The turnover munber of an enzvme is the number of substrale molecules converted inlo product by an enzyme molecule in a unit line when the enzyme is fully saturated with substrate. It is equal to the kinetic constant k]. The maximal rate Vmax reveals the turnover number of an enzyme if the con- centration of active sites [Er] is known, because Vmax =kler] (33 For example, a 10-6m solution of carbonic anhydrase catalyzes the for- mation of 0.6 M HoCO3 per second when it is fully saturated with sub- strate. Hence, ks is 6 X 105 s-l. This turnover number is one of the larg- est known. Each round of catalysis occurs in a time equal to l/k3, which is 1.7 us for carbonic anhydrase. The turnover numbers of most enzymes ith their physiological substrates fall in the range from 1 to 10 per second(Table 8-3)