Past exam of the biological physics course of the University of Oxford / 2015 / 2 / c / Solution

For two-state ion-channel gating, let ${\alpha }_n(V)$ and ${\beta }_n(V)$ be the rates of entering and leaving the permissive state. Gain minus loss givesAt fixed membrane potential, these coefficients are constant. The deviation $w=n-n_{\infty }$ obeys $\dot{w}=-w/{\tau }_n$, so the initial condition $n(0)=n_0$ yieldsThis approaches the fixed-voltage limit monotonically and stays between $n_0$ and $n_{\infty }$, as a valid probability must.

Convention qualification: the displayed single exponential function describes a voltage-clamp step, not an arbitrary time-dependent action potential. For example, a second imposed voltage step changes both coefficients and requires a new exponential segment. For general $V(t)$ the integrating-factor solution is $n(t)=e^{-A(t)}[n_0+{\int }_0^t{\alpha }_n(V(s))e^{A(s)}ds]$, with $A(t)={\int }_0^t[{\alpha }_n(V(s))+{\beta }_n(V(s))]ds$.