Posted by **Amelie** on Saturday, November 10, 2007 at 5:41pm.

Consider the following kinetic parameter for a given enzyme:

Km=4.7x10-5 M

Vmax= 22 nmol/min/mg

[i]=5x10-1 mM

[s]=2x10-4 M

Ki=3x10-4 M

Calculate the rate of product formation in the presence of a competitive inhibitor.

I know that I have to solve for Vo. I also know that I need to convert the concentrations so that they are all alike so they will work in the equation, but the Vmax concentration he gave me is confusing. On every problem we did he gave us moles/(volume)/(time) which is what it should be, and in this problem it is almost like he is giving the me the specific activity of the enzyme. How would i convert Vmax into something useable for the equation for inhibitors?

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