Question

Enzyme kinetics: derive the "reversible" Michaelis Menten reaction E+S<=>ES<=>E+P with 1/Kp=K-2/(K-1+K2) and Keq=K2K1/(K-2K-1). I have the derived irreversible reaction. I found an explanation using E+S<=>ES<=>EP<=>E+P, but not one without EP, and I'm not sure how to start on this.

Answer 1

So far I have

\[k_{1}: [E]+[S] -> [ES]\]
\[k_{-1}: [E]+[S] <- [ES]\]
\[k_{2}: [ES] -> [E] + [P]\]
\[k_{-2}: [ES] <- [E] + [P]\]
\[\frac{ 1 }{ K_{p} }= \frac{ k_{-2} }{(k_{-1} + k_{2)} } \]
\[=> K _{p} = \frac{ (k _{-1} + k _{2})}{ k _{-2} }\]

\[v_{1} + v_{-2} = v_{-1} + v_{2}\]
<=>
\[k _{1}[E][S] +k _{-2} [E][P] = k _{-1} [ES] + k _{2} [ES]\]
<=>
\[k _{1}[E][S] +k _{-2} [E][P] = [ES] (k _{-1} + k _{2} )\]

After that I am unsure what to do. I also don't know whether to express [E]t = [ES] + [E] or [ES] + 2 [E]

Please help

Answer 2

okay. enzyme kinetics is a loong way back for me ^^

but I think the answer lies within the fact that [ES] and [EP] are virtually the same. you could use that explaination then.

I don't get, though, why [ES] -> E + P should be a reversible reaction. it usually isn't. I will try to find a way for you to solve this one.

Answer 3

I know. The course gives a derivation for the normal "irreversible" Michaelis Menten reaction. The "reversible" is only a theoretical one, because indeed in practice the -2 reaction is negligible. But for the exercise we'll have to derive the "reversible" reaction and thus include the -2 arrow. This pdf file mentions the reversible rate law and that the derivation is in appendix b, but of course I can't find the appendix b :( http://sys-bio.org/wp-content/uploads/downloads/2012/03/CB_Chapter3.pdf

Answer 4

also, in an enzymatic reaction, you can do the following thing:

\[v = k _{+2} * [ES]\]
\[v = k_{+2} * \frac{ k_{-1} + k_{+2} }{ k_{+1} }\]

this only works for michaelis menten kinetics under the assumption that the second reaction diminishes the concentration of [ES]

Answer 5

under reversible conditions, you would get a steady state, where K(eq) = 0. maybe try that?

Answer 6

But Keq is defined as k2k1/(k-25-1)

Answer 7

sorry that last must be k-1 instead of 5-1

Answer 8

BTW What's the K(eq) in the irreversible reaction?

Answer 9

since there is no steady state in the irreversible reaction, K(eq) should be irrelevant

Answer 10

Yeah it's not mentioned in there.

Answer 11

luckily, you will NEVER need that again ^^

I can tell you, though, that [E]t = [ES] + [E]

Answer 12

what should your final equation look like?

Answer 13

I found a presentation about the reversible derivation, unfortunately for me it uses the following reaction

[E][S] <=> [ES] <=> [EP] <=> [E] [P]

but there the presentation says that E(t) = E + ES + EP

So it takes the extra E in account

Answer 14

well not in your case, since you don't have [EP].

lemme calculate real quick.

Answer 15

This pdf http://sys-bio.org/wp-content/uploads/downloads/2012/03/CB_Chapter3.pdf says that (on page 60) that

Vp = k2 ES - k(-2) E.P
and that "The expression that describes the steady-state concentration of the enzyme
substrate complex also has an additional term from the product binding
(k(-2) EP. Using these one can derive everything to the general reversible rate expression

\[v=\frac{ vf S/Ks - Vr P/Kp }{ 1+ S/Ks + P/Kp }\]

And at equilibrium the rate is 0 and so it becomes

\[v=vf S/Ks - Vr P/Kp \]

this yields the Haldane relationship: expression for K(eq)

Answer 16

I have never heard of half the terms they are using :/

Answer 17

So both sources I found on this use an extra EP reaction. Perhaps that's the trick? To alter the reversible reaction to include formation of EP before it splits into E and P

Answer 18

well, it's not wrong. if they published it that way...and you CAN calculate it that way, obviously

Answer 19

Anyway, thanx for trying to help me with this :-) Just talking or discussing it sometimes helps to figure out which direction to go.

Answer 20

yeah. sorry I couldn't help you any more =)

Answer 21

Well, I'll be plugging away and try to unravel the problem by tugging at different ends :-) But will keep this open in case any one else has some input or suggestions for me.

Answer 22

take a look at this http://sbw.kgi.edu/sbwwiki/_media/sysbio/labmembers/hsauro/noteskinetics.pdf?id=sysbio:lab..

It's shown pretty straight forward

Answer 23

Yes, aaronq, thank you, it's like the other links I found about it. And I've used it, but I have a sneaking suspicion that in the exercise my prof wants a "simplified" version of that, because in the expressions she defines such as 1/Kp and Keq she never uses a k3. I've asked a question about it to my fellow students. Anyway, thank you for confirming that in literature that is the "reversible" rate.

Answer 24

Hm i see. Why is she using \(K_p\) if there arent any gases?

Answer 25

She doesn't say...She just says "Derive the expression of the so-called "reversible" Michaelis Menten reaction for E + S <=> ES <=> P+E with the definitions

\[\frac{ 1 }{k _{p} } = \frac{ k _{-2} }{ k _{-1}}\]

and

\[\frac{ 1 }{k _{eq} } = \frac{ k _{2}k _{1} }{ k _{-2}k _{-1}}\]

She gives no more information than that.

In the course itself there is a section called near-equilibrium with the scheme

E+S <=> ES <=> EP <=> E+P

and in the section she starts to set up basic formulas and assumptions that you could use to derive a law (but without including these laws) and ends with the simplified scheme

E+S <=> P + E.

I think this is the setup I will have to use to answer the exercise question (the exam will be open book).

And I

Answer 26

well, a reaction of E + S <=> P + E is, of course, of 0th order, and
\[v = \frac{ d[A] }{ dt } = k _{-1} * [P] - k_{1} * [S]\]
you can leave the enzyme completely out of the equation, since it would change both k1 and k(-1) in a similar fashion. I think.