We are going to compare a lineweaver-Burk plot for regular kinetics and competit
ID: 482224 • Letter: W
Question
We are going to compare a lineweaver-Burk plot for regular kinetics and competitive inhibition. a. We showed in class [E]_Total = [E] + [ES] + [EI] and since we have equilibrium k_3[E][I] = k_3 [EI] so we can define the dissociation constant for the inhibitor as K_1 = k_3/k_3 so [EI] = [E][I]/K_1 Now substitute this back in equation (1) and solve for [E] b. As before, we set up steady state for [ES] d[ES]/dt = 0 = k 1 [E][S] - k - 1 [ES] - k2[ES] substitute your expression for [E] from part a. into this equation and solve for [ES] c. Rearrange to show that [ES] can be written as d. Now substitute this is your expression Rate = k2[ES] and compare your result to that of the traditional Lineweaver-Burk plot. In what ways is this plot different?Explanation / Answer
Etotal = E+ES+EI
EI= [E] [I]/K
Etotal = E+ES+[E] [I]/K
Etotal = E(1+I/K)+ES
(Etotal –ES)(1+I/K) = E (1)
d(ES)/dt= K1[E][S]-K-1[ES]-K2[ES]=0
[ES] = K1[E] [S]/(K2+K-1)
From Eq.1 [ES] = K1{ Etotal- ES}(1+I/K)
[ES]= K1[Etotal]*(1+I/K) – ES*(1+I/K)
[ES] (2+I/K) = K1[Etotal] *(1+I/K)
ES = {K1[Etotal]*1+I/K)}/ (2+I/K)
R= K2[ES] = K2K1Etotal *(1+I/K)/(2+I/K)
1/R = (1+1+I/K)/(K2K1[Etotal]*(1+I/K)
1/R= 1/(K2K1[Etotal]*(1+I/K) + 1/K2K1[Etotal]
When a plot of 1/R vs 1/Etotal is drawn, the slope will be 1/K2K1 and intercept will an additional term 1/{K2K1total*(1+I/K)}
The difference is 1+I/K term in the intercept.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.