05 mg of an enzyme gives a Vmax of 10 mMmin The reaction is

0.5 mg of an enzyme gives a Vmax of 10 mM/min. The reaction is carried out in 1 ml and the molecular mass of the enzyme is 40000 g/mol. Find out the specific activity of an enzyme the Turnover number of an enzyme

Solution

B. Turnover number (Kcat) = Number of the substrate molecules converted into the product by an enzyme in unit time, when the enzyme is fully saturated.

Vmax / Et = Kcat (Et = Total enzyme)

10/0.5 = 20

Specific activity = Kcat / Km = 20/05 x 10^-3 = 20000/5 = 4000

Here the Km = MM Constant. It is the substrate concentration where half of the maximum velocity of the enzyme