Biomolecules is a high-yield chapter because it connects core organic/ionic concepts (acids–bases, functional groups, stereochemistry) to biologically important molecules like amino acids, proteins, enzymes, and nucleic acids. For board and competitive exams, the key scoring areas are pI/pKa reasoning, amino acid ionization states at different pH, peptide/protein electrophoresis logic, and enzyme kinetics/free-energy links—these frequently appear as direct numerical or conceptual MCQs.
20
Minutes
15
Questions
1 / -0
Marking
Q1. The pKa values for glycine are and . Using , the isoelectric point () of glycine is nearest to:
4.17
5.97
6.80
7.47
Q2. An enzyme follows Michaelis–Menten kinetics with and . After adding a reversible inhibitor, the Lineweaver–Burk plot slope increases threefold while the y‑intercept is unchanged. At the ratio of reaction rates with inhibitor to without inhibitor, , is closest to (use ):
0.33
0.25
0.67
0.50
Q3. Lysine has , (α‑amino), and . The isoelectric point of lysine is closest to:
9.74
5.56
6.36
8.35
Q4. Consider the pentapeptide NH2–Gly–Lys–Arg–Asp–Glu–COOH. Given , , , , , . In an electrophoresis experiment at pH this peptide will most likely:
Migrate toward the cathode (negative electrode) because it is net positively charged
Migrate toward the anode (positive electrode) because it is net negatively charged
Remain near the origin (no net migration) because its net charge ≈ 0
Migrate slightly toward the cathode due to partial protonation of some residues
Q5. Two proteins X and Y have identical molecular weight () but different isoelectric points: and . Predict their behaviour in native PAGE at and in SDS–PAGE (denaturing):
In native PAGE both migrate toward the anode (pH > both pI); in SDS–PAGE Y migrates faster because it has more charged side chains
In native PAGE X (pI 4.5) will be negatively charged and move toward the anode while Y (pI 9.0) will be positively charged and move toward the cathode; in SDS–PAGE both acquire similar negative charge-to-mass ratio and therefore co‑migrate
In native PAGE both remain at origin (net neutral at pH 8.0); in SDS–PAGE X migrates faster because SDS binds more strongly to acidic proteins
In native PAGE X moves to cathode and Y to anode; in SDS–PAGE they separate according to intrinsic charge rather than mass
Q6. Glycine has and . Using , the isoelectric point of glycine is approximately:
Q7. Consider the tripeptide (free N‑ and C‑termini). Given , , , . At pH , what is the approximate net charge on the peptide? (Treat His as 50% protonated at its .)
Q8. Aspartic acid has , , . Using the Henderson–Hasselbalch relation , estimate the net charge of aspartic acid at pH (two decimal places).
Q9. Consider the peptide with free termini. Given , , , :
(a) Calculate the of the unmodified peptide.
(b) After phosphorylation of Ser an additional acidic group (with ) is introduced (neutral below its , above). What is the approximate after phosphorylation?
Choose the correct pair (unmodified , phosphorylated ).
Q10. Two 20‑base pair DNA duplexes A and B have %GC of and , respectively. Using the simplified melting temperature formula
where is duplex length and is base‑10, calculate the difference at (i) M and (ii) M. Which option is correct?
Difference at 0.05 M and at 0.20 M
Difference at both 0.05 M and 0.20 M
Difference at 0.05 M and at 0.20 M
Difference at both concentrations
Q11. A zwitterionic amino acid X has and . At pH , which ionic form of X predominates?
Fully protonated form
Fully deprotonated form
Zwitterionic form
Neutral uncharged form
Q12. Pure - and -D-glucopyranose have specific rotations and . An aqueous sample shows . Using , the approximate equilibrium composition is:
Q13. A polypeptide contains 4 Lys, 5 Glu and 3 His residues. At pH , take , , . Assume the N-terminal is protonated () and the C-terminal is deprotonated (). Estimate the net charge of the polypeptide at pH (use fraction protonated of His ).
Approximately
Q14. An enzyme-catalysed reaction has and . In the presence of an inhibitor at the observed is unchanged but . If the inhibitor is purely competitive, the inhibition constant is (use ):
Q15. An enzyme increases the rate of a reaction by a factor of compared to the uncatalysed rate at (). Estimate how much the activation free energy is lowered by the enzyme, (use and ).