ps1265A
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#1
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Shouldn't the direction of impulse be the other way around?


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EmmaDalby97
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#2
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#2
I'd think so yeah
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ps1265A
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#3
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(Original post by EmmaDalby97)
I'd think so yeah
It's actually right for some reason, but don't know why :s


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Gnomes&Knights
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#4
This brings back memories from when I was at Swanlea. Not a great school though.
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nexttime
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No its right. Its just presented in a bit of a confusing way.

The right side of the diagram has already depolarised and is now hyper-polarised, so it happened a while ago, yes? Whereas the left side of the diagram is only just depolarising, so the attempted direction illustrated is right to left (opposite to the illustrated flow of time, left to right).

Like I said, its presented weirdly.
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chazwomaq
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#6
The action potential starts at the axon hillock and then proceeds down the axon to the terminal.

Without knowing where the hillock and the terminal are on the diagram, you just have to take their word that it's correct.

As nexttime says, it doens't propagate backwards because of hyperpolarization.

Thought experiment: what would happen if you inserted a micro-electrode into the middle of an axon and depolarised the membrane to start an action potential. Which way would it travel?
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ps1265A
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(Original post by nexttime)
No its right. Its just presented in a bit of a confusing way.

The right side of the diagram has already depolarised and is now hyper-polarised, so it happened a while ago, yes? Whereas the left side of the diagram is only just depolarising, so the attempted direction illustrated is right to left (opposite to the illustrated flow of time, left to right).

Like I said, its presented weirdly.
Thank you so much!


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ps1265A
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(Original post by nexttime)
No its right. Its just presented in a bit of a confusing way.

The right side of the diagram has already depolarised and is now hyper-polarised, so it happened a while ago, yes? Whereas the left side of the diagram is only just depolarising, so the attempted direction illustrated is right to left (opposite to the illustrated flow of time, left to right).

Like I said, its presented weirdly.
I've got one more question if you don't mind me asking?

In terms of a nerve cell, we have the the nodes of Ranvier. It is said that myelinated neurones carry impulses faster than unmyelinated because of saltatory conduction. We know that the myelin sheath acts as an electrical insulator mainly because of the Schwann cell. This means they don't conduct impulses. This therefore means that the nodes of Ranvier conduct impulses?


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Asklepios
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(Original post by ps1265A)
I've got one more question if you don't mind me asking?

In terms of a nerve cell, we have the the nodes of Ranvier. It is said that myelinated neurones carry impulses faster than unmyelinated because of saltatory conduction. We know that the myelin sheath acts as an electrical insulator mainly because of the Schwann cell. This means they don't conduct impulses. This therefore means that the nodes of Ranvier conduct impulses?


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The axon carries current along its length. However, some of it leaks out into the extra cellular space and the membrane potential declines as the current is propagated along the axon. Myelin acts to minimise this leak of current. However there is still some small degree of leak, and at nodes of ranvier there is a high density of voltage-gated sodium channels (which let sodium ions flood into the cell) to re-amplify the membrane potential to its original value. The current is said to 'jump' between nodes of ranvier.

If you compare this to unmyelinated axons, there is a very large amount of leak so the membrane potential needs to be constantly re-amplified and there are sodium channels spread throughout the whole length of the axon membrane to achieve this. Hence propagation is slower in unmyelinated fibres (except for the smallest of axon diameters).


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chazwomaq
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#10
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(Original post by ps1265A)
myelin sheath acts as an electrical insulator mainly because of the Schwann cell. This means they don't conduct impulses. This therefore means that the nodes of Ranvier conduct impulses?
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(Original post by Asklepios)
at nodes of ranvier there is a high density of voltage-gated sodium channels (which let sodium ions flood into the cell) to re-amplify the membrane potential to its original value. The current is said to 'jump' between nodes of ranvier.
SIZE=1]Posted from TSR Mobile[/SIZE]
Asklepios is correct. But to directly answer your question: the myelin sheath does not conduct the impulse. But the axon it surrounds does. The nodes of Ranvier is where the action potential is repropagated. In between the nodes, local current flow occurs causing depolarisation at the next node. So the current doesn't actually jump (hence the inverted commas), it is just periodically repropagated.
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