For the next couple of weeks, I want to discuss with you the electrical properties or nerve cells, and the various types of electrical signals that nerve and muscle cells use to communicate. All cells have an interior milieu that is negatively charged with respect to their environment. That is, if you could take a voltmeter, put one lead inside any cell--plant, animal, bacterial, etc.--and put the other lead in the sap, the bloodstream, or the pond, the voltmeter would give a negative reading. We call this difference in charge between the inside and outside of cells the membrane potential. All cells maintain this electrical charge difference across their membranes by using ion "pumps"--proteins in the cell membrane that use the energy stored in ATP to transport ions across the membrane. Usually we don't think of electrical potentials in all cells--just in neurons--especially in this course. What makes neurons special? The difference between electrically excitable cells like neurons, and other cells, is that neurons have special mechanisms for actively changing their membrane potential, which other cells don’t.

How does the membrane potential arise in the first place? It comes about because there is an asymmetric distribution of ions across the membrane of cells. I mean two things by that. First, the concentration of different ions is different inside the cell (that is, in the cytosol) and outside in the fluid in which the cell resides. Thus K⁺ ions are more concentrated inside cells than outside and Na⁺ ions are more concentrated outside than in. Second, it is the plasma membrane that is responsible for establishing and maintaining this difference in ionic concentrations. The membrane itself is the barrier to free ion flow between the inside and outside of the cell. The resting membrane potential, abbreviated V_m, of a cell that is not undergoing any electrical signaling is typically -60 +/- 20 mV; that is, the voltmeter I described above would read -60 mV, if it were sensitive enough to detect a difference that small. In fact special kinds of voltmeters have been invented that can detect these small difference, as you know from our measurements in lab.

See Figure 2.1 in Purves et.al., Neuroscience

Because large amounts of energy are required to separate positive and negative charges--opposite charges attract and thus it takes energy to keep them apart--a large DV_m can arise from the movement of small numbers of ions through the membrane. The movement of small numbers of ions is very difficult to detect by chemical means, even using radioactive tracers, and only electrical techniques allow us to detect these changes reliably.

Changes in the membrane potential of nerve cells are the basis for the electrical signals that the nervous system uses to communicate. Some signals travel long distances in the form of action potentials and some electrical changes begin and die out locally like ripples in a pond caused by dropping a rock. But all electrical signals arise ultimately from the fact that cells are polarized--i.e., that they have a negative interior charge.

Electrical signaling in the nervous system was first observed over 200 years ago (1791) by Galvani, who showed that isolated frog’s legs would twitch if shocked with an electrical current, and who discovered in turn that the frog could generate its own bioelectric current. Galvani’s conclusion was disputed by another Italian scientist, Volta, who claimed that the current Galvani measured was an artifact of his measuring system; that the electricity was generated by his electrodes, which were made of different metals, rather than the frog’s legs. (Volta was right that junctions between different metals could generate an electrical current; it’s the basis for most thermostats). As a result Volta won that argument, and Galvani retired from science, but history has proved Galvani right. Biological tissues can generate electrical signals.

The membrane potential, i.e., voltage difference, between the inside and outside of the cell arises from two simple "facts":

1. Some ions are differentially distributed across the plasma membrane, so that their concentration inside and outside the cell isn’t the same.

2. The membrane is more permeable to some of these ions than others.

I’m going to spend some time trying to convince you of this. There's nothing magical or mysterious about the resting membrane potential--it's a direct consequence of the properties of biological membranes.

Consider 3 situations: We have a water-filled container with a partition in the middle, and we add K⁺ Cl^- to one side of the partition.

if P_K=P _Cl- = 0, Vm =?

if P_K = P _Cl- = ° , V_m = ?

if P_K = °; P _Cl- = 0, Vm = ?

This is called a semi-permeable membrane because the membrane is selectively permeable to some, but not all, ionic species. (e.g., pH electrode glass).

What will happen in this last case? At first K⁺ will cross the membrane from the more concentrated to the less concentrated side, because there’s a thermodynamic tendency (entropy) for K⁺, or any chemical substance, to have an equal concentration in all regions to which it has access. Will that happen in this case? Why or why not? It won’t because K⁺ has the opposite charge from Cl^- . So as K⁺ diffuses through the membrane, and Cl^- doesn’t, this will create a net charge difference on the two sides of the membrane--and the left side will be negative with respect to the right; that is, there will be more negative than positive charges on the left side and more positive than negative on the right. How long will K⁺ have a net movement from right to left? Until the electrical force that tends to draw it from right to left (charge attraction) = the concentration (diffusion) tendency impelling it to move from left to right. That is, when the electrical energy (measured as potential, in volts) is equaled by the chemical energy (measured as concentration or "activity"). When this occurs the net movement of K⁺ = 0; though individual ions may pass through the membrane in both directions, the movements will balance exactly. Now, a situation in which the net change over time is zero is in equilibrium. So for a given ionic concentration difference, for one species of ion, there will always be a unique electrical potential that exactly balances the chemical tendency of the ion to diffuse from areas higher to lower concentration. This electrical potential is called the equilibrium potential for the ion, written as E_ion, thus the equilibrium potential for K⁺ is defined as E_K. Obviously, the amount of electrical energy required to balance the diffusion of ions depends on the steepness of the gradient, i.e. on the relative concentrations of K⁺ on the two sides of the membrane. We can represent this concentration gradient as a ratio:

[K⁺]_right/ /[K⁺]_left or [K]_outside/[K⁺]_inside.

Thus E_K is proportional to [K]_out/[K]_in. The expression that relates these two variables was developed by tthe German Physical Chemist Nernst, and first applied to biology by Bernstein: It is

it is

where R = 8.3 joules/mole; z = charge of ion; F = 96,500 coulombs/mole of ion^oK; T = temperature in ^oK.
F is called Faraday’s constant; R, the ideal gas constant

If T = 20^oC = 293^oK, which is roughly the temperature of a cool room, then,

Or

In order to determine the Equilibrium Potential for an ion, you need to know the concentration of the ion inside and outside the cell, which can be measured by a variety of chemical means. The values for Mammalian Muscle are:

These data confirm that:

1. Ions do not have the same concentration inside and outside cells, which implies that the cell membrane is selectively permeable to ions. For example, the membrane is essentially impermeable to proteins, ATP, etc.

2. Different ions are distributed differently. e.g. K⁺ has a concentration tendency to diffuse out, but an electrical tendency to diffuse into cells. Cl^- has a concentration tendency to diffuse in, and an electrical tendency to diffuse out. Na⁺ has a concentration and an electrical tendency to diffuse in--thus, sodium ions, at least, are not at equilibrium. We don’t know if potassium or chloride are, but we can calculate this from the Nernst equation.

In mammalian muscle at room T E_K = 58log4/140 = 58 (-1.54) = -90 mV.

That is, if V_m = -90 mV, then there is no net diffusion into or out of the muscle cell.

The German scientist, Julius Bernstein, believed that the membrane of excitable cells like neurons and muscles was permeable only to K⁺, just as in the imaginary example I drew earlier. If that were true, then K⁺ would move in one direction or another across the membrane until it reaches its equilibrium state. At that time the V_m would be identical to E_K. Thus, if you knew the concentration of potassium inside and outside the cell you could predict V_m.. How could you test Bernstein’s hypothesis that the muscle membrane is permeable only to K⁺?. Take a couple minutes. One was is to vary the concentration of K⁺ outside the cell and then plot Vm as a function of [K]_out. If V_m = E_K= 58log [K]_out/[K]_in = 58log [K]_out - 58log [K]_in, then V_m vs. log[K]_o should give a straight line, providing that [K]_in is relatively constant.

See Figure 2.5b in Purves et. al., Neuroscience

In some cells, such as glia, one sees a linear relationship between Vm and [K]o, suggesting the membrane is in fact selectively permeable only to potassium. In neurons, however, the value of Vm deviates from the calculated E_K at in the low range of V_m, <-40mV. For example, the normal resting potential of mammalian muscle is around -75 mV, not -90 mV as predicted from the potassium equilibrium potential. What does this imply about Bernstein’s hypothesis? (Pause).

This result must mean that the hypothesis is false; the membrane cannot be permeable to potassium. This implies that the membrane is permeable to other ions as well. As you can see, the major other ions in mammalian muscle are Na⁺ and Cl^-. Could the membrane also be selectively permeable to one or both of those, so that they also affect the membrane potential? Let’s compute the equilibrium potential for these ions as well.

E_na = 58 log [142]/[12] = 58 x 1.07 = +62 mV.

E_cl = -58 log [120]/[4] = -58 x 1.48 = -86 mV, almost the same as E_K.

Note that at a membrane potential of -75 mV, none of these ions is at equilibrium. Therefore, any ion to which the membrane is permeable is not in equilibrium which means that there must be net movement of the ion across the membrane. For instance, sodium must be moving into the cell, if the membrane is permeable to sodium. Cl- must be moving into the cell, and K⁺ must be moving out at a membrane potential of -75 mV. Because of this complex movement of ions, it’s relatively difficult to predict the membrane potential of the cell from the equilibrium potential of the various ions to which the cell is permeable.

Nevertheless, a theory to describe this situation was developed in the 1940s by Goldman (who called it his constant field theory) and his theory was used by Hodgkin and Katz to derive a way to predict the Vm of neurons. It’s now called the Goldman-Hodgkin-Katz (GHK) equation.

Here P is defined as the permeability of the membrane to the ion. Since this is a pretty complex equation, I first want to consider an imaginary case where the permeability of the membrane to Cl- = 0, in order to simplify the calculation. Then the Cl- terms drops out of the equation. Now let’s define a ration b = P_K/P_Na. In this case then,

Vm = RTln[K]out + b[Na]out
F [K]in + b[Na]in

For mammalian muscle, b ~ 0.02; i.e. Pk = 50 Pna.

Then at 20^o C,

Vm = 58 log{4 + 0.02(142)} = 58 log{4 + 7.8} = 58log 6.8 = 58 x -1.3 = -76 mV
140 + 0.02 (12) 140 + .24 140.24

i.e., according to the GHK equation, introduction of a slight permeability to Na+ in a membrane that’s mostly permeable to K+ has the effect of shifting the Vm away from E_K and toward E_na..

What would be the effect of varying the concentrations of the various ions on Vm? We can use the

GHK equation to compute it.

a. If we increase Ko by tenfold, to 40 mM, then E_k = 58log[40/140] = -31.6 ~ -32 mV. (Vs -90 mV). Note Æ = 58 mV.

Vm = 58 log (40 + 0.02(142) = 58 log 42.8 = 58 x -0.51 = -30 mV
140 + 0.02 (12) 140.24

b. If we reduce Nao by tenfold to 1.2 mM, E_Na = 58 log 142/1.2 = 58 log 118 = 58 x 2.07 = +120 mV (vs +62 mV previously).

Vm = 58 log 4 + 0.02(142) = 58 log6.8 = -76 mV
120 + 0.02(1.2) 140

What these calculations show is that changing [K⁺] will have a large effect on Vm, but changing [Na⁺] has a small, virtually negligible effect. Indeed , over much of the possible physiological range of [K⁺], the Vm ~ E_k. For example in the calculation we just did, -30mV ~ -32mV. In fact, then Bernstein was nearly right. Most excitable cells act as if they are selectively permeable only to K+ except at very negative values of Vm. When Vm is very different from E_Na, then Na leaks into the cell at a fast enough rate that Vm isn't = E_k.

Now if it's true that Vm is determined only by the concentration of ions on the two side of the membrane, and by the permeability of the membrane to ions, then one should be able to alter the Vm predictably by changing the gradients.

This hypothesis was tested experimentally in 1960 by Baker, Hodgkin, and Shaw, using squid giant axon. They knew that the normal resting potential of the axon, in sea water, was -60 mV and they knew the concentration of K⁺, Na⁺ and Cl^- inside the axon and in the sea water. So they took the giant axon and extruded the axoplasm, and washed out all the interior contents. (Draw on board). Now they could experimentally set the concentration of ions inside and outside the axon.

If they put the normal conc. of ions inside and outside, Vm = -60 mV.

If they put identical conc. inside and outside Vm = 0 mV

If they put the normal inside conc. outside and vice versa, Vm = + 60 mV.

This expt provided direct verification that Vm is direct consequence of the unequal distribution of ions, the membrane permeability, and nothing else.

Why is Vm constant? At the resting potential we calculated earlier, none of the three major ions were at their equilibrium potentials. If Vm > Ek, then K+ will leave the cell, tending to reduce Vm.

If Vm < ENa, then Na+ will enter the cell, tending to raise Vm. So at the resting potential, there are some ions that have a net movement into the cell and some that have a net movement out of the cell. Shouldn't that change the charge distribution and thus Vm?

Recall Ohm's Law: V = IR; where V = electrical potential (in V), I = current (in A), and R = resistance (in Ohms).

Thus DV =DIR. If there is a change in current over time, there should be a change in membrane potential as well. But since it has been observed experimentally that DV = 0, then DI = 0. In short there must be no net flow of charge across the membrane; or SIions = 0). If only Na⁺ and K⁺ are carrying charge through the membrane, then I_Na + I_K = 0, or I_Na = -I_K. That is, the net movement of Na⁺ ions into the cell must be exactly balanced by the net rate of movement of K⁺ ions outward. This situation is not a true equilibrium, but a kinetic steady state. In short the small net exit of K⁺ and the small net entry of Na+ won't have a significant effect on [K⁺] and [Na⁺], and thus there will be a negligible effect on Vm, which will remain relatively constant. However, if the net movement of Na⁺ into the cell and K⁺ out continues for a long time, and there is not process that opposes it, then eventually both [K⁺] and [Na⁺] will change substantially, and Vm will also change, according to the GHK equation. Thus the "resting" membrane potential is unstable (non-equilibrium), but is the only "stable" state for a membrane permeable to many ions (except when Vm = 0, because all ions have equal conc. inside and out).

How do these basic features of cells--unequal distribution of ions and selective permeability--come to exist? The unequal distribution arises from two different processes. One that we’ve already briefly touched on is the so-called Donnan equilibrium, which just states that if the membrane is impermeable to some ions (e.g., proteins, ATP), then the ions to which the membrane is permeable will distribute themselves in such a way as to equalize the positive and negative charges on the two sides of the membrane, which will inevitably lead to unequal distribution. However, this alone isn’t enough both because Donnan forces can create a difference in osmolarity that could cause the cell to swell or shrink fatally, and because, as I already mentioned, the resting cell is in disequilibrium, with K+ tending to diffuse out, and Na+ tending to diffuse in. So the cell uses some of its own metabolic energy (ATP) to restore the ion gradients; that is, as ions leak into and out of the cell, the cell moves them back to their original location using the energy stored in ATP. This is accomplished by an integral membrane protein that is both a transporter protein and an enzyme, called the Na⁺/K⁺ ATPase (and sometimes referred to as the "sodium pump"). This protein breaks down ATP to ADP and uses the energy released thereby to move Na⁺ out of the cell, keeping [Na⁺]_in low, and moving K⁺ back into the cell, keeping its concentration high. In both cases Na⁺ and K⁺ are moving against their "electrochemical" gradients, so this would not happen spontaneously without the input of cellular energy, hence the need for ATP.

One can show directly that this protein is essential for maintaining the resting Vm by removing ATP (how?--by poisoning mitochondria), or blocking the Na/K ATPase using ouabain, a drug that binds directly to the enzyme and inhibits it.

In other words all cells convert chemical energy in the bonds of ATP to electrochemical energy (ion gradients) by the use of an enzyme. The activity of this enzyme seems to be exquisitely sensitive to [Na⁺]_in--it works faster when [Na⁺]_n is high, and more slowly when it is low.

Why are biological membranes selectively permeable to ions? Recall the prevailing fluid mosaic model of membrane structure--draw. Membranes consist primarily of a lipid bilayer with imbedded and associated proteins. Are lipids likely to provide selective permeability to ions? No. The lipid interior of the membrane is very hydrophobic while the ions are surrounded by shells of H₂O and are highly charged. They cross artificial lipid bilayers (pure lipid) poorly or not at all. Biology membranes are much more permeable to ions (about 10,000 times) than pure lipid bilayers which suggests that it’s the protein in biological membranes that’s responsible for selective permeability.

Proteins can form hydrophilic "pores" in bilayers that allow ions to pass through and can also act as carriers. These proteins can be pretty specific and selective about which ions they let through and which they exclude, as we’ll see later in the course. So "ion channels" often can discriminate between + and - ions, between +1 (monovalent) and +2 (divalent) ions and even between different +1 ions, such as K⁺ and Na⁺.

Finally, I want to come back to a brief discussion of the role of Cl^- in setting the resting membrane potential. We ignored it when I did the calculation last time, and considered only Na⁺ and K⁺. Though we pretended that Cl^- was irrelevant, real cells do have a differential distribution of and some permeability to Cl^-, so that isn’t really a legitimate assumption. However, for most practical purposes Cl^- can be ignored because it seldom contributes much to the resting potential. Many cells don’t have an active transport system for Cl^-, so Cl^- ions "passively" redistribute themselves until V_m = E_cl. . While K⁺ and Na⁺ are actively distributed by an enzyme that helps in turn to set the membrane potential, that’s often not true for Cl^-. Even when cells do have active Cl^- transport, so that E_Cl is different from V_m, usually E_Cl is still close to V_m, so that the presence of Cl^- permeability doesn’t affect the resting potential by more than a few mV, and is thus a relatively minor component of the system.

To summarize so far:

a. Metabolic energy (ATP) and an unequal distribution of impermeable ions is used by cells to establish an unequal distribution of permeable ions.

b. The properties of biological membranes--relative impermeability to ionic species in general and selective transport of different ions by different proteins give rise to "selective permeability".

c. These two facts and these alone account for the existing of a resting membrane potential in cells. Thus chemical energy (in ATP) is converted into electrochemical energy in the form of an ion gradient. (Do you know of any other circumstances where electrochemical energy and chemical energy can be interconverted?)