The release of transmitter from the motor nerve terminal opens ACh receptor-channels in the muscle membrane at the end-plate, and this action rapidly depolarizes the membrane. The resulting excitatory postsynaptic potential (EPSP), also called the end-plate potential at the nerve-muscle synapse, is very large; stimulation of a single motor cell produces a synaptic potential of approximately 70 mV.
This change in membrane potential usually is large enough to rapidly activate the voltage-gated Na+ channels in the junctional folds, converting the end-plate potential into an action potential, which then propagates along the muscle fiber. In contrast, in the central nervous system most presynaptic neurons produce postsynaptic potentials less than 1 mV in amplitude. As a result, input from many presynaptic neurons is needed to generate an action potential in most neurons.
The End-Plate Potential Is Produced by Ionic Current Through Acetylcholine Receptor-Channels
The end-plate potential was first studied in detail in the 1950s by Paul Fatt and Bernard Katz using intra cellular voltage recordings. Fatt and Katz were able to isolate the end-plate potential by applying the drug curare2 to reduce the amplitude of the postsynaptic potential below the threshold for the action potential (Figure 9–4). They found that the EPSP in muscle cells was largest at the end-plate and decreased progressively with distance (Figure 9–5).
The end-plate potential can be isolated pharmacologically for study.
A. Under normal circumstances stimulation of the motor axon produces an action potential in a skeletal muscle cell. The dashed line shows the inferred time course of the end-plate potential that triggers the action potential.
B. Curare blocks the binding of acetylcholine (ACh) to its receptor and so prevents the end-plate potential from reaching the threshold for an action potential (dashed line). In this way the currents and channels that contribute to the end-plate potential, which are different from those producing an action potential, can be studied. The end-plate potential shown here was recorded in the presence of a low concentration of curare, which blocks only a fraction of the ACh receptors. The values for the resting potential (–90 mV), end-plate potential, and action potential in these intracellular recordings are typical of a vertebrate skeletal muscle.
The end-plate potential decreases with distance as it passively propagates away from the end-plate.
(Adapted, with permission, from Miles 1969.)
A. The amplitude of the postsynaptic potential decreases and the time course of the potential slows with distance from the site of initiation in the end-plate.
B. The decay results from leakiness of the muscle fiber membrane. Because charge must flow in a complete circuit, the inward synaptic current at the end-plate gives rise to a return outward current through resting channels and across the membrane (the capacitor). This return outward flow of positive charge depolarizes the membrane. Because current leaks out all along the membrane, the outward current decreases with distance from the end-plate.
From this, Fatt and Katz concluded that the end-plate potential is generated by an inward ionic current that is confined to the end-plate and then spreads passively away. (Remember, an inward current corresponds to an influx of positive charge, which depolarizes the inside of the membrane.) Inward current is confined to the end-plate because the ACh receptor-channels are concentrated there, opposite the presynaptic terminal from which transmitter is released.
The end-plate potential rises rapidly but decays more slowly. The rapid rise is caused by the sudden release into the synaptic cleft of ACh, which diffuses rapidly to the receptors at the end-plate. (Not all the ACh reaches receptors, however, because ACh is quickly removed from the synaptic cleft by hydrolysis by acetylcholinesterase and diffusion.)
The current that generates the end-plate potential was first studied in voltage-clamp experiments (see Box 7–1). These studies revealed that the end-plate current rises and decays more rapidly than the resultant end-plate potential (Figure 9–6). The time course of the end-plate current is directly determined by the rapid opening and closing of the ACh receptor-channels. Because it takes time for an ionic current to charge or discharge the muscle membrane capacitance, and thus alter the membrane voltage, the EPSP lags behind the synaptic current (see Figure 6–15 and the Postscript at the end of this chapter).
The end-plate current increases and decays more rapidly than the end-plate potential.
A. The membrane at the end-plate is voltage-clamped by inserting two microelectrodes into the muscle at the end-plate. One electrode measures membrane potential (Vm) and the second passes current (Im). Both electrodes are connected to a negative feedback amplifier, which ensures that sufficient current (Im) is delivered so that Vm will remain clamped at the command potential Vc. The synaptic current evoked by stimulating the motor nerve can then be measured at constant Vm, for example −90 mV (see Box 7–1).
B. The end-plate potential (measured when Vm is not clamped) changes relatively slowly and lags behind the more rapid inward synaptic current (measured under voltage-clamp conditions). This is because synaptic current must first alter the charge on the membrane capacitance of the muscle before the muscle membrane is depolarized.
The Ion Channel at the End-Plate Is Permeable to Both Sodium and Potassium
Why does the opening of the ACh receptor-channels lead to an inward current that produces the depolarizing end-plate potential? And which ions move through the ACh-gated channels to produce this inward current? One important means of identifying the ion (or ions) responsible for the synaptic current is to measure the value of the chemical driving force (the chemical battery) propelling ions through the channel. Remember, the current through a set of membrane channels is given by the product of the membrane conductance and the electrochemical driving force on the ions conducted through the channels (see Chapter 6). The end-plate current that underlies the EPSP is determined by:
(9–1) IEPSP = gEPSP × (Vm − EEPSP),
where IEPSP is the end-plate current, gEPSP is the conductance of the ACh receptor-channels, Vm is the membrane potential, and EEPSP is the chemical driving force or battery generated by the transmembrane concentration gradients of the ions conducted through the channels.
The fact that current through the end-plate is inward at the normal resting potential of a muscle cell (−90 mV) indicates that there is an inward (negative) electrochemical driving force on the ions that carry current through the ACh receptor-channels at this potential. Thus, EEPSP must be positive to −90 mV.
The value of EEPSP in Equation 9–1 can be determined by altering Vm in a voltage-clamp experiment and determining its effect on IEPSP. Depolarizing the membrane reduces the net inward electrochemical driving force, thus decreasing the magnitude of the inward end-plate current. If Vm is set equal to EEPSP, there will be no net current through the end-plate channels because the electrical driving force (Vm) will exactly balance the chemical driving force (EEPSP). The potential at which the net ionic current is zero is the reversal potential for current through the synaptic channels; by determining the reversal potential we can experimentally measure the value of EEPSP. If Vm is made more positive than EEPSP, there will be a net outward driving force. In that case stimulation of the motor nerve leads to an outward ionic current (by opening the ACh receptor-channels) that hyperpolarizes the membrane.
If an influx of Na+ were solely responsible for the end-plate potential, the reversal potential for the excitatory postsynaptic potential would be the same as the equilibrium potential for Na+ (ENa = +55 mV). Thus, if Vm is experimentally altered from −100 to +55 mV, the end-plate current should diminish progressively because the electrochemical driving force on Na+ (Vm − ENa) is reduced. At +55 mV the inward current should be abolished, and at potentials more positive than +55 mV the end-plate current should reverse in direction and become outward.
Instead, experiments at the end-plate showed that as Vm is reduced, the inward current rapidly becomes smaller and is abolished at 0 mV. At values more positive than 0 mV the end-plate current reverses direction and becomes outward (Figure 9–7). This reversal potential is not equal to the equilibrium potential for Na+ or any of the other major cations or anions. In fact, this chemical potential is produced not by a single ion species but by a combination of ions: The ligand-gated channels at the end-plate are almost equally permeable to both major cations, Na+ and K+. Thus, during the end-plate potential Na+ flows into the cell and K+ flows out. The reversal potential is at 0 mV because this is a weighted average of the equilibrium potentials for Na+ and K+ (Box 9–1). At the reversal potential the influx of Na+ is balanced by an equal efflux of K+ (Figure 9–7).
The end-plate potential is produced by the simultaneous flow of Na+ and K+ through the same receptor-channels.
The ionic currents responsible for the end-plate potential can be determined by measuring the reversal potential of the end-plate current using a voltage clamp.
A. In the hypothetical case in which Na+ flux alone is responsible for the end-plate current, the reversal potential would occur at +55 mV, the equilibrium potential for Na+ (ENa). The arrow at the right of each current record reflects the magnitude of the net Na+ flux at that membrane potential. (Down arrows = inward; up arrows = outward).
B. In the actual case in which the ACh receptor-channel is permeable to both Na+ and K+, experimental results show that the end-plate current reverses at 0 mV because the ion channel allows Na+ and K+ to move into and out of the cell simultaneously (see Box 9–1). The net current is the sum of the Na+ and K+ fluxes through the end-plate channels. At the reversal potential (EEPSP) the inward Na+ flux is balanced by an outward K+ flux so that no net charge flows.
Box 9–1 Reversal Potential of the End-Plate Potential
The reversal potential of a membrane current carried by more than one ion species, such as the end-plate current through the ACh receptor-channel, is determined by two factors: (1) the relative conductance for the permeant ions (gNa and gK in the case of the end-plate current) and (2) the equilibrium potentials of the ions (ENa and EK).
At the reversal potential for the ACh-gated current, inward current carried by Na+ is balanced by outward current carried by K+:
(9–2) INa + IK = 0.
The individual Na+ and K+ currents can be obtained from
(9–3a) INa = gNa × (Vm − ENa)
(9–3b) IK = gK × (Vm − EK).
Remember that these currents do not result from Na+ and K+ flowing through separate channels (as occurs during the action potential) but through the same ACh receptor-channel. We can substitute Equations 9–3a and 9–3b for INa and IK in Equation 9–2, replacing Vm with EEPSP (because at the reversal potential Vm = EEPSP):
(9–4) gNa × (EEPSP − ENa) + gK × (EEPSP − EK) = 0.
Solving this equation for EEPSP yields
If we divide the top and bottom of the right side of this equation by gK, we obtain
If gNa = gK, then EEPSP = (ENa + EK)/2.
These equations can also be used to solve for the ratio gNa/gK if one knows EEPSP, EK, and ENa. Thus, rearranging Equation 9–6 yields
At the neuromuscular junction, EEPSP = 0 mV, EK = −100 mV, and ENa = +55 mV. Thus, from Equation 9–7, gNa/gK has a value of approximately 1.8, indicating that the conductance of the ACh receptor-channel for Na+ is slightly higher than for K+. A comparable approach can be used to analyze the reversal potential and the movement of ions during excitatory and inhibitory synaptic potentials in central neurons (see Chapter 10).
Why are the ACh receptor-channels at the end-plate not selective for a single ion species like the voltage-gated Na+ or K+ channels? This is because the diameter of the pore of the ACh receptor-channel is substantially larger than that of the voltage-gated channels. Electrophysiological measurements suggest that it may be up to 0.8 nm in diameter, an estimate based on the size of the largest organic cation that can permeate the channel. For example, the permeant cation tetramethylammonium (TMA) is approximately 0.6 nm in diameter. In contrast, the voltage-gated Na+ channel is only permeant to organic cations that are smaller than 0.5 × 0.3 nm in cross section, and voltage-gated K+ channels will only conduct ions less than 0.3 nm in diameter.
The relatively large diameter of the ACh receptor pore is thought to provide a water-filled environment that allows cations to diffuse through the channel relatively unimpeded, much as they would in free solution. This explains why the pore does not discriminate between Na+ and K+. It also explains why even divalent cations, such as Ca2+, can permeate the channel. Anions are excluded, however, by the presence of fixed negative charges in the channel, as described later in this chapter.