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action potential question help-don't understand last point

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I don't understand what they mean by the last point-can someone please help?
Reply 1
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For an action potential to occur and for the membrane to, therefore, become polarised, sodium ions need to diffuse into the cell. So if there are a lot of sodium ions inside the cell then the sodium ion potential gradient will be low and so few sodium ions will continue to diffuse in and hence no action potential will occur.
Reply 3
Original post by elliot-99
For an action potential to occur and for the membrane to, therefore, become polarised, sodium ions need to diffuse into the cell. So if there are a lot of sodium ions inside the cell then the sodium ion potential gradient will be low and so few sodium ions will continue to diffuse in and hence no action potential will occur.

Thanks! But when sodium ions pass a voltage gated sodium ion channel, that voltage gated sodium ion channel is stimulated and so opens up so more sodium ions come in if you see what I mean.
Although I understand the diffusion gradient stuff, whenever sodium ions pass a voltage gated sodium ion channel, then that sodium ion voltage gated channel is triggered to open, right?
So therefore that last point still doesn't quite make sense - could you please let me know your thoughts?
(edited 6 years ago)
Hi,

You have a point here! When the membrane is hyperpolarized, that means it is even more negative inside than the -70mV of normal resting potential. Hyperpolarization is normally produced in a post-synaptic neurone that has been INHIBITED by the [inhibitory] neurotransmitter (e.g. GABA); this action of GABA on the post-synaptic receptor causes an IPSP (Inhibitory Postsynaptic Potential), and one mechanism by which it does this is by reducing Na+ entry, by closing gated sodium channels. The point you are querying would b contrary to this hypothesis (is this an Edexcel Q?:smile:)
Reply 5
Original post by elliot-99
For an action potential to occur and for the membrane to, therefore, become polarised, sodium ions need to diffuse into the cell. So if there are a lot of sodium ions inside the cell then the sodium ion potential gradient will be low and so few sodium ions will continue to diffuse in and hence no action potential will occur.


Firstly, I think you mean depolarise instead of polarise - the resting membrane is already polarised (-70 mV) and the influx of sodium ions into the cell causes depolarisation. Also, that only explains why it is more difficult to fire an action potential, not why it is impossible. It is impossible because when the membrane depolarises, sodium channels become inactivated - they are close and cannot become activated and open again until the membrane potential returns to near its resting value (again, around -70 mV). This is the absolute refractory period. The absolute refractory period is followed by the relative refractory period, during which the cell is hyperpolarised (due to delayed closure of potassium channels). During the relative refractory period, the cell is hyperpolarised and therefore a greater stimulus is required to reach the membrane potential and fire an action potential. Remember that during an action potential, very few ions actually rush into and out of the cell relative to the total number of ions, and therefore you never really abolish the concentration gradients - you don't run out of sodium ions diffusing into the cell or potassium ions diffusing out of the cell.
Reply 6
Original post by Jpw1097
Firstly, I think you mean depolarise instead of polarise - the resting membrane is already polarised (-70 mV) and the influx of sodium ions into the cell causes depolarisation. Also, that only explains why it is more difficult to fire an action potential, not why it is impossible. It is impossible because when the membrane depolarises, sodium channels become inactivated - they are close and cannot become activated and open again until the membrane potential returns to near its resting value (again, around -70 mV). This is the absolute refractory period. The absolute refractory period is followed by the relative refractory period, during which the cell is hyperpolarised (due to delayed closure of potassium channels). During the relative refractory period, the cell is hyperpolarised and therefore a greater stimulus is required to reach the membrane potential and fire an action potential. Remember that during an action potential, very few ions actually rush into and out of the cell relative to the total number of ions, and therefore you never really abolish the concentration gradients - you don't run out of sodium ions diffusing into the cell or potassium ions diffusing out of the cell.

This is great- thank you!
As you advised, the Na+ voltage gated ion channels are closed during hyperpolarisation of the cell membrane, so how does this in the mark scheme make sense "some of the Na+ are still in the cell which would reduce the rate of influx" - how can there even be an influx of sodium ions?
Reply 7
Original post by Jpw1097
Firstly, I think you mean depolarise instead of polarise - the resting membrane is already polarised (-70 mV) and the influx of sodium ions into the cell causes depolarisation. Also, that only explains why it is more difficult to fire an action potential, not why it is impossible. It is impossible because when the membrane depolarises, sodium channels become inactivated - they are close and cannot become activated and open again until the membrane potential returns to near its resting value (again, around -70 mV). This is the absolute refractory period. The absolute refractory period is followed by the relative refractory period, during which the cell is hyperpolarised (due to delayed closure of potassium channels). During the relative refractory period, the cell is hyperpolarised and therefore a greater stimulus is required to reach the membrane potential and fire an action potential. Remember that during an action potential, very few ions actually rush into and out of the cell relative to the total number of ions, and therefore you never really abolish the concentration gradients - you don't run out of sodium ions diffusing into the cell or potassium ions diffusing out of the cell.

This is great- thank you!
As you advised, the Na+ voltage gated ion channels are closed during hyperpolarisation of the cell membrane, so how does this in the mark scheme make sense "some of the Na+ are still in the cell which would reduce the rate of influx" - how can there even be an influx of sodium ions?
Reply 8
Original post by macpatelgh
Hi,

You have a point here! When the membrane is hyperpolarized, that means it is even more negative inside than the -70mV of normal resting potential. Hyperpolarization is normally produced in a post-synaptic neurone that has been INHIBITED by the [inhibitory] neurotransmitter (e.g. GABA); this action of GABA on the post-synaptic receptor causes an IPSP (Inhibitory Postsynaptic Potential), and one mechanism by which it does this is by reducing Na+ entry, by closing gated sodium channels. The point you are querying would b contrary to this hypothesis (is this an Edexcel Q?:smile:)

This is OCR :smile: Although I don't think we need to know about GABA stuff - it just still doesn't make sense.:frown:
Original post by Jpw1097
Firstly, I think you mean depolarise instead of polarise - the resting membrane is already polarised (-70 mV) and the influx of sodium ions into the cell causes depolarisation. Also, that only explains why it is more difficult to fire an action potential, not why it is impossible. It is impossible because when the membrane depolarises, sodium channels become inactivated - they are close and cannot become activated and open again until the membrane potential returns to near its resting value (again, around -70 mV). This is the absolute refractory period. The absolute refractory period is followed by the relative refractory period, during which the cell is hyperpolarised (due to delayed closure of potassium channels). During the relative refractory period, the cell is hyperpolarised and therefore a greater stimulus is required to reach the membrane potential and fire an action potential. Remember that during an action potential, very few ions actually rush into and out of the cell relative to the total number of ions, and therefore you never really abolish the concentration gradients - you don't run out of sodium ions diffusing into the cell or potassium ions diffusing out of the cell.


Yes, sorry I did mean depolarise.
Reply 10
Original post by Jpw1097
Firstly, I think you mean depolarise instead of polarise - the resting membrane is already polarised (-70 mV) and the influx of sodium ions into the cell causes depolarisation. Also, that only explains why it is more difficult to fire an action potential, not why it is impossible. It is impossible because when the membrane depolarises, sodium channels become inactivated - they are close and cannot become activated and open again until the membrane potential returns to near its resting value (again, around -70 mV). This is the absolute refractory period. The absolute refractory period is followed by the relative refractory period, during which the cell is hyperpolarised (due to delayed closure of potassium channels). During the relative refractory period, the cell is hyperpolarised and therefore a greater stimulus is required to reach the membrane potential and fire an action potential. Remember that during an action potential, very few ions actually rush into and out of the cell relative to the total number of ions, and therefore you never really abolish the concentration gradients - you don't run out of sodium ions diffusing into the cell or potassium ions diffusing out of the cell.


This is great- thank you!
As you advised, the Na+ voltage gated ion channels are closed during hyperpolarisation of the cell membrane, so how does this in the mark scheme make sense "some of the Na+ are still in the cell which would reduce the rate of influx" - how can there even be an influx of sodium ions?

Can someone help me please?:frown:
Original post by h26
This is great- thank you!
As you advised, the Na+ voltage gated ion channels are closed during hyperpolarisation of the cell membrane, so how does this in the mark scheme make sense "some of the Na+ are still in the cell which would reduce the rate of influx" - how can there even be an influx of sodium ions?

Can someone help me please?:frown:


The only time an action potential is truly impossible is during the absolute refractory period - so sodium channels are inactivated during this time. I'm surprised that mark scheme does not talk about sodium channel inactivation (different to channel closure) at all. I think what the mark scheme is trying to say is that sodium ions diffused into the cell during depolarisation and while the cell is hyperpolarised, there are still some sodium ions left in the cell that haven't been pumped back outside the cell by the sodium-potassium pump. However, in reality, this wouldn't really make a difference to the sodium concentration gradient as only a small number of sodium ions, relative to the total number of sodium ions, diffuses into the cell. And even still, it doesn't explain why an action potential is impossible, it only explains why a greater stimulus would be needed to fire an action potential.

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