Problem 1 You study kinetics of an enzyme E as a function of substrate concentra

Question

Problem 1 You study kinetics of an enzyme E as a function of substrate concentration first without any inhibitor and then in the presence of 10 M of inhibitor A or inhibitor B (see Table). The enzyme concentration is maintained constant at a level of 1 M (=106 M). The substrate concentration is shown in the Table below. Answer the following questions. 1. From these data determine Vmax, Km, and the turnover number (keat) for the enzyme E in the absence of any inhibitors 2. Determine the type of inhibition for A and B 3. Using your answer to question 2, determine the dissociation constant for each inhibitor Explain your reasoning Comments/hints --Using Lineweaver-Burk plot might be helpful in addressing most of these questions. --You don't have to print out the actual graphs in your answer sheet(s)-you can simply sketclh how they look based on your analysis of the data SI, LIM | Vo, M/min 40 60 90 120 200 300 400 No inhibitor 4.4444 5.4545 6.4286 7.0588 8.0000 8.5714 8.8889 Vo, M/min +Inhibitor AInhibitor B 2.1053 2.8571 3.7500 4.4444 5.7143 6.6667 7.2727 2.3529 2.6087 2.8125 2.9268 3.0769 3.1579 3.2000

Explanation / Answer

Lineweaver-burk 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 is 1/Vmax.

for the case of inhibition, KM becomes Kmapp and Vmax becomes Vmax app.

the three plots of 1/V vs 1/S is drawn and shown below.

from the plot, for the case of no inhibition, 1/Vmax= intercept= 0.1, Vmax= 1/0.1= 10 uM/min

Slope = KM/Vmax= 5.0001, KM= 5.0001*10= 50.001 uM

for the case of inhibitor- A, 1/Vmaxapp = 0.1, Vmaxapp= 1/0.1= 10 uM/min , KMapp/Vmaxapp = 15

Kmapp= 15*10= 150uM

since Vmax remains the same for the no inhibition and inhibition, This is competitive inhibition where the inhibitorp competes with enzyme for binding to the substrate and

KMapp= KM*(1+I/KI), I= concentration of inhibitor

150= 50.001*(1+10/KI)

1+10/KI= 150/50.001= 3

10/Ki= 2

Ki= 5 uM

for the case of inhibitor- B

1/Vmaxapp= 0.3, Vmaxapp= 1/0.3=3.33 uM/min and KMapp/Vmaxapp =5.0009

Kmapp= 5.0009*3.33 uM=16.5 uM

since for this case compared to no inhibition, both KMapp and Vmaxapp are less than Km and Vmax

this is the case of uncompetitive inhibition, where an enzyme inhibitor binds only to the complex formed between the enzyme and the substrate.

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

1+I/KI= Vmax/Vmaxapp= 10/3.33= 3

I/KI= 2

Ki= I/2= 10/2=5

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