The kinetics of an enzyme are measured as a function of substrate concentration

Question

The kinetics of an enzyme are measured as a function of substrate concentration in the presence and in the absence of 2 mM inhibitor (I). What are the values of and in the absence of inhibitor? in its presence? What type of inhibition is it? What is the binding constant of this inhibitor? If |S] = 10 mu M and [I] = 2 mM. what fraction of the enzyme molecules have a bound substrate? A bound inhibitor? If [S] = 30 mu M, what fraction of the enzyme molecules have a bound substrate in the presence and in the absence of 2 mM inhibitor? Compare this ratio with the ratio of the reaction velocities under the same conditions.

Explanation / Answer

Michaelis-Menten kinetics can be written as

V= Vmax[S] /[KM+S]

1/V= (KM/Vmax)*1/S + 1/Vmax

so a plot of 1/V vs 1/S gives a straight line whose slope is KM/Vmax and intercept is 1/Vmax

The plots for with and wiithout inhibitor is shown below

In the absence of inhibitor , intercept = 1/Vmax= 0.0224, Vmax= 44.64 umole/min

KM/Vmax= slope = 0.224

KM= 0.224*44.64 umole =10 umole

in the presecne of inhibitor, Vmax= 1/0.022=44.64 and KMapp= 44.64*0.663=29.74 umole

sicne Vmax is same for both the cases, the inhibition is Competitive inhibition

for competitive inhibition

Kmapp =Km*(1+I/KI) where KI= inhibtion constant

29.74*10-6= 10*10-6 *(1+2*10-3/Ki)

2.974= 1+2*10-3/KI

2*10-3/Ki= 1.974

Ki= 2*10-3/1.974= 1.013*10-3 M

fraction of enzyme bound to substrate= S/ (KM+S) when there is no inhibitor

=10/(10+10)= 0.5

for competiive inhibition , fraction of enzyme bound = S/ {Km*(1+I/Ki)+S}

=10/(10*2.974+10)= 0.31

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