Cable Properties I: Passive Properties

Introduction

• Until this unit, we have treated nerve cells as if they were point entities with no spatial extent.

• In fact, the shapes of neurons play a crucial role in how they function.

• The cylindrical structures that neurons extend over long distances - their axons - are critical for their ability to propagate action potentials rapidly from one place to another.

• The complex shapes of their dendritic trees are also critical for summing up information over space and time. Indeed, the shape of the dendritic tree constitutes a way that a neuron can be "tuned" to be uniquely responsive to particular patterns of inputs over space and time.

• To understand the importance of the shape (the morphology) of neurons for their function, it is essential to study their cable properties.

• We will begin with a qualitative introduction to cable theory. We will then develop the equations necessary to describe change in current and voltage along a single, unbranched uniform cylindrical cable, which is passive, i.e., unable to generate action potentials. Once we have understood this, we will describe how these ideas can be generalized to cables with the regular wrapping known as myelin, to active, voltage-gated ionic channels, and to highly branched structures.

A Qualitative Introduction to Cable Properties

• Cable theory was developed in the mid-nineteenth century, by Lord Kelvin and others, in response to the need to lay long transatlantic cables.

• An example that is closer to home may help you develop an intuition for cable theory that will help you during the more mathematical derivation.

• You are probably all familiar with garden hoses that have small holes all along their lengths; when connected to a water source, uncoiled and placed on the lawn, you can turn on the faucet, and see fine water sprays emerging, which effectively water your lawn.

• If you were asked to design these hoses, what principles would you use to ensure that the fine water sprays were about the same height near the faucet and at the far end of the hose? If they differ greatly in height, this implies that different parts of the lawn are getting more or less water, and ideally one would want to water the lawn uniformly.

• How does one even begin to figure this out?

• It will be helpful to use some symbols to characterize the different components of this system.

• The water faucet provides some initial pressure, $P$. This corresponds to the voltage across the membrane, $V_{m}$.

• Let's refer to the flow of water through the hose using the symbol $J_\text{hose}$, since this is a flux. This is analogous to the current, $I_\text{axon}$, flowing through the center of an axon.

• If the pressure is uniform along the entire hose, there will be no flow through the hose.

• Thus, we can immediately see that the flow of water along the center of the hose is proportional to the difference in pressure along the hose, or (in symbols) $\Delta P \propto J_\text{flow}$.

• What about flow out of the hose through those small holes that generate the fine sprays we want for watering the lawn?

• If the water that flows into a part of the hose near a hole in its side is exactly equal to the water that flows out of that part of the hose at that point, no water will be left over to flow through the hole.

• Let's represent the flux out of one of the holes in the side of the hose by the symbol $J_\text{hole}$.

• The amount of flow through the hole, $J_\text{hole}$, will increase as the flux of water into that location is increasingly greater than the flux of water out of that location. Let us represent the flow of water into the point just before the hole as $J_\text{hose, in}$, and the flow of water out of that region further along the hose as $J_\text{hose, out}$. Then, in symbols, what we have just said is that $J_\text{hole} \propto \Delta (J_\text{hose, in} - J_\text{hose, out})$.

• But there would be no flow of water along the hose into the point where the hole was located unless there was a difference in pressure, and there would be no flow of water along the hose out beyond the point where the hole is located unless there was a difference in pressure beyond the hole in the side of the hose.

• From the definition of flow, this is equivalent to saying, in symbols, $J_\text{hole} \propto \Delta (\Delta P_\text{in} - \Delta P_\text{out})$.

Figure 1: Schematic view of flow into and out of a part of the hose, and its relationship to flow out the side of the hose. [Figure by Yenan Zhu]

• In other words, for any water to flow out of the hose, there has to be a difference in the difference of pressures around the hole. This is the key concept for understanding the cable equation.

• In the axon, for any current to flow out of the membrane, there has to be a difference in the differences of voltages around the point in the membrane where the current will flow out.

• We can now address the question we began with, at least qualitatively. To design a hose that has fine sprays of the same height all along its length, we have to be sure that most of the pressure is not lost quickly. If it is, then the initial fine sprays will be high, but as you proceed along the hose, the loss of pressure will lead to shorter and shorter fine sprays, and perhaps no sprays at all at the end of the hose. So, the holes for the spray must be fairly fine so that most of the water doesn't flow out right near the faucet.

• If we want the hose to respond quickly once the faucet is turned on, we don't want much of the water acting to fill out the sides of the hose. So, we would make the hose out of fairly stiff material, so it doesn't inflate like a balloon when the faucet is turned on. In other words, we would try to lower the capacitance of the hose.

• It is also intuitively clear that the center of the hose should be quite large relative to the size of the holes in the side of the hose, so that most of the water will flow very quickly down the center of the hose. To ensure that the pressure isn't all lost at the end of the hose, it would need to be plugged. Once the water has flowed all the way to the end of the hose, its only place to flow would be through the fine holes, which would effectively water the lawn.

• It turns out that the mathematics describing a linear hydraulic system is the same as that for linear mechanical and electronic systems, so these ideas are not only useful intuitions. They can be turned into exact mathematical statements. Predictions based on these equations will work for hydraulic, mechanical or electronic systems, which is a nice illustration of the power and generality of mathematics.

• With these intuitions in mind, we can now go through a derivation of the equation describing current flow through the membrane for a cylindrical axon subjected to a voltage.

A Discussion of the Analogy Between a Hose and the Axonal Cable

How a hose and an axonal cable are similar

Derivation of the Cable Equation

• The goal of our derivation is an equation that describes how voltage along a cable changes as a function of distance $x$ along the cable, and time $t$.

Figure 2: Schematic view of axon and electrical equivalent circuit

• To represent the spatial differences along the cable, we essentially break it down into equivalent electrical circuits, corresponding to small pieces of the axon membrane, hooked together. In Figure 2, we show three such electrical equivalent circuits; each circuit consists of the membrane capacitance and the membrane resistance in parallel.

• To simplify the derivation, we make several assumptions. We assume that the cable is radially symmetric, that the extracellular resistance is negligible, and that the membrane resistance and capacitance are uniform.

• If we focus on the voltage changes relative to the resting potential, we can treat the batteries as if they are all equal to zero, and thus ignore them. Since we are focusing on the passive responses, we can also ignore the voltage-dependent channels. We also assume that the extracellular resistance is negligible, so we show the three electrical equivalent circuits hooked together with wires. In contrast, the intracellular resistance is not negligible, so we show those parts hooked together with resistors.

• What follows is a derivation of the cable equation in which several parameters are introduced. To ensure that the derivation is correct, it is important to get the units of these parameters correct. You may find it useful to refer to the Cable Theory Parameters and Units page, where we carefully lay out the differences between the parameters and their units.

• Membrane resistance, $r_{m}$, is reported in units that are the reciprocal of conductance per unit length of cylinder, or $\Omega \text{cm}$ (ohm centimeters); thus, the resistance of a 10 cm length of membrane is the resistance of a 1 cm length, measured in $\Omega \text{cm}$, divided by the 10 cm length. The implication is that the resistance of the membrane decreases with length, as there are more channels for current to pass through.

• Membrane capacitance, $c_{m}$, is reported in units of $\text{F}/\text{cm}$ (farads per centimeter); thus, the capacitance of a 10 cm length of membrane will be the capacitance of a 1 cm length (in $\text{F}/\text{cm}$) multiplied by the 10 cm length. The implication is that the capacitance of the membrane increases with length, as there are more and more places for ions to settle on the membrane and charge its capacitance.

• The intracellular resistance (in this context, the resistance of the fluid within the axon), $r_{i}$, is reported in units of $\Omega/\text{cm}$ (ohms per centimeter); thus, the resistance of a 10 cm length of axoplasm is the resistance of a 1 cm length, in $\Omega/\text{cm}$, multiplied by the 10 cm length. The implication is that the intracellular resistance increases along the length of the axon, as ions encounter more and more of the axoplasmic material. Over a short distance of axoplasm, which we can represent using the symbol $\Delta x$, the intracellular resistance would be $r_i \Delta x$.

• We are trying to work out an expression for the flow of current through the membrane per unit length in units of $\text{A}/\text{cm}$ (amperes per centimeter) at location $x$, which will be symbolized by $i_{m} (x)$. Since this is a current per unit length, the total current that flows out over the small region $\Delta x$ is $i_{m} (x)\Delta x$.

• First, since current is neither created or destroyed, we realize that the current that flows out through the membrane at location $x$ has to be equal to the current flowing along the axon into point $x$ minus the current flowing along the axon out of point $x$ (just as above, when we described the water that flows out of a hole in the side of the hose in terms of the water flowing into and out of that point along the hose). If we represent the total current flowing into the point $x$ as $i(x)$, and the current that flows out just beyond the point $x$ as $i(x + \Delta x)$, we can say $$i_{m} \Delta x = i(x) - i (x + \Delta x)). \quad\text{(Equation 1)}$$

• We are now going to convert the expressions for the current flowing into and out of the point $x$ into the voltages and resistances at those locations, using Ohm's law. What is the reason for doing this? As you will see a bit later in the derivation, this will allow us to put everything in terms of the voltage across the membrane, and thus reduce the number of variables we need to analyze. We compute the differences in the potentials following the path of the current flow.

• From Ohm's Law, we can rewrite the first term on the right hand side of Equation 1 as $$i(x) = \frac{V_{m}(x - \Delta x) - V_{m}(x)}{r_{i} \Delta x}. \quad\text{(Equation 2)}$$

• Again applying Ohm's Law, we can rewrite the second term on the right hand side of Equation 1 as $$i(x + \Delta x) = \frac{V_{m}(x) - V_{m}(x + \Delta x)}{r_{i} \Delta x}. \quad\text{(Equation 3)}$$

• Looking back at Figure 2, we see that we can relate all of these quantities to those that were labeled on the figure; this helps to make sure we are following the derivation.

• Now, we are going to rewrite Equation 1 in terms of the voltages and the resistances by substituting Equations 2 and 3 in for the two terms on the right hand side of the equation. This yields $$i_{m} \Delta x = \frac{V_{m}(x - \Delta x) - V_{m}(x)}{r_{i} \Delta x} - \frac{V_{m}(x) - V_{m}(x + \Delta x)}{r_{i} \Delta x}. \quad\text{(Equation 4)}$$

• This can be rearranged so that it can be rewritten this way (you may want to go through the steps yourself, to convince yourself that this is correct): $$i_{m} \Delta x = \frac{(V_{m}(x + \Delta x) - V_{m}(x))- (V_{m}(x) - V_{m}(x - \Delta x))}{r_{i} \Delta x}. \quad\text{(Equation 5)}$$

• It is possible to greatly simplify this equation by using a definition that you almost certainly have seen before in calculus. If you have a function $f(x)$, the difference in values between the function at $x$ and at $x + \Delta x$ can be written in terms of the delta of the original function, or $$\Delta f(x) = f(x + \Delta x) - f(x). \quad\text{(Equation 6)}$$

• Looking carefully at Equation 5, you can see that the first two terms are, by this definition, equal to $\Delta V_{m}(x)$, and that the second two terms are again, by this definition, equal to $\Delta V_{m}(x - \Delta x)$, so that Equation 5 can be rewritten as $$i_{m} \Delta x = \frac{\Delta V_{m}(x) - \Delta V_{m}(x - \Delta x)}{r_{i} \Delta x}. \quad\text{(Equation 7)}$$

• But the remaining difference in the numerator of the right hand side of Equation 7 can be further simplified, again by applying the definition of Equation 6, once we realize that $\Delta V_{m}(x) - \Delta V_{m}(x - \Delta x) = \Delta ( \Delta V_{m}(x - \Delta x))$, which can be written more simply (since we are applying the $\Delta$ operation twice) as $\Delta^2 V_{m}(x - \Delta x)$. If we want an expression for the current that flows out per unit membrane, we can divide both sides of Equation 7 by $\Delta x$, so that Equation 7 can be rewritten as $$i_{m} = \frac{\Delta^2 V_{m}(x - \Delta x)}{r_{i} \Delta x^2}. \quad\text{(Equation 8)}$$

• In words, this equation says that the difference in the difference of the voltages at the point $x$ is what gives rise to the current that flows through the membrane, just as we saw above when we talked about the water flowing out of the side of a hose.

• As we look over smaller and smaller increments of time and space, we can take the limit of Equation 8 as $\Delta x \to 0$. In addition, we must keep in mind that we are taking the rate of change with respect to both time and space, and thus we need to use a partial derivative symbol for the expression, since it is looking only at the change of voltage with respect to the length $x$, and so Equation 8 becomes $$i_{m} = \frac{1}{r_{i}} \frac{\partial^2 V_{m}(x, t)}{\partial x^2}. \quad\text{(Equation 9)}$$

• Another way to understand Equation 9 is to think what would happen if the rate of voltage change with length were identical before and after the location $x$. If they were, then the amount of current flowing into the point $x$ would just equal the amount of current flowing out of that point along the axon, and none would be left over to flow through the membrane.

Figure 3. Flow of water into hose equals flow out, so none flows through the side of the hose. [Figure by Yenan Zhu]

• Now if we could express the current flowing out per unit membrane, $i_{m}$, in terms of the voltage across the membrane, we would have a partial differential equation that was expressed purely in terms of the voltage, which would be easier to solve.

• We can make another assumption, and identify the voltage across the membrane at any location $x$ with the potential along the axon at that point. As long as the external voltage of the medium is not changing, this is a reasonable assumption.

• Referring again to Figure 2, we see that any current flowing out through the membrane has two paths it can take: it can "flow through" the capacitance of the membrane (i.e., charges can land on the lipid bilayer of the membrane, and repel like charges across the membrane), or it can flow through the resistance of the membrane.

• Thus, the total current flowing through the membrane is the sum of the capacitative and resistive currents, or $$i_{m} = i_{c} + i_{r}. \quad\text{(Equation 10)}$$

• Both of these terms can be rewritten in terms of the membrane voltage, because $i_{c} = c_{m}\ \partial V_{m}(x, t) / \partial t$, and $i_{r} = V_{m}(x, t) / r_{m}$, so that Equation 10 becomes $$i_{m} = c_{m} \frac{\partial V_{m}(x, t)}{\partial t} + \frac{V_{m}(x, t)}{r_{m}}. \quad\text{(Equation 11)}$$

• If we set Equation 9 equal to Equation 11, we obtain a single expression for the change in the membrane voltage with space and time: $$\frac{1}{r_{i}} \frac{\partial^2 V_{m}(x, t)}{\partial x^2} = c_{m} \frac{\partial V_{m}(x, t)}{\partial t} + \frac{V_{m}(x, t)}{r_{m}}. \quad\text{(Equation 12)}$$

• Let's simplify this a bit. Multiply the entire equation by $r_{m}$, and move all the terms over to the left side to obtain $$\frac{r_{m}}{r_{i}} \frac{\partial^2 V_{m}(x, t)}{\partial x^2} - r_{m} c_{m} \frac{\partial V_{m}(x, t)}{\partial t} - V_{m}(x, t)= 0. \quad\text{(Equation 13)}$$

• One more step, and we're done. Let's define the constant multiplying the first term as $\lambda^2 = r_{m} / r_{i}$, and let's define the constant multiplying the second term as $\tau = r_{m} c_{m}$. The symbol $\lambda$ is usually referred to as the length constant or the space constant, and the symbol $\tau$ is usually referred to as the time constant, and we will have more to say about each of these constants in the next sections of this unit.

• Using these definitions, we can rewrite Equation 13 as $$\lambda^2 \frac{\partial^2 V_{m}(x, t)}{\partial x^2} - \tau \frac{\partial V_{m}(x, t)}{\partial t} - V_{m}(x, t)= 0. \quad\text{(Equation 14)}$$

• This is the cable equation, and understanding some of its implications will take us the rest of this unit and the next unit.

• Checking the measurement units is a good way of seeing whether the equation is correct. $\lambda^2 = r_{m} / r_{i} = (\Omega \text{cm})/(\Omega/\text{cm}) = \text{cm}^2$. This term is multiplied by a term whose units are $\text{V}/\text{cm}^2$, so the units for the first term are $\text{V}$ (volts). For the second term, $\tau = r_{m} c_{m} = (\Omega \text{cm})(\text{F}/\text{cm}) = \Omega\text{F} = \text{s}$ (seconds), which is multiplied by a term whose units are $\text{V}/\text{s}$, so the second term has the units $\text{V}$; and the third term is already in $\text{V}$, so the units for the equation are correct.

A Discussion of the Derivation of the Cable Equation

Deriving the cable equation from the electrical equivalent circuit

Longterm Behavior and the Space Constant

• The cable equation is complex because it is a partial differential equation describing how voltage changes with both space and time.

• One way to understand a complex equation is to look at simpler cases.

• For example, if we could ignore the role of time, the equation would simplify to an ordinary differential equation.

• How could we do this?

• If we injected a current into the axon, and waited a long time for the transient responses to die down (i.e., for the capacitor of the membrane to charge or discharge), then we could look at the steady state behavior of the cable.

• If no current is flowing through the capacitor, the capacitative current is zero, so the second term of the cable equation goes to zero, i.e., $\partial V_{m}(x, t)/\partial t = 0$, and so the cable equation simplifies to an ordinary differential equation: $$\lambda^2 \frac{d^2 V_{m}(x)}{d x^2} - V_{m}(x)= 0. \quad\text{(Equation 15)}$$

• A function that solves this differential equation is $$V_{m}(x) = V_{0} e^{-x/\lambda}. \quad\text{(Equation 16)}$$

• The solution is readily checked. Take the first derivative of Equation 16 with respect to $x$: $$\frac{dV_{m}(x)}{dx} = -\frac{1}{\lambda} V_{0} e^{-x/\lambda}.$$ Take the derivative again: $$\frac{dV_{m}^2(x)}{dx^2} = \frac{1}{\lambda^2} V_{0} e^{-x/\lambda}.$$ If you substitute this into Equation 15, and Equation 16 for $V_{m}(x)$, the two terms cancel out so that the equation is satisfied.

Figure 4: Fall in voltage along cable given a length constant of 1 mm (blue) and 2 mm (red).

• What does the equation imply? It says that as the length constant $\lambda$ increases, the rate of fall of voltage from the point of current injection will be slower, so that the voltage change will be smaller as you move away from the point of injection.

• Recall that the length constant, $\lambda$, is defined as $\sqrt{r_{m}/r_{i}}$, the square root of the ratio of the membrane resistance to the intracellular resistance. If we think back to the hose analogy, we can understand intuitively how this works. If the resistance of the holes in the sides of the hose are very high (i.e., the holes are very small) relative to the size of the center of the hose, the rate at which the fine sprays of the hose will decrease from the faucet will be very small. Similarly, if the resistance of the nerve cell membrane is much higher than the resistance along the axoplasm, the voltage change as one moves away from the point of current injection will be small.

• This also provides us with an intuitive understanding of the importance of axon diameter for the decay of voltage away from the point of current injection. As the diameter $d$ of the axon increases, the circumference of the membrane will increase as $\pi d$, so that $r_{m}$ will decrease linearly; at the same time, the cross-sectional area of the axoplasm will increase as $\pi d^2/4$, which will cause the intracellular resistance $r_{i}$ to decrease much faster than the membrane resistance decreases with the increase in axonal diameter. In turn, this explains the reason that large diameter axons have much larger length constants than smaller diameter axons.

• Figure 4 illustrates the rate of fall of voltage from the center of the cable (the point at which current is injected) in a cable that has a length constant of 1 mm versus one that has a length constant of 2 mm. The axon with the larger length constant shows a consistently higher voltage at every point along the cable except at the point of injection:

A Discussion of the Length Constant

Deriving and understanding the length constant

A Spherical Cell and the Time Constant

• By allowing transients to die out, we were able to understand the longterm behavior of a cable in response to steady current injection.

• Is there an equivalent simplification that would allow us to understand the temporal response of the system, without having to be concerned with the spatial dimension of the cable?

• One way to eliminate the spatial dimension is to simplify the cable into a spherical cell. Currents injected into a spherical cell will travel essentially instantaneously to all parts of the cell membrane, and so the spatial term in the cable equation will be zero, i.e., $\partial^2 V_{m}(x, t)/\partial x^2 = 0$. Once again, the cable equation simplifies to an ordinary differential equation: $$- \tau \frac{dV_{m}(t)}{dt} - V_{m}(t)= 0. \quad\text{(Equation 17)}$$

• A solution to this equation, showing how the cell charges from an initial voltage $V_{0}$ to a final voltage $V_{\infty}$ is given by $$V_{m}(t) = V_{0} + (V_{\infty} - V_{0})(1 - e^{-t/\tau}). \quad\text{(Equation 18)}$$ Note that at time $t = 0$, the exponential term is 1, so the term in parentheses is just $1 - 1 = 0$, and thus $V_{m}(0) = V_{0}$, i.e., the membrane voltage is just the initial voltage. As time increases, note that the exponential term heads to zero, so the right hand side becomes $V_{0} + (V_{\infty} - V_{0})$, which simplifies to $V_{\infty}$, as it should. The derivation for the solution is given here.

• Note that if the cell starts at a membrane potential of 0 mV, i.e., $V_{0} = 0$, Equation 18 simplifies even further to $$V_{m}(t) = V_{\infty}(1 - e^{-t/\tau}). \quad\text{(Equation 19)}$$

• Furthermore, if the final value, $V_{\infty}$, is equal to zero, the righthand side of Equation 18 becomes $V_{0} + (0 - V_{0})(1 - e^{-t/\tau})$, which simplifies to $$V_{m}(t) = V_{0}(e^{-t/\tau}). \quad\text{(Equation 20)}$$

• What could alter the time constant of a cell? Recall that the time constant, $\tau$, is defined as $r_{m} c_{m}$, i.e., the membrane resistance times the membrane capacitance. Since the capacitance of the membrane is essentially a constant, this suggests that controlling the input resistance of the cell can greatly affect its time constant. We already saw this in operation, qualitatively, when you explored the passive properties of the membrane.

• It is now clear that the time constant, $\tau$, plays the same role as a "measuring stick" for time as the length constant $\lambda$ does for space. The larger the time constant, the more slowly the cell will charge up to its final value; the smaller the time constant, the more rapidly the cell responds to an input.

• Figure 5 illustrates the much more rapid rise of voltage in a spherical cell with a smaller time constant.

Figure 5. Voltage rise in two spherical cells, one with a time constant of 5 msec (red), and one with a time constant of 10 msec (blue).

• Note that in situations where geometry is simple (e.g., a small spherical cell or a short cylinder of axon with a cytoplasmic resistance that is negligible in comparison to the membrane resistance), the voltage as a function of space will be essentially uniform, and the voltage change will occur across the resistance of the membrane. Thus, the input resistance of these structures (measured as the voltage change in response to currents injected into them, in units of $\Omega$) is equal to the membrane resistance of a unit area (in units of $\Omega \text{cm}^2$) divided by the surface area of the membrane.

A Discussion of the Time Constant

Deriving and understanding the time constant

Exploring the Effects of Passive Cable Properties on Inputs

You now have enough information to begin to explore the functional consequences of the passive cable properties of a neuron. Let's begin by examining the properties of a single compartment, and use the equation we just derived for the spherical cell. Here is the simulation you have used before in the Passive Membranes unit. Once again, you may want to open the simulation in another window as you answer the questions.

• Question 1. Set the number of pulses to 1, and set the pulse duration to 20 ms. Set the stimulus current to 5. Before you run the simulation, write down the appropriate equation (not directly given in this unit) and use it to predict the maximum voltage change that you will obtain, in the correct units; confirm your prediction by running the simulation.

• Question 2. Now focus on the rate of rise towards the final voltage.
  • Which equation that was presented in this unit is appropriate to describe the rate of rise? Please write it down, and use it to predict the time at which the voltage will rise to half its final value. You will need to solve the equation for this time; please show your work. To fill in the parameters for the equation, recognize that you just calculated one of them in Question 1 (the long term voltage), and that the other two that you need for calculating the time constant are in the parameter list of the simulation.
  • Check your prediction by measuring the time at which the voltage rises to half its final value in the simulation. Make sure not to include the delay time; start your measurement from when the voltage begins to change. It may be helpful to zoom in to see the voltage and time more clearly (we show how to do this in the second video on the Course policies page if you've forgotten how to do this; the relevant information is at 12 minutes and 57 seconds). Also make sure to check your units and report your results using appropriate scientific notation.

• Question 3. Before you begin working with the simulation of the extended cable, think carefully about the following questions:
  • If you inject a fixed amount of current at one point into a spherical cell of a fixed diameter, and then into a passive cable that had the identical diameter, how much will the membrane voltage rise at the site of injection in the two different cells? Assume that all membrane properties of the two cells are the same (e.g., membrane conductance and capacitance). You don't need to provide a number. Think about the paths of escape for the injected current away from the electrode in a spherical cell as opposed to a cable, and what the likely consequences are for the steady state membrane potential at the point of injection. Please write down your qualitative prediction.
  • Will increasing the resistance to current flow down the cable cause the change in voltage at the point of injection to be higher, the same, or lower? Please explain your reasoning.

In this next simulation, we have two electrodes in a long dendrite of a neuron separated by a short distance. We are injecting current into one electrode, and looking at the resulting intracellular potential at both that electrode and the other electrode a short distance away.

• Question 4. Let us focus first on the response of the cable at the point at which the current is injected into it (this is shown in the top panel of the simulation, labeled Membrane Potential 1). Mathematical analysis of the passive cable, based on the definitions of the membrane and axial resistance in terms of the cable's geometry, demonstrates that the change in voltage of the cable at the point of injection is equal to the current injected times the input resistance, or $$V_{0} = I r_{input} = I \frac{1}{\pi} \sqrt{\frac{R_{i}}{g_{m} d^3}}, \quad\text{(Equation 21)}$$ where $r_{input}$ is the input resistance of the cable, $R_{i}$ is the intracellular resistivity of the cytoplasm within the cable in $\Omega\text{cm}$ (this is similar to, but different from, the intracellular resistance per unit length, $r_i$, discussed above, because it is independent of geometry), and $g_{m}$ is the membrane conductance in $\text{mS}/\text{cm}^2$.
  • Use this equation to predict the voltage change at the point of current injection in response to a 10 nA current for cable diameters of 1, 2, 3, 4 and 5 microns. For $R_{i}$ and $g_{m}$, use the default values given in the simulation. Converting each parameter to the appropriate units is critical to solving this problem. Think through this carefully, and show all your work. Always indicate units on all quantities. Click here for some guidance in solving this.
  • Plot your results, with cable diameter along the x axis, and voltage along the y axis.
  • Test your predictions by running the model and plotting the actual data along with your predictions (don't forget to change the Stimulus current to 10 nA). (Deviations from your predictions by about 2% is normal, since the simulation is actually a finite cable model, instead of the infinite cable model that Equation 21 assumes.)
  • Explain the implications of your results for how cable diameter can be used to control the depolarization of a neuronal process in response to currents.

• Question 5. Now let's focus on how voltage changes along the length of the passive cable. Before running the simulation, think carefully about the following questions:
  • How quickly does the voltage reach its final value as you go further along the cable? What property of the membrane will be critical for determining how quickly the voltage changes? It may be helpful to think about the hose analogy, and how the pressure that builds up along the hose will be affected by the stiffness of the sides of the hose, and the distance you are along the hose. Please write down your ideas briefly and clearly.
  • How high will the final value of the voltage rise as you go further along the cable? What property of the membrane will be critical for determining how high it rises? Again, the hose analogy will be helpful for answering this question. Please write down your ideas briefly and clearly.

• Question 6. Let us test your ideas. The voltage measured at some distance from the point of injection is the second panel of the simulation, and is labeled Membrane Potential 2. Press Reset to restore the original parameters. Change the Total duration to 1 ms and the Pulse duration to 0.8 ms.
  • Measure the peak voltage in the more distant part of the cable (100 microns away from the point of injection). Measure the time at which it reaches half of this voltage after the current pulse begins.
  • Double the capacitance (from 1 to 2 $\mu\text{F}/\text{cm}^2$). Again, measure the peak height in the more distant part of the cable, and the time at which it reaches half of this voltage.
  • Double the capacitance once more (from 2 to 4 $\mu\text{F}/\text{cm}^2$). Again, measure the peak height in the more distant part of the cable, and the time at which it reaches half of this voltage.
  • Explain the results you've obtained.

• Question 7. As we saw above, mathematical analysis of the passive cable demonstrates that the rate of fall of the peak voltage with distance can be described by the equation $$V_{m}(x) = V_{0} e^{-x/\lambda}, \quad\text{(Equation 22)}$$ where $V_{0}$ is the voltage at the point of injection as defined above in Equation 21, $x$ is the distance from the point of injection, and $\lambda$ is the space constant, which (again using the geometry of the cable and the definitions of axial and membrane resistance in terms of that geometry) can be shown to be equal to $$\lambda = \sqrt{\frac{d}{4 g_{m} R_{i}}}, \quad\text{(Equation 23)}$$ where the symbols are the same as those defined for Equation 21.
  • Using the equation, predict the peak voltage at a second electrode 100 microns away from the point of injection of 10 nA of current, for cable diameters of 1, 2, 3, 4 and 5 $\mu\text{m}$. As before, use the default values given in the simulation for $R_{i}$ and $g_{m}$ in your calculations. Again, converting each parameter to the appropriate units is critical to solving this problem. Think through this carefully, and show all your work. Always indicate units on all quantities.
  • Plot the diameter (on the x axis) versus the peak voltage at the second electrode (on the y axis).
  • Test your predictions by running the model and plotting the actual data along with your predictions (don't forget to press the Reset button and change the Stimulus current to 10 nA). (This time, deviations from your predictions may be as large as 20-40%, since the simulation is actually a finite cable model, instead of the infinite cable model these equations assume.)
  • Explain how changing the diameter of the axon can control the extent to which a depolarization will spread down the cable (i.e. from electrode 1 to electrode 2), and thus how far a signal is likely to travel. To answer this question correctly, compare the values you calculated in question 4 with those you calculated in this question, and note that dividing the value you obtained in this question for a given diameter by the value you obtained in question 4 for that diameter is equivalent to dividing "Equation 22" by "Equation 21". If you do this, and then look carefully at "Equation 23", you have all the information you need to quantitatively predict what the ratios should be. Please do this. In addition, please explain your results qualitatively.