Question

An enzyme has a Km of 4.7X10^-5 M. If the Vmax of the preparation is (22 micromoles X liters^-1 Xmin^-1). What velocity would be observed in the presence of 2X10^-4 M substrate and 5X10^-4M of

a. a competitive inhibitor

b. a noncompetitive inhibitor

c. an uncompetitive inhibiter

Ki in all three cases is 3X10^-4M. What is the degree of inhibition in all three cases?

Answer 1

Km= 4.7*10-5M and Vmax= 22*10^-6 M/min, Ki= 3*10-4 M

in the absence of inhibitor V= Vmax[S]/ (KM+S ) = 22*10^-6*2*10^-4/ (4.7*10^-5+2*10^-4)= 1.78*10-5 M/min

V= Vmax [S]/ [KM(1+I/Ki)+S] = 22*10^-6* 2*10^-4/ 4.7*10^-5(1+5*10^-4/3*10^-4)+2*10^-4)=1.352*10-5 M/min

degres of inhibition = (V0-V)/Vo, Vo = rate in the absence of inhibitor and V= rate in the presence of inhibitor

degress of inhibition = (1.78- 1.352)/1.78 = 0.244

b)

For competitive inhibition

b) non competivie inhibition

Non competitive

V= Vmaxapp [S]/ (KM+S)

Vmax app = Vmax/ (1+I/KI)

= 22*10^-6/ (1+5*10-4/3*10-4)= 8.24*10-6 M/min

V= 8.24*10^-6* 2*10^-4/ (4.7*10^-5 +2*10^-4)=6.672*10- 6M/min

degree of inhibition = (1.78-0.6672)/1.78= 0.625

c)Uncompetitive inhibition

1/V= KM/VmaxS + (1+I/Ki)/ Vmax

1/V= (4.7*10^-5/22*10^-6*2*10^-4 + (1+5*10^-4/3*10^-4)/22*10^-6

V= 7.5*10^-6 M/min'

degree of inhibtion = (1.78-0.75)/1.78 =0.588