Regarding enzyme kinetics

Question

According to the Michaelis Menten equation, at what percentage of its maximum rate is an enzyme operating when [S] = ½ KM?

Answer 1

M-M equation is v = Vmax[S]/Km + [S] and by definition Km = [S] for 1/2 maximal velocity.

 

Set [S] = 1/2Km and solve for v as a fraction of Vmax, i.e. solve for v/Vmax.

From v = Vmax[S]/Km + [S]

v/Vmax = [S]/Km+[S] = 0.5Km/Km+0.5Km = 0.33Km

So, the answer would be that when [S] = 1/2 Km, the velocity will be 33% of the max. velocity.