An enzyme catalyzed reaction obeying Michealis-Menten kinetics has the following
ID: 99260 • Letter: A
Question
An enzyme catalyzed reaction obeying Michealis-Menten kinetics has the following rate constants: k_1 = 3.2 times 10^7 M^-1 s^-1 k_2 = 7.3 times 10^2 s^-1 k_1 = 5.8 times 10^5 s^-1 (A) Calculate K_s for this enzyme. (B) Calculate K_m for this enzyme. (C) Calculate k_cat for this enzyme. (D) Calculate the specificity constant for this enzyme. (E) If the enzyme concentration is 2.3 times 10^-9 M, calculate V_max for this enzyme (F) If the enzyme concentration is 2.3 times 10^-9 M, What concentration of substrate would generate a velocity equal to 0.25V_max? (G) If the enzyme concentration was increased to 6.4 times 10^-9 M, what would be the values of V_max and K_m?Explanation / Answer
Answer:
(E) Vmax = Kcat x Et = (5.8 x 105) x (2.3 x 10-9) Ms-1 = 13.34 x 10-4 Ms-1
(F) V0 = (Vmax x [S]) / Km + [S]
0.25 Vmax = 13.34 x 10-4 x [S] / 0.018 + [S]
0.25 x 13.34 x 10-4 = 13.34 x 10-4 x [S] / 0.018 + [S]
0.25 x 13.34 x 10-4 x (0.018 + [S]) = 13.34 x 10-4 x [S]
0.25 x (0.018 + [S]) = [S]
0.0045 + 0.25 [S] = [S]
[S] - 0.25 [S] = 0.0045
0.75 [S] = 0.0045
[S] = 0.006 M
(G) Vmax = Kcat x Et = (5.8 x 105) x (6.4 x 10-9) Ms-1 = 37.12 x 10-4 Ms-1
Km is independent on the concentration of enzyme if you are doing the proper experiment ([S] being always much greater than [E]).