Given below are two statements: Statement (I): On hydrolysis, oligo peptides give rise to fewer number of alpha-amino acids while proteins give rise to a large number of beta-amino acids. Statement (II): Natural proteins are denatured by acids which convert the water soluble form of fibrous proteins to their water insoluble form. In the light of the above statements, choose the most appropriate answer from the options given below:

Statement (I) is incorrect because the complete hydrolysis of both oligopeptides and proteins yields alpha-amino acids, not beta-amino acids. Statement (II) is incorrect because fibrous proteins are inherently water-insoluble structural materials. Denaturation typically disrupts the tertiary and secondary structures of water-soluble globular proteins, making them insoluble. Since both Statement I and Statement II are false, option (3) is the correct choice. All naturally occurring proteins are polymers of alpha-amino acids, so any statement mentioning beta-amino acids as direct translation products can be confidently ruled out. Fibrous proteins (like keratin or collagen) are structural and always insoluble, unlike globular proteins. Class 12 Chemistry: Biomolecules

Given below are two statements: Statement (I): On hydrolysis, oligo peptides give rise to fewer number of alpha-amino acids while proteins give rise to a large number of beta-amino acids. Statement (II): Natural proteins are denatured by acids which convert the water soluble form of fibrous proteins to their water insoluble form. In the light of the above statements, choose the most appropriate answer from the options given below:

Statement (I) is incorrect because the complete hydrolysis of both oligopeptides and proteins yields alpha-amino acids, not beta-amino acids. Statement (II) is incorrect because fibrous proteins are inherently water-insoluble structural materials. Denaturation typically disrupts the tertiary and secondary structures of water-soluble globular proteins, making them insoluble. Since both Statement I and Statement II are false, option (3) is the correct choice. All naturally occurring proteins are polymers of alpha-amino acids, so any statement mentioning beta-amino acids as direct translation products can be confidently ruled out. Fibrous proteins (like keratin or collagen) are structural and always insoluble, unlike globular proteins. Class 12 Chemistry: Biomolecules

Given below are two statements: Statement (I): On hydrolysis, oligo peptides give rise to fewer number of alpha-amino…

Year	2026	2025	2024	Total
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Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

Protein Hydrolysis Peptides Hydrolysis α-amino acids

Core Logic

Statement (I) is incorrect because the complete hydrolysis of both oligopeptides and proteins yields $α$ $\alpha$-amino acids, not $β$ $\beta$-amino acids.

Statement (II) is incorrect because fibrous proteins are inherently water-insoluble structural materials. Denaturation typically disrupts the tertiary and secondary structures of water-soluble globular proteins, making them insoluble.

Step 1: Final Conclusion

Since both Statement I and Statement II are false, option (3) is the correct choice.

Pattern Recognition

All naturally occurring proteins are polymers of $α$ $\alpha$-amino acids, so any statement mentioning $β$ $\beta$-amino acids as direct translation products can be confidently ruled out. Fibrous proteins (like keratin or collagen) are structural and always insoluble, unlike globular proteins.

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Class 12 Chemistry: Biomolecules

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).

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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 $n!$ $n!$.

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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').

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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

and

. 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.

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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).

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Class 12 Chemistry: Biomolecules

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More Biomolecules Previous-Year Questions — Page 2

Solution

Core Logic

Step 1: Identifying the specific sugar

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Identify the Essential Amino Acid

Final Conclusion

Pattern Recognition

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Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

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Solution

Related Formula

Core Logic

Step 1: Calculate the Permutations

Pattern Recognition

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