Past exam of the mathematical physiology course of the University of Oxford / 2023 / 1

Write $a=[\mathrm{A}]$, $b=[\mathrm{B}]$ and $c=[\mathrm{C}]$ for the three concentrations. The letters denote chemical species; the forward arrow represents association of one molecule of each reactant into one complex, and the reverse arrow represents dissociation. The positive constants $k_{+}$ and $k_{-}$ are the respective reaction rate constants. For an elementary chemical reaction, the law of mass action gives forward and reverse reaction ratesEvery forward event removes one molecule of each reactant and adds one complex; a reverse event reverses those changes. Thus the reaction-rate equations areThe units of $k_{+}$ are inverse concentration per unit time, whereas $k_{-}$ has units of inverse time. Adding the equations gives the independent conservation lawsIn particular $a-b$ is constant, but this is a consequence of those two conservation laws. The total $a+b+c$ need not be constant: association changes the number of separate molecules.

For the general stoichiometry, take $p,q$ to be positive integers. The net reaction rate and stoichiometric vector areConsequentlyThe factors $p,q$ outside $R$ count molecules consumed per event; the powers inside $R$ express the law of mass action. The corresponding conservation laws are $a+pc=\text{constant}$ and $b+qc=\text{constant}$. No factorial is required in this deterministic convention for the reaction rate constant.

There are four reversible transitions of the ion channel, each changing exactly one gate:These are first-order reaction-rate equations: their reaction rate constants are per-gate transition rates. In the continuous-time Markov chain interpretation, simultaneous changes of both gates have probability of order $(dt{)}^2$, so there is no direct diagonal transition in its infinitesimal generator matrix.

Normalize the total channel density to one. For independent permissive gates in an ion channel, the probability that the $M$ gate is open is $m$, and the probability that the $H$ gate is open is $h$. The joint probability is the product of the appropriate open or closed probabilities, givingIn particularThe original PDF really prints $j$, rather than $1-j$, on the final factor. That expression fails normalization of a probability distribution and does not give the closed-$H$ probability at $j=0$; the displayed correction is necessary. If $S_{ij}$ are literal densities with conserved total density $N$, multiply each expression by $N$.

The product additionally assumes initially independent events and independent gate transitions. The transition scheme alone permits initially correlated gates and does not force this factorization at every time. The all-closed initial preparation used below satisfies the required product assumption.

For each state, add its incoming reaction rates and subtract its outgoing reaction rates. This gives the master equationSumming gives ${\sum }_{i,j}{\dot{S}}_{ij}=0$, verifying the conservation law for total channel density. Nonnegative initial densities stay nonnegative because every loss term vanishes when its own density reaches zero.

With total density one, the open probabilities for the individual gates are $m=S_{10}+S_{11}$ and $h=S_{01}+S_{11}$. Adding the relevant master equations cancels transitions of the other gate:These marginal ordinary differential equations hold even for correlated initial gates. Their product gives $S_{11}$ when the independence assumption in part (c) holds. Solving the first-order linear differential equations from the all-closed initial state givesPut $\lambda =\alpha +\beta =\gamma +\delta >0$. ThenFor positive opening rates this increases strictly from zero, so its half-equilibrium time is uniquely determined by $(1-e^{-\lambda t}{)}^2=1/2$:If an opening rate is zero, the all-open equilibrium density is zero and a nontrivial half-rise time is undefined.

The two identical $M$ gates are independent Bernoulli random variables, so their open count has a binomial distribution. Multiplying by the independent $H$ factor gives the normalized channel densitiesExplicitly,The binomial coefficient $\left(\genfrac{}{}{0ex}{}{2}{1}\right)=2$ counts which of the two distinguishable $M$ gates is open. At open count $i$, the total $M$-opening rate is $(2-i)\alpha$ and the total closing rate is $i\beta$. Dividing the expected value of that count by two therefore recovers the same single-gate ordinary differential equation:For the specified initial conditions and common rate sum $\lambda$, the all-open fraction isThus, assuming positive opening rates,Three gates take longer to reach half their final all-open fraction than two gates: all three activation factors must build up simultaneously.