lecture 45 - enzyme inhibition kinetics

Question 1

where does enzyme inhibition typically occur in the equilibrium equation?

Answer 1

E + S = ES = E + P

at =

Question 2

can there be reversible enzyme inhibition?

Answer 2

yes - enzymes that bind non-covalently

Question 3

describe competitive enzyme inhibtion

Answer 3

• inhibitor binds to the active site and competes with the substrate for binding to the enzyme
• the structure of the inhibitor typically resembles the structure of the substrate
• happens at the E + S = ES stage

Question 4

describe the enzyme inhibtion at the E + S stage

Answer 4

• instead of E + S = ES have E + I = EI
• the presence of an inhibitot shifts the E + S = ES equilibrium to the left
• thus, more S is required to outcompete EI and form ES

Question 5

what is the dissociation constant for EI?

Answer 5

Ki (dissociation consant) = [E][I] / [EI]

Question 6

compare the Vo vs [S] graphs for a substrate with a competitve inhibitor vs a substrate with no inhibitor

Answer 6

-Km appears to increase
-Kmapp = Km(α), where α = 1 + [I] / Ki
-as Km gets smaller, alpha gets bigger, and more and more substrate is needed to overcome the inhibitor
-Vmax appears unchanged
(curve increases less abrubtly but reaches same point)

Question 7

compare the lineweaver-bark plot for a substrate with no inhibtion vs a substrate with inhibtion

Answer 7

• Vo = Vmax[S] / Kmα + [S]
• 1/Vo = αKm/Vmax * 1/[S] + 1/Vmax
• slope = αKm/Vmax
• y-intercept = 1/Vmax
• as add inhibitor, the slope changes and the line becomes more and more steep
• the y-intercept is not affected by adding inhibitor

Question 8

why do we say the Km “appears” to increase in the presence of a competitive inhibitor?

Answer 8

because the actual Km, or the inherent affinity of the substrate for the enzyme, does not actually increase, it only appears to because it is affecting the progression of the reaction

Question 9

describe non-competitive enzyme inhibition

Answer 9

• the inhibitor bind the ES complex to form an inactive ESI complex
• as [S] increases, [ES] increases, which shifts ES + I = ESI to form more ESI
• i.e. more [S] cannot out compete I
• and the presence of I shift E + S = ES to the right

Question 10

compare the Vo vs [S] graphs for a substrate with a non-competitve inhibitor vs a substrate with no inhibitor

Answer 10

• Vmax appears to decrease
• Km appears to decrease
• Vmaxapp = Vmax/alpha(prime)
• Kmapp = Km/alpha(prime)
• alpha = 1 + [I] / Ki

Question 11

what is the dissociation constant for the ESI complex?

Answer 11

Ki(prime) = [ES] [I] / [ESI]

Question 12

describe the linewearver-bark plot for a subsrate with non-competitive inhibition

Answer 12

• Vo = Vmax [S] / Km + alpha(prime) [S]
• 1/Vo = Km/Vmax * 1/[S] + alpha(prime)/Vmax
• as add inhibitor, the slope changes and the line becomes more and more steep
• both slope and the intercept are affected my non-competitive inhibitors
• slope = Km/Vmax
• y-intercept = alpha(prime)/Vmax

Question 13

describe mixed inhibition

Answer 13

have inhibition of the first E and of the ES complex

Question 14

does Ki need to equal Ki(prime) for mixed inhibition?

Answer 14

no

Question 15

describe the Vo vs [S] graph for mixed inhibition

Answer 15

• have both Km and Vmax appearing to decrease
• Kmapp = alpha / alpha(prime)
• Vmaxapp = Vmax/alpha(prime)
• alpha = 1 + ([I] / Ki)
• alpha(prime) = 1 + ([I] / Ki(prime))
• Vo = Vmax[S] / alphaKm + alpha(prime)*[S]
• 1/Vo = alpha*Km/Vmax * 1/[S] / alpha(prime)/Vmax
• slope = alphaKm/Vmax
• y-intercept = alpha(prime)/Vmax

Question 16

does the inhibitor need to look like the substrate for mixed inhibition?

Answer 16

no - I is epxected to look like S for competitive inhibition only!!!