Biomolecules (amino acids, peptides, proteins, enzymes, and genetic chemistry) are fundamental to both CBSE and competitive exams because they link structure to function. Questions here test core quantitative ideas like – calculations, peptide/charge behavior, enzyme inhibition kinetics, protein denaturation/disulfide chemistry, and classic DNA mutation logic—exactly the skills needed for high-scoring board and exam performance.
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Minutes
20
Questions
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Marking
Q1. The amino acid glycine has (carboxyl) and (amino). Using , the isoelectric point of glycine is approximately:
Q2. A freshly prepared solution of pure -D-glucose has observed specific rotation , pure -D-glucose has , and the equilibrium rotation is . At equilibrium, the percentage of -anomer present is closest to:
Q3. An enzyme catalyses reactions of two substrates S and S with the following parameters: for S and for S. Which statement correctly describes which substrate is preferred at very low substrate concentration () and which gives higher rate at saturating substrate concentration?
S preferred at low ; S gives higher rate at saturation
S preferred at low ; S gives higher rate at saturation
S preferred at low ; S gives higher rate at saturation
S preferred at low ; S gives higher rate at saturation
Q4. Assertion (A): In a Lineweaver–Burk plot (plot of vs ) a pure noncompetitive inhibitor produces the same x-intercept as the uninhibited reaction.
Reason (R): A noncompetitive inhibitor increases the apparent while decreasing .
Both A and R are true and R is the correct explanation of A.
Both A and R are true but R is not the correct explanation of A.
A is true but R is false.
A is false but R is true.
Q5. Assertion (A): -D-glucopyranose is thermodynamically more stable than -D-glucopyranose in the chair conformation.
Reason (R): In the -anomer the hydroxyl at the anomeric carbon is equatorial, minimizing 1,3-diaxial steric interactions; in the -anomer that hydroxyl is axial.
Both A and R are true and R is the correct explanation of A.
Both A and R are true but R is not the correct explanation of A.
A is true but R is false.
A is false but R is true.
Q6. Consider alanine with and . What is the net electric charge on alanine at ?
Q7. The dipeptide has ionizable groups with , and . Calculate its isoelectric point (pI).
Q8. The peptide sequence is . Trypsin cleaves peptide bonds after Arg and Lys (unless followed by Pro). Which set of fragments will result on complete digestion by trypsin?
and
and
and
, , and
Q9. Assertion (A): A globular protein treated with 8 M urea becomes partially unfolded but may still retain covalent disulfide bonds; subsequent addition of -mercaptoethanol (-ME) leads to complete unfolding into separate polypeptide chains.
Reason (R): 8 M urea disrupts noncovalent interactions (hydrogen bonds and hydrophobic packing) but does not cleave disulfide bonds, whereas -ME chemically reduces S–S bridges to thiol groups, breaking covalent cross-links.
Both A and R are true and R is a correct explanation of A.
Both A and R are true but R is not a correct explanation of A.
A is true but R is false.
A is false but R is true.
Q10. An enzyme follows Michaelis–Menten kinetics with and in absence of inhibitor. In presence of of a reversible inhibitor the apparent becomes while is unchanged. Assuming competitive inhibition, the inhibition constant (in mM) is:
Q11. Glycine has and . Using , the isoelectric point of glycine is closest to:
Q12. Consider the tripeptide with values , , , , . Using Henderson–Hasselbalch where fraction protonated , the approximate net charge of this peptide at is:
Q13. For the dipeptide the ionizable groups have values , , . The isoelectric point of this dipeptide is best approximated by:
Q14. Assertion (A): Acetylation of lysine residues in histone proteins reduces their net positive charge and weakens histone–DNA interaction. Reason (R): Acetylation converts the -amino group of lysine from positively charged to a neutral amide , thereby reducing electrostatic attraction to the negatively charged DNA backbone.
Both A and R are true but R does not explain A.
A is true but R is false.
Both A and R are true and R explains A.
Both A and R are false.
Q15. During DNA replication a rare imino tautomer of cytosine pairs with adenine instead of guanine. After two rounds of replication the original pair is ultimately converted into which fixed base pair?
...and 5 more challenging questions available in the interactive simulator.