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 $\mathrm ds/\mathrm dt$?

  • $\begingroup$ You understand that the ratio of k3/k4 gives you the equilibrium constant for the third reaction? S, c1 and c2 thus follow equilibrium. $\endgroup$
    – MaxW
    Nov 5, 2015 at 21:11
  • $\begingroup$ So if $[C_1]$ is a constant (quasi-steady state hypothesis) then dp/dt is $k2[C_1] $ $\endgroup$
    – MaxW
    Nov 5, 2015 at 21:22
  • $\begingroup$ I'm not really sure what you mean by that, I come from a mathematical background so don't have much chemistry knowledge $\endgroup$ Nov 5, 2015 at 21:26
  • $\begingroup$ Initially there is no [P]. The assumption is that [P] is so slow to form in the quasi-steady state that the mass balance is S (at t=0 ) = $S + C_1 + 2C_2$ $\endgroup$
    – MaxW
    Nov 5, 2015 at 21:35
  • $\begingroup$ Sorry to be annoying. I understand that assumption, but I don't understand how that leads to your second point and what your second point means. $\endgroup$ Nov 5, 2015 at 22:18

1 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.

  1. 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: $$k_4 c_2 = k_3 s c_1 \;\;\; \Rightarrow \;\;\; c_2 = \frac{k_3}{k_4} s c_1$$

  2. 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.

  3. 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.

  • $\begingroup$ This is the same exact method I outlined in the 9th comment to the original question. $\endgroup$ Nov 6, 2015 at 17:32
  • $\begingroup$ Thanks for letting me know. I don't always read comments and didn't in this case. I prefer to write answers instead. $\endgroup$
    – Curt F.
    Nov 6, 2015 at 19:00
  • 1
    $\begingroup$ I also corrected a couple of errors from the equations in the original post that propagated into your answer. $\endgroup$ Nov 6, 2015 at 19:14
  • $\begingroup$ Thank you! I just copied the equations from the original post so I'm glad someone is out there checking this! $\endgroup$
    – Curt F.
    Nov 6, 2015 at 19:20
  • $\begingroup$ I realized that there were errors in the formulation only when I tried to apply our solution method, and the answer didn't work out. $\endgroup$ Nov 6, 2015 at 19:35

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