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 absence and presence of 2.5 mM inhibitor(I) Provide two plots; V versus[S] and I/V versus I/[s] What are the values of Vmax and Km in the absence of inhibitor? What are the values of Vmax and Km in the presence inhibitor? What type of inhibition is this? What is the binding constant of this inhibitor (Ki)? What percentage of enzyme is bound to substrate when the [S] equals 1 5 mM: in the absence of inhibitor? in the presence of inhibitor (2.5 mM)? What percentage of enzyme is bound to substrate when the [S] equals 2.0 mM in the presence of inhibitor (5.0 mM)?

Explanation / Answer

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

Vmax= maximum reaction rate, S= Substrate concentration and KM= constant

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

so a plot of 1/V vs 1/S gives a straight line with slope of KM/Vmax and intercpet of 1/Vmax

The plots for both the cases is drawn and shown below

in case of inhibition Km becomes Kmapp

a)

with inhibition

From the plots 1/Vmax= 19403 for inhibition and Vmax= 1/19403 =5.15*10-5 moles/s ( inhibtion)

KMapp/Vmax= 20.199 and Km= 5.15*10-5*20.199 , KMapp= 0.00104 M

without inhibition

Vmax= 1/19371= 5.15*10-5 and Km= 11.604*5.15*10-5 =0.000599 M

since Vmax is same the inhibition is competitive

V/vmax = [S]/ {Km[1+I/Ki] +S } = 2*10-3/ (0.0010

0.00104= 0.000509*(1+ I/Ki)

I+I/Ki= 1.74

I/Ki= 0.74, I is concentration of Enzyme

KI= I./0.74 =2.5*10-3/0.74=0.00144 M

V/mvax= [S] / [KM+S] = fractino of enzyme bound for no inhibition

in case of   no inhibition V/Vmax= 1.5*10-3/ (1.5*10-3+0.000599) =0.71 i.e 71%

in case of inhibition V/Vmax = S/ {[KM*(1+I/Ki)] +S } = 2.5*10-3/ (0.000599*(1+5*10-3/0.00144)=0.93 i.e 93%

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