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Enzymes and investigations

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Which statement describes induced fit?

  1. The active site remains completely rigid during substrate binding.
  2. The enzyme is permanently consumed by its first substrate.
  3. Substrate binding changes the active-site conformation.
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Substrate binding changes the active-site conformation.

This can improve alignment and stabilise the transition state.

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Why buffer an enzyme pH investigation?

  1. To maintain the intended pH during the reaction
  2. To prevent the substrate binding to the active site
  3. To keep the reaction temperature constant
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To maintain the intended pH during the reaction

This reduces changes caused by the reaction itself.

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Which is NOT a reason why enzyme activity falls when pH moves far from the optimum?

  1. Ionic bonds holding tertiary structure are broken
  2. Enzyme molecules collide less often with substrate
  3. Charges on R groups in the active site change
  4. The substrate itself may change its charge
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Enzyme molecules collide less often with substrate

Collision frequency depends mainly on temperature and concentration, not pH. pH acts on charges and bonds.

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Doubling enzyme concentration doubles rate most plausibly when...

  1. substrate remains in excess.
  2. there is no substrate.
  3. all enzyme molecules have been irreversibly denatured.
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substrate remains in excess.

Additional active sites can then form complexes and catalyse reactions.

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In the simple non-competitive model, adding much more substrate...

  1. displaces the inhibitor from the same active site.
  2. always restores the original maximum rate.
  3. does not restore the original maximum rate.
  4. pushes the inhibitor off its separate binding site.
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does not restore the original maximum rate.

Binding at a separate site reduces catalytic capacity even with abundant substrate.

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At low substrate, inhibitor X reduces the rate by half. At very high substrate, the rate with X equals the rate without it. Which conclusion is best supported?

  1. X denatures the enzyme permanently.
  2. X is a second substrate for the same enzyme.
  3. X competes with the substrate for the active site.
  4. X binds an allosteric site and lowers Vmax.
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X competes with the substrate for the active site.

Excess substrate outcompetes a competitive inhibitor, so the maximum rate is restored, as the data show.

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Question 1 of 8

Explain end-product inhibition

5 marks

In an invented pathway, enzyme 1 converts substance A to B, and later enzymes convert B to the end product Z. Enzyme 1 has four subunits. Its rate against substrate concentration is S-shaped, and the curve shifts to the right when Z is added. (a) Explain the S-shaped curve. (b) Explain how Z controls the pathway and why this benefits the cell. (c) Explain why adding more A does not fully restore the rate when Z is high. [5]

Cooperative binding between subunits; Z at an allosteric site; negative feedback.

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