PLEASE SHOW ME HOW TO CALCULATE STEP BY STEP QUESTION A. THANK-YOU :) 7. A diffe

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

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