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 ay-intercept of 0.129 M-1·min and an x-intercept of -60.0 mM-1. The second set of experiments contained 79 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 Vmax and Km that are obtained from a Lineweaver-Burke (i.e., double-reciprocal) plot are satisfactory.

Explanation / Answer

from 1/V= 1/Vmax+(Km/Vmax) 1/S for the case, y intercpet =1/Vmax= 0.043 uM-1.min

Vmax= 1/0.043=23.25 uM/min

x intercept is obtained by letting 1/V =0

gives 1/Vmax =- KM/Vmax *1/S

1/S =-KM= -20mM-1

Km=20 mM-1

for the secnond case

1/V = (Km/Vmax)1/S+ 1/(1+I/Ki)

Where I =concentration of enzyme and Ki = dissociation constant of the complex

y intercept =1/(1+I/Ki)= 0.129

1+I/Ki= 1/0.129 = 7.75

I/Ki= 6.75

But I= 79nM

Ki= I/6.75= 79/6.75= 11.7037 nM

Vmax= 1/0.129 uM/min=7.75 uM/min

For the first set of experiments

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