Action potentials in neurons

[icepik]

What scenarios are possible, if one consumes a drug that blocks the closing of the voltage-gated Sodium gates?

 

a) Would the action potential fire as usual, but the amplitude of the action potential be higher?

b) Would the action potential be between the value of Ek and Ena?

 

 

Also, if the voltage gates are never closed, there will be no depolarization phase right? Would that still count as an action potential if the membrane never repolarize?

 

-OR-

 

Would the increase in membrane potential, keep activating more and more potassium channels as well, and keep the value between Ek and Ena?

[BranMuffin]

The AP would not be able to propagate at all, your right in that there would be no depolarization because there (eventually) wouldn't be a gradient between the ECF & ICF.

 

Sorry I can't answer your question about the Ek & Ena - I'm not sure what those stand for...

Edited February 6, 2014 by BranMuffin

[Ringer]

There are two 'gates' on neuron's sodium channels. One is closed at ~50mV and one closes at ~-70mV (IIRC) so it would depend on which one you wouldn't allow to close. If neither could close then after the firing of one action potential the membrane potential couldn't be reestablished even with the K+ channels opening because Na+ would constantly be leaking through and the Na+/K+ pump would just keep pumping to no avail. The electrochemical gradient would reach equilibrium and the neuron would end up dying.

 

Also, Ek and Ena are probably used to represent the potential of K+ and potential of Na+

[icepik]

The reason I asked this question is because I got it wrong on my exam. But, the answer choices were poorly given. The question states that a person takes a drug that doesn't close the voltage sodium gates in neurons. The only two possible choices were that 1. The action potential fires normally, but with an increasing action potential or the 2. magnitude of the membrane potential would stay between Ek and ENa.

 

The amplitude of the action potential would increase because sodium will flow in much faster than the na/k pump can pump out, and although as voltage increases, the efflux of potassium due to the increased permeability ( voltage gated potassium would stay open as sodium is constantly depolarizing). I believe both answers are correct. The action potential magnitude would reach close to the Ena value since sodium is rushing in much faster, and the slow closing potassium gates willl not influence this.

 

Used my iphone to type this. Sorry for errors.

[Ringer]

Well, since both are voltage gated neither would close so there would be a constant flow both in and out, I don't have the time to look up the concentration gradient vs. voltage gradient @ equilibrium. But they would eventually reach equilibrium, +/- variation due to Na+/K+ pump. I agree that the choices seem to be worded oddly. An action potential, by definition, couldn't fire normally without returning to resting potential. I would say it would need to be #2, but as I said before I'm not going to figure out the equilibrium of the membrane.

[CPG]

I have to ask an important question about the question in the OP. A typical sodium channel that contributes to action potential conductances has more than 1 "closed state". It can be closed because the channel is inactivated or because it has deactivated (this technicality may sound silly at first but they are mechanically and functionally very different) - so which is the drug preventing? Did the question specify? I'm going to reply as if the channels cannot inactivate because that's what I would guess is meant here.

 

Normally in an action potential it is inactivation that is responsible for closing Na⁺ channels. If the drug is preventing inactivation (there are definitely drugs that do this, and some very effective ones like pyrethroids have been commonly used as insecticides because they do this to insects. Once their nerve cells are excited they can't repolarize and so you've effectively shut down the nervous system).

 

 

If you don't mind a few simple algebraic equations, it may help out a lot here. A brief description and then the math. More sodium channels being open makes the membrane more conductive to sodium and thus would drive the membrane potential towards E_Na (aka the equilibrium potential for sodium). The driving force for an ion, Na⁺ in this case, is the difference between the membrane potential at any given moment the ion's equilibrium potential (V_m - V_eq). Conductance (G) is another way of describing the same property as resistance ® - it's just that with conductance we're concerned with how easy it is to pass current (I) instead of how strong the impediment to current is. Here comes the reason for all that setup.

 

The current flowing across the membrane is equal to the conductance times the driving force.

 

I=G*(V_m-V_eq)

 

So let's say your drug prevents inactivation of sodium channels, depolarization causes a positive feedback leading to further depolarization. However, since the Na⁺ channels stay open G_Na remains high and continues to drive the membrane potential towards E_Na which would typically be a very positive voltage.

Your K⁺ channels also open in response to depolarization, albeit more slowly. As they open, K+ currents drive the membrane potential towards E_K (hyperpolarizing). At this point because the sodium channels have been doing their thing, you're probably very far away from E_K , so the driving force for K+ would start off very strong. These two currents are now opposing one another, and as K+ current drives the membrane potential down, you get farther away from E_Na, and thus Na+ currents increase in magnitude. Likewise, as you approach E_K, K⁺ currents decrease in magnitude (assuming no conductance changes). Picture two of the equation above, but one for each ion:

 

I_Na = G_Na (V_m - E_Na)

I_K = G_K (V_m - E_K)

 

Hopefully with this illustration you can see that as long as both channel types remain open, the membrane potential would equilibrate somewere between these two values, at a membrane voltage where the currents are equal and opposite. This point is likely to be above the voltage where activation / deactivation takes place.

 

 

 

 

 

If that lost you at any point it's good to always keep Ohm's Law in mind. That will remind you what's going to happen to the membrane potential whenever current is flowing.

 

V=IR (or V=I/G)

 

Depending on the level of complexity of your studies, there may be more involved, but my best guess is that this is generally the logic your instructor had in mind.

The electrophysiologist in me wants to add a list of caveats ("ifs ands & buts"), it's probably best to stop here and ask if what I've laid out makes sense so far?

 

 

(The last part of your question does make me wonder in what depth your instructor went, because the activation of channels can be described by probablistic functions and we might start worrying about conductance changes due to the voltage. The typical approximations of this have different functions for activation of Na⁺ channels and K⁺ channels. However, as long as the voltage remains well above the activation threshold, you can probably assume that the fraction of channels that are closed is negligible.)

[icepik]

WOW, thanks for that detailed response! You definitely clarified the question for me. I like how you added the equations along with the actual mechanistic explanation. I'm at med school, and the professor phrases our questions really bad. I knew that the membrane potential would (obviously) never exceed Ena. As voltage increased, the potassium gates would further limit it's Em from reaching Ena.

 

Thanks again!