Given a turnover number of 1 x 103 s1 and Km of 2 x 103 M fo

Given a turnover number of 1 x 10^3 s-1 and Km of 2 x 10^-3 M for an enzyme, how much less efficient would the enzyme be than the best known enzymes, i.e., perfected enzymes?

a. 10 times

b. 10^2 times
c. 10^5 times
d. 10^7 times

Solution

kcat/km measures the efficiency of the enzyme.the higher the kcat/km value the more efficient is the enzyme.maximum kcat/km dictated by the diffusion limit 10⁹M^-1s^-1

If every collision results in formation of an enzyme-substrate complex,diffusion theory predicts that kcat/KM will attain a value of about 10⁸to 10⁹ M^-1s^-1.

In this case it is 10² times less efficient than the best known enzymes.