Quasi-steady state hypothesis application

Consider the reaction scheme:


$$\ce{S + E ->[$k\_1$] C1} \qquad \ce{C1 ->[$k\_2$] E + P} \qquad \ce{S + C1 <=>[$k\_3$][$k\_4$]C2}$$


where $\ce{S}$ is the substrate, $\ce{E}$ is the enzyme, $\ce{P}$ is the product, $\ce{C1}$ and $\ce{C2}$ are enzyme substrate complexes. Let $[\ce{S}] = s$, $[\ce{E}] = e$, $[\ce{C1}] = c\_1$, $[\ce{C2}] = c\_2$ and $[\ce{P}] = p$ be the concentrations of each respective chemical. I have simplified this system down to 


\begin{align}
\frac{\mathrm ds}{\mathrm dt} &= -k\_1se\_0 + (k\_1-k\_3)sc\_1 + (k\_1s+k\_4)c\_2 \\
\frac{\mathrm dc\_1}{\mathrm dt} &= k\_1se\_0 - (k\_1s+k\_2+k\_3s)c\_1+(k\_4-k\_1s)c\_2 \\
\frac{\mathrm dc\_2}{\mathrm dt} &= k\_3sc\_1-k\_4c\_2 \\
\end{align}


using the conservation equation $e=e\_0-c\_1-c\_2$. I have found that $p(t) = k\_2\int c\_1(t) \,\mathrm dt$. Now I need to use the quasi-steady state hypothesis to show that $$\frac{\mathrm ds}{\mathrm dt}= - f(s),\qquad f(s) = \frac{k\_1e\_0s}{1+\frac{k\_1}{k\_2}s + \frac{k\_1k\_3}{k\_2k\_4}s^2}.$$


Now I'm not really given much information on this hypothesis. I have been told it means



> 
> We assume that the initial stage of complex formation is very fast. After which it is essentially at equilibrium.
> 
> 
> 


So how do I apply the hypothesis to transform $\mathrm ds/\mathrm dt$?




Thanks for the interesting question. Before we get into the details, let me note a few things:


1. Your system is a model of [substrate inhibition](http://www.ncbi.nlm.nih.gov/pubmed/20414900), a phenomenon where higher-than-optimal concentrations of a substrate for an enzyme *decrease* its own rate of conversion to product.
2. It sounds like you have a good handle on the enzyme mass balance, i.e. that $e\_0 = e + c\_1 + c\_2$ is a conserved quantity whose time derivative is everywhere zero (no approximations or hypotheses required) in this kinetic scheme. This makes sense because enzymes are [catalysts](https://en.wikipedia.org/wiki/Catalysis) that accelerate reactions without any net participation in them.


OK, onward to your questions.


3. The assumption of fast equilibrium for $c\_2$ is a different assumption than the PSSH hypothesis. Assuming a fast equilibrium for $c\_2$ means that reactions 3 and 4 happen much more quickly than reactions 1 or 2. Thus the concentration of $c\_2$ is determined solely by the thermodynamics of reactions 3 and 4: $$k\_4 c\_2 = k\_3 s c\_1 \;\;\; \Rightarrow \;\;\; c\_2 = \frac{k\_3}{k\_4} s c\_1$$
4. The pseudo-steady-state hypothesis could be applied to $c\_1$. $$\frac{
dc\_1}{
dt} = k\_1se\_0 - (k\_1s+k\_2+k\_3s)c\_1+(k\_4-k\_1s)c\_2 = 0$$ Substituting in the result of the fast equilibrium assumption (point #3) gives $$k\_1se\_0 - (k\_1s+k\_2+k\_3s)c\_1+(k\_4-k\_1s)\frac{k\_3}{k\_4}s c\_1 = 0$$ You can solve this equation for $c\_1$ and plug the result into your $\frac{
ds}{
dt}$ equation.
5. If you are interested in understanding more about the mathematical validity of the PSSH, you should know that it is related to [singular perturbation](https://en.wikipedia.org/wiki/Singular_perturbation). For a treatment of a "simple" Michaelis-Menten system by singular perturbation, see [this book excerpt](https://books.google.com/books?id=kTmQAz6QXckC&pg=PA78&lpg=PA78&dq=%22singular+perturbation%22+michaelis+menten&source=bl&ots=KXUKc-UlSu&sig=srmvulx2myuz1einHrO4Lhbk5Zw&hl=en&sa=X&ved=0CCsQ6AEwAWoVChMI04Te_Pn7yAIVl36ICh2O5QFq#v=onepage&q=%22singular%20perturbation%22%20michaelis%20menten&f=false). The PSSH solution is an "outer" singular perturbation solution.


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