HOME >HOMEWORK 7

Reading (one of the following):
Aidley chapter 4
or KSJ (3rd edition) chapters 7 & 8

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 r_m and capacitance c_m 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, r_a per unit length of nerve. The equivalent circuit is shown below:

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

V(x) = V_oe^-x/L where L = (r_m/r_a)^1/2

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

One way to start this problem is as follows: A current I_a is forced to flow down the cylinder. Because of the resistance of the cell membrane, I_a depends on position because some current leaves the cell through the membrane => I_a = I_a(x). From ohm's law, we can calculate the potential difference dV across an infinitesimal distance dx: dV/dx = -r_aI_a at a given position x. Then define a second current I_m moving through the membrane at position x. I_m = -dI_a/dx. Relate I_m to I_a and determine the voltage dependence as a function of position.

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

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 r_m is increased and c_m (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.