 Solution

For two-state ion-channel gating, write $C{\rightleftarrows }_{k_c}^{k_o}O$. Gain minus loss of open probability gives the ordinary differential equationThus the ratio of the transition rates determines the final open probability, while their sum determines the response speed. Both given channels have $p_{\infty }=1/2$. For an initially closed population, with $t$ measured in milliseconds,The mean individual open and closed lifetimes are $1/k_c$ and $1/k_o$: both are $2\mathrm{ms}$ for A and $0.2\mathrm{ms}$ for B. With the same channel number, single-channel electrical conductance and driving force, the steady macroscopic electric currents are equal, but B reaches that value ten times faster. If both populations were already at $p_O=1/2$, neither would show a mean relaxation under these fixed rates: a response also requires an initial condition or a change in rates.

The stationary variance of the number open is $N/4$ for independent channels. Its autocovariance is $(N/4)e^{-|t|/\tau }$, so B's fluctuations decorrelate ten times faster even though their instantaneous variance is unchanged. Faster gating allows an ion channel population to follow a brief stimulus and contribute to rapid action potential timing. Slower gating integrates over brief changes and can provide a delayed opposing electric current, useful in setting action potential duration or frequency selectivity. Those examples describe consequences of kinetics; the stipulated voltage-independent rates alone do not define a complete excitable neuron.