Suppose you perform a Kinetics assay and obtain the following data for an enzyme
ID: 273093 • Letter: S
Question
Suppose you perform a Kinetics assay and obtain the following data for an enzyme-catalyzed reaction:
KM = .040 micromole/L
Vmax = 1.2 micromole/L min
kcat = 2.0 min-1
(Report the proper significant figures in all three of your answers.)
a .What is the concentration of enzyme used in the reaction?
______________micromoles/L
b. If the concentration of enzyme was doubled what is the Km value then?
______________micromoles/L
c. If the enzyme concentration is halved what is the kcat value then?
______________min-1
Using the double reciprocal plot provided to the right, calculate KM and Vmax:
0.7 0.6 y 0.4843x+0.1951 0.5 0.4 0.3 0.2 0.1 0.6 -0.4-0.2 0 0.2 0.4 0.6 0.8 1/S] (mMA-1)Explanation / Answer
a.
Vmax = Kcat x enzyme concentration
1.2 uM/min = (2.0/min) x enzyme concentration
Enzyme concentration = (1.2/2.0) uM = 0.6 micromole per litre.
b.
Km value remains same as 0.040 micromole/L. Km is independent of enzyme concentration for a single active site of a monomeric enzyme.
c.
Kcat is the intrinsic property of enzyme so will remain the same as 2.0 min-1. The change in enzyme concentration will result in change in Vmax keeping Kcat constant. Vmax is directly proportional to enzyme concentration, Vmax = Kcat x enzyme concentration.
Km = 1/0.4 = 2.5 mM.
Vmax = 1/0.2 = 5 mM/min.
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