Question

Et= 20nm [S]=40microM kcat=600/s V0=9.6microM/s.
calculate km

Answer 1

Answer:

You gave the value of Vmax as a specific activity, which has already been divided by the enzyme concentration, so it is actually Vmax/[E] (the volume units having already been canceled). I just converted mg of enzyme to moles of enzyme.

103 mg/g is just the conversion of mg to g. I did this in order to used the molecular mass, which is in g/mol, to convert g to mol.

Let's start from the beginning, using a hypothetical example. You measure the value of Vmax as rate of product formation per unit time, such as 3 µM/min. Notice that this is 3 µmoles/L/min. The enzyme concentration in the reaction is 10 µg/ml = 10 mg/L = 0.01 g/L. The enzyme has a molecular mass of 50,000 g/mole, so we calculate the enzyme concentration to be (0.01 g/L) / (50,000 g/mole) = 2 x 10-7 mol/L = 0.2 µmoles/L. Now we divide Vmax by the enzyme concentration: (3 µmoles/L/min) / (0.2 µmoles/L) = 15 min-1 = 0.25 s-1.