Enzyme kinetics

BIOCHEMISTRY 304
Enzyme Kinetic Sample Problems #1
September 2004

1 Given the reaction k1 kp
E + S  ES  E + P k-1

where k1 = 1 x 107 M-1 sec-1 k-1 = 1 x 102 sec-1, and kp = 3 x 102 sec-1

a) Calculate Ks
b) Calculate Km

(a) k-1 1 x 102 sec-1
Ks = k1 = 1 x 107 M-1 sec-1 = 1 x 10-5 M

(b) k-1 + kp (1 x 102 sec-1) + (3 x 102 sec-1)
Km = k1 = 1 x 107 M-1 sec-1 = 4 x 10-5 M

2) An enzyme was assayed at an initial substrate concentration of 2 x 10-5 M. In 6 minutes half of the substrate had been consumed. Km for this enzyme’s substrate is 5 x 10 -3 M.

a) Calculate k
b) Calculate Vmax
c) Calculate the concentration of product produced in 15 minutes.

SOLUTIONS

One can by inspection come to the conclusion that this reaction is first-order. This can be seen as follows: [S] [2 x 10-5]
Is Km < 0.01 ? [5 x 10 -3] = .004 Yes, first order

a) Since this is a first order reaction we can use the following equation to determine the rate constant, k.

0.693 0.693 0.693 k = t1/2 k = 6 min k = 6 min k = 0.115 min-1

b) k = Vmax / Km

Vmax = k x Km

Vmax = 0.115 min-1 x 5 x 10 -3 M = 0.575 x 10 -3 M min-1 Vmax = 0.575 x 10 -3 mole liter-1 min-1 = 575 moles liter-1 min-1

c) We know the initial concentration, [S] o, and we have calculated the rate constant, k. We need to determine an arbitrary [S] at 15 minutes after the initial concentration for a first order reaction.

[S] o 2.303 log [S] =kt

[2 x 10-5] 2.303 log [S] =