Determine the value of the Michaelis constant, K_m, including units, from the graph.

Sketch a second graph on the same axes to show how the reaction rate varies when a competitive inhibitor is present.

Outline the significance of the value of K_m.

«K_m = [substrate] at \(\frac{1}{2}\) V_max»

4.2 x 10^–3

mol dm^–3

Accept answers in the range of 3.5 x 10^–3 to 5.0 x 10^–3 mol dm^–3.

M2 can be scored independently.

[2 marks]

graph to right of curve AND finish at same V_max

Do not penalize if curve does not finish exactly at same V_max as long as it is close to it (since drawn curve does not
flatten out completely at V_max = 0.50).

[1 mark]

K_m is inverse measure of affinity of enzyme for a substrate / K_m is inversely proportional to enzyme activity
OR
high value of K_m indicates higher substrate concentration 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.

Accept “high value of K_m indicates low affinity of enzyme for substrate/less stable ES complex/lower enzyme activity”.

Accept “low value of K_m indicates high affinity of enzyme for substrate/stable ES complex/greater enzyme activity”.

[1 mark]