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In addition to competitive inhibition, there are two other common types of enzym

ID: 1066385 • Letter: I

Question

In addition to competitive inhibition, there are two other common types of enzyme inhibition. One type is called uncompetitive inhibition. For this mechanism, the inhibitor can only bind to the ES complex and NOT to the free enzyme: E + S ES rightarrow E+ P ES + I ESI Follow the following steps (similar, but not identical to problem 2) to derive d[P]/dt for this mechanism. Write the equation for d[P]/dt Write the equation for d[ESI]/dt. Apply the steady state approximation to derive an equation with [ESI] on one side and [ES}, [I] and constants on the other side. Write the equation for d[ES]/dt. Apply the steady state approximation and your result from pan (b) to derive an equation with [ES] on one side and [E], [S], and constants on the other side (it should he a relatively simple equation). For this mechanism. [E] = [E]_0- [ES] - [ESI]. Substitute the results from parts (b) and (c) into this equation and rearrange so that [E] is on one side and everything else is on the other side. Now that you have an expression for [E], you can substitute that expression back into the equation from part (b) to get [ES] in terms of measurable quantities and rate constants. Take the results from part (e) and write the final expression for d[P]/dt.

Explanation / Answer

a. dp/dt = k2 (ES)

b. d(ESI)/dt = k4 (ES) (I) – k-4 (ESI)

Under steady state approximation

   d(ESI)/dt = k4 (ES) (I) – k-4 (ESI) = 0

     k4 (ES) (I) – k-4 (ESI) = 0

(ESI)= (k4/ k-4 ) (ES) (I)

c. d(ES)/dt = k1 (E) (S) – k-1 (ES) – k4 (ES)(I) + k-4 (ESI)

Under steady state approximation

d(ES)/dt = k1 (E) (S) – k-1 (ES) – k4 (ES)(I) – k-4 (ESI) = 0

k1 (E) (S) – k-1 (ES) – k4 (ES)(I) – k-4 (ESI) = 0

Rearranging and substituting for (ESI) in part b

(ES)=    k1 (E) (S)/( k-1+k2 – (k4+1) (I))

d. E = (E)0 – (ES) – (ESI

substituting from part b and part c

E = (E)0 – (ESI) – ((kt + k1 (S))/kt)

Where

kt = k-1+k2 – (k4+1)(I)

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