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

For an experiment beginning by mixing unbound enzyme with substrate, take $\boxed{[E](0)=e_0>0,[S](0)=s_0>0,[X](0)=[Y](0)=[E_i](0)=[P](0)=0}$. The enzyme-substrate complexes, inactivated enzyme and product are initially absent. Nonzero pre-existing product could be allowed without changing the dynamics here, since it has no feedback reaction in the specified scheme.

Write concentrations as $E,S,X,Y,E_i,P$ in this calculation. The law of mass action assigns the forward bimolecular rate $k_{1}ES$ and the first-order rates $k_{-1}X,k_{2}X,k_{3}Y,k_{4}Y$. Each species gains the rates of reactions producing it and loses those consuming it:Adding the equations for all enzyme-bearing species cancels every reaction rate, so $\boxed{E+X+Y+E_i=e_0}$. This is conservation of the total enzyme-derived material, including the inactivated form. It does not assert conservation of active enzyme: mechanism-based inhibition transfers some of that conserved material irreversibly into $E_i$.

The conserved total is $e_0=[E](0)$ under the chosen initial conditions. Substitute $E=e_0-X-Y-E_i$ into the equations from the previous part. This closes the subsystem for $S,X,Y,E_i$; product can be recovered afterward by integrating $\dot{P}=k_{3}Y$.

Let $k_{\Sigma }=k_{-1}+k_2$ and $K_m=k_{\Sigma }/k_1$. The required nondimensionalization is completed byIndeed $X=e_0\sigma x/(1+\sigma )$ and the free-enzyme fraction is $1-\sigma x/(1+\sigma )-y-e_i$. Substitution and division by the indicated time scale giveFor example $k_2/k_{\Sigma }=1/(1+\rho )$ explains the coefficient in the third equation. The small multiplier on the intermediate-species derivatives identifies their potentially fast adjustment relative to substrate depletion.

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.