The usefulness of the Goldman equation is limited because it cannot be used to determine how rapidly the membrane potential changes in response to a change in permeability. It is also inconvenient for determining the magnitude of the individual Na+, K+, and Cl− currents. This information can be obtained using a simple mathematical model derived from electrical circuits. The model, called an equivalent circuit, represents all of the important electrical properties of the neuron by a circuit consisting of conductors or resistors, batteries, and capacitors.1
Equivalent circuits provide us with an intuitive understanding as well as a quantitative description of how current caused by the movement of ions generates signals in nerve cells.
The first step in developing an equivalent circuit is to relate the membrane's discrete physical properties to its electrical properties. We start by considering the lipid bilayer, which endows the membrane with electrical capacitance, the ability of an electrical nonconductor (insulator) to separate electrical charges on either side of it. The nonconducting phospholipid bilayer of the membrane separates the cytoplasm and extracellular fluid, both of which are highly conductive environments. The presence of a thin layer of opposing charges on the inside and outside surfaces of the cell membrane, acting as a capacitor, gives rise to the electrical potential difference across the membrane. The electrical potential difference or voltage across a capacitor is
where Q is the net excess positive or negative charge on each side of the capacitor and C is the capacitance. Capacitance is measured in units of farads (F), and charge is measured in coulombs (where 96,500 coulombs of a univalent ion is equivalent to 1 mole of that ion). A charge separation of 1 coulomb across a capacitor of 1 F produces a potential difference of 1 volt. A typical value of membrane capacitance for a nerve cell is approximately 1 μF per cm2 of membrane area. Very few charges are required to produce a large potential difference across such a capacitance. For example, the excess of positive and negative charges separated by the membrane of a spherical cell body with a diameter of 50 μm and a resting potential of −60 mV is 29 × 106 ions. Although this number may seem large, it represents only a tiny fraction (1/200,000) of the total number of positive or negative charges in solution within the cytoplasm. The bulk of the cytoplasm and the bulk of the extracellular fluid are electroneutral.
The membrane is a leaky capacitor because it is studded with ion channels that can conduct charge. Ion channels endow the membrane with conductance and with the ability to generate electromotive force (emf). The lipid bilayer itself has effectively zero conductance or infinite resistance. However, because ion channels are highly conductive, they provide pathways of finite electrical resistance for ions to cross the membrane. Because neurons contain many types of channels selective for different ions, we must consider each class of ion channel separately.
In an equivalent circuit we can represent each K+ channel as a resistor or conductor of ionic current with a single-channel conductance γK (remember, conductance = 1/resistance) (Figure 6–6). If there were no K+ concentration gradient, the current through a single K+ channel would be given by Ohm's law: iK = γK × Vm. However, since there is normally a K+ concentration gradient, there is also a chemical force driving K+ across the membrane, represented in the equivalent circuit by a battery. (A source of electrical potential is called an electromotive force and an electromotive force generated by a difference in chemical potentials is called a battery.) The electromotive force of this battery is given by EK, the Nernst potential for K+ (Figure 6–7).
In the absence of voltage across the membrane, the normal K+ concentration gradient causes an outward K+ current. According to our convention for current, an outward movement of positive charge across the membrane corresponds to a positive current. According to the Nernst equation, when the concentration gradient for a positively charged ion, such as K+, is directed outward (ie, the K+ concentration inside the cell is higher than outside), the equilibrium potential for that ion is negative. Thus the K+ current that flows solely because of its concentration gradient is given by iK = −γK × EK (the negative sign is required because a negative equilibrium potential produces a positive current at 0 mV).
Finally, for a real neuron that has both a membrane voltage and a K+ concentration gradient, the net K+ current is given by the sum of the currents caused by the electrical and chemical driving forces:
iK = (γK × Vm) − (γK × EK) = γK × (Vm − EK).
The term (Vm − EK) is called the electrochemical driving force. It determines the direction of ionic current and (along with the conductance) its magnitude. This equation is a modified form of Ohm's law that takes into account the fact that ionic current through a membrane is determined not only by the voltage across the membrane but also by the ionic concentration gradients.
A cell membrane has many resting K+ channels, all of which can be combined into a single equivalent circuit consisting of a conductor in series with a battery (Figure 6–8). In this equivalent circuit the total conductance of all the K+ channels (gK), ie, the K+ conductance of the cell membrane in its resting state, is equal to the number of resting K+ channels (NK) multiplied by the conductance of an individual K+ channel (γK):
Because the battery in this equivalent circuit depends solely on the concentration gradient for K+ and is independent of the number of K+ channels, its value is the equilibrium potential for K+, EK.
Like the population of resting K+ channels, all the resting Na+ channels can be represented by a single conductor in series with a single battery, as can the resting Cl− channels (Figure 6–9). Because the K+, Na+, and Cl− channels account for the bulk of the passive ionic current through the membrane in the cell at rest, we can calculate the resting potential by incorporating these three pathways into a simple equivalent circuit of a neuron.
To construct this circuit we need only connect the elements representing each type of channel at their two ends with elements representing the extracellular fluid and cytoplasm. The extracellular fluid and cytoplasm are both good conductors (compared with the membrane) because they have relatively large cross sectional areas and many ions available to carry charge. In a small region of a neuron, the extracellular and cytoplasmic resistances can be approximated by a short circuit—a conductor with zero resistance (Figure 6–10). The membrane capacitance (Cm) is determined by the insulating properties of the lipid bilayer, which separates the extracellular and cytoplasmic conductors.
The equivalent circuit can be made more realistic by incorporating the active ion fluxes driven by the Na+-K+ pump, which extrudes three Na+ ions from the cell for every two K+ ions it pumps in. This electrogenic ATP-dependent pump, which keeps the ionic batteries charged, can be added to the equivalent circuit in the form of a current generator (Figure 6–11).
The use of the equivalent circuit to analyze neuronal properties quantitatively is illustrated in Box 6–2. There we see how the equivalent circuit can be used to calculate the resting potential, Vm. We also see how the equivalent circuit can be simplified by combining all of the resting Na+, K+, and Cl− channels into a single resting current pathway, with a conductance gr and a battery Er, which is equal to the resting potential.
Box 6–2 Using the Equivalent Circuit Model to Calculate Resting Membrane Potential
An equivalent circuit model of the resting membrane can be used to calculate the resting potential. To simplify the calculation we shall initially ignore Cl− channels and begin with just two types of resting channels, K+ and Na+, as illustrated in Figure 6–12. Moreover, we ignore the electrogenic influence of the Na+-K+ pump because it is small.
Because we consider only steady-state conditions, where the membrane potential, Vm, is not changing, we can also ignore membrane capacitance. (Membrane capacitance and its delaying effect on changes in Vm are discussed below.)
Because there are more resting channels for K+ than for Na+, the membrane conductance for K+ is much greater than that for Na+. In the equivalent circuit in Figure 6–12, gK (10 × 10−6 S) is 20 times higher than gNa (0.5 × 10−6 S). Given these values and the values of EK and ENa Vm is calculated as follows.
Under the above conditions where the membrane potential is constant, there is no net current across the membrane. Therefore INa is equal and opposite to IK:
−INa = IK
or
(6–1) INa + IK = 0.
We can easily calculate INa and IK in two steps. First, we add up the separate potential differences across the Na+ and K+ branches of the circuit. Going from the inside to the outside across the Na+ branch, the total potential difference is the sum of the potential differences across ENa and across gNa:*
Vm = ENa + INa / gNa.
Similarly, for the K+ conductance branch
Vm = EK + IK / gK.
Next, we rearrange and solve for I:
(6–2a) INa = gNa × (Vm − ENa)
(6–2b) IK = gK × (Vm − EK).
As these equations illustrate, the ionic current through each conductance branch is equal to the conductance of that branch multiplied by the net electrical driving force. For example, with the K+ current the conductance is proportional to the number of open K+ channels, and the driving force is equal to the difference between Vm and EK. If Vm is more positive than EK (−75 mV), the driving force is positive and the current is outward; if Vm is more negative than EK, the driving force is negative and the current is inward.
Similar equations are used in a variety of contexts in this book to relate the magnitude of a particular ionic current to its membrane conductance and driving force.
As we saw in Equation 6–1, INa + IK = 0. If we now substitute Equations 6–2a and 6–2b for INa and IK in Equation 6–1, multiply through, and rearrange, we obtain the following expression:
Vm × (gNa + gK) = (ENa × gNa) + (EK × gK).
Solving for Vm we obtain an equation for the resting membrane potential that is expressed in terms of membrane conductances and batteries:
From this equation, using the values in our equivalent circuit (Figure 6–12), we calculate Vm = −69 mV.
Equation 6–3 states that Vm approaches the value of the ionic battery that has the greater conductance. This principle can be illustrated by considering what happens during the action potential. At the peak of the action potential gK is essentially unchanged from its resting value, but gNa increases as much as 500-fold. This increase in gNa is caused by the opening of voltage-gated Na+ channels. In the equivalent circuit in Figure 6–12, a 500-fold increase would change gNa from 0.5 × 10−6 S to 250 × 10−6 S.
If we substitute this new value of gNa into Equation 6–3 and solve for Vm, we obtain +50 mV. Vm is closer to ENa than to EK at the peak of the action potential because gNa is now 25-fold greater than gK, so that the Na+ battery becomes much more important than the K+ battery in determining Vm.
The real resting membrane has open channels for Cl− as well as for Na+ and K+. One can derive a more general equation for Vm, following the steps outlined above, from an equivalent circuit that includes a conductance pathway for Cl− with its associated Nernst battery (Figure 6–13):
This equation is similar to the Goldman equation in that the contribution to Vm of each ionic battery is weighted in proportion to the conductance of the membrane for that particular ion. In the limit, if the conductance for one ion is much greater than that for the other ions, Vm approaches the value of that ion's Nernst potential.
The contribution of Cl− ions to the resting potential can now be determined by comparing Vm calculated for the Na+ and K+ circuits only (Figure 6–12) and for all three ions (Figure 6–13). For most nerve cells the value of gCl ranges from one-fourth to one-half of gK. In the simplified example in Figure 6–13 Cl− ions flow passively across the membrane, so that ECl is equal to the value of Vm, which is determined by Na+ and K+.
Because ECl = Vm (−69 mV in this case), there is no current through the Cl− channels. As a result when we include gCl and ECl in the circuit (Figure 6–13), the calculated value of Vm does not differ from that when the Cl− conductance is absent (Figure 6–12). However, in most neurons Cl− is actively transported out of the cell so that ECl is more negative than the resting potential. Adding the Cl− pathway to the calculation would shift Vm to a slightly more negative value.
The equivalent circuit can be further simplified by lumping the conductance of all the resting channels that contribute to the resting potential into a single conductance, gr, and replacing the battery for each conductance channel with a single battery whose value, Er, is given by Equation 6–4 (Figure 6–14). Here the subscript r stands for the resting channel pathway. Because the resting channels provide a pathway for the steady leakage of ions across the membrane, they are sometimes referred to as leakage channels (see Chapter 7). This consolidation of resting pathways will prove useful when we consider the effects on membrane voltage of current through voltage-gated and ligand-gated channels in later chapters.