Question

The enzyme-catalyzed conversion of a substrate at 298 K has Km = 0.032 mol dm-3 and vmax = 4.25 x10^-4 mol dm-3 s-1 when the enzyme concentration is 3.60 x10^-9 mol dm-3. Calculate the catalytic efficiency

Answer 1

Solution:

Catalytic efficiency of enzyme is calculated as,

Catalytic efficiency = Kcat / Km

Where, Kcat = Turnover number

Kcat is related with Vmax and enzyme concentration (Et) as,

Kcat = Vmax / Et

Given, Vmax = 4.25 x 10^-4 mol dm^-3 sec^-1

Et = 3.60 x 10^-9 mol dm^-3

Thus,

Kcat = Vmax /Et

= 4.25 x 10^-4 mol dm^-3 sec-1 / 3.60 x 10^-9 mol dm^-3

= 1.18056 x 10^5 sec^-1

Hence,

Catalytic efficiency = Kcat / Km

= 1.18056 x 10^5 sec^-1 / 0.032 mol dm^-3

= 3.68925 x 10^6 mol-1 dm^3 sec-1

= 3.689 x 10^6 mol-1 dm^3 sec-1