Km and Vmax and Turnover calculations! Very confused please help!

Question

For an enzymatically catalyzed reaction in which the measured values for K_M and V_max are, respectively, 15.1 mM and 0.715 mM min^-1, what is the value of the turnover number? The concentration of enzyme used in the reaction is 50.0 μM.



turnover number =____  min^-1

Answer 1

What we're trying to determine is

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

I'm pretty sure the value of K_M is not part of the calculation

Convert V_max Units to μmol/min

V_max of 0.715 mmol/min × (1000 μmol)/(mmol) = 715 μmol/min

(V_max μmol/min) / (μmol of enzyme) = (715 μmol/min) / (50.0 μmol) = 14.3 min^–1

Answer (already rounded to three significant figures)

turnover number = 14.3 min^–1