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In: Biology

(14 marks) The following kinetic data were obtained for an enzyme in the absence of inhibitor...

The following kinetic data were obtained for an enzyme in the absence of inhibitor (1), and in the presence of an inhibitor at 5 mM concentration (2). Assume[ET] is the same in each experiment. [S] (mM) (1) v(µmol/mL sec) (2) v(µmol/mL sec) 1 12 4.3 2 20 8 4 29 14 8 35 21 12 40 26

[S] (mM)	(1) v(µmol/mL sec)	(2) v(µmol/mL sec)
1	12	4.3
2	20	8
4	29	14
8	35	21
12	40	26

a. Using a graphing program (excel or sigmaplot) construct a Lineweaver burke plot representing the uninhibited reaction and the inhibited reaction on the same plot (to submit your graphs please paste into either a word or a pdf document along with your solutions)

b. What is the equation for each of the lines? (on the graph)

c. Determine Vmax and Km for the enzyme.

d. Determine the type of inhibition and the KI for the inhibitor.

Expert Solution

Accoring to the Lineweaver Berk equation-

The y- axis is 1/v (1/rate) and x axis is 1/s (1/substrate)

Best-fit values	1 (without inhibitor)	2 (with inhibitor)
Slope	0.06314 ± 0.001473	0.2117 ± 0.001806
Y-intercept when X=0.0	0.01955 ± 0.0007614	0.02008 ± 0.0009332
X-intercept when Y=0.0	-0.3096	-0.09487

b) Equation of curve without inhibitor (v1)-

y= 0.06314x + 0.01955

Equation of curve with inhibitor (v2)-

y= 0.2117x + 0.02008

c) For 1- Without inhibitor-

Y-intercept= 1/Vmax = 0.01955

Thus Vmax= 1/0.01955 = 51.15 µmol/mL sec

X-intercept= -1/Km = -0.3096

Km= 3.23mM

For 2- With inhibitor-

Y-intercept= 1/Vmax = 0.02008

Thus Vmax= 1/0.02008 = 49.8 µmol/mL sec

X-intercept= -1/Km = -0.09487

Km= 10.54mM

d) Since the Vmax with inhibitor is lower than without inhibitor and the Km with inhibitor is more than Km without inhibitor, thus the type of inhibition is- Mixed inhibition

gladiator answered 3 months ago

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