Question

A particular enzyme has a MW of 40,000 and its substrate a MW of 54. A. If 10 μg of enzyme converts the substrate into product at a rate of 0.32 g of substrate per minute at Vmax, what is the Turnover Number (kcat) of this enzyme in sec-1? B. Given that the KM value is 2.5 mM, has this enzyme reached catalytic perfection? (ANS: kcat = 4.0 105 s-1 ; kcat/KM = 1.6 x 108 M-1 s-1 , Yes, this is close to the diffusion controlled limit.) Please show all your work

Answer 1

One Dalton is one AMU (atomic Mass Unit).

1 Da = 1g/mole

So, number of moles in 10 micro g of protein is

=10 x 10^-6 g / 40000 g mol^-1

= 2.5 x 10^-10 mol

Like wise 0.32 g of substrate has

= 0.32 g / 54 g mol^-1

= 5.926 x 10^-3 mol min^-1 = 5.926 x 10^-3 mol / 60 s =>

9.876 x 10^-5 mol s^-1

If the reaction has several steps and one is clearly rate limiting, k_cat is equivalent to the rate constant for that limiting step.

k_cat = V_max / [E_t]

V_max = 9.876 x 10^-5 mol s^-1

[E_t] = 2.5 x 10^-10 mol

k_cat = 9.876 x 10^-5 mol s^-1 / 2.5 x 10^-10 mol

k_cat = 395080 s = 3.950 x 10⁵ s^-1 ~ 4.0 x x 10⁵ s^-1

The best way to compare the catalytic efficiencies of different enzymes or the turnover of different substrates by the same enzyme is to compare the ratio k_cat /K_m This parameter, sometimes called
the specificity constant, is the rate constant for the conversion of E + S to E + P.

This diffusion controlled limit is 10⁸ to 10⁹ M^-1s^-1 , and many enzymes have a k_cat / K_m near this range.

we found k_cat as 3.950 x 10⁵ s^-1 and Km is given as 2.5 mM

k_cat /K_m

= 3.950 x 10⁵ s^-1 / 2.5 x 10^-3 M

k_cat /K_m = 1.58 x 10^-8 M^-1s^-1

So, this is close to diffusion controlled limit