Turnover Number Calculation! New topic to me! Very lost!

Question

You do an enzyme kinetic experiment and calculate a V_max of 145 μmol per minute. If each assay used 0.20 mL of an enzyme solution that had a concentration of 0.20 mg/mL, what would be the turnover number if the enzyme had a molecular weight of 137,000 g/mol?

(Enter your answer to two significant figures.)

turnover number = ____ sec^-1

Answer 1

What we're trying to determine is

k_cat = turnover number = (V_max μmol/sec) / (μmol of enzyme)

Convert V_max Units to μmol/sec

V_max of 145 μmol/min × (1 min)/(60 sec) = 29/12 μmol/sec

(keeping the value as an exact fraction is convenient for now)

Calculate Enzyme Amount in μmol

0.20 mg/mL × 0.20 mL × (1g)/(1000 mg) = 4 × 10^–5 g

4 × 10^–5 g × 1 mol/(1.37 × 10⁵ g) = 2.920 × 10^–10 mol enzyme

2.920 × 10^–10 mol × 10⁶ μmol/mol = 2.920 × 10^–4 μmol enzyme

(V_max μmol/sec) / (μmol of enzyme) = (29/12 μmol/sec) / (2.920 × 10^–4 μmol) = 8277 sec^–1

Answer (rounded to two significant figures)

turnover number = 8300 sec^–1