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HOME >HOMEWORK #7 SOLUTION

1. Let's first consider the passive properties of a nerve cell and work out 1 dimensional cable theory. As a first approximation, a nerve cell is an infinite cylinder of some radius. Current can flow through the cell membrane, as it has a finite resistance rm and capacitance cm per unit length of nerve cell. Furthermore, the interior of the nerve cell has a resistance to current flowing down the length of the nerve, ra per unit length of nerve.

3rd year physics students & beyond: Show that if the nerve is held at some voltage Vo at location x = 0, the steady state potential a distance x down the length cylinder is:

V(x) = Voe-x/L where L = (rm/ra)1/2

Everyone else, write a qualitative explanation about why you would expect the voltage to decrease with distance

We're in steady state, so we don't need to worry about capacitance at all.  So, let's just think about resistances.

We were given:dV/dx = -raIa at a given position x in response to a current Ia moving down the nerve cell.  However, Ia depends on position x, since some current flows out through the membrane to ground.  This current is Im(x) = -dIa/dx

The goal now is to relate these two:

dV/dx = -raIa

=> Ia= -dV/(dx * ra)

V = -rm (dIa/dx) This is just because V = IR and I = Im

So,

d2Ia/dx2 = ra/rm Ia

=> Ia = Ae-x/L where L = (rm/ra)1/2 since Ia(inf) = 0

=> V = Voe-x/L

A hand-wavy explanation for why the voltage will decay with distance from the site of current injection is as follows: Injected current into the cell can either flow down the length of the cell or it can flow out through the cell membrane to the outside of the cell. The more current flows down the length of the cell (as opposed to flowing out of the cell), the greater the extent of voltage change. Therefore, the ratio ra/rm will set the length over which injected current will change the cell's voltage.

2. For a typical nerve cell, L in the previous problem has a value of ~1 mm. A typical nerve cell is 1-100 cm in length. Thus, if a nerve cell becomes depolarized at one end (due to channels opening in response to synaptic input from another cell), the amount of depolarization on the far end of the nerve cell due to passive propagation of voltage will be negligible (and may be physically undetectable in any reasonable time because of the various noise sources in the cell).

For this reason, cells have evolved action potentials that allow small voltage signals to be propagated efficiently down nerve cells.
For the cell to do this, there are voltage activated Na and K channels in the cell membrane. Recall that if the voltage rises above a certain threshold value, the Na channels open. This causes the cell to become even more positive until the membrane potential approaches that predicted by the Nerdst equation. Then, the Na channels close (an intrinsic property of the channels), K channels open, K rushes out of the cell, and the membrane potential becomes negative again.

Na and K channels are clustered together along nerve cells as shown in the figure below. Imagine that the synapse on the left side of the cell raises the membrane potential such that the leftmost cluster of channels is brought above threshold (the Na channels open). However, rightmost cluster of channels is not brought above threshold by the synaptic input. Explain how active propagation of the action potential allows the rightmost cluster of channels to be brought to threshold

From the result in problem #1, we know that any voltage in a nerve cell will fall off exponentially from a site of current injection (through the membrane like if current was coming in through a bunch of ion channels).  The length constant for nerve cells is ~1 mm, and this approximately the distance between ion channel clusters (called Nodes of Ranvier).

What happens is this:  When the leftmost cluster of channels is brought above the voltage required for Na channel activation (opening), they open and generate an action potential.  The voltage Vo at this site then becomes very positive, and the cluster of ion channels immediately to the right is activated and the process repeats.  The channels far to the right will have to wait until channels nearby them are activated because the voltage signal is too weak to get them to activate.

A logical question you might ask is why the action potential propogation is unidirectional (or is it?).  The answer lies in the K+ channels.  Recall that shortly after the Na channels bring the nerve cell very positive, the K+ channels open and the membrane potential falls for some time below resting potential.  And the K+ channels remain open for a time long compared to the time it would take for an action potential to propogate by.  This low resistance dramatically decreases L behind the action potential, while L remains high in front of the action potential since most of the channels are closed.

3. In the patches of membrane between the clusters of ion channels, another cell wraps around the nerve cell many times. When the nerve cell is wrapped up by another cell, this is referred to as myelination. This has the effect that rm is increased and cm (as defined in problem #1) is reduced in this region of the cell membrane compared to what these values would be in the absence of mylenation.
In terms of the results to #1 & #2, explain what effect, if any, the myelination has on the conduction velocity of action potentials.

If you look at the arrangements of reistances & capacitences in the figure for problem 1, you can see that a voltage signal applied to the right end of the nerve cell will be heavily filtered by the system:  specifically it will be very strongly lowpass filtered.  Therefore, by reducing cm, the system will become more highpass and thus faster (in terms of how quickly a finite current can bring about a voltage change some distance down the length of the nerve cell).  Increasing rm also would allow voltage signals to travel faster down the length of the nerve cell, since more current will be available to charge up the membrane capacitence instead of leaking out.

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