Question

When 10 μg of an enzyme of Mr 50,000 g/mol is added to a solution containing its substrate at a concentration that is much higher than Km, it catalyzes the conversion of 25 μmol of substrate into product in 1 min. Calculate this enzyme's turnover number!

Answer 1

Turnover number (K_cat) = V_max / E_t

E_t = quantity of enzyme or active sites in moles

V_max = maximum reaction rate

Km is the substrte concentration when enzyme achieves half Vmax. Since substrate cooncentration is very high than Km the Vmax can be calculated as

Vmax = 25 μmol / 1min = 25 μmol / 60 s = 0.42 μmol s^-1

E_t = 10 μg / 50,000 g/mol = 2.0 X 10^-4 μ moles

Therefore Turnover number (K_cat) = 0.42 μmol s^-1  / 2.0 X 10^-4 μ moles = 2.1 X 10³ s^-1