Question

In a simple model of an axon conducting a nerve signal, ions move across the cell membrane through open ion channels, which act as purely resistive elements. If a typical current density (current per unit cross-sectional area) in the cell membrane is 5 mA/cm² when the voltage across the membrane (the action potential) is 50 mV, what is the number density of open ion channels in the membrane?

(a) 1/cm²;

(b) 10/cm²;

(c) 10/mm²;

(d) 100/µm².

Short answer

$100/\mu m^2$or (d)

Step-by-step solution

Concept and given data 

Given data:

• $R=1.0\times {10}^{11}\Omega$
  
• The voltage across the membrane is V = 50 mV
• Current density,

The channels are connected in parallel. For n identical resistors R in parallel, the equivalent resistance is, of

$$\begin{gathered} \frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{}{R_2}+.... \\ =\frac{1}{R}+\frac{1}{R}+..... \\ =\frac{n}{R} \end{gathered}$$

or,

$R_{eq}=\frac{R}{n}$ (1)

The current flowing in the cell membrane is given by,

$l=jA$ (2)

Where A is the cross-sectional area in the cell membrane, and from ohm's law is given by

$l=\frac{V}{R_{eq}}$ (3)

Substitute from (1) in the above equation, and we get,

$l=\frac{nV}{R}$

Write for the number density of open ion channels in the membrane n/A, that is,

$jA=\frac{nV}{R}$

Calculation

Substitute with, V = 50 mV/$c{m}^2$,$V=50mV$ and a typical open ion channel spanning an on's membrane has a resistance of $R=1.0\times {10}^{11}\Omega$, to get,

$$\begin{gathered} \frac{n}{A}=\frac{\left(5mA/c{m}^2\right)\left(1.0\times {10}^{11}\Omega \right)}{50mV} \\ ={10}^{10}/c{m}^2 \\ =100/\mu m^2 \\ \frac{n}{A}=100/\mu m^2 \end{gathered}$$

So the answer must be (d).