Membrane Events of the Action Potential and Ion Channel Function

Understand the IV curve for gated ion channels, the concept of rectification and importance of channel opening probability

Yup, we're going back to IV curves again!

Remember, the IV curve is a graph of current against voltage, according to the equation I = V/R or I = VG (basically a rearrangement of Ohm's Law, V = IR). Since I = VG, where G = conductance, the higher the conductance, the higher the current. Conductance increases when ion channels open, so the more ion channels are open, the steeper the slope.

Things get a bit more complicated when voltage-gated ion channels are involved, because an increase in voltage will increase the number of open ion channels, and therefore also increase the slope of the curve. If a whole bunch of ion channels were to open at exactly the same voltage, the curve would look something like this:

In reality, though, channels do not open at once: some will open slightly earlier and some will open slightly later. As voltage increases, the chance of a channel opening increases, until you reach the point at which all channels have a pretty much 100% chance of being open. This results in a graph that looks something like this:

You probably noticed that in the above graph, there is no current into or out of the cell below the membrane potential at which the channels start to open. This is because, if the channels are closed, ions can't pass through the membrane. (Technically, I probably should have done that for the first graph too, but I have the lamest excuse ever: for some reason, the graph on the slides is like that too, so I went all "monkey see, monkey do" on it, and now I can't be bothered changing my Paint diagram. So deal with it.)

Explain experimental methods used to study ionic mechanism of the action potential

Back in my first post, I gave you the equation I_x = G_x(V_m - E_x). Well, now it's going to come in handy! Measuring the current at a known voltage will give us Gx, or the conductance, which is related to the number of open ion channels (remember, conductance is the ease at which an electric current passes, and ion channels make things a helluva lot easier).

So, how do we measure the current? We can use the voltage-clamp technique, which I outlined here. Normally, a voltage clamp record will give us a transient inward current, followed by a delayed outward current. But how do we tease out which parts are caused by which ion channels? Thankfully, scientists have ingenious ways of figuring it out.

The first method is to change the Na⁺ concentration gradient by changing the concentration of Na⁺ in the extracellular fluid. It was found that if the cell was in a Na⁺-free solution of choline, the transient inward current was lost, and was instead replaced with a small outward going current. This is what we would expect to happen if Na⁺ channels opened first and the concentration gradient was such that [Na⁺]_i > [Na⁺]_o.

Another method for testing this is by using selective blockers. When tetrodotoxin (TTX), a Na⁺-channel blocker, was used, the initial inward current was lost, but the outward current remained. When tetraethyl ammonium (TEA), a K⁺-channel blocker, was used instead, the inward current remained, but the outward current was lost.

Altogether, these methods supported what we know today: Na⁺ channels open first, followed by K⁺ channels.

Describe the different behaviour of voltage-gated Na⁺ and K⁺ channels including Na⁺ channel inactivation
Describe recent advances in understanding the mechanism of Na⁺ channel gating

In a nutshell, voltage-gated Na⁺ channels open first, but they are only open transiently. K⁺ channels, on the other hand, take longer to open, but they will remain open until the membrane becomes polarised again.

Na⁺ channels have three major states: closed, open and inactivated. This is because they actually have two "gates": an activation gate and an inactivation gate. This was discovered by studying mutated ion channels which could become active, but, unlike normal Na⁺ channels, they could remain active for a long period of time. In the closed state, the activation gate (m-gate) is closed, but the inactivation gate (h-gate) is open. In the open state, both gates are open. In the inactivated state, the activation gate is open, but the inactivation gate is closed.

At normal resting potential, around 40% of Na⁺ channels are inactivated. This number decreases as the cell becomes more negative (hyperpolarised). It's because of this that a previously hyperpolarised cell can undergo "post-inhibitory rebound firing" (increase in amplitude and frequency of action potentials) once they are depolarised again. Another phenomenon related to this is "nerve accommodation"- if the membrane becomes depolarised at a very slow rate, a greater proportion of Na⁺ channels can become inactive, so that when it is time for an action potential to occur, the cell can hardly do anything (i.e. it is unexcitable).

I'll finish this off with a quick note on why all of this is important. As I just noted, the proportion of inactivated and available channels can affect excitability. If there are issues here, there might be abnormally high or low levels of excitability. Channel mutations might be responsible for issues such as epilepsy.