You do an enzyme kinetic experiment and calculate a p n r of

You do an enzyme kinetic experiment and calculate a p n r of 88.1 ol per minute If each assay used 0.10 mL of an enzyme solution that had a concentration of 0.20 mg/m would be the turnover number if the enzyme had a molecular weight of 128,000 g/mol? (Enter your answer to two significant figures.) turnover number what sec

Solution

given

Vmax = 88.1 umol /min = 88.1 x 10-6 mol / 60 s = 1.46833 x 10-6 mol /s

now

mass of enzyme = 0.1 ml x 0.2 mg / ml = 0.02 mg

mass of enzyme = 0.02 mg x 1 g / 1000 mg = 2 x 10-5 g

now

moles of enzyme = 2 x 10-5 g / 128000 (g/mol)

moles of enzyme = 1.5625 x 10-10 mol

now

turn over number = Vmax / moles of enzyme

turn over number = 1.468333 x 10-6 (mol/s) / (1.5625 x 10-10 mol)

turn over number = 9397 s-1