An enzyme with a Vmax of 100 umol/minute and a Km of 10 uM was reacted...

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Biology

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

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3.9 Ratings (519 Votes)

Ans. Step 1: Initial uninhibited Vo = Vmax [S] / (Km + [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 %


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