The turnover number (Kcat) of an enzyme is 35 s- and Km = 1.3 mM. Please determine the total amount of the enzyme (E) with mol. wt. 50,000 g/mol; when the Vmax is 3.5 micromoles/s. O 500,000g O 0.05g 5 X 104 g O 0.005 g