Biology

1. 1.

   Explain how peptide bonds form and identify the small molecule released.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Joining monomers removes an OH and an H as water and creates a covalent link. Hydrolysis is the reverse process, consuming water to break that link.

   Marking points

   • The amino group of one amino acid reacts with the carboxyl group of another.
   • A condensation reaction forms a peptide bond.
   • Water is released.

   Examiner tip: Do not describe condensation as adding water.

2. 2.

   Explain why a severe change in pH can reduce an enzyme's activity without breaking all peptide bonds.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Activity depends on a correctly folded and chemically suitable active site. Altered ionic interactions can change folding while leaving the primary amino-acid sequence largely intact.

   Marking points

   • pH alters ionisation of amino-acid side chains.
   • Interactions maintaining tertiary structure can be disrupted.
   • The active site's shape/charge changes, reducing substrate binding or catalysis.

   Examiner tip: Denaturation is not synonymous with complete peptide hydrolysis.

3. 3.

   An enzyme produces 18 micromol of product in the first 30 s. Calculate its mean rate in micromol min^-1 and explain why later measurements may underestimate initial rate.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Convert the time unit by multiplying the per-second rate by sixty. The early average is only an estimate of initial rate; a later interval may include a slowing reaction as conditions change.

   Marking points

   • Mean rate = 18/30 = 0.60 micromol s^-1.
   • Rate = 36 micromol min^-1.
   • Substrate depletion/product accumulation can reduce later rate.

   Examiner tip: A mean over 30 seconds is not necessarily the instantaneous rate at zero time.

4. 4.

   Compare competitive inhibition with pure noncompetitive inhibition in terms of substrate binding, Km, Vmax and the effect of very high substrate concentration.

   [5 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Competition reduces substrate occupancy at a given concentration, but enough substrate outcompetes the inhibitor and the original Vmax remains attainable. In pure noncompetitive inhibition, equal inhibitor affinity for E and ES leaves Km unchanged: substrate binding is still possible, but inhibitor-bound enzyme is catalytically ineffective, lowering Vmax even at saturating substrate. Unequal affinity would describe mixed rather than pure noncompetitive inhibition.

   Marking points

   • A competitive inhibitor competes for the active site.
   • High substrate concentration can overcome competitive inhibition; apparent Km rises but Vmax is unchanged.
   • A pure noncompetitive inhibitor binds free enzyme E and the enzyme-substrate complex ES with equal affinity, so substrate can still bind.
   • Km is unchanged in the pure noncompetitive model.
   • Vmax is lower, and increasing substrate cannot restore it.

   Examiner tip: Do not infer failure of substrate binding from lower Vmax; equal affinity for E and ES is the defining pure-model assumption.

5. 5.

   A substitution replaces a charged amino acid with a nonpolar one far from an enzyme's active site. Explain how this could still affect activity and why the sequence change alone cannot predict the exact effect.

   [4 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. An active site is part of a connected three-dimensional protein. Distant residues may stabilise it indirectly, but some substitutions are tolerated, so a plausible mechanism is not a measured effect.

   Marking points

   • Side-chain interactions contribute to folding.
   • Replacing charge can alter ionic interactions and local stability.
   • A folding change can propagate to active-site geometry/dynamics.
   • The effect depends on structural context; activity and structural measurements are needed.

   Examiner tip: Do not assert every amino-acid substitution denatures the whole protein.

6. 6.

   Design an investigation to demonstrate reversible slowing of catalase at low temperature and distinguish it from persistent activity loss after high-temperature exposure. Include both treatment-temperature and common-temperature assays using matched aliquots.

   [5 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Split the same enzyme preparation into matched aliquots so separate assays do not reuse substrate-depleted mixtures. First compare the low-temperature rate with the control-temperature rate to demonstrate actual cold slowing. Then assay unused cold-, heat- and control-pretreated aliquots at a common temperature with fresh substrate. Recovery of the cold-pretreated sample distinguishes reversible kinetic slowing from persistent loss after heating. Common-temperature assays alone show retained activity, not that catalysis was slower while cold; persistent activity loss supports damage but does not by itself prove its molecular mechanism.

   Marking points

   • Pre-incubate matched enzyme aliquots at low, high and control temperatures for equal time.
   • Assay one cold aliquot at low temperature and a matched control aliquot at control temperature to establish cold slowing.
   • Equilibrate unused matched aliquots from each pretreatment to the same control assay temperature and add fresh substrate to test recovery.
   • Compare replicated initial oxygen-production rates with equal enzyme amount, pH and substrate concentration.
   • A lower cold-assay rate followed by recovery at common temperature supports reversible slowing; persistently low heat-treated activity supports lasting damage.

   Examiner tip: Use separate matched aliquots for cold-rate and recovery assays; depleted substrate must not masquerade as irreversible enzyme damage.

Original, unofficial practice questions from SubjectScout — not official past-paper questions.