Biology 304 February 8, 2000

Lecture notes:

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I. What is a resting potential?
 

Neurons and glia, like all other living cells, have a resting membrane potential.

This means that, in a cell at rest, you can measure a potential difference, or a voltage, across the cell membrane.
This potential difference results from a separation of charge across the cell membrane: there is an excess of positive charges on the extracellular face of the membrane and an excess of negative charges on the inside of the cell membrane. Therefore, under resting (unstimulated) conditions, the inside of the cell is electrically negative relative to the outside.

II. Boundary Conditions

BEFORE WE BEGIN TO DISCUSS THE BASIS OF THE CELL RMP, WE NEED TO ESTABLISH THE BASIC BOUNDARY CONDITIONS UNDER WHICH THE SYSTEM IS OPERATING

For a cell at rest, three major requirements must be met in order to maintain a steady state:

A. Intracellular and extracellular solutions must be electrically neutral.
 

In other words, the total number of anions must equal the total number of cations in the fluids on either side of the cell membrane. It should be noted that the small excess of anions at the inner surface of the cell membrane and cations at the outer surface that create the resting potential is negligible relative to the total number of ions in the bulk intracellular and extracellular solutions. (in a cell with a radius of 25 mm with 120 mM total solute and a RMP of -85 mV there are about 100,001 anions for every 100,000 cations.) If this were not the case, electrical neutrality would be violated and forces of electrical repulsion between like charges would blow the solution apart.

B. The cell must be in osmotic balance. The total number of osmotically active particles on either side of the cell membrane must be equal.
 

Otherwise the cell would swell or shrink as water flowed across the cell membrane until osmotic balance was achieved.

C. There must be no net flux for any ion into or out of the cell.
 

III. Genesis of the resting membrane potential.

The resting membrane potential, V_m, arises because there is an uneven distribution of ions inside and outside the cell and because the cell membrane shows differential permeability to those ionic species which are present.

In most cells, K⁺ and impermeant organic anions are present at higher concentrations inside the cell than outside, while Na⁺ and Cl^- are present at higher concentrations outside the cell. This means that there are concentration gradients for all of the major ions present.
 
 

IF THE CELL MEMBRANE WERE JUST A LIPID BILAYER WHICH WAS COMPLETELY IMPERMEANT TO ALL IONIC SPECIES PRESENT, THERE WOULD BE NO V_m NO MATTER HOW LARGE THE CONCENTRATION GRADIENTS WERE

A. Nernst equation
 

1. The cell membrane is composed of a mosaic of lipids and proteins.

The membrane lipids are hydrophobic and ions cross the membrane through specialized proteinaceous channels. Channels may be gated or ungated and may be highly specific for a given type of ion or relatively non-selective.

A CELL EXCLUSIVELY PERMEABLE TO K
 
 

Let's say that we had a cell that was exclusively permeable to K⁺ and to no other ion.

What will happen?

Since the K⁺ concentration is so much higher inside the cell than outside, K⁺ will diffuse out of the cell down its concentration gradient. (Like the color and flavor molecules leaving a tea bag) As K⁺ diffuses out of the cell, it creates a separation of charge. Since like charges repel and unlike charges attract, the build-up of positive charges on the outer surface of the membrane opposes the continuing flux of K⁺. In other words, a potential difference (or an electrical gradient) is established.

The point at which the potential difference opposing potassium efflux exactly balances the concentration gradient which tends to drive K⁺ out of the cell is known as the K⁺ equilibrium potential (E_K). At E_K there is no net passive movement of K⁺ into or out of the cell.

You can get an intuitive feeling for this by imagining the concentrations as hills of different sizes and thinking of the equilibrium potential as the force you'd need to exert to keep a boulder from rolling down the slopes between them:

2. The equilibrium potential for K⁺ (or any permeant ion) is therefore determined by the concentration gradient and is proportional to the difference between the logarithms of the concentrations on either side of the membrane:

E_K = k (ln[K]_out - ln [K]_in)

k is defined by RT/zF. R is the gas constant (8.314 volts coulombs / ⁰K mole), T is the temperature in degrees Kelvin, z is the valance of the ion and F is the Faraday (96,500 coulombs/ mole). So

E_K = (RT/zF) (ln [K]_out - ln [K]_in)

or:

This is the Nernst equation.

Know it.

At room temperature (20⁰ C = 293⁰ K) RT/zF for a monovalent cation is about 25 mV. To work with logs (base 10 logarithms) rather than natural logs, this is multiplied by the conversion factor 2.3, giving a value of 58 mV.

(The logarithm base " a" of a positive number x (log_a of x) is the number to which "a" must be raised to the power of to give x. So the log₁₀ of 10 is 1, and the log₁₀ of 100 is 2, and the log₁₀ of 1000 is 3 and so forth)

So, if [K]_o=50, and [K]_i=5; E_K =58mV log 10 = 58mV

For the sample cell above (similar to a frog neuron):

The Nernst equation can be used to find the equilibrium potential for any permeant ion whose extracellular and intracellular concentrations (actually activities) are known.

For Na⁺, in the cell above,

It should be noted that Cl^- is an anion. Therefore z = -1 and

3. If you displace the membrane potential from E_K, the concentration gradient will no longer be equal to the electrical gradient and K⁺ will move so as to restore the equilibrium.

So, if you depolarize the cell from E_K (make it less negative) and have a smaller potential difference, K⁺ will tend to move out of the cell, down its concentration gradient. And if you hyperpolarize the cell (make it more negative -- i.e. a greater potential difference), K⁺ will follow the electrical gradient into the cell.

4. You can do the experiment of changing [K]_o, while measuring V_m with a microelectrode. For glia, the resting membrane potential shifts with changes in [K]_out as would be predicted by the Nernst potential. i.e. V_m (RMP) = E_K.

However, for neurons there are small, but definite deviations from Nernstian behavior, particularly at low [K]_out. This suggests that, for neurons, the K⁺ concentration gradient is not the only factor contributing to the resting membrane potential.

B. Goldman - Hodgkin - Katz equation.

In fact, most neurons have a resting permeability to Na⁺ and Cl^- as well as to K⁺, and the permeability to Na⁺ is largely responsible for the deviations from Nernstian behavior.

1. Going back to our model cell, what happens if we insert a few channels into the membrane that are permeable to Na⁺? We determined that E_Na for this cell was around +34 mV. That means that at a V_m of -85 mV not only is there a big concentration gradient for Na⁺ into the cell, but an electrical gradient as well. So there will be a strong driving force acting to drive Na into the cell. This will tend to depolarize the cell towards E_Na.

2. As the cell becomes depolarized, however, K⁺ will no longer be at its equilibrium potential, so it will start moving out of the cell. At some intermediate potential, a balance will be struck so that the influx of Na⁺ (inward Na⁺ current) will equal the efflux of K⁺ (outward K⁺ current) and the membrane potential will be in a steady state. For a cell with multiple permeant ions, V_m is determined both by their concentration gradients and their permeability and is described by the Goldman - Hodgkin - Katz equation (constant field):

In many neurons, changes in [Cl]_o have little effect on resting potential, and this term is frequently ignored. In squid giant axon, where permeabilities to Na⁺ and K⁺ have been measured, the ratio of resting K⁺ to Na⁺ permeability is about 1: 0.04 so V_m is much closer to E_K than to E_Na.

This dependence of Vm on the relative permeabilities of Na⁺ and K⁺ means that one can shift V_m to any value between E_K and E_Na simply by changing the relative permeabilities to these two ions, without any change in the concentration gradients
 

3. To understand how the the combination of different ionic permeabilities and concentration gradients influence V_m, it may be helpful to think in electrical terms and to remember Ohm’s law:

i = E/R = Eg

In the present case, i_K is the current carried by K⁺ and i_Na is the current carried by Na⁺. E, voltage, is given by the driving force, the difference between V_m and E_K or E_Na. Conductance, g, which is related to permeability, is 1/R.

So i_Na = g_Na (V_m - E_Na) and i_K = g_K (V_m - E_K).

If the membrane potential is stable, i_Na = -i_K.

If gK >> gNa, for i_Na to equal - i_K, driving force on Na⁺ (ie the displacement of E_Na from V_m) must be greater than the driving force on K⁺ (and in the opposite direction).

Note: Permeability is inherent in the properties of the membrane, whether or not the permeable species is present. But you can only have conductance, a measure of electrical flow of current, for a given ion if that ion is there.

IV. Maintenance of the Na⁺andK⁺concentrationgradients.

Since neither Na⁺ nor K⁺ are at their equilibrium potentials at V_m, there will be a steady leak of K⁺ out of the cell and Na⁺ in. Over time, this will lead to a dissipation of the concentration gradients and a loss of the membrane potential. AND WOULD VIOLATE OUR THIRD BASIC CONDITION FOR CELLS IN A STEADY STATE The ionic concentration gradients are maintained by active transport of Na⁺ and K⁺ by the Na⁺/K⁺ ATPase, more familiarly known as the Na pump.
 
 

A. The Na pump actively transports Na⁺ and K⁺ against their electrochemical gradients. The energy for this is derived from the hydrolysis of ATP into ADP and P_i. Three Na⁺ ions are bound internally and two K⁺ ions are bound externally for every molecule of ATP hydrolyzed. Phosphorylation of the enzyme leads to a conformational change of the enzyme and a change in binding affinity so that the Na⁺ ions are released outside of the cell and the K⁺ ions are released inside of the cell. The enzyme is then dephosphorylated so that its ready to go again.

B. Because 3 Na+ are extruded for every 2 K+ brought in, the pump is electrogenic and Vm is slightly more negative than it would be based upon the GHK equation alone.

C. The pump also plays a crucial role in maintaining the osmotic balance of the cell. As previously mentioned, the cell contains impermeant intracellular anions. Without the pump, intracellular osmolarity would exceed extracellular osmolarity at electrochemical equilibrium, water would rush into the cell and the cell would burst. This is known as a Donnan equilibrium. The pump maintains Na+ as an effectively impermeant extracellular ion which counteracts the effects of the impermeant intracellular anions; since more Na is pumped out than K pumped in osmotic balance is restored. From this perspective, the resting membrane potential and the whole neuronal signalling apparatus can be viewed as a side effect of the cell’s mechanism to avoid osmotic lysis.

Slides

The assigned chapter in your text,

Nicholls JG, Martin AR and Wallace BG (1992) From Neuron to Brain, Sinauer, Sunderland, MA, p66-85

gives an excellent treatment of this theme, and is the recommended reading

Those who feel uncomfortable with electrical circuitry may also want to read the appendix "Current Flow in Electrical Circuits" on p 689 -689

You may also wish to consult:

Koester J (1991) "Membrane potential" in Principles of Neural Science, ER Kandel, JH Schwartz and TM Jessell (Eds), Elsevier, NY, p81-94.

And to really get the flavor:

Hodgkin AL and Horowicz P (1959) "The influence of potassium and chloride ions on the membrane potential of single muscle fibers" J. Physiology 148:126-160.

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