The kinetics of an enzyme and absence of an 2 mM of inhibitor A and 100 micromol
ID: 559687 • Letter: T
Question
The kinetics of an enzyme and absence of an 2 mM of inhibitor A and 100 micromolar of inhibitor B is measured as a function of substrate concentration in the presence Velocity umol /minute 0 Inhibitor Inhibitor Sl uMinhibitor [A 10.4 4.5 22.5 33.8 40.5 45.9 47.1 6.4 11.3 22.6 33.8 40.5 45.9 2.9 2.9 6.8 10 30 90 300 900 9.3 0.9 Plot to determine the following parameters: A) What are the Vm values in the absence and presence of inhibitors A and B? No enzyme: Inhibitor A: Inhibitor B: B) What are the Km values in the absence and presence of inhibitors A and B? No enzyme: Inhibitor A Inhibitor B C) What type of inhibition is being demonstrated? Justify By Inhibitor A? By Inhibitor B?Explanation / Answer
the lineweaverburk plot need to be used for determination of KM and Vmax
1/V= (KM/Vmax)*1/S+ 1/Vmax
so a plot of 1/V vs 1/S gives straight line whose slope is KM/Vmax and intercept is 1/Vmax
in case of inhibition, KM becomes Kmapp and Vmax becomes Vmaxapp. The plots of 1/V vs 1/S along with data points are shown below.
from the plots drawn, when ther is no inhibition, intercept is 1/Vmax=0.022, Vmax=1/0.022= 45.45 umoles/min
slope is KM/Vmax= 0.225, KM=0.225*45.45=10.23 uM
when there is inhibitor A, 1/Vmaxapp=0.022, Vmaxapp =45.45 umoles/min , KMapp/Vmaxapp =0.667
KMapp= 0.667*45.45= 30.32 uM
since Vmax is same in the presence and absence of an enzyme, this is competitive inhibition
for competitive inhibition, KMapp =Km(1+I/KI)
given I= 2mM= 2*10-3M Kmapp= 30.32*10-6M
30.32*10-6 = 10.23*10-6*(1+2*10-3/KI), KI=0.001018 M
when inhibitor B is present, intercept 1/Vmaxapp = 0.122, Vmaxapp =1/0.122=8.2 umol/min
KMapp/Vmaxapp = 1.142, Kmapp =1.142*8.2 uM=9.4 uM
both Vmax app and KMapp are less than Vmax and KM. This is uncompetitive inhibition.
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