Given that the concentration of bovine carbonic anhydrase is 3.3 pmol ⋅ L - 1 and R max (V max) = 222 μmol ⋅ L - 1 ⋅ s - 1, determine the turnover number of the enzyme molecule bovine carbonic anhydrase, which has a single active site.

The turnover number of the enzyme molecule bovine carbonic anhydrase = 67,272,727.27 s^-1.

Explanation:

Given:

The concentration of bovine carbonic anhydrase = total enzyme concentration = Et = 3.3 pmol⋅L^-1 = 3.3 * 10^-12 mol. L^-1

The maximum rate of reaction = Rmax (Vmax) = 222 μmol⋅L^-1⋅s^-1 = 222 * 10^-6 mol. L^-1⋅s^-1

The formula for the turnover number of an enzyme (kcat, or catalytic rate constant) = Rmax : Et = 222 * 10^-6 mol. L^-1⋅s^-1 : 3.3 * 10^-12 mol. L^-1 = 67,272,727.27 s^-1

Therefore, the turnover number of the enzyme molecule bovine carbonic anhydrase = 67,272,727.27 s^-1