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Consider this enzyme reaction enter image description here with initial conditions $c_1(0) = c_2(0) = p(0) = 0,s(0) = s_0, e(0) = e_0, e^{\ast}(0) = e^{\ast}_0.$

I determined this differential equations for the enzyme reactions: $$\begin{align} s'&=-k_1se+(k_{-1}-k_3s)c_1+k_{-3}c_2 \\ c_1'&=k_1se-(k_{-1}+k_2+k_3s)c_1+(k_4+k_{-3})c_2\\ c_2'&=k_3sc_1-(k_{-3}+k_4)c_2\\ e'&=-k_1se+(k_{-1}+k_2)c_1-be+fe^{\ast} \\ (e^{\ast})'&=be-fe^{\ast}\\ p'&=k_4c_2+k_2c_1 \end{align}$$

Suppose $E$ can bind substrate but $E^{\ast}$ cannot. Determine the equation for enzyme conservation.

I don't understand the first sentence. It is $e'+c_1'+c_2'+(e^{\ast})'=0$, so the amount of enzymes does not change in time (enzyme conservation?). So one gets $$e+c_1+c_2+e^{\ast}=e_0+c_1(0)+c_2(0)+e^{\ast}_0 \Rightarrow e=e_0+e^{\ast}_0-(c_1+c_2+e^{\ast})?$$

Reduce the system (to get a system in $c_1, c_2, p$) with the QSS assumption $(e^{\ast})'=0$.

This assumption provides the differential equation $$\begin{align} s'&=-k_1se+(k_{-1}-k_3s)c_1+k_{-3}c_2 \\ c_1'&=k_1se-(k_{-1}+k_2+k_3s)c_1+(k_4+k_{-3})c_2\\ c_2'&=k_3sc_1-(k_{-3}+k_4)c_2\\ e'&=-k_1se+(k_{-1}+k_2)c_1 \\ (e^{\ast})'&=0\\ p'&=k_4c_2+k_2c_1 \end{align}$$

Now it is $$e(t)=e_0-(c_1(t)+c_2(t))$$ and

$$s(t)=s_0-(c_1(t)+2c_2(t)+p(t))$$

My idea was to replace $s$ and $e$ by the terms above in $c_1', c_2' \text{ and } p'$: $$\begin{align} c_1'&=k_1e_0(s_0-p)-(k_{-1}+k_2+k_3(s_0-c_1-2c_2-p)+k_1(s_0 +e_0-c_1-3c_2-p))c_1+(k_4+k_{-3}-k_1(s_0+2c_2-2e_0+p))c_2\\ c_2'&=k_3(s_0-(c_1(t)+2c_2(t)+p(t)))c_1-(k_{-3}+k_4)c_2\\ p'&=k_4c_2+k_2c_1\\ (e^{\ast})'&=0 \end{align}$$

My question is: what do I do with $(e^{\ast})'$? Since $(e^{\ast})'=0 \Rightarrow e^{\ast}=e^{\ast}_0$.

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  • $\begingroup$ Can you please clearly state the final system of ordinary differential equations you wish to solve? If you need help deriving that from the above chemical process, I gently suggest this might not be the best place to ask chemistry questions. $\endgroup$ May 24, 2022 at 14:41
  • $\begingroup$ I don't need help to derive the differential equations but the enzyme conservation in order to derive the reduced one. I already solved a similar task without $E^{\ast}$. This is not a chemistry question but a task from my mathematical modelling class $\endgroup$
    – Uhmm
    May 24, 2022 at 14:46
  • $\begingroup$ Great, so if you clearly list your set of equations and initial conditions at the bottom of your question, myself or someone else should be able to point you in the right direction to solve it. As it is, it isn't clear to me what system of equations I'm looking at. There are 6 unknowns, right (perhaps 5 once you recognize that one of them is constant)? One for the concentration of each chemical/enzyme. Yet I only see two simultaneous equations. $\endgroup$ May 24, 2022 at 14:51
  • $\begingroup$ @Chessnerd321 I made an edit $\endgroup$
    – Uhmm
    May 24, 2022 at 14:55

1 Answer 1

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The sets of conserved species can be found by looking at the left null-space of the stoichiometric matrix $S$ given in this case by

$$S=\begin{bmatrix} - 1 & 1 & 0 & -1 & 1 & 0 & 0 & 0\\ 1 & -1 & -1 & -1 & 1 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & -1 & -1 & 0 & 0\\ -1 & 1 & 1 & 0 & 0 & 0 & -1 & 1\\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & -1\\ 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 \end{bmatrix},$$

where the state is $(s,c_1,c_2,e,e^*,p)$. Simple calculations show that the left-null space is spanned by the row vectors

$$(0,1,1,1,1,0)\ \text{and } (1,1,2,0,0,1).$$

This implies that

$$c_1(t)+c_2(t)+e(t)+e^*(t)=e_0+e_0^*$$ and

$$s(t)+c_1(t)+2c_2(t)+p(t)=s_0$$ for all $t\ge0$.

For the quasi steady-state question, we simply need to replace $fe^*$ by $be$ in the model.

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  • $\begingroup$ Thank you! So $c_1(t)+c_2(t)+e(t)+e^*(t)=e_0+e_0^*$ is the enzyme conservation? What happens if E can bind substrate but $E^{\ast}$ cannot? We did not have stoichiometric matrices in class so I worked with the differential equation system I derived in my question. I need to reduce it with the QSS assumption $(e^{\ast})'=0$. By replacing $fe^{\ast}$ by $be$ I get a similar system as above but without the last two prases in $e'$. How do I reduce this system into a system of $c_1, c_2, p$? My idea was to replace $e$ by $e_0+e^{\ast}_0-(c_2+c_1+e^{\ast})$ and $s$ by $s_0-(c_1+2c_2+p)$? $\endgroup$
    – Uhmm
    May 25, 2022 at 11:37
  • $\begingroup$ Too many questions in one comment. You can figure out most of them by yourself by putting enough work into it. I will just give one comment: the network you gave already considers that $E$ can bind the substrate $S$ but that $E^*$ can not. $\endgroup$
    – KBS
    May 25, 2022 at 12:57
  • $\begingroup$ Thank you for your help. Can you please give me another comment on my idea? I made an edit at the end of my question. $\endgroup$
    – Uhmm
    May 25, 2022 at 15:31

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