Reading passive membranes

Resistance or Conductance versus Permeability, and the Electrical Equivalent Circuit for an Ion Channel

$image: SchematicMembrane.png|300px|thumb|Figure 1: A schematic view of the membrane of a nerve cell]] Figure 1 shows a simple schematic view of a nerve cell membrane, greatly magnified. It is important to focus on two features of the membrane:

• Most of the membrane consists of a lipid bilayer.
  • The phospholipids making up the membrane have one side that is charged (referred to as the "head", and represented by a circle in Figure 1), and thus can interact with the partial charges of water and ions on the inside or the outside of the nerve cell; they are thus referred to as hydrophilic, i.e., "water-loving".
  • The other part of the phospholipid, represented by the two wavy lines in Figure 1, often referred to as the "tails", are non-polar or hydrophobic ("water fearing"); like any other fat, they prefer each other's company to that of water molecules.
• At infrequent intervals, the membrane is interrupted by the insertion of roughly cylindrical proteins that have a hole in their middle, which can thus act as channels for the flow of ions.

These two features of the membrane—the lipid bilayer and the ion channels—can be represented using equivalent electrical components: a capacitor and a resistor. Let us focus first on the resistor (the channel).

• As an ion attempts to flow through a channel, it may encounter a series of obstacles, which may be due to constrictions in the diameter of the channel, protrusions of proteins into the channel, charges on some of the amino acids that make up the channel, and so on. These obstacles will tend to reduce the rate at which the ion can flow through the channel. What impels the ions to go through the channel? The electrical gradient across the membrane generates an electromotive force, i.e., a potential to do work on a charged entity, and this affects the probability that a charged ion will cross the membrane through the channel. The electromotive force is measured in volts, $V$.
• Thus, it should be clear that the magnitude of the voltage across the membrane will affect the amount of current that crosses through the channel, or $V \varpropto I$.
• The constant of proportionality is the resistance, $R$, measured in ohms.
• Thus, one can re-write the equation as $V = I R$, which is known as Ohm's Law, after its discoverer. Let us think about this equation for a bit:
  • It implies that, for a given voltage, as the resistance to flow increases (it becomes harder for ions to move through a channel), the current will decrease.
  • Another way to think about it is this: if one applied a fixed current, then a measure of the resistance would be how much the voltage changed for that fixed current. If the resistance is low, there would be a small voltage change; if the resistance is high, there would be a larger voltage change.

$Image: EquilibriumFluxes.png|250px|thumb|Figure 2: Fluxes of potassium are balanced when the membrane potential equals the Nernst potential for potassium]] Another way to describe the ease with which an ion can move through a channel is to take the reciprocal of the resistance, which is known as the conductance. The conductance, usually represented by the symbol $g$ and measured in units of siemens, is thus equal to $1/R$.

• Qualitatively, the relationship between resistance and conductance is clear: a high conductance implies a low resistance, and a low conductance implies a high resistance.
• To use conductances with Ohm's law, divide both sides of the equation by $R$ to obtain $V/R = I$; since $1/R = g$, Ohm's law becomes $g V = I$.

$image: IonChannelCircuit.png|250px|thumb|Figure 3: An electrical equivalent circuit for potassium ion channels. On the left is shown the symbol for a voltmeter, and on the right the symbols for a resistor and a battery]] Channels that are permeable to a specific ion will be associated with a specific Nernst or equilibrium potential. That is, the current flow through a channel will not only depend on the voltage across the membrane, but will also depend on how different the voltage is from the Nernst or equilibrium potential.

• If the membrane voltage is equal to the Nernst or equilibrium potential, then there should be no net flux of those particular kinds of ions across the membrane. For example, suppose you have a neuron with a high internal concentration of potassium, and the membrane potential is at the equilibrium potential for potassium. Imagine that the flux of potassium ions caused by the concentration gradient (an outward flux) became a bit larger due to a random fluctuation in the movement of the ions. This would cause the inside of the membrane to become more negative and the outside more positive, leading to a hyperpolarization of the cell. This would increase the strength of the electrical gradient; in turn, the flux of potassium ions caused by the electrical gradient (an inward flux) would strengthen, and the random fluctuation would be undone. Thus, there would be a balance between the flux of potassium ions out of the nerve cell (flux due to the concentration gradient) and the flux of potassium ions into the nerve cell (flux due to the electrical gradient), as illustrated in Figure 2.
• To represent all the ion channels of a particular kind (e.g., all potassium ion channels) using an electrical equivalent, one could use a battery whose voltage was at the Nernst or the equilibrium potential, which was in series with the resistance (or conductance) of the channels. This is shown in Figure 3 — the wavy lines are used to represent a resistor (or conductor), which is labeled with the symbol for the potassium conductance, $g_{\text{K}^+}$. In series with the resistor is the battery, labeled with the symbol for the potassium Nernst or equilibrium potential, $E_{\text{K}^+}$.
• This simple representation of the potassium ion channels can already be useful for predicting relationships between the membrane voltage, the conductance to those channels, and the current that will flow through them (assuming that the linear approximations we are making by using these electronic components are reasonably accurate).
• Kirchoff's Voltage Law can be stated as follows: the potential drops in a closed circuit must sum to zero. Applying that principle to the circuit shown in Figure 3, we see that the membrane voltage must equal the voltage drop across the resistor $\left(I/g\right)$ plus the voltage change across the battery $\left(E\right)$, or $

   V_m = \frac{I_{\text{K}^+}}{g_{\text{K}^+}} + E_{\text{K}^+}. \quad\text{(Equation 1)}$

• We can solve this equation for the current through the potassium channels to obtain $

   I_{\text{K}^+} = g_{\text{K}^+} \left(V_m - E_{\text{K}^+}\right). \quad\text{(Equation 2)}$

• This is an extremely important equation for understanding and mathematically modeling the properties of ion channels. In words, it says that the current through all the potassium ion channels is equal to the conductance of all the potassium ion channels multiplied by the driving force, i.e., the difference between the membrane potential and the equilibrium potential for the potassium ions.
• An important implication of this equation is that current through a set of ion channels could drop to zero for two different reasons:
  • The conductance of all of the potassium channels could go to zero (that is, they could effectively be shut to potassium ions), and thus no current would flow through them.
  • The driving force could go to zero, that is, the membrane potential could be equal to the equilibrium potential for potassium. Fluxes of potassium ions would go in and out of the nerve cells, but the net flux (and thus the net current) would be zero, as we described above.

• What is the relationship between permeability and conductance? Permeability refers to the ability of an ion to flow through a channel; conductance describes the relationship between current and voltage. Many channels may be open, and thus have high permeability; but the conductance not only depends on the channel permeability but also on the concentration of ions available to actually carry charges through the channel; if no ions are available to carry charge, the conductance will be zero despite the many open channels.

Capacitance

• The second feature of the membrane illustrated by Figure 1 is the lipid bilayer, which prevents ions from flowing directly through the membrane, as they can when they encounter a channel.
• However, as illustrated by the figure, if a positive charge lands on the inside leaflet of the membrane, it can act to repel a positive charge that was near the outer leaflet of the membrane.
• Thus, rather surprisingly, currents may be observed across a membrane even if ions are not crossing through the membrane!
• Adding charges to one side of a membrane repels charges from the other side, inducing a current.
• If the charges are not added or removed, then the separation between the positive and negative charges across the membrane gives rise to a voltage.
• In other words, the charge separated across the membrane (or across a capacitor) is proportional to the voltage across the membrane, or $V \varpropto Q$, where $Q$ is the charge separated across the membrane.
• To turn this proportionality into an equation, we define a constant of proportionality, the capacitance, measured in Farads and represented by the symbol $C$; the equation for the capacitor becomes $

   Q = C V. \quad\text{(Equation 3)}$

• As we have just pointed out, changes in the number of charges can lead to changes in voltage, or $C \Delta V = \Delta Q.$
• If the change occurs in a small time increment $\Delta t$, then the change in the membrane's voltage and charge during that time will be, on average, $C \frac{\Delta V}{\Delta t} = \frac{\Delta Q}{\Delta t}.$
• As the time interval gets shorter and shorter, the fractions on the left and right side of the equation become the instantaneous rates of change, or derivatives, of the voltage and charge, respectively: $C \frac{dV}{dt} = \frac {dQ}{dt}.$
• Since the instantaneous rate of change of the charge is the definition of the current, this equation becomes $

   C \frac{dV}{dt} = I. \quad\text{(Equation 4)}$

• In circuit diagrams, capacitors are represented by a symbol that shows two plates of equal length with a gap between; charges can be stored on the plates, but cannot directly jump across them. However, as positive charges are added or removed to one plate, positive charges will be removed or added to the other plate (respectively), so as Equation 4 states, changing amounts of charge, or current, will lead to changes in voltage.

$image: Channels&Capacitor.png|200px|thumb|Figure 4: An electrical equivalent circuit for ungated potassium channels and the passive membrane]]

• We can now represent all of the ungated potassium channels in the membrane, and how they will respond to charges injected into the nerve cell, with an electrical equivalent circuit that shows the capacitor in parallel with the ion channels. The capacitor is in parallel because ions can either flow through ion channels, or affect charges on the other side of the membrane if they land on the membrane, and thus there are two parallel paths for the movement of charge. In Figure 4, the capacitance of the membrane is labeled $C_m$.
• An analogy, which can be shown to be mathematically equivalent, is to think about what happens to a hose with flexible sides when water pressure is applied to one end. The water pressure is equivalent to voltage, the center of the hose is equivalent to the resistor, and the flexible sides are equivalent to the capacitor.
• Initially, the hose center will fill, and the sides will begin to expand. If the hose was placed in water, this would cause water to be pushed away from the sides of the hose — a current of water would be generated at right angles to the hose, even though no water could flow through the sides of the hose!
• After some time, the sides of the hose would be fully distended, and whatever water current continued to flow would go through the center of the hose.
• Similarly, if charges are injected on one side of the membrane, charges build up on that side of the nerve cell membrane, causing charges with the same sign to move away from the other side of the membrane; at the same time, some charges begin to flow through the ion channels. When the membrane capacitance is fully occupied by charges, the flow "through" the capacitor stops; the only current that remains is flowing through the ion channels. If the charge injection is turned off, the capacitor discharges, and the membrane returns to its previous state (just as the hose deflates once the water pressure is turned off).

From Ion Channels and the Lipid Bilayer to the Electrical Equivalent Circuit

Relating the ion channels and the lipid bilayer to the electrical equivalent circuit

The Electrical Equivalent Circuit and the Ionic and Capacitative Currents

Relating the ion channels and the lipid bilayer to the electrical equivalent circuit

Effects of Current on the Passive Membrane

Once again, having been given some background on the underlying mechanisms of how the membrane works, it is time to start doing virtual experiments on a model of the membrane, so that you can develop a better intuition for what the theory and equations actually mean.

• Code for these simulations and others is available on the JSNeuroSim project on GitHub. If you like programming, you are welcome to download, modify, and share the code. Like corrections to the wiki, we may give extra credit in $Comments on Course Materials]] for fixes and useful extensions to the simulations. It may be wise to contact the instructors before spending a lot of time developing an extension, however, to make sure we agree that it will be useful for the course.
• Note: Some users have reported problems running the simulations under Internet Explorer 9. If you encounter problems we recommend trying a browser with more complete HTML 5 support, such as Firefox, Chrome, Safari or Opera.

Here is a simulation of the passive membrane whose description we have developed in the previous section:

• You may find it useful to open the simulation in a separate window, and have that window set up side-by-side with this window, so you can work on the simulation while working on the problems.
• Remember - as you take data, make sure to take pictures, and add them to your notebook! You may not need pictures for everything you do, but make sure to document all the key results you get using figures in your notebook.
• To help orient you to working with the simulation, here is a very short "user's manual":
  • The top trace shows the membrane potential in a cell that we've impaled with an intracellular electrode.
  • The bottom trace shows a pulse of current we are injecting into the cell.
  • The middle trace shows the endogenous membrane currents (such as currents due to potassium, sodium, or chloride ions) that are a response to the change in membrane potential.
  • Because we are letting the equipment adjust the electrode potential to "clamp" the current at the values we choose, this is known as a current clamp experiment.
  • You can measure values from the data by moving your cursor over the plots. A tooltip will appear displaying the value of the trace at a given time.
  • If you click and drag on a plot, you will create a box. After you release the mouse button, the plot will zoom into the selected region. You can double-click the plot to zoom out again.
  • You can take measurements from the plots which are stored in a table for easy reference. In the following problems, you will be asked to take several measurements. You may find this feature very useful.
    • PC users can take measurements by right-clicking on one of the plots.
    • Mac users can take a measurement by either Ctrl-clicking, or, if they are using a MacBook, by placing two fingers on the trackpad and clicking.
  • At the bottom of the page, you will see three buttons:
    • Run, which takes the current settings, and simulates the response of the membrane to these inputs;
    • Reset, which restores the simulation to the original values it had when it was first started;
    • Clear Data Points, which removes any measurements that you have taken from the data tables below the simulation controls.
  • In between the graphs and the buttons, you will see many parameter values grouped into three categories:
    • Cell Properties: These describe the intrinsic properties of the model cell (membrane capacitance, membrane conductance, resting potential), each of which should now make sense given the previous sections of this unit. If you're not sure what they are, re-reading the previous sections will help.
    • Current Clamp: These are the values for the experiment you are running.
    • Simulation Settings: These are the parameter values that apply to the entire simulation (not to either the cell or to the inputs you are giving the cell).
  • When using a new piece of equipment in a lab setting, you would, of course, read the manual first, because it may have important warnings for protecting the equipment and yourself.
  • Manuals are sometimes ambiguous or terse on the details of how the equipment works, however, and it's important to be willing to play with the equipment a bit until you're comfortable with how it works.

To help develop your understanding of the properties of the passive membrane, please answer the following questions; make sure to write down your answers!

• First focus on getting a qualitative feel for what the cell parameters do.
  • Question 8: What happens when you double or halve the cell capacitance or conductance? What happens if you increase or decrease the cell's resting potential by 20 mV? Please do these one at a time, leaving the others unchanged (otherwise you have to look at a very large number of combinations, which is not necessary!). Explain your results.
• Next, try to get a qualitative feel for how the current clamp settings work.
  • Question 9: What happens if you double or halve the stimulus delay, stimulus current, pulse duration, inter-stimulus interval, or number of pulses? What happens if you double or halve the total duration? Once again, do these one at a time. Explain your results.
• Now you're ready to refine your intuition to develop a quantitative understanding of the height of the pulses.
  • Question 10: Reset the simulation parameters, and run the simulation with these original values.
    • At what value does the membrane potential plateau? Use the cursor to obtain a value. You may find the feature described above for taking and storing measurements useful for this and following problems.
    • Double the stimulus current to 20 nA. Where does the membrane potential plateau now?
    • Triple the stimulus current to 30 nA. Where does the membrane potential plateau now?
    • Use Ohm's law to calculate the membrane resistance of the cell. How does this compare with the value that is actually listed for the simulation? Recall that conductance is the reciprocal of resistance, and don't forget to track the appropriate powers of ten for micro, nano, and milli, which will all enter into your answer.
    • Reset the simulation. Set the membrane conductance to half its current value. Before running the experiments, predict whether this will make the neuron show larger or smaller responses to the same levels of currents. Now repeat the experiments you did above (injecting 10 nA, 20 nA, and 30 nA), and again use Ohm's law to calculate the membrane resistance of the cell. How does this compare with the actual value you set?
    • Try halving the membrane capacitance. Does this change the maximum height of the membrane response? (Increasing the membrane capacitance may have a more complicated effect - we'll explore that in the next question).
    • [Required for graduate students, optional for undergraduates] How would changing the resting potential of the membrane alter its response to current? Reset the simulation, change the resting potential to -10 mV, and see if you can deduce the equation for describing the peak membrane potential in response to a current.
• Now that you know how to calculate the height of the pulses, you can start to explore how quickly it approaches this height. First look at the shape of the rise.
  • Question 11: Reset the simulation parameters. Increase the membrane capacitance to 6 nF, the stimulus delay to 5 ms, the stimulus current to 20 nA, the pulse duration to 20 ms, and reduce the number of pulses to one. Rerun the simulation.
    • From question 10, what do you expect the height of the pulse to be? Measure it to confirm your prediction.
    • How long does it take to reach half of this height? (Use the cursor to measure the time from the beginning of the pulse to the point where it first reaches half of the peak membrane potential; in words, subtract off the time before the pulse starts!).
    • How long does it take to reach three quarters of this height?
    • How long does it take to reach seven eighth of this height?
    • What is the pattern in the previous numbers? Predict how long it will take to 15/16th of the height. Check your prediction.
• Now examine the effects of membrane capacitance on the rise of the potential.
  • Question 12: Set the simulation parameters as in question 11, but set the membrane capacitance to 2 nF.
    • Predict the maximum pulse height and check it. Measure the time to reach half of maximal height.
    • Double the membrane capacitance to 4 nF. Does this affect the maximum height of the pulse?
    • Keeping the membrane capacitance at 4 nF, how long does it take to reach half of the maximum height? Repeat this measurement with the membrane capacitance set to 6 nF and 8 nF.
    • Look at the times to reach half maximum height you've recorded for membrane capacitances of 2, 4, 6, and 8 nF. Do you see a pattern? Remember that your measurements may have a small amount of error. Try to predict the time to reach half maximal height for with a membrane capacitance of 10 nF, and check your result.
• Next, look at the effects of injected current and membrane conductance.
  • Question 13: Starting with the settings from question 11, set the membrane capacitance to 4 nF.
    • Measure or recall the time to reach half maximal height with the stimulus current set to 20 nA.
    • Change the stimulus current to 10 nA. Predict the height of the pulse and measure it to check your prediction.
    • Measure the time to reach half maximal height with the 10 nA stimulus current and compare it with the time for 20 nA. Does stimulus current affect the time to reach half of maximum membrane potential?
  • Question 14: Starting with the settings from question 11, set the membrane capacitance to 4 nF.
    • Measure or recall the time to reach half maximal height with a membrane conductance of 2 µS.
    • Double the membrane conductance to 4 µS. Calculate the maximum pulse height, and measure the time to reach half the maximum height.
    • Halve the membrane conductance to 1 µS. Calculate the maximum pulse height, and measure the time to reach half the maximum height.
    • Look for a pattern and predict the time to reach half maximal height with a membrane conductance of 3 µS. Check your prediction.
• Examine how the potential returns to its resting value after a pulse.
  • Question 15: Start with the settings from question 11. It may help to increase the simulation time to 60 ms.
    • Measure the time from the end of the current pulse to when the membrane potential reaches half, one quarter, and one eighth of its initial maximal value after it begins to fall. Look for a pattern and predict the time required to fall to one sixteenth of the maximal value.
    • Measure the time required to fall to half maximal with a membrane capacitance of 4, and 8 nF. Look for a pattern, and try to predict the time required to fall to half of the maximal value when the membrane capacitance is 5 nF. Check your result.

The Electrical Equivalent Circuit for the Resting Potential

Using the electrical equivalent circuit to derive an equation for the resting potential

The Electrical Equivalent Circuit for the Resting Membrane

$image: RestingPotentialEquivalentCircuit.png|250px|thumb|Figure 5: Electrical equivalent circuit of the resting potential]]

• We developed the entire electrical equivalent circuit by focusing solely on the potassium ions for simplicity. Since this is the ion to which the nerve membrane is primarily permeable at rest, this is a reasonable first approximation. However, as we saw in the previous unit, real nerve cells are permeable not only to potassium, but also to sodium and chloride ions.
• We can represent the other conductances in the same way that we did for the potassium ion channels: as a conductance for those channels in series with a battery, representing the equilibrium or Nernst potential for each ion. The result is shown in Figure 5.
• We can use the circuit diagram to derive an electrical equivalent equation for the resting potential, which can be as useful as the Goldman-Hodkgin-Katz constant field equation that we learned about in the previous unit, if the assumptions that allow the membrane to be represented using these electrical components are valid.
• The total current flowing across the membrane is the sum of the currents flowing through each of the branches of the equivalent circuit. In symbols, we can state this as $

   I_m = I_{\text{K}^+} + I_{\text{Na}^+} + I_{\text{Cl}^-}. \quad\text{(Equation 5)}$

$image: RestingPotentialFluxes.png|200px|thumb|Figure 6: Balanced fluxes of sodium and potassium ions across the membrane at the resting potential]]

• At the resting potential, the net flux of currents across the membrane is zero, even though the net flux of each of the ions is not zero. Figure 6 shows a schematic diagram for two of the three ions, for simplicity.
• If the fluxes were not in balance, the membrane would depolarize or hyperpolarize, and the change in potential would increase the fluxes of ions further from their Nernst potential, and reduce the fluxes of ions closer to their Nernst potential; the net effect would be to change the membrane potential back to the resting potential.
• If the fluxes are in balance, then the current across the membrane is zero, or $I_m = 0$.
• Repeatedly applying Equation 2 to the right-hand side of Equation 5, we obtain $

   0 = g_{\text{K}^+} \left(V_m - E_{\text{K}^+}\right) + g_{\text{Na}^+} \left(V_m - E_{\text{Na}^+}\right) + g_{\text{Cl}^-} \left(V_m - E_{\text{Cl}^-}\right). \quad\text{(Equation 6)}$

• A little algebra (which you are encouraged to do yourself) leads to a solution for the membrane voltage in terms of the conductances to the three ions and their Nernst or equilibrium potentials: $

   V_m = \frac{g_{\text{K}^+} E_{\text{K}^+} + g_{\text{Na}^+} E_{\text{Na}^+} + g_{\text{Cl}^-} E_{\text{Cl}^-}}{g_{\text{K}^+} + g_{\text{Na}^+} + g_{\text{Cl}^-}}. \quad\text{(Equation 7)}$

• In words, the equation says that the membrane potential is a weighted sum of the equilibrium potentials, in which the weighting factors are the conductance to each of the ions.
• Note that if the conductance to any two ions were to go to zero, the equation would simplify to the Nernst or equilibrium potential for the remaining ion.
• In subsequent units, we will refer to the conductance through these channels as the leak conductance, and we will report their average equilibrium potential as the leak equilibrium potential.

With this understanding of the electrical equivalent circuit for the resting potential, it is now possible for you to do some additional explorations of the properties of the resting membrane and its response to current. Here is a simulation of the passive membrane which incorporates all three conductances:

• Answer these questions using the simulation.
  • Question 16:
    • Try setting the sodium and chloride conductances to 0.01 µS and the potassium conductance to 30 µS. What is the membrane resting potential?
    • Try setting the chloride and potassium conductances to 0.01 µS and the sodium conductance to 30 µS. What is the membrane resting potential?
    • From the previous two results (or using Equation 7), predict the resting potential when the sodium and potassium conductances are 0.01 µS and the chloride conductance is 30 µS. Test your prediction.
    • What determines the resting potential when the membrane is much more permeable to one ion than the others?
  • Question 17:
    • Try setting the sodium conductance to 0.01 µS and the potassium and chloride conductances to 30 µS. What is the membrane resting potential?
    • Try setting the chloride conductance to 0.01 µS and the sodium and potassium conductances to 30 µS. What is the membrane resting potential?
    • From the previous two results (or using Equation 7), predict the resting potential when the potassium conductance is 0.01 µS and the sodium and chloride conductances are 30 µS. Test your prediction.
    • What determines the resting potential when the membrane is equally permeable to two ions?