Biological Physics: Energy, Information, Life

462 Chapter 12. Nerve impulses[[Student version, January 17, 2003]]

+10mV v

v

ab

sodium conductivity

gNa

+

membrane current flux

jq

,r

v 1 v 2

ohmic

positive

feedback

Figure 12.9:(Sketch graphs.) (a)Dashed curve:Sketch of the conductancegNa+of an axon membrane to sodium
ions, showing an increase as the membrane potential increases from its resting value (v=0).Solid curve:Simplified
hypothesis for membrane sodium conductance (Equation 12.19). This form captures the relevant feature of the
dashed curve, namely that it increases asvincreases and is positive. (Even the dashed line is not fully realistic:
Real membrane conductances do not respond instantly to changes in membrane potential, but rather reflect the past
history ofv. See Section 12.3.1.) (b) Current-voltage relation resulting from the conductance model in (a) (see
Equation 12.20). The special valuesv 1 andv 2 are defined in the text.

The total charge flux through the membrane is then the sum of the sodium contribution, plus
Ohmic terms from the other ions:

jq,r=

(∑

i

(V−ViNernst)g^0 i

)

+(V−VNaNernst+ )Bv^2.

As in Your Turn 12a on page 450, the first term above can be rewritten asg^0 totv.LettingHdenote
the constantVNaNernst+ −V^0 ,wecan also rewrite the last term as (v−H)Bv^2 ,toget

jq,r=vgtot^0 +(v−H)Bv^2. (12.20)

Figure 12.9b helps us understand graphically the behavior of our model. There are three impor-
tant points on the curve of current versus depolarization, namely the points where the membrane
currentjq,ris zero. Equation 12.20 shows that these points are the roots of a cubic equation.
Wewrite them asv=0,v 1 ,andv 2 ,wherev 1 andv 2 equal^12 (H∓

√

H^2 − 4 g^0 tot/B). At small
depolarizationv,the sodium permeability stays small and so the last term of Equation 12.20 is
negligible. In this case a small positivevgives small positive (outward) current, as expected: We
are in the Ohmic regime (stage “a” of Figure 12.8). The outward flow of charge tends to reduce
vback toward zero. A further increase ofv,however, opens the voltage-gated sodium channels,
eventually reducingjq,rto zero, and then below zero as we pass the pointv 1 .Nowthe net inward
flow of charge tends toincreasev,giving positive feedback—an avalanche. Instead of returning to
zero,vdrives toward the other root,v 2.^10 Atstill higherv,weonce again get a positive (outward)
current, as the large outward electric force on all the ions finally overcomes the entropic tendency
for sodium to drift inward.

(^10) This bistability is analogous to the one studied in Problem 6.7c.