PLEASE SHOW ME HOW TO CALCULATE STEP BY STEP QUESTION A. THANK-YOU :) 7. A diffe
ID: 1084366 • Letter: P
Question
PLEASE SHOW ME HOW TO CALCULATE STEP BY STEP QUESTION A.
THANK-YOU :)
7. A different mode. The kinetics of the enzyme considered in problem 6 are measured in the presence of a different inhibitor. The concentration of this inhibitor is 100 M. 6 Velocitymol minute-1) No inhibitor 10.4 14.5 22.5 33.8 40.5 Inhibitor 2.1 2.9 4.5 6.8 8.1 S] (uM) 10 30 90 (a) What are the values of Vmax and KM in the presence of this inhibitor? Compare them with those obtained in problem 6 (b) What type of inhibition is it?Explanation / Answer
Lineweaver-burk plot will be used to determine the constants. The equation is
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.
for the case of inhibition, KM becomes KMapp and Vmax becomes Vmaxapp
So a plot of 1/V vs 1/S is generated and the plot along with data points for both the absence and presence of inhibitorr are shown beow.
when there is no inhibitor, the equation of bst fit gives 1/Vmax= intercept = 0.022, Vmax= 1/0.022 =45.45 umoles/min
slope = KM/Vmax=0.224, Km= 0.224*45.45 uM=10.18 uM
when there is inhibitor, 1/Vmaxapp= 0.112, Vmaxapp= 1/0.112= 8.93 umoles/min and Kmapp//Vmaxapp= 1.108
Kmapp= 1.108* 8.93 uM=9.89 uM
since Vmaxapp<Vmax, and Kmapp<Km, this inhibition is the case of mimics uncompetitive binding. in this type of inhibiton, the enzyme is capable enough to bind to the free enzyme and enzyme-substrate complex.
for this inhibiton, 1/V= Km*(1+I/KI)/Vmax + (1+I/KI)/Vmax
hence 1/ Vmaxapp = (1+I/KI)/Vmax
1+I/KI= Vmax/Vmaxapp = 45.45/8.93=5.08, I/KI= 5.08-1=4.08
KI= I/4.08= 100/4.08 uM=24.51 uM