Question

Assume that a typical open ion channel spanning an axon’s membrane has a resistance of 1 * 1011 Ω. We can model this ion channel, with its pore, as a 12-nm-long cylinder of radius 0.3 nm. What is the resistivity of the fluid in the pore? (a) 10 Ω.m; (b) 6 Ω.m; (c) 2 Ω.m; (d) 1 Ω.m

Short answer

The resistivity is$2.4\Omega /nm$

Step-by-step solution

Resistivity of conductor.

Consider a typical open ion channel spanning an axon’s membrane which has a resistance of R = $1\times {10}^{11}\Omega$.The model of this ion channel with its pore can be treated as a l = 12nm long cylinder of radius r = 0.3nm. The resistance of a cylinder shape conductor is given by,

$R=\frac{\rho \mathcal{l}}{A}$

Where A =${\mathrm{\pi r}}^2$ is the cross-sectional area andis the resistivity. Thus,

$\rho =\frac{AR}{\mathcal{l}}=\frac{{\mathrm{\pi r}}^2\mathrm{R}}{\mathcal{l}}$

Substitute with the given we get,

$$\begin{gathered} \rho =\frac{\mathrm{\pi }{\left(0.3\mathrm{nm}\right)}^2\left(1\times {10}^{11}\mathrm{\Omega }\right)}{12nm}=2.4\times {10}^9\Omega /nm=2.4\Omega /nm \\ \rho =2.4\Omega /nm \end{gathered}$$

Result

Hence, the resistivity is$2.4\Omega /nm$