<p>When a sodium (Na) channel is stimulated by a strong-enough stimulus (such as from an electrical stimulus or from an incoming action potential), it will open and let a lot of Na+ ions influx rapidly into the cell at that channel, creating an action potential there. The locally high concentration of Na+ ion inside the cell at that channel will cause Na+ ions there to spread around immediately. This will cause depolarization and openings of the nearby Na channels, resulting in new actions potentials occurring at the nearby channels. This whole process will repeat itself again at these channels, causing more new action potentials to spread out further. In each cycle of an action potential, the potassium channel opens after the sodium channel does about 1 msec. Although the opening of the potassium channel causes repolarization,  it doesn’t have any effects on the concentration of Na+ ion and does not stop the spreading out of the Na+ ions. Therefore, an action potential inevitably spreads out and out.</p>
<p><a href="https://i.sstatic.net/aqduP.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/aqduP.jpg" alt="enter image description here" /></a></p>
<p>If the axon is stimulated at its trunk, the action potential will spread out both ways, up and down the axon. Figure 3.1 A shows a schematic diagram of an axon and the charge distribution along the membrane of that axon. More positive charges exist on the outside of the axon, and more negative charges on the inside. Now consider the consequences of delivering some stimulus to a point in the middle of the axon, Figure 3.1 B. If the depolarization is sufficiently large, voltage-dependent sodium channels will be opened, and an action potential will be initiated, and the sodium ion (Na+) will influx into the cell.</p>
<p>The influx positive charges Na+ will spread to the adjacent regions of the membrane. As the charges move to the adjacent regions of the membrane, the adjacent regions of the membrane will depolarize. If they depolarize sufficiently, voltage-dependent sodium channels in the adjacent regions of the membrane will be opened and &quot;new&quot; action potentials will be initiated, Figure 3.1 C. This perturbance of charge distribution will then spread to the next regions and initiates other &quot;new&quot; action potentials, Figure 3.1 D to 3.1 F.</p>
<p>[The above text and the figure are adapted from Reference 1]</p>
<p><a href="https://i.sstatic.net/xZOMk.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/xZOMk.jpg" alt="enter image description here" /></a></p>
<p>Now, if the action potential spreads down from the axon hillock (which is located to the left side of the axon and is not shown in the picture) as in Figure 3.12 above, it will spread only one way to the distal end of the axon [inset 2, 3, and 5] because the previously depolarized channels will still be inactive (or in refractory period) for a while after the action potential ends [inset 4].</p>
<p>[The above text and the figure are adapted from Reference 2]</p>
<p><strong>References:</strong></p>
<ol>
<li><p>Byrne JH. <a href="https://nba.uth.tmc.edu/neuroscience/m/s1/chapter03.html" rel="nofollow noreferrer">Chapter 3: Propagation of the Action Potential. Neuroscience Online.</a> The University of Texas Health Science Center at Houston (UTHealth). McGovern Medical School.</p>
</li>
<li><p>Augustine GJ. <a href="https://www.hse.ru/data/2011/06/22/1215686482/Neuroscience.pdf" rel="nofollow noreferrer">Chpater 3. Voltage-dependent membrane permeability.</a> In: Purves D, Augustine GJ, David Fitzpatrick D, Hall WC, Lamantia AS,‎ McNamara JO, Williams SM, editors. Neuroscience. 3rd ed. Sunderland, Massachusetts: Sinauer Associates Inc; 2004. ISBN-13: 9780878937257 ISBN-10: 0878937250.</p>
</li>
</ol>

CC BY-SA 4.0
2019-10-17T10:08:34.527Z
2019-10-17T10:08:34.527Z
2020-06-17T09:55:31.633Z
user287279
<p>Hello I am taking a neurophysiology course and learned about the action potential and that it propagates in one direction from axon hillock down along the axon. 
I understand that action potential is simply the voltage difference caused by influx of Na+ ions causing the inside of the cell relatively more positive</p>

<p>My questions is why/how does action potential propagate? <strong>why should it propogate?</strong></p>

<p>If at one site where a sodium channel opens to depolarize, and k+ channel polarize the cell to membrane potential <strong>at that site</strong>, how does this affect the next neighboring ion channels down the axons to open and close? Why doesn't it just cause an action potential at one section of the membrane and stop? What exactly causes the neighboring ion channels to open continuously as well after the previous one open and closes? </p>

<p>Textbooks just says 'it propagates down the axon' but never mentions why it should and why should the previous ion channel configuration affect the next region of the membrane. I guess this might be a rather dumb question but I hope I did not confuse you, I realize my questions is a bit odd?</p>

<neuroscience>
what causes an action potential to propagate exactly and why should it propagate?
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# what causes an action potential to propagate exactly and why should it propagate?

(Asked by: username_0 on 2019-10-15)

Hello I am taking a neurophysiology course and learned about the action potential and that it propagates in one direction from axon hillock down along the axon.
I understand that action potential is simply the voltage difference caused by influx of Na+ ions causing the inside of the cell relatively more positive

My questions is why/how does action potential propagate? **why should it propogate?**

If at one site where a sodium channel opens to depolarize, and k+ channel polarize the cell to membrane potential **at that site**, how does this affect the next neighboring ion channels down the axons to open and close? Why doesn't it just cause an action potential at one section of the membrane and stop? What exactly causes the neighboring ion channels to open continuously as well after the previous one open and closes?

Textbooks just says 'it propagates down the axon' but never mentions why it should and why should the previous ion channel configuration affect the next region of the membrane. I guess this might be a rather dumb question but I hope I did not confuse you, I realize my questions is a bit odd?

**Tags:** neuroscience


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## 1 answers

### Answer by username_1 on 2019-10-17. Score: 1

When a sodium (Na) channel is stimulated by a strong-enough stimulus (such as from an electrical stimulus or from an incoming action potential), it will open and let a lot of Na+ ions influx rapidly into the cell at that channel, creating an action potential there. The locally high concentration of Na+ ion inside the cell at that channel will cause Na+ ions there to spread around immediately. This will cause depolarization and openings of the nearby Na channels, resulting in new actions potentials occurring at the nearby channels. This whole process will repeat itself again at these channels, causing more new action potentials to spread out further. In each cycle of an action potential, the potassium channel opens after the sodium channel does about 1 msec. Although the opening of the potassium channel causes repolarization, it doesn’t have any effects on the concentration of Na+ ion and does not stop the spreading out of the Na+ ions. Therefore, an action potential inevitably spreads out and out.

[![enter image description here](https://i.sstatic.net/aqduP.jpg)](https://i.sstatic.net/aqduP.jpg)

If the axon is stimulated at its trunk, the action potential will spread out both ways, up and down the axon. Figure 3.1 A shows a schematic diagram of an axon and the charge distribution along the membrane of that axon. More positive charges exist on the outside of the axon, and more negative charges on the inside. Now consider the consequences of delivering some stimulus to a point in the middle of the axon, Figure 3.1 B. If the depolarization is sufficiently large, voltage-dependent sodium channels will be opened, and an action potential will be initiated, and the sodium ion (Na+) will influx into the cell.

The influx positive charges Na+ will spread to the adjacent regions of the membrane. As the charges move to the adjacent regions of the membrane, the adjacent regions of the membrane will depolarize. If they depolarize sufficiently, voltage-dependent sodium channels in the adjacent regions of the membrane will be opened and "new" action potentials will be initiated, Figure 3.1 C. This perturbance of charge distribution will then spread to the next regions and initiates other "new" action potentials, Figure 3.1 D to 3.1 F.

[The above text and the figure are adapted from Reference 1]

[![enter image description here](https://i.sstatic.net/xZOMk.jpg)](https://i.sstatic.net/xZOMk.jpg)

Now, if the action potential spreads down from the axon hillock (which is located to the left side of the axon and is not shown in the picture) as in Figure 3.12 above, it will spread only one way to the distal end of the axon [inset 2, 3, and 5] because the previously depolarized channels will still be inactive (or in refractory period) for a while after the action potential ends [inset 4].

[The above text and the figure are adapted from Reference 2]

**References:**

1. Byrne JH. [Chapter 3: Propagation of the Action Potential. Neuroscience Online.](https://nba.uth.tmc.edu/neuroscience/m/s1/chapter03.html) The University of Texas Health Science Center at Houston (UTHealth). McGovern Medical School.
2. Augustine GJ. [Chpater 3. Voltage-dependent membrane permeability.](https://www.hse.ru/data/2011/06/22/1215686482/Neuroscience.pdf) In: Purves D, Augustine GJ, David Fitzpatrick D, Hall WC, Lamantia AS,‎ McNamara JO, Williams SM, editors. Neuroscience. 3rd ed. Sunderland, Massachusetts: Sinauer Associates Inc; 2004. ISBN-13: 9780878937257 ISBN-10: 0878937250.


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