# At which voltage will the conductance decrease due to the narrow pore in a voltage-gated potassium channel?

If you have a voltage-gated potassium channel that do not inactivate, at which voltage would you see decrease in ion conductance related to the narrow path in the pore?

With that said I know that in a solutions with other ions there is a possibility that the high driving force eventually will lead to blockage of the pore by other ions. This is more of a theoretical question.

## 1 Answer

There is a point where diffusion into the pore can be a limiting factor in the channel's conductance. If the trans-membrane voltage and the recording solution viscosity are high enough, unitary ionic current departs the Ohms law; approaching a plateau. This asymptotic saturation amplitude represents the diffusion limited rate with which ions approach the channel entrance. The bigger the pore entrance, or the higher the ion concentration, or the larger the diffusion coefficient, the higher is the limiting asymptotic current. The pore entrance is measured as the capture radius.

The diffusion limited maximum unitary current (iDL) is directly proportional to the capture radius of the channel ($$r_c$$):

$$iDL=2π z e_o r_c D C$$

z is the valence of the ion, $$e_o$$ is the elementary charge, C is the bulk concentration of the ion, and D is its diffusion coefficient.

Here you can find a case in which single channel current is not linear with voltage and an example calculation of the capture radius of shaker $$K^+$$ channel.