2. The following kinetic data were obtained for an enzyme in the absence of inhi

Question

2. The following kinetic data were obtained for an enzyme in the absence of inhibitor (1), and in the presence of two different inhibitors (2) and (3). Concentration for each inhibitor was at 5 mM. Assume that [ET] is the same in each experiment IS] mM (mol/mL sec) (mol/mL sec) (mol/mL sec) 12 20 29 35 40 4.3 5.5 13 16 18 4 21 26 12 inhibition(s) that has occurred. (Attach graph) b. Does the inhibitors 2 and 3 combine with E, with ES, or with both? Explain. c. Calculate the KM for the substrate with and without the inhibitors. d. What are the Vmax for the substrate with and without the inhibitors? What does this tell you? e. If [ET] was 10 mg, what would the kcat be for the substrate in the absence of the inhibitors (1)? f. What would be the Ki for inhibitor 3?

Explanation / Answer

double reciprocal plot 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 of 1/Vmax.

in case of inhibiton, KM becomes Kmapp and Vmax becomes Vmaxapp.

So the data on substrate concentration (S) and rate (V) are converted to 1/S and 1/V.

From no inhibition, 1/V = 0.065*1/S+0.017 , from the equation, the intercept 1/Vmax= 0.017, Vmax=1/0.017 =58.82 umol/ml.sec, KM/Vmax=0.065, KM= 0.065*Vmax=3.82 mM

for the case of inhibition-1. 1/V=0.214*1/S+0.017, Vmaxapp =1/0.017= 58.82 umol/ml.sec and KMapp= 58.82*0.214 mM= 12.6 mM

for the case of inhibition-2,1/V= 0.139*1/S+0.042, Vmaxapp =1/0.042=23.81 umol/ml.sec and KMapp=23.81*0.139=3.3mM

The Vamxapp for inhibition-1 is same as the Vmax of no inhibitor. This is the case of competitive inhibition. In case of this type of inhibirtion, the Inhibitor binds to the Enzyme.

KMapp= KM*(1+I/KI)

12.6= 3.82*(1+5/KI), 5/KI= 2.3, Ki= 5/2.3 =2.17 mM. Kcat= Vmax/ET= 58.82/10 =5.882 umol/ml.sec.mg

for the case of inhibiton-2, both the KMapp and Vmaxapp are less than that of Km and Vamx. So this is an example of uncometitive inhibition. The inhibitior binds to ES complex formed.

Kcat= Vmax/ET= 23.81/5= 4.762 umol/ml.sec.mg

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