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?

To calculate the rate of product formation (Vo) in the presence of a competitive inhibitor, you will need to use the Michaelis-Menten equation. However, before we proceed, let's clarify the units for Vmax.

In this problem, Vmax is given as 22 nmol/min/mg. This means that it represents the maximum rate of product formation per unit enzyme mass (mg) in the absence of any inhibitor.

To convert Vmax into something usable for the equation for inhibitors, we need to consider the units of Km and inhibitor concentration (Ki). Since Km is specified in M (moles per liter) and Ki is given in M as well, we should convert Vmax into moles/(volume)/(time).

Given:
Km = 4.7 x 10^-5 M
Vmax = 22 nmol/min/mg
[i] = 5 x 10^-1 mM (convert to M) = 5 x 10^-4 M
[s] = 2 x 10^-4 M
Ki = 3 x 10^-4 M

To convert Vmax to moles/(volume)/(time), you need to multiply by the mass of the enzyme. You can assume a convenient enzyme mass, such as 1 mg, to calculate the rate.

Let's proceed with the calculation now:

1. Convert [i] to M (since Km and Ki are given in M).
[i] = 5 x 10^-4 M

2. Plug the values into the Michaelis-Menten equation:
Vo = Vmax * [s] / (Km * (1 + [i]/Ki))

Vo = (22 nmol/min/mg) * (2 x 10^-4 M) / (4.7 x 10^-5 M * (1 + (5 x 10^-4 M)/(3 x 10^-4 M)))

Simplifying the equation:
Vo = (22 nmol/min/mg) * (2 x 10^-4 M) / (4.7 x 10^-5 M * (1 + 5/3))

Vo = (22 nmol/min/mg) * (2 x 10^-4 M) / (4.7 x 10^-5 M * (8/3))

Vo = (22 nmol/min/mg) * (2 x 10^-4 M * 3/ (4.7 x 10^-5 M * 8)

Vo = (22 nmol/min/mg) * (6 x 10^-4 M / (37.6 x 10^-5 M)

Vo ≈ (22 nmol/min/mg) * 0.159

Vo ≈ 3.498 nmol/min/mg

Therefore, the rate of product formation (Vo) in the presence of a competitive inhibitor is approximately 3.498 nmol/min/mg.