Derivation of rate law for random bi bi kinetic reaction
504 views
Skip to first unread message
Padma Priya
Jun 15, 2021, 11:07:55 PM6/15/21
to COPASI User Forum
Hi.
I am an amateur and have started exploring COPASI for my research.
I was working with an enzyme, citramalate synthase which was reported to follow
nonrapid equilibrium random bi-bi kinetic mechanism, with a preferred pathway to
the ternary complex.
The generalized rate law for the equation has been reported to be
(V/Vmax)= (A.B)/ (KAs . KAB + KAB [A] + KBA [B] + [A][B]) where Vmax= kcat ET
Previously I understood that for an ordered sequential bi bi reaction, the generalized rate law was (V/Vmax)= (A.B)/ (KAs . KAB + KAB [A] + [A][B]) where Vmax= kcat ET
but that in the model was a very complex one and I have no clue how this was derived.
(Vf∗(substratea∗substrateb-productp∗productq/Keq))/█(substratea∗substrateb∗(1+productp/Kip)+Kma∗substrateb+Kmb∗(substratea+Kia)@+Vf/(Vr∗Keq)∗(Kmq∗productp∗(1+substratea/Kia)+productq∗(Kmp∗(1+Kma∗substrateb/(Kia∗Kmb))+productp∗(1+substrateb/Kib))) )
It would be great if some one could just help me out in deriving this complex rate equation from the generalized rate law as it would be a great eye opener for my work.
sven....@gmail.com
Jun 16, 2021, 9:59:02 AM6/16/21
to COPASI User Forum
Hello, I do not know the model and cannot immedeately see how this specific rate law was derived, but one basic difference between the simpler rate laws and the complex one is that the simpler ones are for irreversible reactions (they only describe the rate of the forward reaction) while the more complicated rate law descibes a reversible reaction with a given equilibrium constant.
Sven
Mendes,Pedro
Jun 16, 2021, 10:24:16 AM6/16/21
to copasi-u...@googlegroups.com
Hello,
Summary: the difference is thermodynamics (and reversibility).
Longer explanation:
The rate laws that you express are for irreversible reactions. In that case you don't have to worry about thermodynamics because the statement "reaction is irreversible" means that there are no parameters for the reverse reaction (and also that Keq=infinty). (Now I don't want to go to the sideline issue whether irreversible reactions are ever legitimate in models; it all depends on the purpose of the model and the strength of the assumptions).
For reversible reactions, i.e. those with 0 < Keq < infinity, if one does not include the Keq in the rate law then we can easily change parameters such that they end up with very different thermodynamics (ie different values of Keq). One has to consider the so-called Haldane Relationships, which are constraints on the kinetic parameters given microscopic reversibility.
The simplest case to explain is the reversible Michaelis-Menten equation (which should really be called the Michaelis-Menten-Haldane equation). It is usually written as:
v = ( Vmaxf * S/Kms - Vmaxr * P/Kmp) / (1 + S/Kms + P/Kmp) (1)
this equation appears to have four kinetic parameters: Vmaxf, Vmaxr, Kms, and Kmp . But Haldane showed that they are not independent and to satisfy microscopic reversibility the following relationship (the Haldane relationship) exists:
Keq = Vmaxf*Kmp / Vmaxr*Kms (2)
This means that if you know the equilibrium constant of the reaction, then there are only 3 independent kinetic paramters. You can re-write the reversible Michaelis-Menten equation as (for example):
v = ( Vmaxf/Kms * ( S -P/Keq) ) / (1 +S/Kms + P/Kmp ) (3)
which now has 3 kinetic parameters and the equilibrium constant. This equation is preferable to use in modeling when you know the Keq. If you use this, you would only fit Vmaxf, Kms and Kmp (Vmaxr depends on these and Keq). Also, if you were to do a parameter scan and you wanted to keep the reaction equilibrium constant fixed (ie do a parameter scan that is consistent with thermodynamic constraints), you would only change Vmaxf, Kms and Kmp and keep Keq constant. There are other benefits from writing the rate law using (3) but for brevity I will not discuss them here.
(Note that I removed Vmaxr from the equation, but you can instead remove any of the other 3 parameters, and arrive at different equations, which nevertheless are all equivalent in thermodynamic terms. It is really a modeler's choice which of the 4 parameters to factor out).
So the above explains why sometimes you may see Eq (1) being used, and other times Eq (3) being used. The latter looks more complicated, but is not really.
Now, for more complicated reactions, with several substrates, things get more complicated. It turns out that if you have more than one substrate and product there are more thermodynamic constraints (ie several Haldane relationships), as discussed in the classic series of 3 papers by W.W. Cleland (the most important being [1] but see also [2]. This means that to write a thermodynamically consistent equation for those reactions, requires factoring out more than one kinetic parameter. This ends up with equations that look a bit ugly (like the one in COPASI for Ordered Uni Bi mechanism), but that is just because of factoring out some parameters using the Haldane relationships. (Additionally we re-write the equations in a form that minimizes divisions because a) divisions are slow operations, b) this avoids accidental division-by-zero exceptions)
References
[1] Cleland WW (1963) The kinetics of enzyme-catalyzed reactions with two or more substrates or products. I. Nomenclature and rate equations. Biochimica et Biophysica Acta 67:104–137 Cleland WW (1963) The kinetics of enzyme-catalyzed reactions with two or more substrates or products. I. Nomenclature and rate equations. Biochimica et Biophysica Acta 67:104–137
[2] Cleland WW (1982) An analysis of Haldane relationships. Methods Enzymol. 87:366–369
Pedro
Summary: the difference is thermodynamics (and reversibility).
Longer explanation:
The rate laws that you express are for irreversible reactions. In that case you don't have to worry about thermodynamics because the statement "reaction is irreversible" means that there are no parameters for the reverse reaction (and also that Keq=infinty). (Now I don't want to go to the sideline issue whether irreversible reactions are ever legitimate in models; it all depends on the purpose of the model and the strength of the assumptions).
For reversible reactions, i.e. those with 0 < Keq < infinity, if one does not include the Keq in the rate law then we can easily change parameters such that they end up with very different thermodynamics (ie different values of Keq). One has to consider the so-called Haldane Relationships, which are constraints on the kinetic parameters given microscopic reversibility.
The simplest case to explain is the reversible Michaelis-Menten equation (which should really be called the Michaelis-Menten-Haldane equation). It is usually written as:
v = ( Vmaxf * S/Kms - Vmaxr * P/Kmp) / (1 + S/Kms + P/Kmp) (1)
this equation appears to have four kinetic parameters: Vmaxf, Vmaxr, Kms, and Kmp . But Haldane showed that they are not independent and to satisfy microscopic reversibility the following relationship (the Haldane relationship) exists:
Keq = Vmaxf*Kmp / Vmaxr*Kms (2)
This means that if you know the equilibrium constant of the reaction, then there are only 3 independent kinetic paramters. You can re-write the reversible Michaelis-Menten equation as (for example):
v = ( Vmaxf/Kms * ( S -P/Keq) ) / (1 +S/Kms + P/Kmp ) (3)
which now has 3 kinetic parameters and the equilibrium constant. This equation is preferable to use in modeling when you know the Keq. If you use this, you would only fit Vmaxf, Kms and Kmp (Vmaxr depends on these and Keq). Also, if you were to do a parameter scan and you wanted to keep the reaction equilibrium constant fixed (ie do a parameter scan that is consistent with thermodynamic constraints), you would only change Vmaxf, Kms and Kmp and keep Keq constant. There are other benefits from writing the rate law using (3) but for brevity I will not discuss them here.
(Note that I removed Vmaxr from the equation, but you can instead remove any of the other 3 parameters, and arrive at different equations, which nevertheless are all equivalent in thermodynamic terms. It is really a modeler's choice which of the 4 parameters to factor out).
So the above explains why sometimes you may see Eq (1) being used, and other times Eq (3) being used. The latter looks more complicated, but is not really.
Now, for more complicated reactions, with several substrates, things get more complicated. It turns out that if you have more than one substrate and product there are more thermodynamic constraints (ie several Haldane relationships), as discussed in the classic series of 3 papers by W.W. Cleland (the most important being [1] but see also [2]. This means that to write a thermodynamically consistent equation for those reactions, requires factoring out more than one kinetic parameter. This ends up with equations that look a bit ugly (like the one in COPASI for Ordered Uni Bi mechanism), but that is just because of factoring out some parameters using the Haldane relationships. (Additionally we re-write the equations in a form that minimizes divisions because a) divisions are slow operations, b) this avoids accidental division-by-zero exceptions)
References
[1] Cleland WW (1963) The kinetics of enzyme-catalyzed reactions with two or more substrates or products. I. Nomenclature and rate equations. Biochimica et Biophysica Acta 67:104–137 Cleland WW (1963) The kinetics of enzyme-catalyzed reactions with two or more substrates or products. I. Nomenclature and rate equations. Biochimica et Biophysica Acta 67:104–137
Pedro
--
You received this message because you are subscribed to the Google Groups "COPASI User Forum" group.
To unsubscribe from this group and stop receiving emails from it, send an email to copasi-user-fo...@googlegroups.com.
To view this discussion on the web visit https://groups.google.com/d/msgid/copasi-user-forum/00d53894-78d3-47dc-a05b-665b439f3160n%40googlegroups.com.
-- Pedro Mendes, PhD Professor and Director, Richard D. Berlin Center for Cell Analysis and Modeling University of Connecticut School of Medicine group website: http://www.comp-sys-bio.org
Padma Priya
Jun 16, 2021, 12:48:30 PM6/16/21
to copasi-u...@googlegroups.com
Thank you very much Dr. Pedro and Dr. Sven. I have one more doubt. Is it possible to achieve a steady state in case of a linear pathway? I understand that for a cyclic pathway, that is mentioned in the copasi tutorials, the steady state was achieved and in turn the metabolic control analysis could be performed. However, in myc case which involves a linear pathway, the model says that the steady state was not found and hence it gives on elasticity coefficient. Is there a way that the model achieves a steady state with a linear pathway?
To view this discussion on the web visit https://groups.google.com/d/msgid/copasi-user-forum/ae45654e-4dc0-e425-b177-0f1eb9b732b3%40uchc.edu.
Padma Priya
Jun 17, 2021, 4:38:32 AM6/17/21
to copasi-u...@googlegroups.com
Hi
I am not able to determine the rate law for a few of the enzymes that are involved in my pathway of research. Could anyone help with this?
1. Isopropylmalate isomerase (also known as isopropyl malate dehydratase)
2. Pyruvate decarboxylase
Also can the ordered bi bi rate law be used for random bi bi kinetic reaction? Will that make a huge difference?
Pedro Mendes
Jun 17, 2021, 8:31:41 AM6/17/21
to copasi-u...@googlegroups.com
You should be able to reach a steady state with a linear pathway or any
other network topology. The only necessary conditions for establishing a
steady state is that the "environment" be fixed. The environment is the
outside of your model. For example in a small linear pathway liket this:
X1 = X2 = X3 = X4
(where the "=" means reversible reaction)
the environment is X1 and X4. These two have to be made constant in
order to achieve a steady state for X2 and X3 (which are the system).
The most common way to achieve this in COPASI is to set X1 and X4 as
"fixed" (instead of the default "reactions"). You do this in the Species
part of the model.
Another way in which this could be achieved is to set the kinetics for
reaction 1 (X1 = X2) and reaction 3 (X3 = X4) as "constant flux" and set
them exactly to the same flux value.
If you don't set X1 and X4 as fixed, nor set reactions 1 and 3 as
constant flux, then you have a closed system (and in this case X1 and X4
are considered part of the system). You will still achieve a steady
state but in this case it is garanteed to exactly the chemical equilibrium.
Hope this helps
Pedro
> equations. /Biochimica et Biophysica Acta/ 67:104–137 Cleland WW
>> <mailto:copasi-user-fo...@googlegroups.com>.
> <https://groups.google.com/d/msgid/copasi-user-forum/ae45654e-4dc0-e425-b177-0f1eb9b732b3%40uchc.edu?utm_medium=email&utm_source=footer>.
> <https://groups.google.com/d/msgid/copasi-user-forum/CABG8L5nuAxjM0PAxL-kacBciSYiLNG4sFPeofT4vCZ0NPsa5gQ%40mail.gmail.com?utm_medium=email&utm_source=footer>.
other network topology. The only necessary conditions for establishing a
steady state is that the "environment" be fixed. The environment is the
outside of your model. For example in a small linear pathway liket this:
X1 = X2 = X3 = X4
(where the "=" means reversible reaction)
the environment is X1 and X4. These two have to be made constant in
order to achieve a steady state for X2 and X3 (which are the system).
The most common way to achieve this in COPASI is to set X1 and X4 as
"fixed" (instead of the default "reactions"). You do this in the Species
part of the model.
Another way in which this could be achieved is to set the kinetics for
reaction 1 (X1 = X2) and reaction 3 (X3 = X4) as "constant flux" and set
them exactly to the same flux value.
If you don't set X1 and X4 as fixed, nor set reactions 1 and 3 as
constant flux, then you have a closed system (and in this case X1 and X4
are considered part of the system). You will still achieve a steady
state but in this case it is garanteed to exactly the chemical equilibrium.
Hope this helps
Pedro
>> To view this discussion on the web visit
>> https://groups.google.com/d/msgid/copasi-user-forum/00d53894-78d3-47dc-a05b-665b439f3160n%40googlegroups.com
>> <https://groups.google.com/d/msgid/copasi-user-forum/00d53894-78d3-47dc-a05b-665b439f3160n%40googlegroups.com?utm_medium=email&utm_source=footer>.
>> https://groups.google.com/d/msgid/copasi-user-forum/00d53894-78d3-47dc-a05b-665b439f3160n%40googlegroups.com
>
> --
> Pedro Mendes, PhD
> Professor and Director,
> Richard D. Berlin Center for Cell Analysis and Modeling
> University of Connecticut School of Medicine
> group website:http://www.comp-sys-bio.org
>
> --
> You received this message because you are subscribed to the Google
> Groups "COPASI User Forum" group.
> To unsubscribe from this group and stop receiving emails from it,
> send an email to copasi-user-fo...@googlegroups.com
> <mailto:copasi-user-fo...@googlegroups.com>.
> --
> Pedro Mendes, PhD
> Professor and Director,
> Richard D. Berlin Center for Cell Analysis and Modeling
> University of Connecticut School of Medicine
> group website:http://www.comp-sys-bio.org
>
> --
> You received this message because you are subscribed to the Google
> Groups "COPASI User Forum" group.
> To unsubscribe from this group and stop receiving emails from it,
> send an email to copasi-user-fo...@googlegroups.com
> To view this discussion on the web visit
> https://groups.google.com/d/msgid/copasi-user-forum/ae45654e-4dc0-e425-b177-0f1eb9b732b3%40uchc.edu
> <https://groups.google.com/d/msgid/copasi-user-forum/ae45654e-4dc0-e425-b177-0f1eb9b732b3%40uchc.edu?utm_medium=email&utm_source=footer>.
>
> --
> You received this message because you are subscribed to the Google
> Groups "COPASI User Forum" group.
> To unsubscribe from this group and stop receiving emails from it, send
> an email to copasi-user-fo...@googlegroups.com
> <mailto:copasi-user-fo...@googlegroups.com>.
> --
> You received this message because you are subscribed to the Google
> Groups "COPASI User Forum" group.
> To unsubscribe from this group and stop receiving emails from it, send
> an email to copasi-user-fo...@googlegroups.com
> To view this discussion on the web visit
> https://groups.google.com/d/msgid/copasi-user-forum/CABG8L5nuAxjM0PAxL-kacBciSYiLNG4sFPeofT4vCZ0NPsa5gQ%40mail.gmail.com
> <https://groups.google.com/d/msgid/copasi-user-forum/CABG8L5nuAxjM0PAxL-kacBciSYiLNG4sFPeofT4vCZ0NPsa5gQ%40mail.gmail.com?utm_medium=email&utm_source=footer>.
Pedro Mendes
Jun 17, 2021, 8:40:12 AM6/17/21
to copasi-u...@googlegroups.com
I don't know specifics of those two enzymes, but you can look them up in
two online databases: Brenda and Sabio-RK, both have kinetics for
several enzymes.
In any case, there is a big difference between random bi bi and ordered
bi bi. The first one has interaction terms between all the substrates
and products, while the second one has less interaction terms. However
in practice it may be difficult to distinguish them when embedded in a
pathway (instead of in vitro kinetics of the purified enzymes). You can
also try to use one of the "generic rate laws" that have been developed
for this case when you don't know much details. See for example [1] and [2]
[1] Liebermeister W, Klipp E (2006) Bringing metabolic networks to life:
convenience rate law and thermodynamic constraints. Theoretical Biology
& Medical Modelling 3:41 https://doi.org/10.1186/1742-4682-3-41
[2] Liebermeister W, Uhlendorf J, Klipp E (2010) Modular rate laws for
enzymatic reactions: thermodynamics, elasticities and implementation.
Bioinformatics (Oxford, England) 26:1528–1534
https://doi.org/10.1093/bioinformatics/btq141
> Nomenclature and rate equations. /Biochimica et Biophysica Acta/
>> <mailto:copasi-user-fo...@googlegroups.com>.
> <https://groups.google.com/d/msgid/copasi-user-forum/ae45654e-4dc0-e425-b177-0f1eb9b732b3%40uchc.edu?utm_medium=email&utm_source=footer>.
> <https://groups.google.com/d/msgid/copasi-user-forum/CABG8L5ku4sbdX0U6rN9tSn%3Dd%2By9qs8Ui%3DU7%3Dfx1dNC1mLU1hBQ%40mail.gmail.com?utm_medium=email&utm_source=footer>.
two online databases: Brenda and Sabio-RK, both have kinetics for
several enzymes.
In any case, there is a big difference between random bi bi and ordered
bi bi. The first one has interaction terms between all the substrates
and products, while the second one has less interaction terms. However
in practice it may be difficult to distinguish them when embedded in a
pathway (instead of in vitro kinetics of the purified enzymes). You can
also try to use one of the "generic rate laws" that have been developed
for this case when you don't know much details. See for example [1] and [2]
[1] Liebermeister W, Klipp E (2006) Bringing metabolic networks to life:
convenience rate law and thermodynamic constraints. Theoretical Biology
& Medical Modelling 3:41 https://doi.org/10.1186/1742-4682-3-41
[2] Liebermeister W, Uhlendorf J, Klipp E (2010) Modular rate laws for
enzymatic reactions: thermodynamics, elasticities and implementation.
Bioinformatics (Oxford, England) 26:1528–1534
https://doi.org/10.1093/bioinformatics/btq141
>> To view this discussion on the web visit
>> https://groups.google.com/d/msgid/copasi-user-forum/00d53894-78d3-47dc-a05b-665b439f3160n%40googlegroups.com
>> <https://groups.google.com/d/msgid/copasi-user-forum/00d53894-78d3-47dc-a05b-665b439f3160n%40googlegroups.com?utm_medium=email&utm_source=footer>.
>> https://groups.google.com/d/msgid/copasi-user-forum/00d53894-78d3-47dc-a05b-665b439f3160n%40googlegroups.com
>
> --
> Pedro Mendes, PhD
> Professor and Director,
> Richard D. Berlin Center for Cell Analysis and Modeling
> University of Connecticut School of Medicine
> group website:http://www.comp-sys-bio.org
>
> --
> You received this message because you are subscribed to the
> Google Groups "COPASI User Forum" group.
> To unsubscribe from this group and stop receiving emails from
> it, send an email to
> copasi-user-fo...@googlegroups.com
> <mailto:copasi-user-fo...@googlegroups.com>.
> --
> Pedro Mendes, PhD
> Professor and Director,
> Richard D. Berlin Center for Cell Analysis and Modeling
> University of Connecticut School of Medicine
> group website:http://www.comp-sys-bio.org
>
> --
> You received this message because you are subscribed to the
> Google Groups "COPASI User Forum" group.
> To unsubscribe from this group and stop receiving emails from
> it, send an email to
> copasi-user-fo...@googlegroups.com
> To view this discussion on the web visit
> https://groups.google.com/d/msgid/copasi-user-forum/ae45654e-4dc0-e425-b177-0f1eb9b732b3%40uchc.edu
> <https://groups.google.com/d/msgid/copasi-user-forum/ae45654e-4dc0-e425-b177-0f1eb9b732b3%40uchc.edu?utm_medium=email&utm_source=footer>.
>
> --
> You received this message because you are subscribed to the Google
> Groups "COPASI User Forum" group.
> To unsubscribe from this group and stop receiving emails from it, send
> an email to copasi-user-fo...@googlegroups.com
> <mailto:copasi-user-fo...@googlegroups.com>.
> --
> You received this message because you are subscribed to the Google
> Groups "COPASI User Forum" group.
> To unsubscribe from this group and stop receiving emails from it, send
> an email to copasi-user-fo...@googlegroups.com
> To view this discussion on the web visit
> https://groups.google.com/d/msgid/copasi-user-forum/CABG8L5ku4sbdX0U6rN9tSn%3Dd%2By9qs8Ui%3DU7%3Dfx1dNC1mLU1hBQ%40mail.gmail.com
> <https://groups.google.com/d/msgid/copasi-user-forum/CABG8L5ku4sbdX0U6rN9tSn%3Dd%2By9qs8Ui%3DU7%3Dfx1dNC1mLU1hBQ%40mail.gmail.com?utm_medium=email&utm_source=footer>.
Padma Priya
Jun 18, 2021, 7:25:59 AM6/18/21
to copasi-u...@googlegroups.com
Thank you very much Dr. Pedro. Since i am working with the enzymatic biotransformation in cell free system using multi enzymatic steps, I always design my cell free experiments in such a way that there is no accumulation of intermediates (which overrules inhibition of enzymes by intermediate compounds), use a fed batch strategy to feed my substrate if there is substrate inhibition and try to use enzymes which lack the product inhibition. In such a case, I felt that developing a model without any inhibition constants would be the best way to mimic my multi- enzymatic cell free system.
![]()
![]()
In connection to that, taking your inputs into consideration, I derived a rate equation for random bi bi mechanism, by starting with a generalised rate law of
V=[Vf.(A/KmA).(B/Kmb)-Vr.(P/Kmp).(Q/Kmq)]/([1+(A/Kma)+ (B/Kmb)+(AB/KmaKmb)+(P/Kmp)+(Q/Kmq)+(PQ/KmpKmq)]
and a haldane relationship without inhibition term by modifying the haldane of Michaelis menten equation where I assumed
Keq= (Vf.Kmp.Kmq)/(Vr.Kma.Kmb)
Further by following the steps suggested by you, I have obtained a rate law as follows for random bi bi mechanism,
V=Vf[A.B-(P.Q)/Keq] /[Kma.Kmb +A.Kmb +B.Kma + A.B+ {Vf/(Vr.Keq)}.{PKmq+QKmp+PQ}] which actually corroborated with the format for generalized rate law for random bi bi mechanism.
Could you please let me know if this rate law and the assumptions sound correct and I could proceed further by feeding this rate equation into the model?
Your inputs are opening great insights for me and it is very helpful. Thank you so much for the valuable inputs.
To unsubscribe from this group and stop receiving emails from it, send an email to copasi-user-fo...@googlegroups.com.
To view this discussion on the web visit https://groups.google.com/d/msgid/copasi-user-forum/a1eadf66-d3fa-6d0a-bce4-5a9b14137bdc%40copasi.org.
Padma Priya
Jun 20, 2021, 12:05:20 AM6/20/21
to copasi-u...@googlegroups.com
Hello Dr. Pedro,
As suggested by you, I have been working with a pathway which involves the recycling of X2 and X4 making the model cyclic
X1+ X2= X3+X4
X3=X5
X5+X6=X7
X7+X4=X2+X8
Now, usually to run a time course simulation for the conversion of 10mM X1, I supply 10mM X6 and 0.2mM of X2 (Since X2 gets recycled)
To achieve a steady state, I tried both the options that you suggested.
1. In the first case, I set X1, X2 and X8 as fixed with 5mM as the initial concentration. I had the other metabolites as "reaction" with X6 alone set as 5mM and concentrations of other metabolites as 0. I was not able to achieve a proper steady state and only equilibrium steady state, particularly the region after the completion of the reaction, was achieved.
2. When the second suggestion given by you was executed by assuming the constant flux of 2umol/min for Reaction 1 and Reaction 4 while retaining the original rate laws for the two reactions in between, I was not able to achieve a steady state.
I don't know what mistake I am making and I am sure it has something to do with the initial concentration of metabolites that I assume in the species tab. Even though I make the assumptions that you suggested I am not able to ascertain the initial concentrations of the metabolites that should be fed into the model while performing steady state analysis. Also, should there be any changes in the kinetic parameters of the reactions, like Vf Vr, to obtain a proper steady state. Could you please help me out by identifying the things that I am missing while doing a steady state analysis. I am in need of figuring out this problem for my research paper. It would be great if you could suggest a way out
Thanks in advance
To unsubscribe from this group and stop receiving emails from it, send an email to copasi-user-fo...@googlegroups.com.
To view this discussion on the web visit https://groups.google.com/d/msgid/copasi-user-forum/8c43ce6d-9994-f7cb-dc8e-294a5a0fe038%40copasi.org.
Frank Bergmann
Jun 21, 2021, 6:37:47 AM6/21/21
to copasi-u...@googlegroups.com
Hello Padma,
maybe a visualization of the network helps (you can do that in COPASI, selecting diagrams under the model tree. It will give you for your reaction network something like this:
In order to obtain a steady state, you want to 'fix' the borders of your system. So in this case it would be X1, X6 and X8. You can either do this by setting the reaction type to fixed or by adding inflow / outflow reactions that would indicate that the species are constantly produced, or removed from the system.
I hope this helps,
best
Frank
To view this discussion on the web visit https://groups.google.com/d/msgid/copasi-user-forum/CABG8L5kwjW%2BJGugpbN5_CBVQspvt0%2BcfSuFS0sfSrqN2nT_bxA%40mail.gmail.com.
Padma Priya
Jun 23, 2021, 4:25:40 AM6/23/21
to copasi-u...@googlegroups.com
Hi frank
I understand that I am supposed to fix certain metabolites. In that case what should be the initial concentration of the metabolites that is to be given?
Also what are the initial concentrations that I should provide for the other substrates in the species tab which have been selected as "reactions" instead of fixed.
Is there any particular principle behind setting the initial concentration of the metabolites before doing the steady state analysis?
The model states that the equilibrium steady state has been reached. But when I take a close look, it points out the portion after the reaction has got completely converted as steady state.
To view this discussion on the web visit https://groups.google.com/d/msgid/copasi-user-forum/CAEnSYuLA0UtFvRWAmKWU0ytkCRMJhgmQ_r3br%2BZg2iqES4D2qg%40mail.gmail.com.
Reply all
Reply to author
Forward
0 new messages