We have seen that the sodium-potassium
pumps control the osmotic
pressure of a cell and maintain the concentrations of sodium,
potassium, and chloride at about 50, 400, and 52 mM inside the cell
when it is immersed in a bath in which the Na^{+}, K^{+},
and Cl^{-} concentrations are 440, 20, and 560 mM. The
resulting voltage across the membrane is

ΔV = V_{2} - V_{1} = V_{inside} - V_{outside}
= -60 mV,

and the electric field *E* points into the cell.

A nerve cell, or *neuron*, consists of a cell body with
branching *dendrites* and a long *axon*, which may
range from 1 mm to 1 m in length, with its terminal branches which make
*synapses* with target
cells. Axon diameters run from less than a micron to more than a
millimeter.

The dendrites receive incoming
signals; the axon and its branches carry
signals to other cells.

Nerve signals are called *action potentials*.
They are pulses of about 1 ms in which ΔV is + 40 mV instead of the
resting potential of -60 or -70 mV.
Thus the action potential is a change
in ΔV of about 100 mV.
These pulses travel with little change in shape at constant speeds in
the range of
0.1 to 120 m/s.

For about 1 ms after an action
potential
passes a given part of an axon,
that part of the axon is hard to stimulate.
This pause is called the *refractory
period*. After the
action potential passes a given part of an axon, ΔV recovers,
overshoots, and then returns to its
resting value in about 5 ms.
This behavior is called *afterhyperpolarization*
.

Neural contacts occur at synapses, most often when a branch of an
axon touches a dendrite or the body of a target cell.

In this micrograph of the cell body
of a motor neuron of a spinal cord,
the axon terminals are red and the cytoskeleton is green:

The gap between the end of an axon branch and a dendrite (or a spine on
a dendrite) of a target neuron is between about 10 and 30 nm; it is
called the synaptic cleft.
Here are cartoons of a resting and
an active synapse:

Neurotransmitters, such as
acetylcholine, serotonin, GABA, glycine, and glutamate, can open transmitter-gated sodium, potassium, and
chloride channels. The excitatory
neurotransmitters open sodium
channels, and the resulting influx of sodium depolarizes the membrane, making ΔV = V_{2} - V_{1}
= V_{inside} - V_{outside }more positive and

possibly leading to an action potential. The inhibitory neurotransmitters
open potassium or chloride channels. The efflux of
potassium further polarizes the membrane, making ΔV = V_{2} - V_{1} _{
}more negative. The
concentration of Cl^{-} is 10 times higher outside the cell,
and so when the chloride
channels open, the influx of Cl^{-}
tends to further polarize
the membrane. So both potassium
and chloride
transmitter-gated channels tend to stop the generation of action
potentials.

Mammalian brains add extra non-NMDA
glutamate-receptor sodium channels after repeated
entries of Ca^{2+} thru the
NMDA receptor:

Here is a scanning electron
micrograph of a neuromuscular junction in a
frog:

Glial cells (Schwann cells and
oligodendrocytes) wrap their plasma membranes about the axons of many
of the neurons of vertebrates
forming *myelin sheaths*
.
The voltage-gated sodium channels
are concentrated in gaps in the myelin,
the *nodes of Ranvier*,
about 1 mm apart. The action potential
jumps from node to node, traveling faster
than in unmyelated axons.
The disease *multiple sclerosis*
destroys some of the myelin in the
central nervous system, slowing down action potentials with devastating
results.

The human brain has about 100
billion (10^{11}) neurons.
Here is a drawing by Golgi of a small part
of the olfactory bulb (sense of smell) of a dog's brain:

What happens if we electrically
stimulate a nerve cell? If we
slightly decrease ΔV throughout a small patch of the membrane of a
nerve cell
and then measure ΔV at points several mm away, we find a reduced
decrease ΔV that fades
with time and distance.
The cell resets its ΔV to - 60 mV, and at distances
of 100 mm, the change in ΔV is very small.
A neuron recovers from a slight *depolarization*.
If we slightly increase Δ V in a small patch
of membrane, then at points several mm away
ΔV will show an increase that fades
with distance and time.
The cell resets its ΔV to -60 or -70mV.
A neuron recovers from a slight *hyperpolarization*.
This passive spread of electrical signals
is called *electrotonus*.
In both depolarization and hyperpolarization,
the magnitude of the spreading change in
ΔV is proportional to the stimulus:
the response is *graded*.

But if we depolarize the membrane
by more
than about 10 mV, then we can set in motion
an action potential.

The huge difference in the response
is
due to *voltage-gated Na ^{+}
channels*:

How do we describe action potentials
mathematically? We note
that the voltage drop across
the membrane ΔV = V_{2} - V_{1} is the same for
all three species, Na^{+}, K^{+}, and Cl^{-}.
This voltage is the sum

ΔV = *I _{i}R_{i}*
+

of the Ohmic drop

Here

Next, we will ignore the myelin
and the sodium-potassium pumps. Instead, we'll just assume the
resting values
of the bulk concentrations and of ΔV mentioned above. In any case,
these values
remain fairly constant for about 6 10^{5}
s
after the pumps are stopped. We'll denote the quasi-steady value
of ΔV by *V*^{0}.
If g_{i} is the
conductance of
species i,
then the current of that species
will be (*V*^{0}
- *V _{i}*

∑

Let g

g

Then the quasi-steady membrane voltage is

which is called the

which is not very different from the value of -72 mV found in Eq.(11.12).

The axon's membrane is a
capacitor.
When a voltage V is applied across it,
opposite charges q and -q appear within
a Debye length of the membrane.
If C is the capacitance of the membrane,
then the charge q and the voltage V are related by

q = CV.
(12.4)

If the voltage is changing with
time,
then by differentiating the preceding equation,
we find

(dq/dt) = I = C (dV/dt).
(12.5)

This current is called a *capacitive current*.
The capacitance C is proportional to
the area A of the axon,
C = A *C*.
Experimentally,

*C*
= 10^{-2} F m^{-2}
= 1 µ F cm^{-2}.

We still have not modeled the
voltage-gated
sodium channels, so we don't expect
to get much of an action potential
with the present model, but let's push
ahead anyway.
Consider a cylindrical slice of radius a
and thickness dx of an axon
that points in the x-direction.
Let *I _{x}(x)* be the current
flowing into the cylindrical slice at x.
Then -

We can relate the current *I _{x}*
to the voltage difference

and the resistance of the axoplasm (the axon's cytoplasm)

by the formula

Thus we get

If we now put Eq.(12.6.5) into (12.6),
then we find

*
π a ^{2}κ (*d

or

a

which is the *cable equation*.

Now *V ^{0}* is the
quasi-steady voltage across the
membrane
— the voltage that gives no net
radial flux

Now let

*
v(x,t) = V(x,t) -V ^{0} *.

And let's define the axon's

and its

With these definitions and assumptions, the cable equation becomes the

Letting *v(x,t)* = exp(*-t/τ*)
*w(x,t)*,
we find that (12.9) becomes

(*λ ^{2}/τ*)
(d

which is the diffusion equation we studied in Sec.(4.6.5). Using its solutions, we have with N a normalization constant

which is the passive-spread solution. With

and

Here is a cartoon that describes both voltage-gated
sodium channels

and (slightly better) voltage-gated potassium
channels:

When the membrane is depolarized by more than about 10
mV, the voltage-gated
sodium channels open. When enough of them
are open on a given patch of the axon's membrane, the
sodium conductance momentarily jumps from the value of Table 11.1,
which was g_{Na+
}= 0.04_{ }g_{K+
}to

g_{K+} = 0.05 g_{Na+}
= 2g_{Cl-}
(12.14)

that is, the sodium conductance
jumps from

g_{Na+
}= 0.04_{ }g_{K+
}to g_{Na+ }= 20_{ }g_{K+
}

which is to say by a
factor of 25 x 20 =
500. Sodium ions come flooding in when the sodium
channels are open.

The voltage-gated potassium
channels are slower than the voltage-gated
sodium
channels to respond to changes in the voltage
across the membrane. They open only after this voltage V = V_{2}
- V_{1} has reached its maximum of about +40 mV.

The action potentials of some muscle, egg, and endocrine cells depend upon voltage-gated calcium channels.

We seek a traveling-wave
solution, that is, one of the form

V(x,t) = V(t - x/s)

where s is the speed of the
wave. The text has a tilde on the second V and uses a Greek theta for the
speed, but I am not good enough at html to be so fancy. Now the
spatial derivative of this traveling wave is

dV(x,t)/dx = dV(t - x/s)/dx = (-1/s) dV(t - x/s)/dt
(12.145)

and its second spatial derivative is

d^{2}V(x,t)/dx^{2
}= (1/s^{2})
d^{2}V(t - x/s)/dt^{2}
. (12.146)

If we put this traveling-wave solution
into the cable
equation (12.7b), then we find

which is an ordinary, not
a partial, differential equation.

Our simplified
voltage-gating hypothesis is that the outward radial current j_{q,r }is

j_{q,r
}= _{
}∑_{i} ( V - V_{i}^{N
}) g_{i}(V).
(12.17)

g

Nelson then finds an action potential that rises from 0 to a voltage he calls v

But an action potential rises and then falls slightly below 0. Also, the voltage-gated sodium channels stop the sodium current once the voltage v has risen to about 100 mV (V rises from -60 mV to + 40 mV). So Eq.(12.19) is a very bad approximation.

So let's try a more general expression for the sodium conductance:

g

Then the outward radial current is from (12.17)

j

or by (12.2-3)

j

= v g

where H = V

(a

or with v(t) = V(t - x/s) - V

(a

Now v = V - V

Let's simplify (A.6). First, we drop the first term (aκ/(2s

Next, we take the term f(v) to be

f(v) = f θ(v - 0.01) θ(v - 0.1) (A.7)

in which the Heaviside functionθ(x) = 1 if x is positive and 0 otherwise and f is a constant, which we'll take to be

f = g

So we now have

0 = v g

Now we imagine that at time t = 0 the voltage v = v

We write this first-order, linear, inhomogeneous differential equation as

dv/[ H f - ( g

which we may integrate to

- (1/(g

or

[ Hf - ( g

A little high-school algebra now gives

v(t) = Hf [ 1 - e

This function starts at v

As v(t) approaches Hf /( g

C

To integrate this simpler equation, we write it as

dv/v = - g

Integrating both sides, we find

ln ( v/v

Thus we find

v(t) = v

Here's a plot of v(t) as given by (A.13) followed by (A.17):

We may improve this model by including the voltage-gated potassium channels. After the voltage-gated sodium channels close and the voltage-gated potassium channels open, instead of (A.1), we have

g

And instead of (A.4), we have

j

in which K = V

which we may integrate as we did (A.9). We get

v(t) = k K [ 1 - e

Here's a plot of the resulting action potential (A.13, 21, & 17):

The voltage v(t) rests at 0 for the first 0.5 ms; then the sodium channels open and (A.13) raises v above 0.1 V in 0.5 ms; then the sodium channels close, and the potassium channels open, and (A.21) drops v drops, overshooting; then the potassium channels close, and (A.17) returns v to its resting value of 0. The input values were: k = 50 m

Let's now return to the full Eq.(A.6) with the Heaviside representation (A.7) of the flow thru the voltage-gated sodium channels but without the potassium channels. Before depolarization, the voltage v is lying at 0, and the channels are closed, so (A.6) is

(a

This is a linear, homogeneous, second-order differential equation.

We may solve it by letting v = exp(pt). Since dv/dt

d

(aκ/(2s

which is a quadratic equation for p

(aκ/(2s

There are two solutions

p

and

p

The general solution to Eq.(A.22) is then

v = c

There are three time zones here: before the v-g sodium channels open, while they are open,

and after they close. Since the voltage v relaxes to 0 in the far past and in the far future, the solution before the v-g sodium channels open must be

v(t) = c

and

v(t) = c

after they close.

While they are open, Eqs.(A.6&7) give

(aκ/(2s

which is a linear, second-order, inhomogeneous DEQ.

It is easy to find a particular solution of the inhomogeneous equation: the constant

v(t) = H f/( g

will do. To find the general solution of the homogeneous equation, we try v(t) = exp(qt).

Inserting this into (A.26), we find that q must satisfy the quadratic equation

(aκ/(2s

The solutions are

q

and

q

So the general solution to the inhomogeneous equation (A.26) is

v(t) = d

The voltage-gated sodium channels open when v = v

So the solutions (A.24&25) must be

v(t) = v

and

v(t) = v

Let's allow a discontinuity in the time derivative dv/dt of the voltage

at t = t

v(t) = d

d

Here is a plot of (A.33-35) for s = 10 m/s, t

This plot has two defects: the time-derivative dv/dt is discontinuous at t

First, we use (A.18 & 19) to model the voltage-gated potassium channels after the voltage-gated sodium channels have shut down. So instead of (A.1), we have

g

And instead of (A.4), we have

j

in which K = V

The general solution v(t) to this linear, inhomogeneous, second-order ordinary DEQ is the sum of a particular solution to the inhomogeneous equation, such as

v = k K/(g

and the general solution of the homogeneous equation

(aκ/(2s

The general solution to this linear, homogeneous equation is

v(t) = e

in which

r

and

r

So the general solution v(t) to Eq.(A.44) is

v(t) = e

We now stitch together the solution v(t) = 0 while the neuron is resting, followed by the solution (A.35) for 0.5 ms while the voltage-gated sodium channels are open (and the K ones closed), followed by the solution (A.48) for 0.5 ms while the voltage-gated potassium channels are open (and the Na ones closed), followed by the solution (A.29) for 1 ms after all the voltage-gated ion channels are closed. We match the coefficients to keep the time-derivative dv/dt continuous except at t = 0.0005 s when the voltage v jumps due to the opening of the transmitter-gated sodium channels, which start the action potential. It is interesting that for k = 50 m

Here is a plot for speed s = 75 m/s:

The voltage v stays at zero until the initial depolarization opens the sodium channels at t = 0.5 ms. The influx of Na

In Eq.(A.42), the speed s enters only thru the combination (aκ/(2s