12.1. The problem of nerve impulses[[Student version, January 17, 2003]] 451
the interior fluid (representing the axon’s cytoplasm, oraxoplasm,represented by the upper hor-
izontal line), or through the surrounding extracellular fluid (represented by the lower horizontal
line). Taking dx→ 0 amounts to describing the membrane as a chain of infinitesimal elements, a
distributed networkof resistors, capacitors, and batteries.
Toexplore the behavior of such a network under the sort of stimuli sketched in Figure 12.1, we
now take four steps:
a. Find numerical values for all the circuit elements in Figure 12.4, then
b. Translate the figure into an equation;
c. Solve the equation; and
d. Interpret the solution.
a. Values
Tofind the interior axial resistance dRx,recall that the resistance of a cylinder of fluid to axial
current flow is proportional to the cylinder’s length divided by its cross-sectional area, or dRx=
dx/(κπa^2 ), whereκis the fluid’s electrical conductivity (see Section 4.6.4). The conductivity of
axoplasm can be measured directly in the lab. For squid axon its numerical value isκ≈ 3 Ω−^1 m−^1 ,
roughly what we would expect for the corresponding salt solution (see Problem 12.5).
Tosimplify the math, we will set the electrical resistance of the exterior fluid equal to zero:
dR′x=0.This approximation is reasonable, since the cross-sectional area available for carrying
current outside the cylindrical axon is much larger than the areaπa^2 of the interior. Thus we have
the very convenient feature that the entire exterior of the axon is “short-circuited,” and so is at a
uniform potential, which we take to be zero:V 1 (x)≡0. The membrane potential difference is then
∆V(x)=V 2 (x); to simplify the notation we will call this quantityV(x).
The resistanceRrof the membrane surrounding the axon slice is just the reciprocal of its total
conductance; according to Your Turn 12a, this equals (gtot× 2 πadx)−^1 ,wheregtotis the sum of
thegi’s. As mentioned in Section 11.2.2, a typical value forgtotin squid axon is≈ 5 m−^2 Ω−^1.
Finally, Section 12.1.2 says that the membrane capacitance is dC=(2πadx)×C.Section 12.1.2
quoted a typical value ofC≈ 10 −^2 Fm−^2.
b. Equation
Toget the equation for the spread of an external stimulus, we write down the condition of charge
neutrality for one cylindrical slice of the axon (Figure 12.4a). This condition says that the net
current into the ends of the slice,Ix(x)−Ix(x+dx), must balance the total rate at which charge
flows radially out of the axoplasm. The radial current equals the sum of the charge permeating
throughthe membrane, or 2πadx×jq,r,plus the rate at which chargepiles up atthe membrane,
(2πadx)×CddVt (see Equation 12.5). Thus
Ix(x)−Ix(x+dx)=−
dIx
dx
×dx=2πa(
jq,r(x)+C
dV
dt)
dx. (12.6)This equation is a good start, but we can’t solve it yet: It’s one differential equation in three
unknown functions, namelyV(x, t),Ix(x, t), andjq,r(x, t). First let’s eliminateIx.
The axial current at a pointxof our axon just equals the potential drop along a short distance,
divided by the axial resistance dRx:
Ix(x)=−
V(x+^12 dx)−V(x−^12 dx)
dx/(πa^2 κ
)=−πa^2 κ
dV
dx