Question

An enzyme has a Km for substrate of 10 mM and Vmax of 5 mol L-1 sec-1 at a total enzyme
concentration of 1 nM. At [S] = 10 mM, kcat is:
A) 2500 per M per sec.
B) 5000 per M per sec.
C) 1250 per M per sec.
D) 2500 per sec.
E) 5000 per sec.

Answer 1

We can use the following equation

Vmax = Kcat x [E]

Therefore ,

Kcat = Vmax / [E]

= 5 x 10 -6 / 10-9

= 5000 per sec.

Here it should be noted that the unit of Kcat is per sec.