The Ion Channel at the End-Plate Is Permeable to Both Sodium and Potassium Ions
Once a receptor-channel opens, which ions flowthrough the channel, and how does this lead to depolarization of the muscle membrane? 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 single open channel is given by the product of the single-channel conductance and the electrochemical driving force on the ions conducted through the channel (Chapter 9). Thus, the current generated by a single ACh receptor-channel is given by:
IEPSP = EPSP . (Vm − EEPSP), (12–1) where IEPSP is the amplitude of current through one channel, EPSP is the conductance of a single open channel, EEPSP is membrane potential at which the net flux of ions through the channel is zero, and Vm − EEPSP is the electrochemical driving force for ion flux. The current steps change in size as the membrane potential changes because of the change in driving force. For the ACh receptor-channels, the relationship between IEPSP and membrane voltage is linear, indicating that the single-channel conductance is constant and does not depend on membrane voltage; that is, the channel behaves as a simple ohmic resistor. From the slope of this relation, the channel is found to have a conductance of 30 pS (Figure 12–7B). As we saw in Chapter 9, the total conductance, g, due to the opening of a number of receptor-channels (n) is given by: g n .
The current–voltage relation for a single channel shows that the reversal potential for ionic current through ACh receptor-channels, obtained from the intercept of the membrane voltage axis, is 0 mV, which is not equal to the equilibrium potential for Na+ or any of the other major cations or anions. This is due to the fact that this chemical potential is produced not by a single ion species but by a combination of two species:
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 12–1). At the reversal potential, the influx of Na+ is balanced by an equal efflux of K+ (Figure 12–7A).