= Solution
For <two-state ion-channel gating>, write $C\mathrel{\mathop{\rightleftarrows}^{k_o}_{k_c}}O$. Gain minus loss of open <probability> gives the <ordinary differential equation>
$$
\dot p_O=k_o(1-p_O)-k_cp_O,\qquad
p_O(t)=p_\infty+[p_O(0)-p_\infty]e^{-t/\tau},\quad
p_\infty=\frac{k_o}{k_o+k_c},\quad \tau=\frac1{k_o+k_c}.
$$
Thus 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,
$$
\boxed{p_A(t)=\tfrac12(1-e^{-t}),\quad \tau_A=1\,\mathrm{ms};\qquad
p_B(t)=\tfrac12(1-e^{-10t}),\quad \tau_B=0.1\,\mathrm{ms}.}
$$
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>.
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