Saltatory Conduction: single AP or not?

In summary, Saltatory conduction is a phenomenon that occurs in myelinated axons, where action potentials do not propagate as waves but instead recur at successive nodes of Ranvier. This allows for faster conduction than in unmyelinated axons. The process involves the passive spread of charge between nodes, triggering action potentials in each successive node. This was discovered by Ichiji Tasaki and Andrew Huxley. The cable theory, which is used to explain this phenomenon, shows that at any given time, there are multiple action potentials at different points along the axon, each at a different stage in its time course. The equation for cable theory includes parameters such as the space constant and time constant, which determine the speed and shape
  • #71
somasimple said:
Hmmm, no: It is better to lock something to ensure the transmission. Locking something before the transmission will be the best source of problem and no transmission.

the idea is to prevent misfirings. so yes the idea is to create no transmission at all until everything is ready and the first few nodes are unlocked.

at anyone time no more than 3 nodes would be unlocked
 
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  • #72
every ap is initiated at one node (which had just locked itself so the ap con only go in one direction) passes without delay through the next node (which then gets ready to fire) and finally the by now much weakened ap strikes the third node which is locked so the signal ends there. but in striking the third node it unlocks it so the next ap can pass.

when each node fires it first locks itself so the normal state of all nodes is to be locked
 
  • #73
Suppose a node remains locked => no transmission.
We have seen that APs are overlapping.
If a locking mechanism exists (high probability) it may function when a node is already firing and initiating an AP at the next node. It may avoid node interactions.
 
  • #74
granpa said:
when each node fires it first locks itself so the normal state of all nodes is to be locked
contradiction: nodes are locked at firing and at rest.
 
  • #75
somasimple said:
contradiction: nodes are locked at firing and at rest.
to signals coming from one direction.

each node is unlocked by one signal
it then passes a second signal
at then locks itself and then fires. it doest need to be unlocked to fire because it isn't passing a signal. its creating a signal which only goes one way from the node.

here is a very good picture of what I am talking about:
http://www.pubmedcentral.nih.gov/pagerender.fcgi?artid=1392492&pageindex=8
 
  • #76
granpa said:
here is a very good picture of what I am talking about:
http://www.pubmedcentral.nih.gov/pagerender.fcgi?artid=1392492&pageindex=8
Elaborate!
 
  • #77
somasimple said:
contradiction: nodes are locked at firing and at rest.
locked means it won't pass a signal. not that it can't initiate one. but if it does initiate one then the ap can only go in one direction.
 
  • #78
curve B after the delay at the node fires. this is the firing of the node. it travels to the next node (which does not yet fire) and passes straight through without delay becoming the horizontal part of curve A and finally ending at the second node which is locked. while that node is being unlocked there is a delay. then the next node fires and repeats the process. as each node fires a backward propagating anti-action-potential (curve C) restores the previous internode to its resting state. that is why the node must lock itself before firing.
 
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  • #79
so each node receives 2 signals. a weaker one that unlocks it and a stronger one that causes it after a short delay to fire.
a single signal alone isn't enough to cause it to fire. this also helps to prevent misfiring.
 
  • #80
granpa said:
curve B after the delay at the node fires. this is the firing of the node. it travels to the next node and passes straight through without delay becoming the horizontal part of curve A and finally ending at the second node which is locked. while that node is being unlocked there is a delay. then the next node fires and repeats the process. as each node fires a backward propagating anti-action-potential (curve C) restores the previous internode to its resting state. that is why the node must lock itself before firing.

The A curve shows rising phases of APs
The B curve shows peaks
The C one: falling phases.

Since there is decay in internode then all is OK.
Your process is too complicated and threshold + delay explains fully the process.
 
  • #81
granpa said:
a single signal alone isn't enough to cause it to fire.
No, a single is largely sufficient.
 
  • #82
somasimple said:
No, a single is largely sufficient.

the first signal is small (decayed). and in fact is usually just considered to be part of the rising phase of a single ap which is in fact 2 aps. the older decayed one followed rapidly by a newer more intense one. so it would appear that one signal was sufficient to trigger it but its actually 2
 
  • #83
somasimple said:
The A curve shows rising phases of APs
The B curve shows peaks
The C one: falling phases.

Since there is decay in internode then all is OK.
Your process is too complicated and threshold + delay explains fully the process.

I know what ABand C are supposed to show. but I think they are wrong. A is the older decayed signal from further away. B is the actual signal from a freshly opend gate. and C is the backward propagating anti-action-potential.

I don't understand why you think that's complicated. is it any more complicated than a human eye? or the human brain? seems well within the capability of evolution. moreover it explains the seemingly needless delay at each node. it has to wait for the node down the line to become unlocked. and the whole process is to prevent noise and misfirings.

seems pretty reasonable to me.

not to mention supported visually by this:
http://www.pubmedcentral.nih.gov/pagerender.fcgi?artid=1392492&pageindex=8
 
  • #84
taking two signals implies a delay and AP that is larger in duration.
Since there is a delay, it would be very easy to verify your hypothesis: axon elongation => enlarges or shrinks the AP duration...
I keep the single signal processing.
 
  • #85
I know what ABand C are supposed to show. but I think they are wrong.
Facts are facts: They were recorded and these curves are results of simple computations.
 
  • #86
somasimple said:
taking two signals implies a delay and AP that is larger in duration.
Since there is a delay, it would be very easy to verify your hypothesis: axon elongation => enlarges or shrinks the AP duration...
I keep the single signal processing.
none of that makes any sense. axon elongation wouldn't have any effect on the ap duration. you have apparently misunderstood something i said. and what delay are you talking about. the only delay is the one that is apparent in the graph. about 0.1 ms at each node.
 
  • #87
somasimple said:
Facts are facts: They were recorded and these curves are results of simple computations.

I don't contest the facts. those curves are the basis of everything i am saying. why would i contest them?

look at one node on the axon. first it receives the A signal then the B signal then then it fires then receives the C signal. i think the A signal is the ap from 2 nodes away. the B signal before the delay is the ap from 1 node away. the B signal after the delay is the node itself firing. the C signal is the backward propagating anti-action-potential which is resetting the previous internode to its resting state.
 
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  • #88
somasimple said:
taking two signals implies a delay and AP that is larger in duration.
Since there is a delay, it would be very easy to verify your hypothesis: axon elongation => enlarges or shrinks the AP duration...
I keep the single signal processing.
i'm not sure i understand you but you appear to be forgetting that each ap passes through 2 internodes before being stopped by a locked node. so a single node con receive 2 signals just as fast as before. there is no delay.

the first signal is decayed and doest cause the node to fire. it just unlocks it. the second one causes it to fire (after a short delay).
 
  • #89
somasimple said:
Facts are facts: These curves are results of simple computations.
Computations may distort the facts in that case. But I cant' how you compute your two signals => They must be added and there is a delay.
 
  • #90
somasimple said:
Computations may distort the facts in that case. But I cant' how you compute your two signals => They must be added and there is a delay.

i already told you in post 87. the only delay is the one everyone already agrees on. there are 2 signals because the ap passes through the first node it reaches (which after a short delay fires) and continues on without delay to the next node which it unlocks.

i don't understand what it is that you don't understand. please ask more specific questions.
 
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  • #91
this is just an idea that I am presenting. it isn't established fact. the facts are here:
http://www.pubmedcentral.nih.gov/pagerender.fcgi?artid=1392492&pageindex=8

nodes 1234. nodes 3 and 4 and all further nodes are locked. meaning that they will not pass signals.
ap=action potential

t0 node 1 fires producing ap1 which moves at the speed of sound in water. 1500 m/s
t1 ap1 almost instantly reaches node 2 and passes through WITHOUT DELAY
t2 ap1 almost instantly reaches and ends at node 3 and unlocks node 3 (which takes some time)
t3 after 0.1 ms node 2 fires producing ap2 which moves at the speed of sound in water. 1500 m/s
t4 ap2 almost instantly reaches node 3 and passes through WITHOUT DELAY
t5 ap2 almost instantly reaches and ends at node 4 and unlocks node 4 (which takes some time)
t6 after 0.1 ms node3 fires producing ap3 which moves at the speed of sound in water. 1500 m/s

t0 node 1 fires producing ap1
t1 ap1 almost instantly reaches node 2
t1-t3 delay of 0.1 ms at node 2 before it fires
t3 node 2 fires producing ap2
t4 ap2 almost instantly reaches node 3
t4-t6 delay of 0.1 ms at node 3 before it fires

there is therefore only one delay and it is the 0.1 ms one that everyone already agrees on. so it takes 0.1 ms for an action potential at one node to create an action potential at the next node which is typically 1 or 2 mm away. that gives a net speed of 10-20 m/s. if there were no delay at each node then the signal would move at 1500 m/s. the speed of sound in water. (thats just a guess but its certainly at least a good fraction of that speed)

just before each node fires its ap it relocks itself so the ap can only go in one direction and the previous internode can immediately begin to return to its resting state. this is seen in curve C:
http://www.pubmedcentral.nih.gov/pagerender.fcgi?artid=1392492&pageindex=8
 
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  • #92
https://www.physicsforums.com/showpost.php?p=1897614&postcount=66

This spread takes place with a finite velocity (not necessarily constant) so that graph B becomes later, and graph C earlier towards the distal end of each internode.
finite => delay
+ delay to initiate the next AP since there is a decay in internode.

So I reject, one more time, your point of view. Sorry.
 
  • #93
i thought we agreed that the speed of the ap through the internode was around 1000 m/s and all the delay came at the node as is suggested by the data here:
http://www.pubmedcentral.nih.gov/pagerender.fcgi?artid=1392492&pageindex=8
 
  • #94
granpa said:
i thought we agreed that the speed of the ap through the internode was around 1000 m/s and all the delay came at the node.
It gives 1~2µs for the internode and since the decay is quite 1/3 => total delay around 20 µs.
 
  • #95
according to this:
http://www.pubmedcentral.nih.gov/pagerender.fcgi?artid=1392492&pageindex=8
the delay at the node is about 0.1 ms or 100 microseconds. a 20 microsecond delay would be almost negligible
 
  • #96
travelling at 1000 m/s an ap will travel the 1-2 mm from node to node in about 1 or 2 microseconds. so i guess we agree on that. whether decay results in a delay i don't know but i don't see what it matters. the inherent delay at the node swamps it out anyway.

i fail to see what any of this has to do with the idea that nodes lock and unlock
 
  • #97
granpa said:
whether decay results in a delay i don't know but i don't see what it matters.
A lot, since the threshold will be reached later at next node. o:)
 
  • #98
HH model for unmyelinated axon:
-Describes AP at a point and its propagation.
-Considers active and passive circuit components uniformly distributed along the axon.
-Is a wave equation with a well-defined propagation velocity which matches experiement.
-Reduces in a certain limit to the linear passive cable equation which does not have a well defined velocity.

HS model for myelinated axon
-Nodes active, internodes passive
-Internodes considered as resistor and capacitor in parallel (I think I know what they mean, but agree with somasimple it's not obvious), and apparently equivalently with an equation that resembles the linear passive cable equation.
-Expected backwards propagation from node into internode is apparently seen in the data and discussed.
-No explicit calculation of internode velocity, but heuristic and dimensional arguments are given for its form.

FH model data for myelinated axon
-FH model is standard reference for myelinated axon
-Only FH 1964 seems to be available to me, and does not describe propagation, but there may be other FH papers.
 
  • #99
not much later. not enough to make any difference. the delay at the node is already 100 microseconds anyway. i don't see what internode delay has to do with anything at all much less whether nodes lock or unlock.
 
  • #100
atyy said:
-Expected backwards propagation from node into internode is apparently seen in the data and discussed..
http://www.pubmedcentral.nih.gov/pagerender.fcgi?artid=1392492&pageindex=8
but the backward propagation that is seen (at least in the data i saw) isn't a backward propagating ap. its an anti-ap. it doesn't depolorize the axon. it returns it to its resting state.

but that's not the issue at the moment.
 
  • #101
somasimple said:
Why the capacity is omitted since it is 40 time greater than at node?

Capacity is not omitted - they are discussing resistor and capacitor in parallel as a model for the internode.
 
  • #102
atyy said:
HH model for unmyelinated axon:
-Describes AP at a point and its propagation.
-Considers active and passive circuit components uniformly distributed along the axon.
-Is a wave equation with a well-defined propagation velocity which matches experiement.
-Reduces in a certain limit to the linear passive cable equation which does not have a well defined velocity.
Hi atyy, do you by any chance have a link for this? None of the variants of the HH models that I have seen have any spatial terms, but it has been years since I studied this stuff. I would be very interested to see a single model that includes the voltage-gated channels and spatial terms.
 
  • #103
DaleSpam said:
Hi atyy, do you by any chance have a link for this? None of the variants of the HH models that I have seen have any spatial terms, but it has been years since I studied this stuff. I would be very interested to see a single model that includes the voltage-gated channels and spatial terms.

The HH paper discussing AP propagation in an unmyelinated axon doesn't seem to be free online, unlike the others. I learned about this from somasimple, haven't read it, but looks sensible on a quick scan: http://butler.cc.tut.fi/~malmivuo/bem/bembook/ .

I'll summarise the argument presented by Koch (Biophysics of Computation, OUP 1999) [Vxx is second partial of V wrt x, I haven't bothered about correct signs]:

1. im~Vxx

2. im~Vt+F(V), where F(V) represents the HH model for the AP at a point, including terms that look like dp/dt~f(p)

3. So Vt~Vxx+F(V)

"no general analytical solution is known ... Hodgkin and Huxley only had access to a very primitive hand calculator ... Instead they considered a particular solution to these equations ... postulated the existence of a wave solution ... Vxx~Vtt ... [more steps until an ordinary DE is also obtained] ... Hodgkin and Huxley iteratively solved this equation until they found a value of u leading to a stable propagating wave solution. In a truly remarkable test of the power of their model, they estimated 18.8 m/s at (18.3oC) ... a value within 10% of the experimental value of 21.2 m/s ..."

" ... more than 10 years later that Cooley, Dodge and Cohen solved the full partial differential equation numerically ..."

It boggles my mind they did that with a "primitive hand calculator"?! :smile:
 
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  • #104
I think I may finally understand somasimple's "discontinuity" objection - it makes sense to me if "discontinuous" means "non-analytic".

Linear passive cable equation: Vxx~Vt, which is a linear parabolic partial differential equation.

HH equation: Vxx~Vt+F(V,p,dp/dt), where p are the HH point conductance parameters. The considerations in its derivation are the same as in deriving the cable equation, but it is not parabolic. This is usually called the HH equation only if p is not a function of x, but I will refer to it as the HH equation even for p(x).

For an unmyelinated axon, some parameter like the density of sodium channels pn is spatially constant.

For a myelinated axon, the spatial distribution of sodium channels can presumably be modeled by pn(x), which if analytic will approach zero only asymptotically, and the equation will not be exactly parabolic for any axon segment, and we cannot do an exact separation into "active" and "passive" compartments (HS discuss this, but in different language, they say the internode may be active, but not active enough for current to lead voltage).

If pn(x) is smooth but not analytic, then it can be exactly zero over some internode segment, and the equation will reduce exactly to the cable equation. In this case we can do an exact separation into "active" and "passive" compartments.

Presumably since the full analytical solution is not known, whether one chooses the parameter to be smooth and analytic, or smooth but not analytic, will be a matter of numerical convenience, since the difference will probably not be experimentally detectable.
 
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  • #105
atyy said:
3. So Vt~Vxx+F(V)
OK, that makes sense. The Vt~Vxx part is the cable equation and the F(V) part is what I knew as the HH model. I just hadn't seen them put together like that, but it is pretty obvious when someone else points it out for you :smile:
 

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