From a kinetics experiment, Kcat was determined to be 55sec^-1. For the kinetic assay, 0.05mL of a 0.05mg/mL solution of enzyme was used, and the enzyme has a molecular weight of 30,000g/mole. Assume a reaction volume of 3mL. Calculate Vmax (um*min^-1) for the enzyme and catalytic efficiency (in  M^-1sec^-1) for the enzyme. The Km for the enzyme was determined to be 8.3*10^-2M.

Biochemistry
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From a kinetics experiment, Kcat was determined to be 55sec^-1. For the kinetic assay, 0.05mL of a 0.05mg/mL solution of enzyme was used, and the enzyme has a molecular weight of 30,000g/mole. Assume a reaction volume of 3mL. Calculate Vmax (um*min^-1) for the enzyme and catalytic efficiency (in  M^-1sec^-1) for the enzyme. The Km for the enzyme was determined to be 8.3*10^-2M. 

Expert Solution
Initial calculations

The concentration of enzyme stock solution was 0.05mg/ml.

i.e. 1ml contain 0.05mg of enzyme.

We took 0.05ml of this enzyme stock solution.

0.05ml contains; 0.05×0.05=2.5×10-3 mg = 2.5×10-6g of enzyme.

Molecular weight of the enzyme is 30,000 g/mole.

1g of enzyme contain 130,000moles of enzyme.

So; 2.5 x 10-6 g of enzyme contain, 2.5×10-630,000=8.333×10-11 moles of enzyme.

Reaction volume =3mL=3×10-3LTotal enzyme concentration, [E]t =Number of moles of enzymetotal volume                                                           =8.333×10-11moles3×10-3L                                                           =2.78×10-8M

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