law-bug
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I don't get how the action potential is transmitted along the neurone by local circuits.
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The Illuminati
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me neither :bump:
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racheatworld
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(Original post by law-bug)
I don't get how the action potential is transmitted along the neurone by local circuits.
Here's me testing my own revision skills.

Excuse my awful diagram of a neurone:

Image

When an action potential in one part of the neurone occurs, Na+ ions diffuse across the plasma membrane into the neurone. Therefore, Na+ concentration increases in the neurone where the channels are open. (area A on the diagram)

Na+ ions diffuse sideways down a concentration gradient, from an area of high concentration (area A) where the action potential is, to the areas of low concentration (B and C). The Na+ ions diffuse to an area ahead of the action potential (area B) and to the area behind the previous action potential (area C), as both of these "resting" areas have lower concentration of Na+ ions than area A. This movement of the Na+ ions is called the local circuit.

When this local circuit is set up, this depolarises the resting regions (B and C) which causes voltage-gated sodium ion channels to open. Na+ ions flood into the cell and K+ ions flood out of the cell, which generates a new action potential.

As both the area infront of the previous action potential (area B) and the area behind the previous action potential (area C) are depolarised, then surely an action potential should be generated in both directions? However, an action potential is only generated to the area INFRONT of the previous one (area B), because the area behind (area C) is in the refractory period, and therefore cannot be re-stimulated immediately.
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The Illuminati
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(Original post by racheatworld)
Here's me testing my own revision skills.

Excuse my awful diagram of a neurone:

Image

When an action potential in one part of the neurone occurs, Na+ ions diffuse across the plasma membrane into the neurone. Therefore, Na+ concentration increases in the neurone where the channels are open. (area A on the diagram)

Na+ ions diffuse sideways down a concentration gradient, from an area of high concentration (area A) where the action potential is, to the areas of low concentration (B and C). The Na+ ions diffuse to an area ahead of the action potential (area B) and to the area behind the previous action potential (area C), as both of these "resting" areas have lower concentration of Na+ ions than area A. This movement of the Na+ ions is called the local circuit.

When this local circuit is set up, this depolarises the resting regions (B and C) which causes voltage-gated sodium ion channels to open. Na+ ions flood into the cell and K+ ions flood out of the cell, which generates a new action potential.

As both the area infront of the previous action potential (area B) and the area behind the previous action potential (area C) are depolarised, then surely an action potential should be generated in both directions? However, an action potential is only generated to the area INFRONT of the previous one (area B), because the area behind (area C) is in the refractory period, and therefore cannot be re-stimulated immediately.
where do K+ channels come into this?
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racheatworld
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(Original post by The Illuminati)
where do K+ channels come into this?
As with any action potential (I assume, this isn't in my notes), Na+ channels have opened and Na+ ions have moved in to the cell. This causes depolarisation, and once a potential difference of +40mv has been reached, Na+ channels close, K+ channels open and K+ ions move out.

EDIT: Just re-read the way I wrote about K+ channels in my first post, it was written like that in my notes but it isn't so clear now. I think it is as I said above in this post, would be helpful if someone else could clarify this though.
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law-bug
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(Original post by racheatworld)
As with any action potential (I assume, this isn't in my notes), Na+ channels have opened and Na+ ions have moved in to the cell. This causes depolarisation, and once a potential difference of +40mv has been reached, Na+ channels close, K+ channels open and K+ ions move out.

EDIT: Just re-read the way I wrote about K+ channels in my first post, it was written like that in my notes but it isn't so clear now. I think it is as I said above in this post, would be helpful if someone else could clarify this though.
What you've said in both posts is correct i finally got it

fanks anyway
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Fatma Adel
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Dude thank so much for this simple and easy way of explaining local circuits 😄
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macpatgh-Sheldon
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(Original post by racheatworld)
As with any action potential (I assume, this isn't in my notes), Na+ channels have opened and Na+ ions have moved in to the cell. This causes depolarisation, and once a potential difference of +40mv has been reached, Na+ channels close, K+ channels open and K+ ions move out.

EDIT: Just re-read the way I wrote about K+ channels in my first post, it was written like that in my notes but it isn't so clear now. I think it is as I said above in this post, would be helpful if someone else could clarify this though.

Hi boys and girls,

Just saw this Q, and was quite impressed with the explanation of the propagation of the action potential by Illuminati and by OP.

I felt I should try to do my bit to explain the role of K+ ions:-

One crucial baseline fact to remember [often overlooked as it seems too obvious] is that in ALL cells in the body, the sodium concentration is high extracellularly (135-150 mM/l in plasma) whereas the potassium conc is high intracellularly (therefore low in e.g. plasma = 3.5-5.1 mM/l [these figures were the reference values of the labs of the hospital where I trained, and do vary slightly from hospital to hospital]). This difference has a fundamental role in the generation of the resting potential, in tandem with the different permeabilities of cell membranes to Na+ and K+, as well as the existence of the Na+-K+ ATPase pump.

Looking at the action potential, at around the time this reaches its peak [about +40mV as stated], the direction of the electrical gradient for Na+ is reversed, and this reduces sodium influx; in addition, the voltage-gated K+ channels open, so that potassium moves out of the neurone. These events explain the repolarization phase (later downward part of the action potential chart) which is, of course, an increase in negativity of the inside of the membrane.

The movement of potassium is much slower than that of Na+, which explains the shallowness of the hyperpolarization part of the curve.

The role of K+ in propagation of the impulse is not generally described, but I would predict that it is deducible from the same principles as that of Na+.

One other important point that should not be missed out is the propagation of the action potential in myelinated neurones. You will remember that myelin sheaths are separated by a notch at the nodes of Ranvier. So the explanation by both the above-quoted TSR members needs to be extended to the fact that the lengthwise movement of Na+ [and presumably of K+] along the inside of the membrane involves a jump from one node of Ranvier to the next, so the propagation is much quicker - it has been estimated that the neuronal impulse in a myelinated nerve is 50 times faster than in an unmyelinated nerve [this explains the drastic and profound derangement of function in MS, where there is a defect in myelination].

M (ex medic)
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