Question

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Question 6 O out of 10 points An enzyme catalyzed reaction was performed in the presence and in the absence of an inhibitor. The Lineweaver Burk plot showed competitive kinetics with X- intercepts of -10mM -1 and -3.5mM-1 in the presence and absence of the inhibitor respectively. If the inhibitor concentration used was 2micro molar (UM), calculate KI for the inhibitor enzyme binding? Question 7 O out of 10 points An enzyme catalyzed reaction was performed in the presence and in the absence of an inhibitor. The Lineweaver Burk plot showed competitive kinetics with x- intercepts of -10mM-1 and -3.5mm -1 in the presence and absence of the inhibitor respectively. If the inhibitor concentration used was 0.25nM, calculate KI for the inhibitor enzyme binding?

Answer 1

In Lineweaver Berk plot-

Y- intercept = 1/Vmax

X intercept = -1/Km

6) Without inhibitor-

X- intercept = -10/mM

Thus Km without inhibitor = - (-1/10) = 0.1 mM

With inhibitor-

X intercept = -3.5/mM

Thus K'm with inhibitor = -(-1/3.5) = 0.286 mM

We know that, K'm = alpha * Km

0.286 = alpha * 0.1

Thus alpha = 2.86

Since alpha = 1 + [I]/ Ki

2.86 = 1 + 2/Ki

1.86 = 2/ Ki

Thus Ki = 1.075 uM

7)

Without inhibitor-

X- intercept = -10/mM

Thus Km without inhibitor = - (-1/10) = 0.1 mM

With inhibitor-

X intercept = -3.5/mM

Thus K'm with inhibitor = -(-1/3.5) = 0.286 mM

We know that, K'm = alpha * Km

0.286 = alpha * 0.1

Thus alpha = 2.86

Since alpha = 1 + [I]/ Ki

2.86 = 1 + 0.25/Ki

1.86 = 0.25/ Ki

Thus Ki = 0.134 nM

8) Without inhibitor-

Y intercept = 15

Thus Vmax = 1/15 = 0.067

With inhibitor-

Y intercept = 5

Thus V'max = 1/5= 0.2

SInce it is non competitive inhibitor-

V'max =Vmax/alpha

0.067 = 0.2/alpha

alpha = 2.985

Since alpha = 1 + [I]/Ki

2.986 = 1 + 0.25/Ki

1.986 = 0.25/Ki

Ki = 0.125 nM

Thus the correct answer is-

B. 0.125nM