You measure the initial rate of an enzyme reaction as a function of substrate co

Question

You measure the initial rate of an enzyme reaction as a function of substrate concentration in the presence and absence of an inhibitor. The following data are obtained: 8. Vo -Inhibitor +Inhibitor 0.0001 0.0002 0.0005 0.001 0.002 0.005 0.01 0.02 0.05 0.1 0.2 50 71 83 91 96 98 17 29 50 67 80 91 95 98 100 100 100 100 100 a) What is the Vmax in the absence of inhibitor? b) What is the Km in the absence of inhibitor? c) When [S]- 0.0004, what will Vo be in the absence of inhibitor? d) when [S] =0.0004, what will Vo be in the presence of inhibitor? e) What kind of inhibitor is it likely to be?

Explanation / Answer

a) Vmax in the absence of inhibitor:

Slope= KM/Vmax and intercept= 1/Vmax

The intercept of the plot in the in the absence of the inhibitor= 0.01

Vmax= 1/0.01= 100

b) Km in the absence of inhibitor

Slope of the plot= 2E-06

Km/Vmax= 2E-06

=2E-6*100= 2E-4

Km= 2E-4

c) when [S]= 0.0004 Vo in the absence of inhibitor:

the linear equation from the plot for 1/vo Vs 1/[S] is

1/[Vo]=2E-6 (1/[S])+0.01

for [S]= 0.0004, on evaluating,

1/Vo=2*E-6*2500+0.01

Vo= 66.2

d) 1/Vo= 5E-6*2500+0.01

Vo= 44

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