The KM of a Michaelis-Menten enzyme for a substrate is 1.0 x 10-4 M. At a substrate concentration of 0.2 M, Vo=43x10^-6 M for a certain enzyme concentration. However, with a substrate concentration of 0.02 M, Vo has the same value.
(A) Using numerical calculations show that this observation is accurate.
(B) What is the best range of [S] for measuring KM?
Please see the attachment.(A) Km = 1.0 x 10-4M
[S] = 2 x 10-1M
Vo = 43 x 10-6M
Using Michaelis-Menten equation:
Vo = Vmax[S]
Km + [S]
At [S] = 2 x 10-1M,
43 x ...
This solution involves step by step calculations for finding the KM for a Michaelis-Menten enzyme
Understanding the Michaelis-Menten Equation, Enzyme Kinetics
For an enzyme which obeys Henri-Michaelis-Menten kinetics,
(1) At what substrate concentration will an enzyme characterized by a kcat of 30 s-1 and Km of 0.005 M show one quarter of its maximal rate?
(2) Calculate the fraction of Vmax that would be found when [S]=0.5 Km, [S]=3Km, and [S]=10KmView Full Posting Details