Here's the problem:
A student who performed the experiments in this lab exercise* found the Vmax of alkaline phosphatase to be 0.073 umole substrate hydrolyzed per minute. The concentration of alkaline phosphatase in the stock solution used in the experiment was 0.7 mg protein/mL. Calculate the activity of the enzyme at Vmax as umole substrate hydrolyzed per minute per mg of enzyme protein.
*We basically did a Pierce 660 assay to measure the rate of a enzyme-catalyzed reaction with & without the presence of an inhibitor. This problem refers to the procedure without the inhibitor. I think all the info you'd need to know to answer is that the total volume of each cuvette (containing the reaction mix) was 3mL. Each cuvette contained 0.1mL alkaline phosphatase. The amount of water and substrate varied between cuvettes but iirc we shouldn't need those numbers for this calculation.
Here's I've done so far:
(0.7 mg/mL)(0.1 mL) = (C2)(3 mL)
C2 = .0233 mg/mL
At first I thought I could just divide 0.073 umol/min by that number... but then I realized that the units would be wrong. It'd give me umol*mL/min*mg, and the problem wants the answer in umol/min*mg.
I'm not sure what volume I would have to multiply C2 by though, because doesn't that calculation already account for the volume of the cuvette & the stock soln?
I feel like I'm misunderstanding or overlooking something simple. Any help would be appreciated.