= Solution
The average <electric current> is the number of channels times their <ion-channel open probability> times the <electric current> carried by an open channel:
$$
I=N P_O\gamma_K(V_m-V_K),\qquad
\frac{I}{NP_O}=\frac{9\ \mathrm{nA}}{3000\times0.6}=5.0\ \mathrm{pA}.
$$
Consequently the identifiable answer is
$$
\boxed{\gamma_K=\frac{5.0\ \mathrm{pA}}{(-50\ \mathrm{mV})-V_K}.}
$$
\b[TODO (missing input):] The source does not supply $V_K$ or the <potassium> <thermodynamic activities> for this experiment. A unique numerical <electrical conductance> cannot follow from the listed <electric current>, count and <ion-channel open probability> alone. For the explicitly additional choice $V_K=-90$ mV, $\gamma_K=125$ pS. Taking $V_K=-100$ mV instead gives $100$ pS: both match the measured <electric current>. The table in question 3 describes a skeletal-muscle resting state, and cannot silently supply the conditions of this separate recording.
Rapid permeation is possible because the hydrated pore and coordinating groups of the <selectivity filter> provide a favorable free-energy landscape for repeated <ion> transit. Several <ions> can occupy the filter and their interactions favor onward passage, while the channel remains open for many successive transits. Selectivity need not mean that each transported <ion> remains tightly trapped for a long time. This is passive downhill transport; it does not consume one <ATP> <molecule> per <ion> crossing.
Back to article page
