We need to identify the key structural difference between cellulose and starch.

Step 1: Structure of starch

Starch exists in two forms:

Step 2: Structure of cellulose

Cellulose is a linear polymer of $\text{D}$-glucose units connected by $\beta (1\to 4)$ glycosidic linkages. The $\beta$ configuration allows extended straight chains that can align and form strong intermolecular hydrogen bonds, giving cellulose its fibrous, rigid nature.

Step 3: Why the linkage type matters

The $\alpha$ linkage in starch creates a helical structure (digestible by humans). The $\beta$ linkage in cellulose creates straight chains (not digestible by humans, as we lack the enzyme $\beta$-glucosidase/cellulase).

Answer:

Starch contains $\alpha (1\to 4)$ glycosidic linkages while cellulose contains $\beta (1\to 4)$ linkages.

Quick Tip:

"$\alpha$ links make starch (edible); $\beta$ links make cellulose (structural)." This single difference explains all digestibility and physical property contrasts.

Understanding:

We need to identify the correct description of an enzyme as a biochemical catalyst.

Step 1: Fundamental properties of enzymes

Step 2: Eliminate incorrect options

Step 3: Confirm correct option

Option B correctly states that enzymes lower activation energy and are substrate-specific globular proteins.

Answer:

Enzymes lower the activation energy and are specific globular protein catalysts that are not consumed in the reaction.

Quick Tip:

A catalyst (including an enzyme) can never change $\Delta G$ or $K_{eq}$ — it only provides an alternative pathway with lower $E_a$.

Understanding:

We need to identify which amino acid among the options has a sulfur-containing side chain (R-group).

Step 1: Examine each option

Valine has the side chain $-\text{CH}({\text{CH}}_3{)}_2$, which contains only carbon and hydrogen.

Leucine has the side chain $-{\text{CH}}_2\text{CH}({\text{CH}}_3{)}_2$, also only C and H.

Threonine has the side chain $-\text{CH}(\text{OH}){\text{CH}}_3$, containing oxygen but no sulfur.

Cysteine has the side chain $-{\text{CH}}_2\text{SH}$, which contains a thiol ($-\text{SH}$) group with a sulfur atom.

Step 2: Recall sulfur-containing amino acids

The two common sulfur-containing amino acids are cysteine ($-{\text{CH}}_2\text{SH}$) and methionine ($-{\text{CH}}_2{\text{CH}}_2{\text{SCH}}_3$). Among the given options, only cysteine qualifies.

Answer:

Cysteine is the amino acid with a sulfur-containing side chain due to its thiol ($-\text{SH}$) group.

Quick Tip:

Cysteine's thiol group plays a critical role in forming disulfide bonds ($-\text{S}-\text{S}-$) between protein chains, which are essential for tertiary and quaternary protein structure.

Understanding:

We need to determine the orientation of the $-\text{OH}$ group at $\text{C}-1$ (anomeric carbon) in the Haworth projection of $\alpha$-$\text{D}$-glucose.

Step 1: Recall the convention for Haworth projections

In a Haworth projection of a $\text{D}$-sugar (pyranose ring), groups that are written on the right in the Fischer projection point downward (below the ring plane), and groups on the left point upward (above the ring plane).

Step 2: Define $\alpha$ and $\beta$ anomers

In the $\alpha$-anomer of a $\text{D}$-sugar, the $-\text{OH}$ at the anomeric carbon ($\text{C}-1$) is on the same side as the reference group (the $-{\text{CH}}_2\text{OH}$ at $\text{C}-5$ determines the $\text{D}$-configuration, which points upward). The $\alpha$ designation means the $-\text{OH}$ at $\text{C}-1$ is trans to the $-{\text{CH}}_2\text{OH}$ group, placing it below the ring plane.

Step 3: Confirm for $\alpha$-$\text{D}$-glucose

For $\alpha$-$\text{D}$-glucopyranose:

Answer:

In the Haworth projection of $\alpha$-$\text{D}$-glucose, the $-\text{OH}$ at $\text{C}-1$ is below the ring plane.

Quick Tip:

A simple rule: in $\alpha$-$\text{D}$-sugars, the $-\text{OH}$ at $\text{C}-1$ is down (below the ring); in $\beta$-$\text{D}$-sugars it is up (above the ring).

Understanding:

We need to identify the correct statement about the disaccharide sucrose.

Step 1: Analyse the structure of sucrose

Sucrose is a disaccharide formed by the condensation of $\alpha$-$\text{D}$-glucose and $\beta$-$\text{D}$-fructose. The glycosidic bond is formed between the anomeric carbon of glucose ($\text{C}-1$) and the anomeric carbon of fructose ($\text{C}-2$), giving an $\alpha$-$1,\beta$-$2$-glycosidic linkage, not a $\beta$-$1,4$ linkage.

Step 2: Check reducing sugar property

Because both anomeric carbons are involved in the glycosidic bond, there is no free anomeric $-\text{OH}$ group. Therefore, sucrose cannot open to give an aldehyde or ketone form, and it is a non-reducing sugar. It gives neither a positive Tollens' test nor a positive Fehling's test.

Step 3: Confirm the hydrolysis products

Hydrolysis of sucrose yields one molecule of $\text{D}$-glucose and one molecule of $\text{D}$-fructose. This is the correct statement.

Answer:

Sucrose hydrolyzes to give one molecule of glucose and one molecule of fructose.

Quick Tip:

The mixture of glucose and fructose obtained from sucrose hydrolysis is called invert sugar because the optical rotation changes from positive (sucrose, dextrorotatory) to negative (fructose dominates, levorotatory).

Understanding:

We need to identify which levels of protein structure are lost during denaturation.

Step 1: Define denaturation

Denaturation is the process by which a protein loses its native three-dimensional conformation due to disruption of non-covalent interactions (hydrogen bonds, hydrophobic interactions, electrostatic interactions) and disulfide bonds, without breaking the peptide bonds.

Step 2: Identify what is preserved

The primary structure — the sequence of amino acids linked by peptide bonds — is maintained during denaturation because peptide bonds are covalent bonds that are not broken by typical denaturing agents such as heat, urea, or $\text{pH}$ changes.

Step 3: Identify what is lost

All higher-order structures that depend on non-covalent interactions are disrupted:

All three are lost upon denaturation.

Answer:

Denaturation disrupts secondary, tertiary, and quaternary structures while the primary structure (peptide bond sequence) remains intact.

Quick Tip:

Renaturation (refolding) is possible in some cases if the denaturing agent is removed gently, showing that the primary sequence contains all the information needed to regain the native fold.

Understanding:

We need to identify which nitrogenous bases found in nucleic acids are classified as purines.

Step 1: Recall the classification of nitrogenous bases

Nitrogenous bases in nucleic acids are classified as:

Step 2: Assign each base to its class

Purines: Adenine ($\text{A}$) and Guanine ($\text{G}$) — both have a double-ring system.

Pyrimidines: Cytosine ($\text{C}$), Thymine ($\text{T}$, in DNA), and Uracil ($\text{U}$, in RNA) — all have a single ring.

Step 3: Match to the options

Among the choices, only Adenine and Guanine are purines.

Answer:

The purine bases in nucleic acids are Adenine and Guanine.

Quick Tip:

A useful mnemonic: "Pure As Gold" — Purines are Adenine and Guanine. The word 'purine' itself is longer (like the double ring), while 'pyrimidine' names three single-ring bases.

Understanding:

We need to identify the correct description of the peptide bond ($-\text{CO}-\text{NH}-$) that links amino acids in a protein chain.

Step 1: Recall the formation of the peptide bond

The peptide bond is a covalent amide bond formed between the carboxyl group ($-\text{COOH}$) of one amino acid and the amino group ($-{\text{NH}}_2$) of the next, with the loss of water:

Step 2: Understand the resonance structure

The lone pair on nitrogen delocalizes into the $\text{C}=\text{O}$ bond:

This resonance gives the peptide bond approximately 40% double bond character.

Step 3: Consequences of partial double bond character

Because of this partial double bond character:

Answer:

The peptide bond has partial double bond character due to resonance and is therefore planar and rigid.

Quick Tip:

In the trans configuration (most common), the two $\alpha$-carbons flanking the peptide bond are on opposite sides, minimising steric clashes between side chains.

Understanding:

We need to identify the monosaccharide products obtained when lactose is completely hydrolyzed.

Step 1: Recall the structure of lactose

Lactose is a disaccharide found in milk. It is formed by a $\beta$-$1,4$-glycosidic linkage between:

Step 2: Write the hydrolysis reaction

Step 3: Note the reducing nature

In lactose, the anomeric $-\text{OH}$ of glucose is free (only galactose's anomeric carbon is used in the bond), so lactose is a reducing sugar. Upon hydrolysis, one molecule each of galactose and glucose are produced.

Answer:

Complete hydrolysis of lactose gives one molecule of glucose and one molecule of galactose.

Quick Tip:

Lactose intolerance arises from deficiency of the enzyme lactase ($\beta$-galactosidase), which cleaves the $\beta$-$1,4$-glycosidic bond in lactose.

Understanding:

We need to classify the given proteins and identify which one is a fibrous protein.

Step 1: Recall the two major structural categories of proteins

Step 2: Classify each option

Step 3: Identify the answer

Only collagen is a fibrous protein among the given options.

Answer:

Collagen is a fibrous protein, characterized by its triple-helix structure and structural role in connective tissues.

Quick Tip:

Collagen is the most abundant protein in the human body. Its triple helix is stabilized by hydrogen bonds and requires hydroxyproline (formed by post-translational hydroxylation of proline, requiring Vitamin C). Deficiency of Vitamin C disrupts collagen synthesis, causing scurvy.