biochem Each student must turn in a separate set of answers, on paper (not elect

Question

biochem

Each student must turn in a separate set of answers, on paper (not electronically) even if they worked with other students on this assignment. Answers without any work being shown to support those answers will be awarded partial or no credit Answers must be legible For a normal Michaelis-Menton enzyme that obeys this mechanism: E + S reversiblearrow ES rightarrow E + P When [E_r] = 1.0 nM, the following kinetic values were obtained k= 2 times 10^3 M^-1 s^-1 K_1= 1 times 10^3 s^-1, k_2 = 5 times 10^3 s^-1 Calculate K_m Calculate K_s (equilibrium constant for the dissociation of the ES complex) Calculate max. Calculate. Calculate/Km Does the catalytic efficiency approach the limits of diffusion If initial velocity was measured using 4 n moles of enzyme per ml and saturating amounts of substrate, what would V_max and K_M equal under these conditions?

Explanation / Answer

a) Km = (k-1 + k2)/k1 = (1 x 103 X 5 X 103 )/2X108 = 2.5 X 10-5

b) Rate of formation of [ES] = k1[E][S].

Rate of dissociation of [ES] = k-1[ES] + k2 [ES].

So in the steady state,
k-1[ES] + kcat[ES] = k1[E][S]
ks = (k-1 + k2)/k1 = [E][S] /[ES]

c) The maximum velocity Vmax occurs when the enzyme is saturated, that is, when all enzyme molecules are tied up with S, or [ES] = [E]total
Thus
Vmax = k2 X [E]total = 5 X 103 X 1 = 5 X 103

d) kcat = 5 X 103 M-1 S-1

e) kcat / km= 5 X 103 / 2.5 X 10-5 = 2.5 X 10-5

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.