An enzyme was obtained in the lab. To study the kinetics of this enzyme, a gradu

Question

An enzyme was obtained in the lab. To study the kinetics of this enzyme, a graduate student measured the velocity of the reaction at a substrate concentration of 0.1 mM substrate. He found that after 1 mM competitive inhibitor was added, the initial rate decreased to 1/4 of the original rate. KM, the Michaelis constant, is 0.5 mM. What is the KI, the dissociation constant for the inhibitor binding to the enzyme?

(Hint: in the presence of the competitive inhibitor,

A. 0.33 mM?

B. 0.45 mM?

C. 0.14 mM

D. 0.28 mM

E. 0.9 mM

Explanation / Answer

[S] = 0.1 mM

[I] = 1 mM

Vo = 1/4 * Vmax

KM = 0.5 mM

We have Vo = Vmax * [S] / { (1 + [I] / KI)KM + [S] }

1/4 * Vmax = Vmax * 0.1 / { (1 + 1 / KI) * 0.5 + 0.1 }

1/4 = 0.1 / { (1 + 1 / KI) * 0.5 + 0.1 }

{ (1 + 1 / KI) * 0.5 + 0.1 } = 0.4

(1 + 1 / KI) * 0.5 = 0.3

1 + 1 / KI = 3/5

1 / KI = 3/5 - 1

1 / KI = - 2/5

KI = - 2.5

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