Question

Why do we use kcat (also called turnover number) and not Vmax when we characterize the properties of an enzymecatalysed reaction?

Please explain to me in details. Thank you!

Answer 1

Turnover number (K_cat) = V_max / [E_T]         ; where, [E_T] = concentration of enzyme.

Turnover number is the rate of product formation per unit time per unit enzyme concentration.

Under ideal conditions (1 binding site per enzyme, sufficient [S] and other conditions), Kcat is in fact the same rate of conversion of [E-S] complex into E and P- it gives the rate of formation of product.

The Kcat value of enzyme is practically more informative-

1. Suppose you procured alpha-amylase with labelled Vmax = X mmol s^-1

Case: A: Suppose there is only “1” molecule of enzyme in the solution, it’s V_max = X

Case: B: Suppose there are 10 molecules of the enzyme, the V_max still remains the same, i.e. V_max = X

In both cases, V_max is the same. However, overall rate of catalysis in case B shall be around 10 times greater than that of Case A. Because, V_max does not tells you about the concentration of enzyme in the solution.

Therefore, Kcat = V_max / [E_T] is preferred.

Case: A1, Kcat =1

Case B1, kcat = 10

Thus, Kcat is more informative and accurate about rate of catalysis per unit time than Vmax.