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Biomolecules appeared 35 times across 3 years — 4.1% of Chemistry. This question is from Structure of Nucleic Acids.

Year	2026	2025	2024	Total
Questions	9	18	8	35

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Match List-I with List-II tracking nitrogenous bases structures:
List-I (Base) List-II (Chemical Structure Diagram)
(A) Adenine (I)
The Match List contains nitrogenous bases aligned with their corresponding biochemical molecular ring structures.
(B) Cytosine (II)
The Match List contains nitrogenous bases aligned with their corresponding biochemical molecular ring structures.
(C) Thymine (III)
The Match List contains nitrogenous bases aligned with their corresponding biochemical molecular ring structures.
(D) Uracil (IV)
The Match List contains nitrogenous bases aligned with their corresponding biochemical molecular ring structures.
Choose the correct answer from the options given below :

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

Solution & Explanation

Core Logic

Let's identify the chemical structures of the nitrogenous bases used in nucleic acids:

* (A) Adenine: A purine derivative featuring a characteristic fused bicyclic ring system with an amino substituent at position 6 (6-aminopurine) $arrow$ $\rightarrow$ (III). * (B) Cytosine: A pyrimidine monocyclic derivative with an amino group at position 4 and a carbonyl group at position 2 (4-amino-2-oxo-pyrimidine) $arrow$ $\rightarrow$ (IV). * (C) Thymine: Found in DNA, this pyrimidine derivative features a methyl substituent at position 5 along with carbonyl groups at positions 2 and 4 (5-methyl-2,4-dioxo-pyrimidine) $arrow$ $\rightarrow$ (II). * (D) Uracil: Found in RNA, this pyrimidine derivative lacks the methyl group found in thymine, featuring just carbonyl groups at positions 2 and 4 (2,4-dioxo-pyrimidine) $arrow$ $\rightarrow$ (I).

Matching these structures gives the sequence: (A)-(III), (B)-(IV), (C)-(II), (D)-(I).

Step-by-Step Structural Validation

The biochemical structures correspond to the following configurations:

Structure of Nucleic Acids solution diagram for Q44 - JEE Main 2025 Evening — The Match List contains nitrogenous bases aligned with their corresponding biochemical molecular ring structures.

Structure of Nucleic Acids solution diagram for Q44 - JEE Main 2025 Evening — The Match List contains nitrogenous bases aligned with their corresponding biochemical molecular ring structures.

Structure of Nucleic Acids solution diagram for Q44 - JEE Main 2025 Evening — The Match List contains nitrogenous bases aligned with their corresponding biochemical molecular ring structures.

Structure of Nucleic Acids solution diagram for Q44 - JEE Main 2025 Evening — The Match List contains nitrogenous bases aligned with their corresponding biochemical molecular ring structures.

Pattern Recognition

Quick identification keys:

Bicyclic ring = Adenine
Monocyclic ring with a $-CH₃$ $-\mathrm{CH}_3$ group = Thymine
Monocyclic ring without a $-CH₃$ $-\mathrm{CH}_3$ group = Uracil
Monocyclic ring with an amino group ( $-NH₂$ $-\mathrm{NH}_2$) = Cytosine

Chapter Mix

Class 12 Chemistry: Biomolecules

Reference Study Guides

More Biomolecules Previous-Year Questions — Page 2

Q54 jee_main_2026_23_january_evening Nucleic Acids

Both human DNA and RNA are chiral molecules. The chirality in DNA and RNA arises due to the presence of

A. $Base unit$ $\text{Base unit}$
B. $Chiral phosphate unit$ $\text{Chiral phosphate unit}$
C. $D-sugar component$ $\text{D-sugar component}$
D. $L-sugar component$ $\text{L-sugar component}$

Solution

Core Logic

Nucleic acids (DNA and RNA) are composed of three fundamental building blocks: nitrogenous bases, a pentose sugar, and phosphate groups. The nitrogenous bases are planar and generally achiral. The phosphate groups are also achiral. The chirality in both DNA and RNA originates strictly from their pentose sugar components, which have multiple chiral carbon centers.

Step 1: Identifying the specific sugar

DNA contains $β$ $\beta$-D-2-deoxyribose, and RNA contains $β$ $\beta$-D-ribose. Both of these are chiral D-sugar components.

Pattern Recognition

Any question asking about the source of chirality in nucleic acid backbones always points to the pentose sugar ring (D-sugar component).

Chapter Mix

Class 12 Chemistry: Biomolecules

Q53 jee_main_2026_24_january_evening Proteins and Amino Acids

The number of possible tripeptides formed involving alanine (ala), glycine (gly) and valine (val), where no amino acid has been used more than once is:

A. 6
B. 3
C. 4
D. 8

Solution

Core Logic

Since we have 3 distinct amino acids and each must be used exactly once, the number of unique tripeptides corresponds to the number of permutations of these 3 items. Possible sequences:

Gly - Ala - Val
Gly - Val - Ala
Val - Gly - Ala
Val - Ala - Gly
Ala - Val - Gly
Ala - Gly - Val

Total tripeptides $= 3! = 6$ $= 3! = 6$

Step 1: Final Conclusion

Total possible tripeptides without repetition is 6.

Pattern Recognition

For $n$ $n$ distinct amino acids forming a peptide chain of length $n$ $n$ without repetition, the number of distinct isomers is simply $n!$ $n!$.

Chapter Mix

Class 12 Chemistry: Biomolecules Class 11 Maths: Permutations and Combinations

Q64 jee_main_2026_28_january_morning Peptide Sequencing

In the given pentapeptide, find out an essential amino acid (Y) and the sequence present in the pentapeptide:

Chemical structure of a pentapeptide — Full structural formula of a polypeptide chain for sequence deduction.

Choose the correct answer from the options given below:

A. $(Y) Threonine, (Sequence) Ser-Thr-Asp-Gly-Ala$
B. $(Y) Serine, (Sequence) Thr-Ser-Asp-Ala-Gly$
C. $(Y) Threonine, (Sequence) Thr-Ser-Asp-Gly-Ala$
D. $(Y) Serine, (Sequence) Ser-Asp-Thr-Ala-Gly$

Solution

Core Logic

Breaking down the peptide sequence from the N-terminal (left) to C-terminal (right):

First amino acid side chain is $-CH(OH)CH₃$ $-CH(OH)CH_3$. This matches Threonine (Thr).
Second amino acid side chain is $-CH₂OH$ $-CH_2OH$. This matches Serine (Ser).
Third amino acid side chain is $-CH₂COOH$ $-CH_2COOH$. This matches Aspartic Acid (Asp).
Fourth amino acid side chain is just $-H$ $-H$. This matches Glycine (Gly).
Fifth amino acid side chain is $-CH₃$ $-CH_3$. This matches Alanine (Ala).

Sequence: Thr - Ser - Asp - Gly - Ala.

Hydrolysis of pentapeptide into constituent amino acids — Full structural formula of a polypeptide chain for sequence deduction.

Step 1: Identify the Essential Amino Acid

Among the constituents (Thr, Ser, Asp, Gly, Ala), Threonine is an essential amino acid (cannot be synthesized by the human body).

Final Conclusion

Essential Amino acid (Y) = Threonine. Sequence = Thr-Ser-Asp-Gly-Ala.

Pattern Recognition

N-terminal reading is standard left-to-right. Memory aid for essential amino acids: 'PVT TIM HALL' (Threonine is the 'T').

Chapter Mix

Class 12 Chemistry: Biomolecules

Q61 jee_main_2026_28_january_evening Reducing And Non Reducing Sugars

Structures of four disaccharides are given below. Among the given disaccharides, the non-reducing sugar is :

A. $(1)$ $(1)$
B. $(2)$ $(2)$
C. $(3)$ $(3)$
D. $(4)$ $(4)$

Solution

Core Logic

For a sugar to be non-reducing, it must NOT possess a free hemiacetal or hemiketal group. The linkage connecting the monosaccharide units must consume both anomeric carbons via an acetal/ketal linkage. Structure (1) represents Sucrose, which has an $α, β -1,2$ $\alpha, \beta -1,2$-glycosidic bond. Both anomeric carbons are locked in the glycosidic linkage

Reducing And Non Reducing Sugars

and

Reducing And Non Reducing Sugars

. Structures (2), (3), and (4) are maltose or lactose-type structures containing a free anomeric -OH, thus acting as reducing sugars.

Step 1: Final Conclusion

Structure (1) is sucrose, which lacks a hemiacetal linkage, making it a non-reducing sugar.

Pattern Recognition

Whenever asked for a non-reducing sugar among standard disaccharides, look for Sucrose (linked at C1 of glucose and C2 of fructose). Both anomeric carbons are tied up.

Chapter Mix

Class 12 Chemistry: Biomolecules

Q44 jee_main_2025_02_april_evening Proteins and Amino Acid Sequences

A tetrapeptide "x" on complete hydrolysis produced glycine (Gly), alanine (Ala), valine (Val), leucine (Leu) in equimolar proportion each. The number of tetrapeptides (sequences) possible involving each of these amino acids is

A. 16
B. 32
C. 8
D. 24

Solution

Related Formula

Number of unique sequences = n!

Core Logic

A tetrapeptide is formed by connecting four amino acids through three peptide linkages.

Since the problem specifies that complete hydrolysis of the tetrapeptide produces Gly, Ala, Val, and Leu in equimolar proportions, the peptide must contain exactly one molecule of each of these four distinct amino acids.

Step 1: Calculate the Permutations

The number of unique peptide sequences corresponds to the number of ways we can arrange these 4 distinct amino acids:

Number of permutations = 4! = 4 \times 3 \times 2 \times 1 = 24

Pattern Recognition

Combinatorics in Chemistry: If we have $n$ $n$ unique, non-repeating amino acids, the number of linear isomeric peptides is $n!$ $n!$. If repetition were permitted, the number of possible peptides would be $nⁿ$ $n^n$ (which would be $4⁴ = 256$ $4^4 = 256$ in this case).

Chapter Mix

Class 12 Chemistry: Biomolecules

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