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Enzyme kinetics: de… - QuestionCove
OpenStudy (anonymous):

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.

3 years ago
OpenStudy (anonymous):

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

3 years ago
OpenStudy (anonymous):

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.

3 years ago
OpenStudy (anonymous):

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

3 years ago
OpenStudy (anonymous):

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]

3 years ago
OpenStudy (anonymous):

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

3 years ago
OpenStudy (anonymous):

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

3 years ago
OpenStudy (anonymous):

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

3 years ago
OpenStudy (anonymous):

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

3 years ago
OpenStudy (anonymous):

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

3 years ago
OpenStudy (anonymous):

Yeah it's not mentioned in there.

3 years ago
OpenStudy (anonymous):

luckily, you will NEVER need that again ^^ I can tell you, though, that [E]t = [ES] + [E]

3 years ago
OpenStudy (anonymous):

what should your final equation look like?

3 years ago
OpenStudy (anonymous):

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

3 years ago
OpenStudy (anonymous):

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

3 years ago
OpenStudy (anonymous):

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)

3 years ago
OpenStudy (anonymous):

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

3 years ago
OpenStudy (anonymous):

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

3 years ago
OpenStudy (anonymous):

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

3 years ago
OpenStudy (anonymous):

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

3 years ago
OpenStudy (anonymous):

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

3 years ago
OpenStudy (anonymous):

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.

3 years ago
OpenStudy (aaronq):

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

3 years ago
OpenStudy (anonymous):

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.

3 years ago
OpenStudy (aaronq):

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

3 years ago
OpenStudy (anonymous):

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

3 years ago
OpenStudy (anonymous):

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.

3 years ago
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