Question

Question 28 1 pts Determine the type of inhibition that has occurred in a first order Michaelis-Menten enzyme catalyzed reaction that has yielded the following data. Vi is the velocity in the presence of inhibitor, V is the velocity when run without inhibitor and [S] refers to the substrate concentration s V 4.00 0.28 0.28 .33 0.27 0.25 0.56 0.24 0.20 0.20 0.18 0.13 0.10 0.13 0.08 O Non-competitive O Competitive O None of the above No inhibition at all O Uncompetitive

Answer 1

Vi 0.28 0.25 0.2 0.13 0.08 1/1/V1/Vi 3.57 3.70 4.17 5.56 7.69 14.00 0.28 0.27 0.24 0.18 0.13 0.25 0.75 1.79 5.00 10.00 3.57 4.00 5.00 7.69 12.50 0.9114x+3.3103 1.33 0.56 0.2 0.1 12.00 R 0.9992 10.00 8.00 -Without Inhibitor With Inhibitor 6.00 ...Linear (Without Inhibitor) y 0.427x+3.4188 R-0.9997 4.00 .Linear (With Inhibitor) 2.00 0.00 0.00 2.00 4.00 6.00 8.00 10.00 12.00 1/[S]

In absence of Inhibitor,

$\frac{1}{V}=\frac{K_{M}}{V_{max}}\frac{1}{[S]}+\frac{1}{V_{max}}$

$\frac{1}{V_{max}}=3.4188$

$V_{max}=0.2925$

$\frac{K_{M}}{V_{max}}=0.427$

$K_{M}=0.1249$

In presence of Inhibitor,

$\frac{1}{V_{i}}=\frac{K_{M}}{V_{i,max}}\frac{1}{[S]}+\frac{1}{V_{i,max}}$

$\frac{1}{V_{i,max}}=3.3103$

$V_{i,max}=0.3021$

$\frac{K_{M}}{V_{i,max}}=0.9114$

$K_{i,M}=0.082$

Now on compairing,

$V_{i,max}>V_{max}$

$K_{M}>K_{i,M}$

It can be uncompetitve inhibition.