Kinetics for reversible reactions

[Deepa Maheshvare]

Dear All,

I'm trying to collect the kinetic information of this particular reaction -KEGG REACTIONID:R00754, EC Number:1.1.1.1.

Ethanol + NAD+ <=> Acetaldehyde + NADH + H+

I am looking for the rate expression that corresponds to the above reaction.

For instance, in SABIO EntryID : 23824  the kinetic parameters(Km and Vmax) are reported for Michaelis-Menten equation with NAD+ as the substrate.

I would like to know whether the rate law, mentioned in EntryID:23824, will be applicable just for the forward reaction? 

I would also like to get some advise on how to proceed, if I wish to make use of the parameters in a reversible Michaelis-Menten rate equation.

v=vmaxfS/Km,1−vmaxbP/Km,2(1+S/Km,1+P/Km,1)

For the above reaction , suppose there are two SABIO EntryID's: A and B

 Entry A contains the parameters for forward reaction with NAD+ as the substrate.B contains the parameters for backward reaction with Acetaldehyde as the substrate.Would it be right if I plug both these in the above rate law?

I would be grateful for any help you can offer.

Thanks a lot for your time

Deepa

[Andreas Dräger]

Dear Deepa,

The rate law you found for your reaction of interest applies to a bidirectional, i.e., reversible reaction. The rate law is meant to describe the forward and reverse direction simultaneously. You can see this because in the numerator there is a difference composed of two terms with two v_max constants, one of which is named v_max_f (the forward limiting rate) and the other is v_max_b (the backward limiting rate). The forward limiting rate is multiplied with a fraction of the substrate over its corresponding Km value and the limiting rate for the backward reaction is multiplied with a similar fraction but composed of the product over its corresponding Km value.

The term v_max should (according to A. Cornish-Bowden's book on "Enzyme Kinetics", the red one) no longer be used, because v_max is not a maximum in a mathematical sense, but a limit, an upper bound if you will, for the enzyme's rate.

If you decompose the reaction into separate steps, i.e., one separate reaction for the forward direction and another reaction for the reverse case, you also need to have separate (and different) rate laws for the forward and reverse reaction. If you do this, you will need to apply a classical Henry-Michaelis-Menten reaction to both reactions.

I hope this helps!

With best regards

Dr. Andreas Draeger
Principal Investigator
---
University of Tübingen
Center for Bioinformatics Tübingen (ZBIT)
Applied Bioinformatics Group
Sand 14 · Office #C320 · 72076 Tübingen · Germany
Phone: +49-7071-29-70459 · Fax: +49-7071-29-5152
Web: http://draeger-lab.org · Twitter: @dr_drae
YouTube: https://www.youtube.com/channel/UCp7fWtXGFOIjV35u7ONiVbg

[Andreas Dräger]

Thanks, Deepa,

for this clarification! My first personal impression would be that you should pick the rate law whose substrate and product molecules match the ones in your reaction, i.e., the rate law should follow the same direction as the reaction.

The best way to figure out which rate law meets your criteria is to create a toy example. Tools such as Tellurium provide a nice Python interface where you can quickly generate small models using the scripting language Antimony (see http://tellurium.analogmachine.org for details).

Just create two models:
1) one that has only one reaction and one rate law that you like.
2) another one that has two reactions with separate rate laws.
Set initial values and the parameters and run a simulation and compare the differences.

It would certainly be helpful to read and study a good book about modeling, such as

Palsson, Bernhard Ø. 2011. Systems Biology: Simulation of Dynamic Network States.

I hope this helps!

Cheers
Andreas



> Am 31.08.2017 um 04:22 schrieb Deepa Maheshvare <deepama...@gmail.com>:
>
> Dear Andreas ,
> I'm sorry if I was not clear in the above mail.
> The rate law given in SABIO for EntryID : 23824 is the Henry-Michaelis-Menten expression.
>
> Since the reaction is bidirectional,I would like to use v=Vforward[(S/Km,1)−(P/(Km,2*Keq))]/[(1+S/Km,1+P/Km,2)]
>
> In order to do this, I found two entries -ENTRIID:16888 with ethanol as the substrate and ENTRYID 16885 with acetaldehyde as the substrate.Can we use these Km values(Km of ethanol given in ENTRIID:16888 and Km of acetaldehyde given in ENTRYID 16885 ) in place of Km,1 and Km,2 ,the Vmax(/Vforward) which is given in 16888?
>
> Or is it required to form two different rate laws?
>
> Thanks a lot for all your suggestions and clarifications.
>
> Deepa