Question

An enzyme with a Vmax of 100 umol/minute and a Km of 10 uM was reacted with a irreversible active site specific inhibitor. After reaction with the inhibitor, the enzyme was assayed using a 2 mM concentration of substrate, and it gave a reaction rate of 20 umol/min. What percentage of the enzyme did the inhibitor inactivate?

Answer 1

Ans. Step 1: Initial uninhibited Vo = V_max [S] / (K_m + [S])       - MM equation

            Or, Vo = (100 umol min^-1) x 2000 uM / (10 uM + 2000 uM)

            Hence, Vo = 99.5025 umol min^-1

# Step 2: Given, reaction velocity in presence of inhibitor, Voi = 20 umol min^-1

Now,  

Fraction of inhibited enzymes = (Vo - Voi) / Vo

                                    = (99.5025 umol min^-1 – 20.0 umol min^-1) / 99.5025 umol min^-1      

                                   = 0.7990

% of inhibited enzymes = fraction of inhibited enzymes x 100

                                    = 0.7990 x 100

                                    = 79.90 %