Question

In: Chemistry

An experimenter performs two sets of enzyme kinetics experiments. One set of experiments, when plotted as...

An experimenter performs two sets of enzyme kinetics experiments. One set of experiments, when plotted as a double-reciprocal plot, yielded a y-intercept of 0.043 μM-1·min and an x-intercept of -20.0 mM-1. A second set of experiments, when similarly plotted, yielded a y-intercept of 0.129 μM-1·min and an x-intercept of -60.0 mM-1. The second set of experiments contained 92 nM of an uncompetitive inhibitor. Given these data, what is the Ki' (in nM to the nearest tenths) for this inhibitor? Hint: If you do not recall what the x-intercept represents, you can write the double-reciprocal equation and set 1/v equal to zero. Also, for this problem, you can assume that the enzyme exhibits Michaelis-Menten kinetics and that the estimates for Vmaxand Km that are obtained from a Lineweaver-Burke (i.e., double-reciprocal) plot are satisfactory. (Recall that I advocated in class for the use of nonlinear fitting algorithms.)

Solutions

Expert Solution

The reciprocal plot of enzyme kinetics is

1/V= (KM+S)VmaxS = (KM/Vmax )*1/S +1/Vmax

y= mx+C, where x= 1/S and y= 1/V, when y intercept is given x=0 whicis 1/Vmax,C = 0.043min/uM Vmax= 1/0.043 uM/min, Vmax= 23.25 uM/min

x= (y-C)/m, when x intercept is given y=0    -20 = -C/m,

m= -C/20 = 0.043/20 = 0.00215

KM/Vmax =slope = 0.00215

KM= 0.00215*1/0.043=0.005 uM

so a plot of 1/V vs 1/S gives slope of KM/Vmax and intercept of 1/Vmax

the reciprocal plot for uncompetitive inhibition

1/V= (KM/Vmax)*1/S +(1+I/Ki)/Vmax

y= mx+C where C=( 1+I/Ki)/ Vmax= 0.129

1+I/Ki= 0.129*23.25=3

I/Ki=2, KI=1/2= 92* nM/2= 46n M


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