Quasi-steady state hypothesis application

Question

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 $\mathrm ds/\mathrm dt$?

Answer 1

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, 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 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. For a treatment of a "simple" Michaelis-Menten system by singular perturbation, see this book excerpt. The PSSH solution is an "outer" singular perturbation solution.