An experimenter performs two sets of enzyme kinetics experiments. One set of exp

Question

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 M-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 M-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.)

Explanation / Answer

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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