Past exam of the mathematics and the environment and mathematical physiology course of the University of Oxford / 2007 / B / 5 / iv / Solution

For quasi-steady enzyme kinetics at low substrate saturation, the quasi-steady-state approximation is controlled here by $\epsilon =e_0/(s_0+K_m)$, rather than by $e_0/s_0$ alone. Even with $e_0\approx s_0$, taking $K_m\gg s_0$ makes $\epsilon \ll 1$ and $\sigma =s_0/K_m=O(\epsilon )$.

After the initial fast layer $T=O(\epsilon )$, the third equation givesSince $\varphi =O(\epsilon )$, the last equation gives $(e_i{)}_T=(\varphi /\epsilon )y=O(\epsilon )$. Starting from $e_i(0)=0$, this implies $e_i=O(\epsilon )$ on bounded intervals of the slow time. The quasi-steady second equation then gives $x=s+O(\epsilon )$. In the first equation,Matching to the initial substrate value $s(0)=1$ therefore yields $\boxed{s(T)\sim e^{-T/(1+\rho )}}$. The relation $x\sim s$ is an outer approximation: it does not replace the initial layer needed to reconcile $x(0)=0$ with $s(0)=1$. Likewise the smallness of accumulated inactivated enzyme has been justified for bounded slow times, not arbitrarily long times.