Past exam of the mathematics and the environment and mathematical physiology course of the University of Oxford / 2008 / B / 5

For an assay begun by mixing free enzyme and substrate, take $\boxed{[S](0)=S_0>0,[E](0)=E_0>0,[C](0)=[P](0)=0}$. The complex has not yet formed and no product has been made. The later small-enzyme limit assumes $E_0\ll S_0$. Other preincubation protocols could require nonzero initial complex, but these initial conditions are the ones used by the stated inner solution.

Mass action givesThus $E+C=E_0$ and $S+C+P=S_0$ are conserved. Eliminating free enzyme yields the stated two-equation system.

With $s=S/S_0$, $c=C/E_0$ and $\tau =t/T$, substitution givesThe dimensionless initial data are $\boxed{s(0)=1,c(0)=0}$. The parameter named $\delta$ here is an enzyme kinetic coefficient and is unrelated to the loss parameter used in the later blood-cell question.

Set $T_f=(k_1{S}_0{)}^{-1}$, $ϵ=E_0/S_0$, $K=(k_{-1}+k_2)/(k_1{S}_0)$ and $\lambda =k_2/(k_1{S}_0)$. On fast time ${\tau }_f=t/T_f$,For ${\tau }_f=O(1)$, leading substrate remains $s=1$. The initial complex buildup therefore solves $c^{\prime }=1-(1+K)c$, $c(0)=0$, givingTime belongs inside the exponential. Its placement outside the exponential bracket in the printed expression is a typographical error; the derived form satisfies both the kinetic equation and zero initial complex concentration.

Use the longer scale $T_s=(k_1{E}_0{)}^{-1}=T_f/ϵ$ and ${\tau }_s=t/T_s=ϵ{\tau }_f$. The equations becomeAt leading order the fast complex variable has equilibrated to the quasi-steady manifold. ThereforeMatching from the initial layer gives $s(0)=1$ for this outer problem; its implicit solution is $\boxed{s+K\log s=1-\lambda {\tau }_s}$. The dimensional substrate loss is the Michaelis-Menten rate with maximum rate $k_2{E}_0$ and Michaelis constant $(k_{-1}+k_2)/k_1$.

The fast scale resolves the rapid complex buildup that is compressed into a thin initial layer on the substrate-depletion scale. The outer scale follows depletion over a much longer time and cannot satisfy $c(0)=0$ by itself.

In the overlap $1\ll {\tau }_f\ll {ϵ}^{-1}$, the inner solution tends to $(s,c)=(1,1/(1+K))$. The outer solution as ${\tau }_s\downarrow 0$ has the same limit. Subtracting this common part produces a leading composite approximation,It satisfies the leading initial values and tends to the outer quasi-steady solution after the initial layer. Corrections of order $ϵ$ describe the small substrate loss during complex buildup.

In physical concentration coordinates, the two nullclines areSince $K_S<K_M$, the substrate nullcline lies above the complex nullcline for $S>0$. Starting at $(S_0,0)$, the complex rises and substrate falls; with $E_0/S_0\ll 1$, this fast segment is nearly vertical when the axes use their respective concentration scales. The trajectory then follows the complex quasi-steady manifold toward the origin. On its slow descending portion it is slightly above the complex nullcline so that $\dot{C}<0$; the leading outer curve itself sets that derivative to zero only at zeroth order.