Cable Properties II: Temporal Characteristics and Myelination

Introduction

• In the previous unit, we began to characterize the cable properties of a neuron, and we developed a partial differential equation that provided detailed predictions about how a cable structure's voltage would vary with time and space.

• To understand the implications of that equation, we examined two simplified cases: the longterm response to current injection, which allowed us to ignore the transient, time-dependent changes; and the temporal response to current injection in a spherical cell, which allowed us to ignore the spatial extent of the cable.

• In this unit, we will begin to explore some of the implications of cable theory for more complex behaviors: propagation of action potentials, myelination, and branching.

• An entire book is devoted primarily to this topic: "Electric current flow in excitable cells", by Jack, J.J.B., Noble, D., and Tsien, R. W. (Clarendon Press, Oxford, 1975, 1983); a copy is available in our library. This provides an extensive introduction and a great deal of detail about approaches to solving the cable equation. A more recent reference that also provides up to date references is "Biophysics of Computation: Information Processing in Single Neurons", by Christof Koch, Oxford University Press, 1999, which is also available in our library. We will only examine a few of the many examples developed at length in those books.

Response to Transient Input

Figure 1: The response of a passive cable to an instantaneous pulse of charge delivered to its middle.

• How does the cable transform signals injected into it in both space and time? We will look at one special case.

• We will not go through the derivation of the solution, which is provided in Jack, Noble and Tsien (see pp. 28 - 29, and 46). By using Laplace transforms, and treating the instantaneous pulse as an impulse or delta function, they obtain the following analytical solution: $$V(x, t) = \frac{Q_{0}}{2 c_{m} \lambda \sqrt{\pi T}} e^\tfrac{-X^2 - 4T^2}{4 T}, \quad\text{(Equation 1)}$$ where $\lambda$ is the length constant, as defined in the previous unit, $\tau$ is the time constant, as defined in the previous unit, $c_{m}$ is the membrane capacitance, $X = x/\lambda$, $T = t/\tau$, and $Q_{0}$ is the total charge transferred instantaneously to the cable at the middle of the cable, which by convention is referred to as $X = 0$.

• The response of a passive cable to the instantaneous current pulse is shown in Figure 1.

• The left back corner shows the response of the cable at the point where the current is injected at $X = 0$ and at time $t = 0$. The front axis shows how the pulse changes with time, and the axis along the right top shows how the response changes over space. As one proceeds further and further down the cable, the time of the peak response is more and more delayed, and the response significantly spreads out in time, so that it rises more and more slowly, and decays even more slowly at any given point along the cable.

• This figure provides helpful intuition for how initially strong signals that are very brief can end up diffused over time and space, and thus provides insight into the way a cable will respond to a single input from another neuron (e.g., the input from an electrical or a chemical synapse).

Cable Properties and Conduction Velocity

• A very important aspect of the paper in which Hodgkin and Huxley presented their quantitative model of the action potential is that they used the model to make predictions that went well beyond the original data they had used to construct the model in the first place.

• They used the model to predict the conduction velocity of the action potential, which was certainly not a part of their original measurements. Indeed, because they space-clamped the axon, the entire membrane generated a single instantaneous response to voltage commands, and there was no way in which the action potential could propagate.

• The key assumption they made was that the action potential would propagate at a fixed velocity along the axon; let us refer to that velocity as $v$.

• In turn, this implied that the solution to this constant velocity wave would provide a fixed relationship between space and time; that is, if one looked at the solution a distance $\Delta x$ from the current location, the potential across the membrane would be identical, after a time equal to the time it took the wave to propagate that distance, i.e., $\Delta x / v$. By applying this transform to the original differential equation, it was possible to convert it to an ordinary differential equation, with the velocity as an explicit parameter.

• Then, by iteratively guessing the value of that velocity parameter, and seeing whether the resulting solution was stable, Hodgkin and Huxley were able to obtain an estimate of the conduction velocity of about 18.8 meters per second (at a temperature of 18.3 C), which was within about 10% of the experimentally measured value (21.2 m/s).

• There are two difficulties with this approach: first, unless the estimates are close, the system becomes numerically unstable. Second, the relationship between the changing conductances of the membrane and the conduction velocity are very complex.

• Recently, Tasaki published a derivation for the conduction velocity which provided validation for an equation that he had developed based on empirical measurements. The paper describing the detailed derivation is here. I will briefly describe the approach he took and the key result that he obtained.

• Assuming that there are two regions, active and resting, that move relative to one another at a constant velocity, one can construct two ordinary differential equations describing the propagation of the potential along the nerve fiber. Making the reasonable assumption that the voltage profiles along the membrane on either side of the "active front" are symmetrical, one can then derive the following equation for the conduction velocity, assuming that the ratio of the membrane resistance at the peak of the action potential to the resting membrane resistance is much less than one, and the capacitance during that action potential is essentially identical to the capacitance of the resting membrane: $$v = \frac{1}{c_{m}\sqrt{2 r_{i} r_{m}^*}} \quad\text{(Equation 2)}$$ where $c_{m}$ and $r_{i}$ are the capacitance of the resting membrane and intracellular resistance, respectively, and $r_{m}^*$ is the membrane resistance at the peak of excitation. Again, you may find it useful to refer to the Cable Theory Parameters and Units page, where we carefully describe each parameter and specify its units.

• These quantities can be converted to the more readily measured properties of the axon by using the following relationships:
  • The membrane capacitance per unit area, $C_m$, if $d$ is the diameter of the axon, is defined (in $\mu\text{F}/\text{cm}^2$) as $$C_m = \frac{c_{m}}{\pi d}. \quad\text{(Equation 3)}$$
  • The intracellular resistivity, $R_i$ is defined (in $\Omega \text{cm}$) as $$R_i = \frac{r_{i} \pi d^2}{4}. \quad\text{(Equation 4)}$$
  • The membrane resistance of a unit area at the peak of excitation, $R_m^*$, is defined (in $\Omega \text{cm}^2$) as $$R_m^* = r_{m}^* \pi d. \quad\text{(Equation 5)}$$
  • For a more thorough description of the differences among these quantities, refer to the Cable Theory Parameters and Units page.

• After substituting these quantities into Equation 2, the resulting equation is $$v \simeq \frac{1}{\sqrt{8}} \frac{1}{C_m} \sqrt{\frac{d}{R_m^* R_i}}. \quad\text{(Equation 6)}$$

• For the squid giant axon, the value of $C_m$ is approximately $1\ \mu\text{F}/\text{cm}^{2}$, the diameter $d$ is $0.4\ \text{mm}$, the intracellular resistivity $R_i$ is approximately $36\ \Omega \text{cm}$, and $R_m^*$ is approximately $22\ \Omega \text{cm}^2$.

• Question 1. Use the given data and Equation 6 to compute the conduction velocity in the squid giant axon. By compute, we mean calculate and report a specific number with correct units. Getting the units right will be essential for obtaining the correct answer. Recall that 1 ohm times 1 Farad is equal to one second, or $\Omega \text{F} = \text{s}$. Note that most dendrites have diameters measured in microns (micrometers, $\mu\text{m}$), so use that as your unit, and make sure to convert the parameters of the squid giant axon into microns. Conduction velocity is typically reported for neurons in meters per second; please make sure to report your final velocity in meters per second.

• Question 2. Here is a simulation that allows you to explore the properties of the active cable. As usual, it may be helpful to open it in a separate window.
  • To ensure you are using the initial parameters for the Active Cable Simulation, press the Reset button, and then press the Run button. Estimate the conduction velocity by measuring the time of the peak of the action potential at Membrane Potential 1 and the time of the peak of the action potential at Membrane Potential 2. Note the distance between the electrodes. Use your measurements to estimate the conduction velocity. Make sure to convert your measurement into meters/second, based on the analysis of the units you used for Question 1.
  • Now use the equation to predict the conduction velocity, using the $R_m^{*}$ for the squid giant axon. The value of the intracellular resistivity and the capacitance given in the simulation Cell Properties list should be used for $R_i$ and $C_m$, respectively, and the dendrite diameter for $d$. How well does the equation predict the conduction velocity? What does this imply about the value of $R_m^{*}$ for this system?

• Question 3. Find a value of $R_m^{*}$ that provides a better match to the data. Use the values for the conduction velocity that you measured, the parameter values for the simulation, and solve Equation 6 for $R_m^{*}$.

• Question 4. Use the revised value of $R_m^{*}$ to predict the conduction velocity when the dendrite diameter is $1\ \mu\text{m}$ and $3\ \mu\text{m}$. Plot the function over the range of 1 to 3 $\mu\text{m}$, and superimpose the measured values at 1, 2 and 3 $\mu\text{m}$. How well do they match?

• Question 5. If an animal needs to rapidly escape from a predator, it will need to increase the conduction velocity of action potentials. Assuming that the animal uses unmyelinated axons for its escape response, predict what will happen to the axon diameter of the neurons that generate the escape response over the course of evolution (i.e., over many generations) to increase the likelihood that an animal can successfully escape a predator.

Cable Properties and Myelination

• From Equation 6, it is clear that the velocity of an unmyelinated axon is proportional to the square root of its diameter.

• Is there any way of increasing conduction velocity without having to increase axon diameter in this way?

• If we go back to the hose analogy that we used to understand cable properties in the previous unit, we can gain an intuitive understanding of an alternative way of speeding up conduction velocity.

• Imagine that you wanted the fine sprays to extend out to a very long distance, far longer than the ordinary garden (say, you inherited a massive estate - wouldn't that be nice!)

• How would you change the hose so that it could continue to provide water out to a much greater distance? Assume that you no longer were worried about watering every patch of grass along the length of the hose; you assumed that a fine spray every so often would be enough.

• One way to do this would be to insert into the hose long pipes that had no holes in the side, and were made of much stiffer material.

• Let's refer to the parts of the hose that have holes in them as the "nodes", and the parts that have the stiff pipe as the "internodes".

• The internodes have much higher resistance, ensuring that the only flow of fine sprays will be through the holes in the nodes.

• The internodes, being much stiffer, also have much lower capacitance; they bulge out much less when the faucet is turned on, and so most of the central flow goes right past them into the nodes.

• Qualitatively, we have just described the effects of providing myelination for a nerve axon.

• Myelin is an insulating layer made by special support cells in the nervous system, known as glial cells. Glia perform a whole variety of maintenance and support functions for neuronal cells, and there are dedicated glia just for making myelin. In the peripheral nervous system, the glial cells that make myelin are called Schwann cells, after their discoverer. In the central nervous system, myelin is made by oligodendrocytes. The glial cells wrap part of their membrane multiple times around the axon, creating a very thick, high resistance, low capacitance layer.

• The regions that are not covered by myelin are referred to as the nodes of Ranvier (again, after their discoverer); the regions covered by myelin are referred to as the inter-nodal regions.

• At the nodes, in frog peripheral neurons, there are both voltage-dependent sodium and potassium channels, which can serve as the basis for regular action potential generation. In mammalian myelinated nerves, it appears that there are voltage-dependent sodium channels, and repolarization may be due to leak channels.

• In all myelinated nerves, the action potential appears to "jump" from one node to the next, because it cannot be generated in the internodal region, and the currents from one node rapidly traverse the internodal region and induce an action potential in the next node. This is referred to as saltatory conduction.

• In myelinated nerves, the speed of conduction is proportional to the diameter of the axon, i.e., $v \varpropto d$, rather than being proportional to the square root of the axon diameter, as is true for unmyelinated axons.

• Here is a simulation of a cable with three segments; the middle segment is myelinated:

• Question 6. In the Myelinated cable simulation, compare the capacitance, leak conductance, fast transient sodium conductance, delayed rectifier potassium conductance, and the intracellular resistivity for the node of Ranvier versus the myelinated segment. How do they differ? Based on your understanding of active and passive cable properties, how should this affect transmission in these segments?

• Question 7. Compute the conduction velocity between node 1 and node 2 in the simulation. Note that the first electrode is placed at the beginning of the first node of Ranvier, and the second electrode is placed at the end of the second node of Ranvier; make sure to note the lengths of the nodes and of the intervening myelinated section to ensure that you measure the right length over which the action potential propagates.
  • Now simulate the effects of a demyelinating disease by changing all of the myelinated segment properties to be identical to those of the node of Ranvier (i.e., change the membrane capacitance, leak conductance, fast transient sodium conductance and delayed rectifier potassium conductance to be the same). Also, increase the duration of the simulation from 6 to 15 ms. Again measure the conduction velocity. What is the ratio of the two conduction velocities? What are the implications for a patient?
  • In making these changes, we assumed that the voltage-dependent sodium and potassium channels would be expressed in the demyelinated region. What happens if the demyelinated region does not express these channels? Change the fast transient sodium conductance and delayed rectifier potassium conductance back to 0. Run the simulation again. Explain the results. What would the implications be for a patient under these circumstances?