JEE Main · Chemistry ↓ Falling

Biomolecules appeared 36 times across 3 years — 4.2% of Chemistry. This question is from Proteins and Enzymes.

Year 2026 2025 2024 Total
Questions 9 19 8 36

The correct statements are: A. Activation energy for enzyme catalysed hydrolysis of sucrose is lower than that of acid catalysed hydrolysis. B. During denaturation, secondary and tertiary structures of a protein are destroyed but primary structure remains intact. C. Nucleotides are joined together by glycosidic linkage between C₁ and C₄ carbons of the pentose sugar. D. Quaternary structure of proteins represents overall folding of the polypeptide chain. Choose the correct answer from the options given below:

Solution & Explanation

Core Logic
  • A is correct: Enzymes lower activation energy significantly.
  • B is correct: Denaturation affects secondary and tertiary structures while primary structure remains intact.
  • C is incorrect: Nucleotides are joined by phosphodiester linkages, and glycosidic linkages connect bases to sugars.
  • D is incorrect: Quaternary structure represents aggregation of multiple polypeptide subunits, not single chain folding.
Step 1: Final Conclusion

Only statements A and B are correct, matching option (3).

Pattern Recognition

Sees: Biomolecules structural and enzymatic properties. Trap: Confusing primary structure disruption with denaturation.

Chapter Mix

Class 12 Chemistry: Biomolecules

Reference Study Guides

More Biomolecules Previous-Year Questions — Page 8

Q78 jee_main_2024_31_jan_morning Reactions of Glucose
Match List I with List II
LIST-ILIST-II
A. Glucose/NaHCO₃/ΔI. Gluconic acid
B. Glucose/HNO₃II. No reaction
C. Glucose/HI/ΔIII. n-hexane
D. Glucose/Bromine waterIV. Saccharic acid
Choose the correct answer from the options given below:
  • A. A-IV, B-I, C-III, D-II
  • B. A-II, B-IV, C-III, D-I
  • C. A-III, B-II, C-I, D-IV
  • D. A-I, B-IV, C-III, D-II

Solution

Core Logic

Matching the reactions of glucose: (A) Glucose does not react with NaHCO₃, so there is no reaction. (A arrow II) (B) Oxidation of glucose with strong oxidizing agents like HNO₃ yields a dicarboxylic acid called saccharic acid. (B arrow IV) (C) Prolonged heating of glucose with HI forms n-hexane, indicating a straight chain of six carbon atoms. (C arrow III) (D) Oxidation with mild agents like bromine water converts glucose to gluconic acid. (D arrow I)

Chapter Mix

Class 12 Chemistry: Biomolecules

More Biomolecules Questions — jee_main_2026_21_jan_evening

Practice all Biomolecules previous-year questions →

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)