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Date	November 2020	Marks available	1	Reference code	20N.3.hl.TZ0.10
Level	HL	Paper	3	Time zone	TZ0
Command term	Identify	Question number	10	Adapted from	N/A

Question

The kinetics of an enzyme-catalysed reaction are studied in the absence and presence of an inhibitor. The graph represents the initial rate as a function of substrate concentration.

Identify the type of inhibition shown in the graph.

[1]

Determine the value of $V_{\max}$ and $K_{m}$ in the absence and presence of the inhibitor.

[3]

b(i).

Outline the significance of the value of the Michaelis constant, $K_{m}$ .

[1]

b(ii).

Markscheme

non-competitive «inhibition» ✔

✔✔✔

Award [3] for four values correct.
Award [2] for three values correct.
Award [1] for two values correct.
Ignore units.
Accept $\pm 0.1$ for $K_{m}$ and $V_{m a x}$ . No ECF applied.

b(i).

$K_{m}$ is an inverse measure of affinity of substrate for enzyme
OR
higher $K_{m}$ indicates higher substrate concentration is needed for enzyme saturation
OR
low value of $K_{m}$ means reaction is fast at low substrate concentration ✔

Idea of “inverse relationship” must be conveyed.

b(ii).

Examiners report

Most scored the one mark here for identifying the correct type of inhibition, namely, non-competitive inhibition.

The stronger candidates scored all three marks here for the four correct values. Most scored at least one mark for the V_max values in the absence and presence of the inhibitor. The Km values sometimes were outside the permitted ±0.1 range.

b(i).

b(ii).

Syllabus sections

Options » B: Biochemistry » B.7 Proteins and enzymes (HL only)

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