Biology

1. 1.

   Explain why a competitive inhibitor increases apparent Km but does not change Vmax in an enzyme-catalysed reaction.

   [3 marks] · no calculator

   Answer explanation

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

   1. Compare the enzyme's rate at low and very high substrate concentrations while keeping enzyme concentration fixed.
   2. The inhibitor competes with substrate for the active site.
   3. More substrate is needed to reach half of Vmax, so apparent Km rises.
   4. At sufficiently high substrate concentration, substrate outcompetes inhibitor and Vmax is unchanged.

   Marking points

   • The inhibitor competes with substrate for the active site.
   • More substrate is needed to reach half of Vmax, so apparent Km rises.
   • At sufficiently high substrate concentration, substrate outcompetes inhibitor and Vmax is unchanged.

   Examiner tip: Higher apparent Km does not mean a lower Vmax. This kinetic treatment may exceed some specifications.

2. 2.

   Contrast facilitated diffusion and active transport across a cell membrane.

   [3 marks] · no calculator

   Answer explanation

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

   1. Compare direction relative to the gradient and energy requirements, then identify the shared use of transport proteins.
   2. Both involve membrane transport proteins.
   3. Facilitated diffusion is down a concentration or electrochemical gradient and does not directly require metabolic energy.
   4. Active transport can move substances against a gradient and requires energy, directly or indirectly.

   Marking points

   • Both involve membrane transport proteins.
   • Facilitated diffusion is down a concentration or electrochemical gradient and does not directly require metabolic energy.
   • Active transport can move substances against a gradient and requires energy, directly or indirectly.

   Examiner tip: Facilitated diffusion uses proteins but remains passive; a protein does not automatically imply active transport.

3. 3.

   Two heterozygous parents (Aa x Aa) have a child. A recessive condition appears only in aa. Calculate the probability of an affected child and explain why an unaffected child can still carry the allele.

   [3 marks] · no calculator

   Answer explanation

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

   1. Each parent supplies A or a with equal probability. Combine the possibilities in a two-by-two genetic cross.
   2. The possible genotypes are AA, Aa, Aa and aa with equal probabilities.
   3. The probability of aa is 1/4.
   4. Aa is unaffected because A is dominant, but carries and can transmit a.

   Marking points

   • The possible genotypes are AA, Aa, Aa and aa with equal probabilities.
   • The probability of aa is 1/4.
   • Aa is unaffected because A is dominant, but carries and can transmit a.

   Examiner tip: An unaffected phenotype does not exclude a recessive allele; distinguish phenotype from genotype.

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