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Enzymes: Question 10

Syllabus 3.2

Multiple choice AS 1 mark

Compared with the same enzyme-catalysed reaction with no inhibitor present, a competitive inhibitor and a non-competitive inhibitor have different effects on Vmax and on Km.

Which option correctly states these effects?

Choose an answer to check it, then compare with the worked solution below.
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Worked solution

Step 1: Recall where each inhibitor binds

A competitive inhibitor has a shape similar enough to the substrate to bind reversibly at the enzyme’s active site, directly blocking substrate from binding there. A non-competitive inhibitor binds at a separate site elsewhere on the enzyme, causing a conformational change that distorts the active site indirectly.

Step 2: Work out the effect on Vmax

Because a competitive inhibitor only occupies active sites some of the time, adding enough substrate lets substrate molecules out-compete the inhibitor for those active sites. Given a sufficiently high substrate concentration, the same maximum rate can still be reached, so Vmax is unchanged.

A non-competitive inhibitor, once bound, permanently disables the active sites of the enzyme molecules it affects (for as long as it remains bound), and no amount of extra substrate can displace it from its separate binding site. With a smaller pool of catalytically active enzyme available, the maximum rate that can ever be reached is lower, so Vmax is decreased.

Step 3: Work out the effect on Km

With a competitive inhibitor present, more substrate is needed to reach half of Vmax, because some of it is “used up” out-competing the inhibitor rather than immediately forming enzyme-substrate complexes. So the apparent Km is increased.

With a non-competitive inhibitor present, the active sites that remain functional are entirely unaffected, they still bind substrate exactly as readily as before. So the substrate concentration needed to reach half of the (now lower) Vmax is the same as without the inhibitor, meaning Km is unchanged.

Step 4: Evaluate each option

  • A: correct, competitive: Vmax unchanged, Km increased; non-competitive: Vmax decreased, Km unchanged.
  • B: incorrect, this swaps the two inhibitors’ effects around.
  • C: incorrect. A competitive inhibitor does raise Km; it is not unchanged.
  • D: incorrect. A competitive inhibitor does not lower Vmax, and it raises (not lowers) Km.

Final answer

  • Competitive inhibitor: Vmax unchanged, Km increased. Non-competitive inhibitor: Vmax decreased, Km unchanged, option A.