Biomolecules is an important Class 12 Chemistry chapter because it links chemical concepts (acids–bases, peptides/proteins, kinetics, electrophoresis) with biological structures (amino acids, peptides, DNA). Board and competitive exams frequently test the ability to calculate ionization/protonation states, isoelectric points, net charges at given pH, hydrogen bonding in DNA, and to apply Michaelis–Menten kinetics—so mastery of this chapter is high-impact for scoring well.
15
Minutes
10
Questions
1 / -0
Marking
Q1. Glycine has and . The isoelectric point of glycine is
Q2. Consider the peptide . Given values: , , , . The net charge of this peptide at pH is
Q3. A double-stranded DNA fragment contains base pairs and has GC content . If each A–T pair forms H‑bonds and each G–C pair forms H‑bonds, the total number of hydrogen bonds in the duplex is
Q4. Assertion (A): A lysine residue buried in the hydrophobic core of a folded protein has a lower than the same lysine exposed to aqueous solvent.
Reason (R): Burial in a nonpolar environment destabilizes the positively charged (protonated) form because of the large desolvation penalty and lack of stabilizing solvent interactions; therefore the neutral (deprotonated) form is favoured and its decreases.
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. The peptide has values: , , , , , . The isoelectric point of this peptide is approximately
Q6. An enzyme follows Michaelis–Menten kinetics with and . Using , calculate the initial rate when .
Q7. A mixture contains glycine, lysine and glutamic acid at pH . Given values: glycine (), lysine (), glutamic acid (). Which amino acid will migrate towards the anode (positive electrode) in electrophoresis at pH ?
Glutamic acid
Glycine
Lysine
Both glycine and glutamic acid
Q8. An enzyme has and . In presence of a competitive inhibitor the apparent increases to while remains unchanged. If the initial rate at without inhibitor is to be restored in the presence of the inhibitor, what substrate concentration is required? (Use .)
Q9. Consider the dipeptide H–Glu–Lys–OH (N‑terminus = Glu, C‑terminus = Lys). Use these values: N‑terminal –NH , C‑terminal –COOH , Glu side‑chain COOH , Lys side‑chain NH . What is the net charge of this dipeptide at pH ?
+2
0
−2
−1
Q10. A catalytic histidine in an enzyme acts as a general base; its in the wild‑type enzyme is . A mutation changes the microenvironment and raises the histidine to . At pH , the fraction of deprotonated histidine is given by . Assuming the catalytic rate is directly proportional to the fraction of deprotonated histidine, what approximate fraction of the original catalytic activity remains after the mutation?
of the original activity
of the original activity
of the original activity
Activity remains essentially unchanged