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==Introduction==
* We will focus briefly on electrical synapses, whose functional roles are important in many neural circuits.
* Nerve cells may be connected together via tight gap junctions. These are formed by proteins called '''connexons''', creating a barrel-shaped opening in the membrane which is apposed to a similar opening in the membrane of the second neuron. The staves of the barrel are made of a protein called '''connexin'''.
* If two cells (let us call them $<math>A$ </math> and $<math>B$</math>) are connected by an electrical synapse, then injecting depolarizing current (positive charges) into $<math>A$ </math> (the bottom trace in Figure 1) could induce a depolarization in neuron $<math>B$</math>.* In contrast, injecting hyperpolarizing current (negative charges) into $<math>A$ </math> can induce a hyper polarization in neuron $<math>B$</math>, as shown in Figure 2.
* If one neuron can both depolarize and hyperpolarize another neuron, that is usually a very strong indicator that the two neurons are electrically coupled to one another.
* Interactions between electrically coupled cells can be characterized using the electrical equivalent circuit shown in Figure 3.
* If a current is injected into the lefthand neuron ($<math>A$</math>), setting it to voltage $<math>V_1$</math>, then the voltage to which neuron $<math>B$ </math> will be set will be determined by how much current flows directly through the electrical synapse, and how much instead flows out through the membrane of neuron <math>B</math>. Once the transients due to the capacitance of the membrane settle out, the voltage $<math>V_2$ </math> to which the righthand neuron ($<math>B$</math>) goes will be $$<math display="block"> \frac{V_{2}}{V_{1}} = \frac{R_{2}}{R_{2} + R_{e}} = K_{1,2}, \eqnumquad\text{(Equation 1)}$$ </math> where $<math>R_2$ </math> is the membrane resistance of neuron $<math>B$</math>, and $<math>R_e$ </math> is the resistance of the electrical synapse.* Similarly, if a current is injected into the righthand neuron ($<math>B$</math>), setting it to voltage $<math>V_2$</math>, then the voltage to which neuron <math>A</math> will be set will be determined by how much current flows directly through the electrical synapse, and how much instead flows through the membrane of neuron <math>A</math>. The final voltage <math>V_{1}</math> of neuron <math>A</math> will be $$<math display="block"> \frac{V_{1}}{V_{2}} = \frac{R_{1}}{R_{1} + R_{e}} = K_{2,1}, \eqnumquad\text{(Equation 2)}$$ </math> where $<math>R_1$ </math> is the membrane resistance of neuron $<math>A$</math>.* The terms for $<math>K_{1,2}$ </math> and $<math>K_{2,1}$ </math> are ''coupling coefficients''.
* Although the picture shows two neurons whose sizes are identical, which would likely lead them to have identical coupling coefficients for currents injected into either neuron, if one neuron is much larger than the other, its membrane resistance is likely to be much lower. As a consequence, current injected from the smaller neuron, which can generate much less current, would have very little effect on the larger neuron; in contrast, the larger process will almost certainly be able to generate a larger current, and because of the larger input resistance of the smaller neuron, its voltage might change much more significantly. So even though electrical connections are bi-directional in principle, they can in practice effectively work primarily in one direction.
* Similarly, although this is less common, the gap junctions can sometimes show voltage dependence, so that the amount of current carried by the electrical synapse will vary with the membrane voltage of the two neurons.
* At the peak of the action potential, the membrane is in a steady state, and <math>I_{EPSP} + I_{L} = 0, </math> which allows us to determine the actual membrane voltage that will be reached by the nerve cell due to the synaptic input.
* From the channel equation, we know that <math>I_{EPSP} = g_{EPSP}(V - E_{EPSP})</math>, and <math>I_{L} = g_{L}(V - E_{L})</math>. Combining these equations, we can solve for the membrane voltage at the peak of the synaptic potential: $$<math display="block"> V_{m} = \frac{g_{EPSP} E_{EPSP} + g_{L} E_{L}}{g_{EPSP} + g_{L}}. \eqnumquad\text{(Equation 3)} $$</math>
* For simplicity, we focused on a single kind of postsynaptic potential, one that opened ion channels (induced an '''increased''' conductance), and had a reversal potential that was significantly more depolarized than the resting potential of a neuron, and thus would not only depolarize the neuron, but will also tend to increase the likelihood that the postsynaptic neuron will fire an action potential. This is an example of an '''excitatory postsynaptic potential''', usually abbreviated as '''EPSP'''.
