Past exam of the mathematics and the environment and mathematical physiology course of the University of Oxford / 2010 / B / 4

Use $S,C_i,P$ for concentrations and define net complex-formation rates $R_i=k_i{C}_{i-1}S-(k_{-i}+k_i^{+})C_i$, with $R_{n+1}=0$. The law of mass action givesEach binding event removes one free substrate molecule; each unbinding returns one; each catalytic event releases one reaction product and reduces the number of occupied sites by one. Suitable initially unbound enzyme and substrate data are $S(0)=S_0$, $P(0)=0$, $C_0(0)=E_0$, $C_i(0)=0$ for $i\ge 1$. All rates are nonnegative on the boundary of the nonnegative concentration region in the direction required to preserve it. These equations also conserve $S+P+{\sum }_{i=1}^{n}i{C}_i=S_0$, accounting for bound substrate as well as free substrate.

Adding the ordinary differential equations for all enzyme states makes the net rates telescope:With initially unbound enzyme, this sum is $E_0$ for all time. HenceThis is enzyme conservation, not conservation of each complex individually. The reaction changes site occupancy and makes reaction product, but no event creates or destroys an enzyme molecule.

Besides the stated scales, set $P=S_{0}p$, $ϵ=E_0/S_0$, ${\alpha }_i=k_i/k_1$, ${\beta }_i=(k_{-i}+k_i^{+})/(k_1{S}_0)$, and ${\kappa }_i=k_i^{+}/(k_1{S}_0)$. Then $r_i={\alpha }_i{c}_{i-1}s-{\beta }_i{c}_i$, and direct substitution, remembering $d/dt=k_1{E}_{0}d/d\tau$, givesThe exact initial values are $s(0)=1$, $p(0)=0$, $c_0(0)=1$, $c_i(0)=0$ for $1\le i\le n$. The enzyme-state equations carry the small parameter because the enzyme concentration scale is small relative to the initial substrate concentration; the substrate evolves on the chosen slower depletion time scale.

For an outer solution with bounded derivatives as $ϵ\to 0$, the final complex equation gives $r_n=0$. Working backwards in $0=r_i-r_{i+1}$ then gives every $r_i=0$, including $r_1$; setting only their sum to zero would not suffice. This is the quasi-steady-state approximation, valid after the initial complex-formation layer with the relevant dimensionless rate parameters held fixed.

The recurrence is $C_i=S{C}_{i-1}/K_i$, where $K_i=(k_{-i}+k_i^{+})/k_i$. Thus $C_j={\varphi }_j{S}^j{C}_0$, ${\varphi }_j={\prod }_{i=1}^j{K}_i^{-1}$. Substitution into enzyme conservation yieldsEach state $j$ produces reaction product at rate $k_j^{+}{C}_j$, soThe source's uppercase $N$ in the index range is read as the number $n$ of binding sites. This approximation preserves the total enzyme population and the different catalytic rates of the occupied states.

The sequential-enzyme Hill limit uses the intended distinguished limit for $n\ge 2$ holds the other rates fixed while $k_1\to 0$, $k_n\to \infty$, and $k_1{k}_n$ tends to a positive finite constant. Then for every $1\le j<n$, ${\varphi }_j={\prod }_{i=1}^j{k}_i/(k_{-i}+k_i^{+})=O(k_1)\to 0$, whereas ${\varphi }_n$ remains finite and nonzero. Hence intermediate complexes have negligible occupancy relative to the unbound and fully bound states.

In the quasi-steady-state approximation rate, all intermediate numerator and denominator terms disappear. Multiplying by $1/{\varphi }_n={\prod }_{i=1}^n{K}_i$ givesThis is the Hill equation with maximal rate $E_0{k}_n^{+}$ and effective half-saturation concentration $(\prod K_i{)}^{1/n}$. Here its power $n$ follows from the sequential binding mechanism and the suppression of intermediate states.

The phrase “$k_1{k}_n$ finite” needs this scaling interpretation for a nontrivial limiting rate. Merely bounding that product, while permitting unrelated rates to vary, does not ensure that all intermediate weights vanish relative to the terminal one. The derivation explicitly supplies the usual fixed-other-rates, positive-product assumptions. For $n=1$, the rate in the box is already the exact one-state quasi-steady-state approximation rate, without this two-endpoint limiting argument.

Write $c=c_1$, $\beta =(k_{-1}+k_1^{+})/(k_1{S}_0)$ and $\kappa =k_1^{+}/(k_1{S}_0)$. Since $c_0=1-c$, the exact equations areAt the prescribed initial data $s=1,c=0$, $r_1=1$, rather than zero. The quasi-steady-state approximation would instead demand $c=s/(s+\beta )$, giving $c(0^{+})=1/(1+\beta )$. This mismatch signals an initial layer.

Set $\sigma =\tau /ϵ$ and expand $s=s_0+ϵs_1+\cdots$, $c=c_0+\cdots$. At leading order $s_0^{\prime }=0$, so $s_0=1$, while $c_0^{\prime }=1-(1+\beta )c_0$ and $c_0(0)=0$. ThusFor a more informative substrate approximation, integration of $s_1^{\prime }=-1+(1+\beta -\kappa )c_0$ with $s_1(0)=0$ givesOn the slow scale $c_{\mathrm{out}}=s_{\mathrm{out}}/(s_{\mathrm{out}}+\beta )$ and $s_{\mathrm{out},\tau }=-\kappa s_{\mathrm{out}}/(s_{\mathrm{out}}+\beta )$, with matched leading value $s_{\mathrm{out}}(0)=1$. Integrating gives $s_{\mathrm{out}}+\beta \log s_{\mathrm{out}}=1-\kappa \tau$. In the overlap $1\ll \sigma \ll 1/ϵ$, the inner complex tends to $1/(1+\beta )$ and its substrate slope matches the expansion of this outer solution. A leading composite asymptotic expansion isIt satisfies both initial values at leading order and smoothly joins the initial complex build-up to substrate depletion. The displayed $O(ϵ)$ inner substrate term records the small binding-induced initial slip if greater accuracy is wanted.