Fig. 3-1. Orientation of lipid molecules in an aqueous medium. When placed in water, lipid molecules line-up on the surface with the polar ends in the water and the nonpolar ends out of the water.
Fig. 3-2. Lipid bilayer between two aqueous phases. Lipids place in a solution containing two phases (like intra- and extracellular fluids) form a bilayer with polar ends away from each other and nearer the aqueous phases and nonpolar ends toward each other and away from the aqueous phases.
Fig. 3-3. Fluid mosaic model of membrane. Lipid bilayer forms a fluid sea in which protein molecules "float" either on the surface, as extrinisic or structural proteins, or with the fluid, as intrinsic proteins. Channels are thought to be intrinsic proteins or aggregates of such proteins, in contact with both intra- and extracellular fluids, surrounding a water- filled pore through which ions pass. Voltage-gated channels open when the transmembrane potential changes, and chemically gated channels open when they are contacted by a transmitter molecule, indicated here as acetylcholine. (de Robertis E: Science 171:963-971, 1971)
What distinguishes one membrane from another is the nature of the proteins in the membrane. In nerve, there appear to be five functional types of membrane proteins. These are: structural elements, enzymes, receptors, channels or pores, and pumps. The structural proteins may be involved in holding cells together at junctions, stabilizing other proteins within the membrane, or maintaining some aspects of subcellular structure. Enzyme proteins facilitate chemical reactions, and presumably those located in or on the membrane are involved in chemical reactions involved in membrane function. The membrane of the cell must be able to recognize particular molecules (e.g., transmitter substances) and bind selectively with those molecules. Receptor proteins provide binding sites, some of high affinity and selectivity. Charged molecules do not pass readily through lipid media; thus, the membrane must contain specific channels through which ions can move to enter or leave the cell. These channels are provided by certain protein molecules or groups of molecules. As we shall see shortly, the concentrations of different ions are not the same inside and outside the cell; yet the membrane is leaky to these ions. In order to maintain these differing ion concentrations that are fundamental to neuronal function, pumps expend energy to expel certain ions from the cell and bring other ions into the cell. The pump proteins are intimately involved in this process.
It is not necessary that a given protein molecule serves only one of the functions above. In fact, at the neuromuscular junction the same protein is thought to act as a receptor for acetylcholine and a channel for both Na+ and K+ ions. We shall have more to say about channels and pumps later.
The nerve membrane is permeable to a large number of compounds and ions. In general, lipid-soluble substances move through the membrane more rapidly than lipid-insoluble substances, and smaller molecules move through more rapidly than larger ones. Medium and large molecules move through the membrane only if they are lipid-soluble, but small molecules move through whether they are lipid-soluble or not. The existence of pores or channels through the membrane was postulated to explain how lipid-insoluble substances could diffuse through a lipid membrane in which they cannot dissolve. No one has yet seen a pore under the microscope, but it seems likely that the pores are tortuous pathways through protein molecules or between them. It appears that there is a spectrum of pore diameters and permeabilities, some admitting most ions less than a certain maximum size and others selectively permitting only particular ions to pass through. Divalent ions usually pass through less readily than monovalent ions. For some cells, the permeability of the membrane to monovalent ions is
Cs+ > Rb+ > K+ > Na+ >Li+ and I- > Br- > Cl- > F-;
for other cells, the sequence is different. These sequences are not determined strictly by molecular weight or diameter; Na+ ions are 30% smaller in diameter than K+, yet the membrane of most cells is 20-30 times more permeable to K+.
This surprising result is explained in part by the tendency of ions to be hydrated. There is an attractive force between the positive charge of the ion and the negative end of the water dipole, so that water molecules accumulate around the ion, forming a shell. The number of water molecules attracted to an ion is a function of the ion's charge density. Both sodium and potassium have the same charge, but because sodium is smaller it has a larger charge density. That means sodium attracts more water molecules. Therefore, the radius of the hydration shell for sodium is larger than that for potassium; it has an effectively larger diameter. As we shall see, the two ions share some channels and also have separate channels in the membrane, and the ease with which they pass through the channels is a function of interactions with parts of the channel structure and the positions of water molecules within the channel.
|Electrical Properties of Membranes|
The nerve cell membrane behaves like a simple electrical circuit containing resistance and capacitance. To understand its behavior it is important to review the behavior of electrical circuits. First, let's recall a few definitions. A voltage or potential difference is a separation of unlike charge in space; the greater the amount of charge separated, the larger the voltage, V, and the greater the tendency for the charges to flow toward each other. Voltage is always measured at one point with respect to another point. There cannot be a voltage at one point in space. The unit of voltage is a volt, V. A flow of electrical charges is a current, i, measured in coulombs/sec or amperes, A. Resistance is a measure of the difficulty with which current flows in a circuit; the greater the difficulty, the greater the resistance, R. The unit of resistance is the ohm, upper-case omega. The reciprocal of the resistance is called the conductance, g, a measure of the ease with which a current flows in a circuit. The unit of conductance is the siemen, S (the older term was mho).
Current, voltage, and resistance are related to each other by Ohm's law: i = V/R. This relationship for simple resistive circuits says that for a given resistance, the current increases linearly with the voltage or, alternatively, that the current driven by a particular voltage depends upon the resistance. For those not familiar with electrical concepts, a useful analogy might be carrying rocks up a hill. Current would then be like the number of rocks you could carry per trip, resistance like the steepness of the hill, and voltage like the amount of energy you have available to use in carrying the rocks. Clearly, the less steep the hill (the smaller the resistance) and the higher your energy level (the greater the voltage), the greater will be the number of rocks you can carry up the hill per trip. Of course, Ohm's law can be rewritten i = gV, where g = 1/R. In this form of the equation, current is a direct function of both the voltage and the conductance. We will have occasion to use this form later.
Fig. 3-4. A simple resistance circuit. A battery, V, is used to apply a voltage of 10 V across a resistance, R, of 10 ohms. The voltage and current in the circuit are measured using a voltmeter, Mv, in parallel with R and an ammeter, Mi, in series with R. In this case, the voltmeter reads 10 V, with polarity as indicated by +,- signs, and the ammeter reads 1 A.
That current can be calculated from Ohm's law as follows:
Let V = 10 V.
And let R = 10 ohms.
Then, i = V/R = 1 A.
The current flowing through the resistor produces a voltage drop (a difference in electrical potential) across the resistor that can be measured by the meter, Mv, a voltmeter. We can assume that the meter is perfect (real meters seldom are), i.e., no current flows through it, and it does not influence the operation of the circuit; thus, the entire voltage, V, is imposed across R. The measured voltage drop across the resistor, R, will be 10 V with the polarity of the voltage as shown. The wires of the circuit have a very low resistance, so there will be essentially no voltage drop due to the wires (from Ohm's law again, V = iR; if R = 0, then V = 0), and the potential will be the same at the upper ends of the resistor and the battery. The same argument applies to the lower ends.
The direction of current flow in the external circuit, that part of the circuit outside the battery, is taken, by convention, to be the direction a positive particle would move (single, semi-circular, clockwise arrow). Actually, current in this type of circuit is carried by electrons, but the direction of current flow is opposite to their movement, i.e., from + to -. In order to complete the circuit, current flows from - to + within the battery (the internal circuit; wide, upward arrow). We will see that most charge carriers for membrane events are positively charged ions, or cations; currents, by convention, will flow the way they move. Bear in mind though, that when an anion, or negatively charged ion, moves, the current is said to be in the direction opposite its motion. This fact frequently gives students problems when considering chloride ions, Cl-.
Also note that when the current enters a resistor (the top in Figure 3-4), that end of the resistor becomes positive with respect to the other end. As we will see later, current passing outward through a membrane at rest will make the membrane more positive on the inside with respect to the outside. The reverse is true for inwardly directed currents.This observation holds for current flow through all passive elements like resistors and "resting" cell membranes. Refer back to the figure to see that current in the source or active element (the battery in this case) flows from negative to positive (again, the wide upward arrow). Therefore, an inward current through active membrane will make the membrane more positive inside with respect to outside, i.e., it will hypopolarize it. We will soon see that an inward sodium current hypopolarizes the active nerve or muscle membrane, whereas an outward current in resting membrane (passive) hypopolarizes it.(1)
A capacitor is a device that is capable of separating and storing charge. Usually a capacitor is made of two parallel conducting plates separated by a nonconductor or dielectric material. The capacitance of a capacitor is symbolized by C and related to voltage by C = Q/V, where Q is the charge. The unit of capacitance is the farad. The larger the capacitance, the larger is the amount of charge that must be added to the plates to bring the voltage between them to a given voltage. Looked at in another way, the larger the capacitance, the larger will be the amount of charge stored on the plates for a given voltage between them. We speak of increasing the charge stored on a capacitor as "charging the capacitor" and decreasing the charge stored as "discharging the capacitor."
Fig. 3-5. A simple resistance-capacitance circuit. The same circuit as in Figure 3-4, but with a capacitor, C, placed in parallel with the resistance, R. The voltages across the resistance and the capacitance are the same as the battery voltage, V, and the same as read by the voltmeter, Mv. The polarity of the voltages is as indicated by +,- signs.
ic = C dV/dt,
where dV/dt is the rate of change of voltage between the plates. Before the battery is hooked up, the voltage is zero and unchanging; therefore, dV/dt = 0 and ic = 0. When the battery is hooked up, the plates start to charge and continue until the voltage between the plates equals V, the battery voltage. At this time, dV/dt again becomes zero and ic = 0. Thus, there will only be a capacitative current when the voltage is actually changing.
If there were no resistance in the circuit of Figure 3-5, the battery could be disconnected and the voltage between the plates of the capacitor would remain at V (at least for a perfect capacitor). The resistor provides a current pathway through which the capacitor can discharge, with current flowing from the positive plate onto the negative plate. The larger R is the less current will flow per unit of time and the less rapidly the capacitor voltage will fall. In fact, the time for discharge varies directly with both R and C. The time required for the voltage to attain (1-1/e) V or about 2/3 V (V is the total voltage change) is called the time constant, tau, and is given by the equation tau = R x C. If R is expressed in ohms and C in farads, then tau will be expressed in seconds. In circuits, such as that in Figure 3-5, without rectification(2), the charging and discharging time constants are equal. As we shall see, cell membranes have rectifying properties which make them unequal.
Fig. 3-6. A simple parallel resistance circuit. The two resistance, R1 and R2, are in parallel, that is, they share the same input and combine their outputs. The voltmeter, Mv, still reads the value of the battery, V.
iR1 = V/R1
and that through R2 will be
iR2 = V/R2.
The total current will be as follows:
i = V/Rtotal = iR1 + iR2 = V/R1 + V/R2 .
Dividing by V gives:
1/Rtotal = 1/R1 + 1/R2 .
Thus, in a parallel resistive circuit, resistances add reciprocally. Capacitors in parallel each store charge; therefore, more charges are stored for the same voltage. In parallel, Ctotal = C1 + C2. When capacitors are in series, each sees a voltage that is smaller and thus stores fewer charges. In series:
1/Ctotal = 1/C1 + 1/C2 .
Fig. 3-7. Effects of varying resistance on circuit voltage. A. A battery and a resistor are shown in each of two parallel circuits. With the configuration shown, the voltmeter, M, will read 85 V, point b positive with respect to point d. B. The same circuit as in A, but with the resistors interchanged, leading to a voltmeter reading of 35 V, point b negative with respect to point d. The polarity and voltage have been changed by simply changing the resistors.
Figure 3-7B is the same as 3-7A except that the values of the resistors are interchanged. The current in the circuit is still 1.5 A, and the potential drop across 10 ohms is still 15 V, but this time the meter will read 50 V - 15 V or 35 V, with point b negative with respect to point d. Again, by symmetry, the calculation for the 90-ohm resistor will give the same value. Note that the potential difference between points b and d, is changed in magnitude and polarity simply by changing the resistance (or conductance), without altering the battery voltages. This observation will become important later when we consider the origin of the action potential.
Electrically, the nerve axon behaves like a very poor conductor of electricity. The axoplasm, the cytoplasm in the axon, has a very high resistance. For the squid axon, the longitudinal resistivity is 30 to 60 ohm-cm. This is extremely high compared with a copper wire of the same diameter, which has a resistivity of only 1.8 x 10-6 ohm-cm. This difference in resistance comes about because the density of charge carriers is smaller (ions in cytoplasm are less dense than electrons in copper wire) and their mobility is less (ions in solution move much less readily than electrons in copper wire). The voltage drop (loss) along 1 cm of cytoplasm is about 107 times greater than that occurring along 1 cm of wire. In other words, you wouldn't want to replace the cord on your toaster with the same length of nerve axon.
The membrane of a neuron behaves like a resistor, that is, when current is passed through the membrane, there is a voltage drop that is predictable from Ohm's law. Values of membrane resistance are usually expressed as area specific resistance in ohm-cm2. (Membrane resistance decreases with increases in membrane area; therefore it is measured in ohms x cm2, whereas membrane capacitance increases with increases in area and is expressed as microFarads per cm2.) Motoneurons of the lobster (a large neuron in which such measurements are easily made) have a membrane resistance of 2300 ohm-cm2.
The neural membrane also behaves like a capacitor, that is, it is capable of separating and storing charge. The capacitance of most neural membranes is of the order of 1 microFarad/cm2, meaning that the membrane can separate and store a charge of 1 x 10-6 coulomb/volt of potential across the membrane per cm2. Thus, the membrane can separate charges due to 10-12 mole of univalent ions/volt (Faraday's constant: 96,516 coulombs/mole of univalent ions). We will see shortly that the membrane does in fact separate considerable charge.
Fig. 3-8. Effects of membrane resistance and capacitance on membrane voltage. Recording setup shows two micropipette electrodes penetrating an axon membrane, one used for stimulation (right) and the other for recording (left). The transmembrane potential is recorded between the micropipette inside the axon and another electrode outside the axon, using amplifier A and an oscilloscope, M. The stimulating current flows from the second micropipette, out through the membrane, and back, through another external electrode, to the battery. The distance between the micropipette electrodes can be varied. On the right are the injected current in the upper trace and the transmembrane voltage in succesively lower traces when the electrodes are 0, 0.5, 1.0 and 2.0 mm apart. The gain of the amplifier is the same in all voltage traces.
Fig. 3-9. Resistance-capacitance model of the membrane. The outside of the membrane is at the top, the inside at the bottom of the figure. A single short segment of membrane is shown at the left and two consecutive segments at the right. Membrane resistance is indicated by Rm, membrane capacitance by Cm, and the longitudinal resistance of the intracellular fluid by Ri.
The distance along the membrane at which the imposed voltage, V, has been attenuated to
l/e x V is called the space constant, lambda. Because e=2.72 or about 3, lambda is the distance along the membrane at which V has fallen to about 1/3 of its initial value. The space constant is related to axon resistances as:
lambda = sq. rt.(Rm/(Ro + Ri)),
where Rm is the transmembrane resistance, and Ro and Ri are external and internal longitudinal resistances, respectively. For most neurons, lambda is about 2 mm. Thus, if information to be transmitted is expressed as a membrane voltage and this voltage is conducted passively or electrotonically along the neuron, distances over 5 mm would be well outside the conduction ability of the membrane; the voltage would be essentially zero (less than 1/10 V) 5 mm away from its source.
What happens in transatlantic telephone cables is similar, but the space constant there is longer. Still, the telephone signal must be amplified at intervals across the Atlantic Ocean in order for any telephone message sent from New York to be heard in London.
The membrane capacitances shown in Figure 3-9 are in parallel, so the farther away from the current source the greater will be the capacitance. The greater the capacitance and resistance of a circuit the greater the time constant and the more slowly voltage changes occur. It is not surprising, in view of these drastic attenuations and distortions over short distances, that neural transmission occurs by a specialized process, the self-regenerative action potential.
|Membrane Ionic Environment|
The nerve membrane is capable of storing, transmitting, and releasing electrical energy and yet it contains no metallic conductors. Neither are such metallic conductors found in the intra- or extracellular fluids. It is, therefore, unlikely that membrane currents represent flows of free electrons, but it is to the electrolytes, the ions, that we must look for carriers of membrane currents.
The intracellular fluid consists of an aqueous solution containing relatively large amounts of potassium, but small amounts of chloride, sodium, calcium, and magnesium. In addition, it contains some organic anions (negatively charged molecules) to which the membrane is impermeable, i.e., they cannot leave the cell. Table 3-1 shows both intracellular and extracellular concentrations of ions for the giant axon of the squid mantle (a cylinder of membrane of about 0.5-mm diameter), frog muscle fibers, and motoneurons in the cat spinal cord. Relative to intracellular fluid, extracellular fluid is rich in sodium, chloride, calcium, and magnesium, but poor in potassium.
Ionic Environments of the Membrane
(Vr = -70 mV)
(-100 <Vr< -70 mV)
(-80 <Vr< -60 mV)
Fig. 3-10A. Diffusion and diffusion potentials. A container, divided by a membrane permeable to sodium but not chloride, is filled with water. NaCl is added to side 1. Na+ diffuses to side 2, but chloride cannot.
Fig. 3-10B. Diffusion and diffusion potentials. An excess of positive charge, carried by Na+, builds up on side 2, leaving an excess of negative charge, carried by Cl-, on side 1, that is, a voltage builds up between the two sides.
Equilibrium potentialsThus far, we have not considered the effect of ionic charge on movement of the ions in solution. In Figure 3-10A, the compartments of the vessel are again separated by a membrane, but this time let us suppose that the membrane is permeable only to sodium. If we now add salt to one compartment as before, there will be a diffusion of sodium ions (recall that NaCl ionizes almost completely in water), from side 1 to side 2 (as before) because of the concentration gradient. Chloride would move with sodium to preserve electrical neutrality, but it cannot penetrate this membrane. As sodium crosses the membrane, an excess of positive charge (sodium is a positively charged ion or cation) builds up on side 2, leaving an excess of negative charge on side 1 due to unpaired chloride ions. This separation of charge is by definition a voltage. We speak of this voltage as being an electrical gradient. The excess of positive charge acts as a repellent to further movement of sodium across the membrane (like charges repel) or, alternatively, the excess negative charge on side 1 attracts sodium ions back across the membrane (unlike charges attract). The greater the accumulation of sodium ions and excess positive charge on side 2, the greater will be the repulsive force against further diffusion. This is the principle of operation of a concentration cell or battery; voltage is generated by differences in concentration across a semi-permeable membrane.
|The value of the membrane potential at|
electrochemical equilibrium for a particular ion
is called the equilibrium potential for that ion.
In other words, an ion will be in electrochemical
equilibrium if Vm = Vion.
At some point, the concentration force promoting diffusion of sodium ions will be exactly balanced by the electrical force preventing diffusion (Fig. 3-10B), and a condition of equilibrium will result, in which JN = 0. This is called electrochemical equilibrium. The value of the membrane potential at electrochemical equilibrium for a particular ion is called the equilibrium potential for that ion and is abbreviated, VX. The subscript is added to show the ion at equilibrium, in the case of sodium, Na+. The symbol for the equilibrium potential for sodium ions is VNa+.
Fig. 3-11. The equilibrium condition of the experiment in Figure 3-10B and C. Electrochemical equilibrium develops in which the Na+ concentration gradient, indicated by the upper arrow, is just balanced by the elctrical gradient, indicated by the lower arrow. At this point, there is no net flux of Na+ to either side.
Wt = Wc + We .
At equilibrium, Wt=0 because there is no net ion movement. Thus,
Wc = -We(1)
Wc = RT ln( [Na+]1 / [Na+]2 ),
where R is the gas constant; T, the absolute temperature; [Na+]1 is the sodium concentration on side 1; and [Na+]2 is the sodium concentration on side 2. This is the work required to maintain the concentration gradient. At chemical equilibrium [Na+]1 = [Na+]2 and Wc=0; no work need be done to maintain equal concentrations.
The work necessary to move ions against the electrical gradient is
We = zFV,
where z is the valence of the ion, F is Faraday's constant and V is the potential difference between the two compartments. Substitution for We and Wc in
equation 1 yields
Wc = -We = -RT ln( [Na+]1 / [Na+]2 ) = zFVNa.
VNa+ = (-RT/zF) ln( [Na+]1 / [Na+]2 ) .
This equation is known as the Nernst equation. Substitution of values for the gas constant, Faraday's constant and physiological temperature allows the Nernst equation to be simplified, using logarithms to base 10, to the form:
VNa+ = (-58/z) log10( [Na+]1 / [Na+]2 ) .
The Nernst equation not only specifies the transmembrane voltage necessary to maintain a given concentration difference, it also specifies the voltage that will result from maintenance of a given concentration difference, i.e., it specifies the voltage that will result from a concentration cell. In its most general form, the Nernst equation is written: .
VX = (-58/z) log10 ( [X]1 / [X]2 ),(2)
where X is any ion.
|The Nernst equation specifies the|
transmembrane voltage necessary to maintain
a given concentration difference or the
voltage that will result from maintenance of a
given concentration difference.
|The Membrane Potential|
These concentration gradients should form concentration cells and one would expect to find a potential difference (or voltage) between the inside and outside of a living cell. But, what should be the value and polarity of the potential? These can be calculated from Ohm's law and the Nernst equation. Any current flowing across the membrane will result in a voltage drop, predictable from Ohm's law:
i = gV.
That current will have to be carried by ions because the membrane contains no metallic conductors, so the total current will be equal to the sum of the currents carried by each ion:
i = iK+ + iNa+ + iCl- + iHCO3- + iMg2+ + iCa2+ . . .
The current carried by any ion will be determined by its driving force and the conductance of the membrane to it. The driving force is the electrical force experienced by the ion, and it is determined by the deviation of the membrane voltage from the equilibrium potential for the ion. The conductance, as you remember, is a measure of how easily the ion passes through the membrane(6). Thus:
iK+ = gK+(Vm - VK+)
iNa+ = gNa+(Vm - VNa+)
iCl- = gCl-(Vm - VCl-),
and so on, where Vm is the membrane potential. When Vm=VK+, there is no driving force on K+ and no K+ current. Any potassium current will be influenced by both the driving force and the membrane conductance for potassium. This is true for any ion.
Ohm's law can now be rewritten to the form:
i = gK+(Vm-VK+) + gNa+(Vm-VNa+) + gCl-(Vm-VCl-) + . . .(3)
Because of their very low concentrations, Ca2+ and Mg2+ ions contribute only slightly to the membrane potential, though they play a significant role in the functioning of both nerve and muscle. We shall ignore these ions for the present.
|The driving force is the electrical force|
experienced by the ion, and it is determined
by the deviation of the membrane voltage
from the equilibrium potential for the ion.
We want to determine the value of Vm in equation 3 above. To do this, we have to know the values of the equilibrium potentials for the various ions. Let's choose the squid giant axon for our example and calculate the equilibrium potentials from values in Table 3-1. It is customary in neurophysiology to express membrane potentials as potential inside the cell with respect to outside. Therefore, the Nernst equation takes the form:
VX = (-58/z) log10( [X]inside / [X]outside ).(4)
Substituting values for K+, we get:
VK+ = (-58/+1) log10 (400/20) = -58 log10 (20) = -75.5 mV.
Thus, K+ will be in electrochemical equilibrium at a 20/1 concentration ratio (inside/outside) when the inside is 75.5 mV negative with respect to the outside. At any potential less negative than this value, K+ will leave the cell; at any potential more negative than this value, K+ will enter the cell. We can calculate the equilibrium potentials for both Na+ and Cl- in a similar way:
VNa+ = (-58/+1) log10 (50/440)
= +58 log10 (440/50)
= +58 log10 (8.8) = +54.8 mV
VCl- = (-58/-1) log10 (40/560)
= -58 log10 (560/40)
= -58 log10 (14) = -66.5 mV.
|It is customary to express membrane|
potential as potential inside with respect to
In cell membranes, the conductance to chloride ions is high. Nerve membrane conductance to chloride is slightly lower than for potassium, but muscle membrane has a higher conductance to chloride than to potassium. Because of the high conductance, chloride is able to redistribute across the membrane by simple diffusion in response to changes in the membrane voltage. Thus, (at least in the squid axon) the chloride ion will always be at equilibrium in the vicinity of the resting membrane potential. (This might not be true for every cell, but it's true for the one we are considering here.) The contribution of chloride to the membrane potential can be ignored because of its tendency to redistribute itself.
In order to maintain electrical neutrality (the net charge in any solution must be zero), some positive ions must be present inside to balance the charge of the large anions that cannot leave the cell. The low sodium conductance of the resting membrane and the presence of the sodium pump prevent sodium ions from redistributing to provide the positive charge to balance the large intracellular anions; so potassium ions must remain inside the cell in high concentration.
That leaves us with only K+ and Na+ to contribute to the membrane potential. At rest, the membrane is at equilibrium, i.e., there is no net transmembrane current flow, so i = 0, therefore
0 = i = gK+(Vr - VK+) + gNa+(Vr - VNa+)
gK+(Vr - VK+) = - gNa+(Vr - VNa+),
where Vr is the value of Vm at equilibrium and is called the resting membrane potential. Rearranging terms:
The nerve membrane is permeable to both Na+ and K+, but the permeability to K+ is much higher. The ratio of the conductance of the membrane to K+ to the conductance to Na+ varies from 10 to 30 in different cells. If the ratio is taken to be 20, then
20/1 = -(Vr - VNa+) / (Vr - VK+) .
Solving for Vr, we see that
Vr = 20/21 VK+ + 1/21 VNa+.
|The value of the membrane potential is due|
mainly to the concentration difference for
potassium and is close to VK+.
Substituting values for VK+ and VNa+ from Table 3-1, we see that Vr = -69.3 mV. That is, we expect the membrane potential to be 69.3 mV negative inside with respect to outside. Taking the two extremes of the ratio of conductances, namely 30/1 and 10/1, yields predicted membrane potentials of -71.3 mV and -63.6 mV. The greater the ratio, i.e., the (relatively) more impermeable the membrane is to Na+, the closer Vr will be to VK+. Thus, the membrane potential is due mainly to the concentration difference for potassium and is close to VK+. The greater the relative permeability to Na+, the farther Vr will deviate from VK+ or the closer it will deviate toward VNa+. If the conductances to Na+ and K+ were equal, Vr would lie half way between VNa+ and VK+.
Fig. 3-12. The membrane potential. A graph of the voltage recorded between a movable micropipette electrode and a fixed electrode in the extracellular fluid (ordinate) against time (abscissa). At the origin, both the pipette and the fixed electrode are in the extracellular fluid, and the voltage between them is zero (A). When the micropipette penetrates the membrane, the voltage changes to -70 mV, inside with respect to outside (B). When the electrode is backed out of the cell, the potential returns to zero (C). The positions of VNa+ and VK+ are indicated on the ordinate.
Not all cells, not even all neurons, have the same resting membrane potential. In all cells, however, the value of the resting membrane potential depends upon the values of VK+, VNa+ and gK+/gNa+ (for some cells chloride and other ions may also play an important role). The values of VK+ for the three types of cells in Table 3-1 vary from -75 to -99 mV, and these values are reflected in their membrane potentials. Frog muscle cells have resting potentials of -70 to -100 mV, whereas cat motoneurons vary from -60 to -80 mV. Neurons in the cerebral cortex typically have resting membrane potentials around -50 mV, presumably reflecting either lower values of VK+ or gK+/gNa+ or both.
Fig. 3-13. A plot of the measured membrane potential (ordinate) for frog muscle fibers against the extracellular concentration of potassium (abscissa, log scale). Note that the membrane potential becomes more positive as the extracellular K+ increases. The solid line is a plot of the Nernst equation for potassium. The dasked line is the relationship predicted for a small contribution of sodium ions to the membrane potential (Hodgkin and Horowicz, J Physiol (Lond) 145:405-432, 1959).
Fig. 3-14. An electrical equivalent circuit model for the resting membrane. Outside of the membrane is at the top, inside at the bottom. Shown are the membrane capacitance, Cm; sodium and potassium resistance, RNa+ and RK+; internal longitudinal resistance of the intracellular fluid, Ri; and the sodium and potassium batteries, VNa+ and VK+.
|Recall that an ion is in equilibrium if its|
equilibrium potential equals the membrane
There exists, within the membrane, a mechanism for expelling Na+ and taking up K+ against their concentration gradients. This mechanism is called the sodium-potassium pump or, simply, the sodium pump. Obviously, if the ions have a natural tendency to move in one direction, causing them to move in the opposite direction requires the expenditure of energy. The tendency of ions to move down their concentration gradients constitutes a potential energy (as opposed to kinetic energy) that can be counteracted only by expending at least an equal amount of energy. That this process of pumping ions is, indeed, an active process is indicated by its sensitivity to changes in temperature, i.e., cells accumulate more sodium and lose more potassium at low temperature, and by the blockade of the pumping and decline of the membrane potential over a prolonged period when metabolic activity, involving splitting the high-energy phosphate bonds of adenosine triphosphate (ATP), is blocked by the inhibitor ouabain or by cyanide. The extrusion of 3 sodium ions requires the hydrolyzation of 1 molecule of ATP.
The rate of extrusion of Na+ by the sodium pump is proportional to the internal sodium concentration. For example, at an internal concentration of Na+ of 50 mM, the rate of pumping of Na+ ions is about 30 pmol/cm2/sec (pmol = picomoles) in the squid axon, but the rate rises to about 150 pmol/cm2/sec at an internal concentration of 230 mM. The relationship between pumping rate and the internal Na+ concentration appears to be linear for some cells (e.g., squid axon) but not for others (e.g., frog muscle).
In most cells, the pumping of sodium ions is linked to the pumping of potassium. In fact, extrusion of sodium can be reduced by as much as 70% in the absence of potassium in the extracellular fluid. A pump that exchanges one Na+ for one K+ will not produce a net charge transfer, and therefore there will be no change in membrane potential as a result of pump activity. This is called a non-electrogenic pump. However, if the amount of Na+ transported exceeds the amount of K+ transported, a potential will develop as a result of the net transfer of charge. Such a pump is called an electrogenic pump. An electrogenic pump can add to potentials set up by diffusing ions and cause the ions to redistribute themselves in order to restore equilibrium. The activity of electrogenic pumps can therefore affect ion distributions and membrane potentials either through action on concentration gradients or by altering the membrane potential. In any event, at equilibrium the net fluxes due to pump activity, or any other forces, must still be zero. Active fluxes (pump-related) must exactly balance passive fluxes (diffusional).
The sodium pump of erythrocytes has a Na+/K+ transport ratio that is nearly one, i.e., it is essentially non-electrogenic. In the squid axon, the ratio can be as high as 3 or as low as 1, depending upon whether the intracellular Na+ concentration is high or low. Thus, the transport ratio need not be fixed, but can vary, providing a mechanism for a relatively constant K+ transport rate and a variable sodium transport rate.
There is evidence that the transport system involves Na+-K+-activated ATPases (enzymes that hydrolyze ATP) within the membrane. These have a greater affinity for Na+ and a lesser affinity for K+ at the inner surface of the membrane and a lesser affinity for Na+ and greater affinity for K+ at the outer surface. They hydrolyze ATP at a rate dependent upon the Na+ and K+ concentrations. The actual physical mechanism by which the ions are moved is unknown, but the suggestion has been made that the ATPase is an intrinsic protein that perhaps rotates or otherwise changes shape after picking up intracellular Na+ and extracellular K+. In this new orientation or conformation, the affinity for the transported substance is reduced, and Na+ is released outside and K+ inside the cell. Finally, the protein rotates back to its original orientation or changes to its original conformation and affinities, and the process repeats.
It has been estimated that each sodium pump can transport up to 200 Na+ and 130 K+ per sec, the actual rate being determined by ion concentrations. Other estimates suggest that, on the average, sodium pumps have a density of 100-200 per mm2 of membrane, but, in some parts of the cell, they can be up to 10 times more dense. Thus, a typical neuron might have a million pumps with a maximum capacity of 200 million Na+ ions/sec.
The importance of the membrane potentialThe importance of the resting membrane potential is seldom stated explicitly, but its importance cannot be over-emphasized. All living cells, not just nerves and muscles, have a resting membrane potential. It is a ubiquitous property of living matter. The work of nerve and muscle cells depends upon the membrane potential. It acts as an energy store, a source of potential energy from which these cells draw to initiate their characteristic activities, the nerve and muscle action potentials and muscle contraction. The secretion of chemical substances by nerve terminals and glands also depends upon the existence of the membrane potential. We will see presently, how the membrane potential results in action potentials, and we shall see later, how the membrane potential is involved in muscle contraction.
1. It might be good to commit to memory the phrase: "Outward currents hypopolarize passive membrane; inward currents hypopolarize active membrane."
2. Here rectification is used in the engineer's sense of a lower resistance to current flow in one direction than the other through a circuit element.
3. Kuffler SW, Nichols JG: From Neuron to Brain, Sunderland, MA: Sinauer Assoc., 1976
4. Conway EJ: Physiol. Rev. 37:84-132, 1957
5. Coombs JS, Curtis, DR, Eccles, JC: J. Physiol. (Lond.) 130:326-373, 1955
6. You should always keep in mind that conductance and driving force are independent quantities. Driving force depends upon the membrane potential and the equilibrium potential. The equilibrium potential depends, in turn, upon the concentration difference for the ion. Conductance is a measure of the qualities of the membrane itself, how easily it admits the ion. Driving force does not depend in any way upon the qualities of the membrane.
7. Keep in mind that a concentration gradient is a source of potential energy, energy released if the gradient is allowed to dissipate.
8. This is true in the cell we have chosen for our examples, but it may not be true in the majority of cells. In particular, it is not true in cardiac muscle cells.