Question

The capacitance of the cell membrane is estimated to be 89 pF, and approximately 10000 Na+ ions flow through an ion channel when it opens.

1. Based on this information and what you learned in this chapter about the action potential, estimate the total number of sodium ion channels in the membrane of a nerve cell.

Express your answer using two significant figures. (N = ???)

Answer 1

The value of capacitance = (charge stored) / (potential difference across capacitance). This is usually written as C = Q / V.

The charge on one Na+ ion is 1.6x10^-19C ((atom is missing 1 electron).
The charge on 10000 Na+ ions = 10000 x 1.6x10^-19 = 1.6x10^-15C.
If there are N channels, the total charge pumped per firing is N x 1.6x10^-15C

The potential difference across the membrane of a nerve cell is around 50mV, but you should use whatever value your book tells you, as the is no single 'correct; value. I'll use 50mV as an example here.

C = 89pF = 89x10^-12F
V = 50mV = 50x10^-3V
Q = N x 1.6x10^-15

Since C=Q/V
89x10^-12 = N x 1.6x10^-15 /50x10^-3

N = 89x10^-12 x 50x10^-3 / (1.6x10^-15)
= 2781