Action Potential: Passive Spread Current

1. Sep 6, 2008

somasimple

Hi All,

The Action Potential propagation involves a passive event called: Passive Spread or Electrotonic conduction.

Here is some references:
http://www.ncbi.nlm.nih.gov/books/bv.fcgi?rid=mcb.figgrp.6138
http://www.ncbi.nlm.nih.gov/books/bv.fcgi?rid=.0zyfzkapx787Lxyk2TNcPpbCOnVmwIAZMxK6R2 [Broken]
http://www.ncbi.nlm.nih.gov/books/bv.fcgi?rid=mcb.figgrp.6145
http://butler.cc.tut.fi/~malmivuo/bem/bembook/03/03.htm [Broken]

It is defined a Constant Length that enables this passive event.
What are the normalized values of $$\lambda$$ for unmeylinated fibers that have 1 m/s and 20 m/s (such as the giant squid axon)?

Last edited by a moderator: May 3, 2017
2. Sep 9, 2008

somasimple

I brought these examples because they are the rare that have temporal values.
http://hawk.med.uottawa.ca/public/reprints/PropagationAP.pdf [Broken]
http://www.ncbi.nlm.nih.gov/books/bv.fcgi?rid=mcb.figgrp.6145
This one says that the AP (of 2 ms duration) travels/runs at 1 mm/ms. So it must be a thin fiber since it runs at 1m/s.
The space constant of such a fiber is around 0.1/0.2 mm but the "length" of the AP is 2mm. Centered on the peak value of the AP, the space constant is too short to activate any next patch of membrane.
BTW, this example is perhaps not very good. The famous squid axon has a conduction velocity around 20/25 m/s. If we take an AP of the same duration, we get a length that is 40 mm. 4 cm. All experiments recorded a space constant around 5/6 mm => 0.5 cm. It is just 10 fold less the necessary value.

And it is a fact. The facts exclude a passive spread at such a distance. The positive loop involved in the HH model may be discussed. The cable proterties, too.

Last edited by a moderator: May 3, 2017
3. Sep 9, 2008

somasimple

A second disproof comes from myelinated fibers.
The propagation is an extension of the previous model. Unfortunately there is no room for any travel at the node of Ranvier. It is just a few µ large and can't stand a complete AP "length".
If the AP doesn't move in myelinated fibers, it is not the same phenomenon as in unmyelinated fibers.

4. Sep 12, 2008

somasimple

An AP (Action Potential) is traveling all along the axon for an unmyelinated fiber.
Its speed (conduction velocity) is variable and depends of the diameter of the fiber.
A Squid axon has a common speed around 20ms-1.
The duration of an AP is variable ans it is around 2/4 ms for the giant squid axon.
A phenomenon, which has a duration and travels at a known speed, occupies a length that is the result of duration*speed.
In our example we have: 40 mm (0.04 m).

If we record the AP at location Z, we get have at time t1, the value A, t2 => B and so on.
But at time t0, all values exist on the axon but located at x1, x2...
This means its exist sequences ABC, DEF or GHJ.
These sequences may be explained by the positive loop of the HH model. The HH models fits only the second one. And all exist at the same time.

The length of the phenomenon excludes also the local current process. the space constant is too short to enable any next patch.

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5. Sep 13, 2008

somasimple

If we take, now, a common myelinated fiber (a cat one: values come from Bullock).
A fiber that has a 75ms-1 average speed. Its AP duration is measured: 1 ms.

The HH model expects the second AP (Node 2) is initiated when the AP, recorded at Node 1, reaches the A value.
This means a new AP is instantiated every 0.3 ms.

Since we know that an internode length is comprised between 1 and 2 mm, there is 500/1000 per meter.

In the worst case, we get:
1000*0.3E-3 => 0.3sm-1 => 3.3 ms-1
That is far from the expected 75ms-1

In the best case, we get:
500*0.3E-3 => 0.15sm-1 => 6.67 ms-1
That is far from the expected 75ms-1

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6. Sep 13, 2008

somasimple

It is easy to find a solution that fits the measured values (facts).
If speed is 75ms-1 => AP travels at 75mm/ms.
If internode length is 1.5 mm, then a new AP appears at Node 51, just after 1ms.
Thus, it is mandatory that 50 AP exist at the same time.
Since the internode length is presumed constant then each AP is separated by 20 µs and the initiating value A must be much below the expected value.

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7. Sep 14, 2008

Staff: Mentor

Hi somasimple,

If you have a patch of membrane that is in the "resting state" what event is required to initiate a rising potential cascade at that patch?

Can you use the cable conduction theory rigorously to demonstrate that the required event will not occur at a resting patch next to a rising patch? (I am willing to accept numerical solutions to the partial differential equation)

8. Sep 15, 2008

somasimple

I can't because my little computation is based upon registered facts. These facts are only facts and I can't contest them.
In my first example: If a AP "takes" 40 mm, what is a good length for a membrane patch? 1.3 cm? We must admit that a good patch size is shorter because a patch of 1.3 exhibits too many values.
Since a complete AP exists and is 40 mm long, how do you solve the sequences ABC, then CDE then DEF, then FGH and finally GHJ with a single positive loop?

I'll bring another example for the myelinated fiber based on the http://www.ncbi.nlm.nih.gov/pubmed/...el.Pubmed_DefaultReportPanel.Pubmed_RVDocSum"r.

Last edited by a moderator: Apr 23, 2017
9. Sep 15, 2008

Cincinnatus

Somasimple, you CAN perform the calculation that DaleSpam suggested. If you did, then you would see that cable theory predicts that action potentials do propagate down an axon. Of course, you wouldn't accept this anyway since we all know that you don't accept cable theory.

For anyone following along with these silly discussions, the best reference that I know of on cable theory in neuroscience is Cristof Koch's book: Biophysics of Computation.

10. Sep 15, 2008

somasimple

We will round the 23.2 to 23 ms-1.
From figure 6, the AP duration is around 0.5 ms.
The apparent length/duration of the AP is thus 11.5 mm.
And because the AP began at time 0, at time +1ms, an AP must begin at a distance of 23 mm.

If the internode is 2 mm => 11 nodes
1.5 => 15
1 mm => 23

Then divide an AP that is 0.5 ms by our 11 equidistant intervals.
Make a graph and solve the differential equation by readind the values...

11. Sep 15, 2008

somasimple

12. Sep 15, 2008

somasimple

13. Sep 15, 2008

somasimple

I made a mistake because I didn't take account of the first spike where there is a latency. You're right. This delay enables a propagation with higher values. BTW, it doesn't change anything about space constant.

14. Sep 15, 2008

somasimple

The cable model is a simple cascade of low pass filters.
It is normal that each cell fills itself with some virtual delay (at rising phase).
But if the current is shut the cells are all emptied at the same time.
It means that the duration of the initial pulse is reduced with each added cell.
This does not reflect the real situation since AP duration remains unchanged.
And, and there is no propagated delay!!!

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15. Sep 15, 2008

Staff: Mentor

I agree that experimental facts can't be contested, but when you are trying to disprove a theory you must also demonstrate that the theory fails to predict a given experimental fact.

Here, you have compared the length constant with the AP length and determined that the length constant is shorter. So what? As far as I know the cable theory does not predict that the length constant will be the same size as the AP length. Your cited observation therefore does not appear to be in contradiction of the theory.

By the way, the required event in the question I posed above is "a depolarization greater than the threshold". So, in order to demonstrate that the cable model fails to predict AP propagation, you need to rigorously use the full cable theory to demonstrate that the next patch never gets a supra-threshold depolarization.

16. Sep 15, 2008

somasimple

http://hawk.med.uottawa.ca/public/reprints/PropagationAP.pdf [Broken]
So what?
A passive spread that is under theses conditions is unable to promote a propagation.
The given graphs come from the cable theory. The electric schema fails to produce any delay.
So what?

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17. Sep 15, 2008

Staff: Mentor

That is precisely what you have not yet demonstrated rigorously using the cable theory.

18. Sep 15, 2008

atyy

Maybe the space constant measurement is wrong?

And anyway, the internode distance is less than the space constant, so it's not obvious there's a problem.

The point of the node of Ranvier is that it increases the "space constant" of the axon, because it is an "active" process.

All that is needed is that the AP can travel passively from one node to the next, and then it is regenerated "actively" at each node. The problem with passive travel is not so much the speed, but rather that the amplitude of the action potential dies down during passive travel.

Last edited: Sep 15, 2008
19. Sep 15, 2008

atyy

A single action potential can travel down the axon. It does not exist at multiple points on the axon at the same time. The "active" part of the HH model describes the generation of an action potential at a single point. The "passive" part describes how the action potential travels down the axon. The main problem of passive travel is that the action potential dies down (after one space constant, the action potential will be very small). So nodes of Ranvier have to be spaced much shorter than the space constant in order to regenerate the amplitude of the action potential.

20. Sep 15, 2008

atyy

OK, I see your point. But now it seems that the AP is travelling too fast, not too slow?

So first it seemed the AP was too slow. Now its too fast. So maybe that indicates there's no problem - it seems that some adjustment of parameters would produce a value between the slowest and fastest estimates, and our calculation is just too simple?

Last edited: Sep 15, 2008
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