From a kinetics experiment Kcat was determined to be 35 sec1

From a kinetics experiment, K_cat was determined to be 35 sec^-1. For the kinetic assay, 0.3mL of a 0.25mg/mL solution of enzyme was used, and enzyme has a molecular weight of 105.000 g/mole. Assume a reaction volume of 3mL. Calculate Vmax (mu M middot min^-1) for the enzyme and catalytic efficiency (in M^-1sec^-1) for the enzyme. The K_m for the enzyme was determined to be 2.5 times 10^-2M. Consider the following peptide: Lys-Tyr-Glu-His-Cys-Glu-Ala-Arg-Cyc-Arg, is the net charge of this peptide at pH = 14 9 you must show your work for full credit!)?

Solution

Answer:

Given:

Molar mass of the enzyme

1mL---->0.25mg

0.3 mL---> 0.25 x 0.3 = 0.07mg added into total 3 mL

So 0.023mg/mL

FROM (mg/ml) TO molarity (M): Divide the concentration (mg/ml) by the molecular weight.

=0.023 mL/105,000 = 0.00021 x 10^-3M

Formula for Kcat

Kcat=Vmax/ [E]

Rearranging the formula

Vmax= Kcat x [E]

Vmax= 35 x 0.00021 x 10^-3M

Vmax=0.0073 x 10^-3M/sec or

Vmax = 7.3 uM/sec

Catalytic efficiency= Kcat/Km

=35/2.5 x 10^-2 M

Catalytic efficiency=14 x 10^2 M