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

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

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

Â Â Â Â Â Â Â Â Â Â Â Hence, Vo = 99.5025 umol min^-1

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

Now,Â Â

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

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = (99.5025 umol min^-1 â€“ 20.0 umol min^-1) / 99.5025 umol min^-1Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.7990

% of inhibited enzymes = fraction of inhibited enzymes x 100

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.7990 x 100

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 79.90 %

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