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I understand that after depolarisation, repolarisation and then hyperpolarisaiton occurs and that an area in hyperpolarisation is in its "refractory period".

Why does this prevent Na+ ions diffusing "backwards"? Since an area of the neurone membrane in hyperpolarisation will have a very negative charge so surely positive Na+ ions would diffuse towards them rather than away from them - "forward".

Similarly, why does this prevent another action potential firing too soon?

(I know this is kind of 2 questions in one, but as they are so very related I thought it more prudent to combine them).

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  • $\begingroup$ Possible duplicate of: biology.stackexchange.com/questions/56621/… $\endgroup$ – Bryan Krause Apr 5 at 19:45
  • $\begingroup$ Yes, that answer kind of answers my first part of my question (about being uni-directional) but doesn't answer the part about another ap firing too soon. Although I see why you might think it's a duplicate :) $\endgroup$ – Ben Hughes Apr 5 at 19:47
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I'll borrow a diagram from another answer to a similar question, but not quite a duplicate (I hope @AliceD won't mind my laziness):

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Fig. 1. Refractoriness. source: University of British Columbia

One important concept to understand about the refractory period is that it has two components. One is the hyperpolarizing part, which is due to open voltage-gated potassium channels. Because there is more potassium conductance than at rest, the membrane voltage gets closer to the reversal potential for potassium.

If you were to open a sodium conductance during this period, you would indeed get a larger sodium current than you would get at rest (because you are further from the sodium reversal potential), however it will also be accompanied by potassium leaving the cell (because there is more potassium conductance), such that the net current and net change in voltage may not be much larger than at rest. Voltage-gated sodium channels that open to cause an action potential are gated by voltage, not by sodium. However, this type of refractory period is referred to as a relative refractory period because you can overcome it if you open a large enough sodium conductance (or inject a bunch of current with a patch clamp amplifier).

However, there is also a second part to the refractory period, the "absolute" part, which is shown in the diagram above. The voltage gated sodium channels themselves are not only closed but they are also inactivated. While inactivated, those channels will not open even in response to depolarizing voltage. It takes some time for the channels to return to the 'closed' state where they can be opened by voltage.

Summary

To speak to your question more directly, nothing prevents sodium from diffusing 'backwards', but action potentials travel quickly, sodium channel inactivation prevents them from traveling into axon segments where the sodium channels were just recently open, and the open potassium channels that cause hyperpolarization will 'shunt' current out of the cell while they remain open. Lastly, voltage-gated sodium channels are gated by voltage, not sodium.

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  • $\begingroup$ "or inject a bunch of current with a patch clamp amplifier" --> Current is the flow of electrons, would hence current injection not make the membrane more negative (and make it hence harder for sodium channels to open)? $\endgroup$ – Pugl Jul 27 at 20:44
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    $\begingroup$ @Pugl The convention is to refer to positive charge as positive current. In neuroscience, it is not in fact electrons that are moving but ions of positive or negative charge. You can hyperpolarize or depolarize a cell freely when you have access to a cell via patch clamp, it doesn't matter what is flowing in the circuit of the amplifier. $\endgroup$ – Bryan Krause Jul 27 at 22:14
  • $\begingroup$ So, one would in fact inject cations (= positive current) for instance to depolarize a cell? $\endgroup$ – Pugl Jul 28 at 1:12
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    $\begingroup$ @Pugl Physically with an electrode it's probably more accurate to say you are attracting electrons/anions towards the cathode but it really doesn't matter mathematically. $\endgroup$ – Bryan Krause Jul 28 at 1:24

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