Question

HUUR UUU UUILUUITUL LUIUILINU 2 PV, 10 pul, posta 10. Applying the Michaelis-Menten Equation II An en- zyme catalyzes the reaction M = N. The enzyme is present at a concentration of 1 nm, and the Vmax is 2 um s-1. The Kn for substrate M is 4 um. (a) Calculate kcat. (b) What values of Vmax and Km would be observed in the presence of sufficient amounts of an uncompetitive inhibitor to generate an a' of 2.0?

Answer 1

Q10. Concentration of enzyme = 1 nM = 1 x 10^-9 M

V_max = 2 uM s^-1 = 2 x 10^-6 M s^-1

(a) Turnover number, k_cat = (V_max) / (Concentration of enzyme)

k_cat = (2 x 10^-6 M s^-1) / (1 x 10^-9 M)

k_cat = 2 x 10³ s^-1

(b) For uncompetitive inhibition,

observed V_max = (V_max) / $\alpha$'

observed V_max = (2 uM s^-1) / 2

observed V_max = 1 uM s^-1

observed K_m = (K_m) / $\alpha$'

observed K_m = (4 uM) / 2

observed K_m = 2 uM