Biological Physics: Energy, Information, Life

12.1. The problem of nerve impulses[[Student version, January 17, 2003]] 447

horizontal wires imply (by point (3) above) that all three legs have the same value of

∆V=VNernsti +IiRi. (12.1)

Here as always ∆V =V 2 −V 1 is the interior potential relative to the outside,Ri=1/(giA)isthe
resistance, andIi=jq,iAare the currents through a patch of membrane of areaA,considered
positive if the ion current flows from inside to outside.

Quasi-steady approximation Figure 12.3b includes the effect of diffusive ion transport through
the cell membrane, driven by entropic and electrostatic forces. However, the figure omits two
important features of membrane physiology. One of these features (gated ion conductances), is not
needed yet; it will be added in later sections. The other omitted feature is active ion pumping. This
omission is a simplification which will be used throughout the rest of this chapter, so let’s pause to
justify it.
The situation sketched in Figure 12.3b cannot be a true steady state (see Sections 11.2.2– 11.2.3).
The dissipative flow of ions, shown by the dashed arrow in the figure, will eventually change the
sodium and potassium concentrations, until all three species come to Donnan equilibrium, obeying
Equation 12.1 with all currents equal to zero.^3 Tofind a true steady state, we had to posit an
additional element, the sodium–potassium pump. Setting the inward diffusive flux of sodium into
the cell equal to the pumped flux out of the cell, and similarly with potassium, gave us the steady
state (Equation 11.12 on page 423).
But imagine that we begin in the steady state, then suddenly shut down the pumps. The ion
concentrations will begin to drift toward their Donnan-equilibrium values, but rather slowly (see
Problem 11.4). In fact, the immediate effect on the membrane potential turns out to be rather
small. We will denote the potential difference across the membrane shortly after shutting down the
pumps (that is, the quasi-steady value) by the symbolV^0 .Tofind it, note that while the charge
fluxesjq,ifor each ion species need not separately be zero, as they must be in the true steady state,
still they must add up to zero, to avoid net charge pileup inside the cell. The Ohmic hypothesis
(Equation 12.1) then gives ∑

i

(V^0 −ViNernst)gi=0. (12.2)

Example Find the valueV^0 of ∆Vshortly after shutting off the pumps, assuming the initial

ion concentrations in Table 11.1 on page 416 and the relative conductances per

area given in Equation 11.9 on page 420. Compare to the estimated steady-state

potential found in Section 11.2.3.

Solution: Collecting terms in Equation 12.2 and dividing bygtot≡

∑

igigives the

chord conductance formula:

V^0 =

∑

i

gi

gtot

VNernsti. (12.3)

Evaluating everything gives that V^0 =− 66 mV,only a few millivolts different from

the true steady-state potential− 72 mVfound from Equation 11.12.

In fact, the ion pumpscanbeselectively turned off, using drugs like oubain. The immediate effect
of oubain treatment on the resting potential is usually small (less than 5mV), just as found in the

(^3) In this case Equation 12.1 reduces to the Gibbs–Donnan relation, Equation 11.4.