= Solution
An open <ion channel> carries the <electric current>
$$
i=g(V-E_K)=(90\,\mathrm{pS})(60\,\mathrm{mV})=5.4\,\mathrm{pA}.
$$
The macroscopic mean is $I=NP_oi$, so $I/i\simeq204$ is the mean number open, not the total number present. Reading $P_o(-20\,\mathrm{mV})\simeq0.40$ from the original graph gives
$$
\boxed{N=\frac{1.1\,\mathrm{nA}}{(0.40)(5.4\,\mathrm{pA})}\simeq5.1\times10^2}.
$$
The graph reading is approximate, so a precision of hundreds is appropriate. Only a few hundred active <ion channels> generate a nanoampere-scale cellular <electric current>; molecular discreteness and stochastic gating therefore matter. For independent identical <ion channels>, the open-count <standard deviation> is $\sqrt{NP_o(1-P_o)}\simeq11$, corresponding to a <coefficient of variation> in <electric current> of about $5.4\%$. This is an instantaneous ideal independent-gating estimate, before temporal filtering and other noise sources, and requires the <single-channel recording> $P_o$ to apply to the cellular recording conditions.
Back to article page
