The Resting Potential and Spike Generation

The resting potential and spike generation

The Resting Potential and Spike Generation

This lecture concerns the computational language of the nervous system--the electrical potentials that are used by the nervous system to decide whether or not information should be relayed further along, and finally the signals that carry that information.
1. We talk about the ability of the nervous system to perform logical addition, subtraction, multiplication, and division, and this is entirely accomplished via electrical potentials.
2. Before describing these potentials a word or two should be said about how they are recorded.
  1. Most of the potentials that will be discussed early in the course are measured across the membrane separating the inside of the neuron from the outside (trans-membrane potential).
  2. An electrode filled with a solution that conducts current is lowered into the neuron and the voltage difference between the inside of the cell and the outside of the cell is measured.
    1. This is done by comparing what is measured by the inserted electrode to what is measured by a reference electrode outside of the cell--the potential that we record is always derived by comparing the internal and external state of the neuron.
    2. The circuit is shown in Figure 5.1 (a) of the text.
  3. As one lowers the electrode into an axon or cell body when the cell is in the resting state, it becomes apparent that the inside of the cell is negative relative to the outside.
    1. The potential is approximately 70 mv, although its magnitude varies somewhat from species to species and from neuron to neuron (its magnitude is less that 1/10 volt). -see Figure 5.1(b)
  4. This potential is the resting potential of the cell.
    1. The importance of this potential rests in the fact that it serves as a battery, a source of stored energy, from which the computational neural potentials are generated.
    2. Since it is the starting point of the neuron, all other neural potentials have a -70 mv baseline.

From where does this -70mv battery come? To answer we must examine the behavior of charged particles (ions).

There are two tendencies that govern the movement of ions in a solution or in a gas, and these two tendencies are responsible for the generation of the membrane potential.

First, molecules tend to diffuse so that they are evenly distributed in space.
1. As such, increasing the concentration of some local region causes a net movement of these molecules into neighboring regions until they are evenly distributed.
  - We speak of the ion as running down its concentration gradient (osmotic pressure if there's a membrane) towards chemical equilibrium.
  - There is nothing magical nor volitional about this--it occurs through purely random motion of particles that are in constant motion.
  - The probability of an ion moving from a region of high concentration to one of low concentration is greater simply because there are more ions in the region of high concentration to begin with.
  - Once equilibrium is reached, the random movement of ions maintains it (no net movement).
The second force is electrostatic pressure--compounds placed into solution tend to dissociate into cations (positive) and anions (negative).
1. Ions of opposite charge tend to be attracted to one another, while those of the same charge are repelled.
  - This repulsion/attraction is the basis for electrostatic pressure.
Often electrostatic and osmotic forces are placed in opposition with one another, so that for a given ion there is one force drawing it one way and another force drawing it the other
1. For example, let us imagine some sort of semi- permeable membrane with high concentrations of K+ on the inside and high concentrations of large anions also inside.
2. The membrane is semi-permeable in the sense that it allows the passage of certain ions through channels but not others.
3. Let's assume that the membrane is freely permeable to potassium, but that the anions are too large to traverse the membrane.
4. Electrostatic forces drive the K+ inward because of the negative charge of the anions, while the concentration gradient will tend to move K+ outward.
5. As such, the concentration of K+ inside the cell will tend to be elevated (cannot run down its concentration gradient), but the + and - charges will not be able to balance out either (electrostatic equilibrium cannot be achieved).
6. Electrostatic pressure forces K+ inward and the concentration gradient (laws of diffusion) force it outward.
While the two forces themselves cannot achieve equilibrium, we can speak of electrostatic pressure and the concentration gradient as counteracting one another (electrochemical equilibrium is reached).
1. At this point there is an equal likelihood that ions to which the membrane is permeable will move inward and outward, but a voltage will be recorded across the membrane.
  - Without the semi-permeable membrane, each of the two forces could reach equilibrium and no membrane potential would develope.
If one examines the relative concentrations of ions on the inside and outside of neurons, large differences are observed:
9-10 times more Na+ outside.
20-40 times more K+ inside.
14 times more Cl- outside.
All of the large protein A- inside.

The Nernst equation is an expression that allows one to calculate the membrane potential that exactly counteracts an observed concentration gradient (i.e., when diffusion and electrostatic pressure are in equilibrium).

The theoretical value should equal the value observed when one empirically measures it with electrodes.

V_mv=k log [(Conc. OUT)/(Conc. IN)]; k=58

Cl-	K+	Na+
-58 log 14	58 log (.05)	58 log (9.2)
-58 (1.15)	58 (-1.3)	58 (.96)
-66.5 mv	-75 mv	56 mv

If the cell membrane was permeable to all three ions, then they should all yield the same estimated voltage.

The Nernst equation assumes that electrostatic and osmotic forces are the only factors involved, and that the membrane is permeable to the ions for which the measurements are made.
1. When electrical measurements differ from the theoretical values, some aspect of the assumptions must have been violated.
  - The Cl- equilibrium potential is close to the resting potential, so the assumptions seem valid.
  - The sodium equilibrium potential would be +56 mv, which is greatly different from the resting potential, and this is due to the fact that the membrane is impermeable to Na+.
    - Indeed both osmotic and electrostatic forces are acting to drive sodium inward.
2. The disagreement between K+'s equilibrium potential and the resting potential arises from its role in the active exclusion of Na+ from the inside of the cell.
  - This sodium pump requires metabolic activity (its active rather than passive), and exchanges 3 Na+ for every 2 K+.
  - As such, part of the K+ concentration gradient is due to the pump and not due to counteracting electrostatic pressure--a violation of the Nernst assumptions.
  - The concentration gradient of K+ is larger that that necessary to counteract the electrostatic pressure, so that the theoretical solution overestimates the actual equilibrium potential.
3. Others argue that the discrepancy between the RP and E_k is due to some slight residual permeability to Na+, resulting in a slight inward, positive current

We have shown how the Nernst equation tells us how large of a membrane potential is needed to support an observed concentration gradient.
1. There is a single membrane potential, so the ions for which the assumptions of the Nernst equation are appropriate should all aline themselves on the two sides of the membrane such that osmotic pressure and electrostatic pressure are in equilibrium.
  1. If the concentration gradient does not support the measured membrane potential, then one of the assumptions-- equilibrium, only osmotic and electrostatic forces, and permeability--is violated by the ion in question.
    - For example, the membrane is not permeable to sodium, and the concentration gradient for potassium is due to electrostatic pressure AND the Na+/K+ pump.
2. Usually the permeabilities of membranes to particular ions is not "all or none," so computation of the membrane potential (rather than the equilibrium potential) takes into account the relative permeabilities of the ions in question (Goldman):
  
             P_K[K+]out + P_Na[Na+]out + P_Cl[Cl-] in
  58 log ---------------------------------------
            P_K[K+]in + P_Na[Na+]in + P_Cl[Cl-] out
  or
  
            P_Na E_Na + P_K E_K + P_Cl E_Cl
  V_m----------------------------------------------
            P_Na + P_K + P_Cl
  
  At rest, P_K:P_Na:P_Cl = 1.0:0.04:0.45.
  At the peak of the Action Potential, try 1.0:20:0.45.
All of the principles discussed so far are appropriate for discussion of the other potentials found in the nervous system.
1. If one takes a stimulating electrode and inserts it into an axon, the inside of the cell can either be made more positive or more negative by passing positive or negative current.
  1. We speak of depolarizing the cell when we make it less negative and hyperpolarizing the cell when we make it more negative.
  2. Over a range of current magnitudes, the membrane behaves very reasonably (according to Ohm's Law: E=iR; E=voltage, i=current, and R=resistance)
    1. Over this range we speak of the system as being passive--it responds to the input without supplying any additional voltage besides what is supplied by passing current.
    2. For brief pulses it takes the membrane a short period of time to charge up and charge down, and this is consistent with passive cable properties. (See figure 5.2)
2. As we continue to depolarize the membrane to about -60 mv, something rather spectacular occurs--the resulting voltage change far exceeds what one would expect from the amount of current passed.
  1. The system is no longer passive.
  2. As a matter of fact, the inside of the cell becomes 40 mv positive relative to the outside of the cell.
  3. If we plot the change in membrane potential as a function of time, the resulting waveform is referred to as the action potential, a neural discharge or firing, or a spike
  4. This potential is active, with the energy supplied by the resting potential
  5. Once threshold is reached, the spike always has the same waveform regardless of the amount of current applied.
3. What is the ionic basis of the action potential?
  1. First of all, it can be shown that decreasing the amount of extracellular Na+ can abolish the action potential or reduce its amplitude.
  2. Secondly, by radioactively tagging the Na+ it can be shown that the amount of Na+ increases following repeated discharges.
  3. Tetrodotoxin (TTX), a poison known to block entry of sodium into cells, eliminated the ability of axons to spike.
4. It appears that the depolarization induces the opening of Na+ gates, and this results in sodium running down its concentration and electrostatic gradients.
  1. As a result, the membrane potential moves towards the sodium equilibrium potential (+56 mv), not quite getting there because the gates are opened for only about 1 msec.
    - Sodium activation is followed by sodium inactivation.
  2. Because the inside of the cell is now positive (and permeability to K+ remains high), potassium runs down its concentration and electrostatic gradients and leaves the cell.
    - This is termed potassium activation.
    - This helps to restore the resting potential.
    - There also appears to be a decreased membrane resistance to potassium.
  3. The slight undershot at the end of the discharge is due to the fact that there was excess K+ in the cell to begin with (because of the Na+/K+ pump), so the exit of K+ drives the membrane potential towards the K+ equilibrium potential (which is more negative than the resting potential).
    - Although the membrane is quite permeable to K+ when in the resting state, there appears to be some residual membrane resistance.
    - This allows the Na/K pump to accumulate potassium, otherwise it would simply exit the cell again immediately.
    - During potassium activation there is a decrease in resistance to K+ (relative to that measured in the resting state), so the decreased membrane resistance allows the membrane potential to approach the potassium equilibrium potential.
  4. Finally the membrane potential returns to -70 mv; over a longer time course the pump removes the sodium that has accumulated internally.
5. During Na+ inactivation it is impossible to fire the cell again regardless of how far the membrane potential is driven.
  1. This is the absolute refractory period
  2. During potassium activation, the voltage needed for the axon to reach threshold is elevated--this is the relative refractory period.
6. The action potential only occurs beyond the axon hillock in vertebrates, while some invertebrates show somal spikes.
7. Often one represents the membrane potential in terms of its electrical analogue (a metaphor), with each equilibrium potential represented by a separate battery across the membrane which has the appropriate voltage (E) and polarity for that ion.
  1. In series with each battery is a resistance (R), which is related to the permeability of the membrane to the ion in question (actually we want conductance, G, which is inversely related to resistance--R=1/G).
    - G_K . P_K([K+]out/[K+]in).
    - Thus, the permeability to potassium can be quite high, yet the conductance can be 0 if there is no external potassium.
  2. Lastly, there is a membrane capacitance (C), which is due to the fact that the lipids of the membrane are poor conductors, so they tend to charge and discharge somewhat slowly.
    See Figure from Class showing the circuit diagram.
How the action potential travels along the axon.
1. If the axon were like a simple cable or uninsulated wire, the potential would decrease in magnitude as it traveled down the axon because of the resistance of the axon and because of current leakage through the membrane into the surrounding medium.
  1. The predicted potential (V) at any point along the axon is given by Vo exp[-x / (r_m/r_i)^0.5], where Vo is the starting voltage, r_m is the radial membrane resistance X cm, r_i is the longitudinal core resistance/cm, and x is distance.
    - As such, the potential goes to 1/e at (r_m/r_i)^0.5;the higher the membrane resistance and the lower the core resistance, the further the potential spreads down the cable.
  2. Yet, when we actually record the magnitude of the waveform near the terminal of an axon we see that it has not diminished.
2. In unmyelinated fibers, the action potential invades the neighboring region of the axon, causing it to reach threshold and a new spike to be generated (a "regenerative" process).
  1. As such the spike one sees at the axon hillock is not really the same electrical potential (at least in some sense) as the one at the end of the axon.
3. For myelinated fibers, the action potentials travel with a small decrement (obeying cable properties) between the Nodes of Ranvier.
  1. Only at the nodes is there exposure of the membrane to extracellular fluid, so Na+ can only enter the cell to generate a new discharge at the Nodes.
  2. This is termed saltatory conduction, and it greatly increases the conduction velocity of the axon because new spikes do not have to be continuously generated.
  3. The insulation of the myelin is sufficient that the threshold for spike generation is always reached at each Node of Ranvier as long as an initial spike is produced at the hillock.
    - The myelin sheaths have a high resistance and low capacitance that makes the electrotonic current tend to flow down the fiber to the next node rather than leak back across the membrane.
  4. Saltatory conduction is both faster and preserves metabolic energy because there is less need for the Na/K pump.
4. Conduction is generally in one direction along the axon because of the refractoriness of the previously depolarized membrane.
The membrane of the neuron consists of two lipid layers with protein molecules spanning them.
1. Some of the protein molecules appear to be the cites of the Na+/K+ pump.
  1. These membrane proteins are called Na-K-ATPase, and they react with ATP to cause the movement of Na+ outward and K+ inward (3 for 2).
    - The reaction only occurs if Na+ resides near the inner cell membrane.
2. In addition, proteins in the membrane are thought to compose the sodium and potassium channels.
  1. Although they have not been isolated, the two channels do appear to be separate.
    - TTX poisons Na+ channels only, eliminating action potentials (the early current).
    - TEA (tetra-ethyl-ammonium) slows the return to the resting potential by blocking the K+ channels.
  2. The assumption is that the proteins change configuration dependent upon the membrane potential, so that there are opened and closed states.
  3. At least for sodium, there may be a limited number of carrier molecules at the pores so that sodium activation is of a limited duration.