Myelin increases resistance across the cell membrane

[granpa]

according to wikipedia:
Myelin increases resistance across the cell membrane by a factor of 5,000 and decreases capacitance by a factor of 50.[citation needed]


why would it decrease capacitance? I thought dielectrics were added between capacitor plates to increase capacitance.
https://www.physicsforums.com/showthread.php?t=257719&highlight=capacitance


does this change in capacitance explain the increase of he speed of the action potential? what effect would capacitance be expected to have on signal velocity? it would reduce the number of ions that would need to move acress the membrane to produce a given voltage so I just assume it would increase the speed of the signal.a myelintated axon might transmit AP's at 75 m/s.
if the same axon is demyelinated then how fast would the AP travel? (as determined by experiment not theory)

 

[atyy]

Not sure, would have to look at the right equations and numbers. But off the top of my head, C~kA/d (the exact formula depends on the geometry) where A is the area of the plates, k is related to the dielectric constant and d is the distance between the plates. So it may be that myelin increases d much more than it changes k.

 

[Moonbear]

I'll start at the end of your question, because that's the easy answer. In an unmyelinated giant squid axon (~500 micrometers in diameter), an action potential is transmitted at about 25 m/s; that has been directly measured experimentally. In an unmyelinated human nerve of about 10 micrometers diameter, the conduction velocity is calculated then to be about 0.5 m/s. Addition of myelin increases conduction velocity by a factor of about 100 (so that 10 micrometer human nerve fiber would have a conduction velocity around 50 m/s).

As for how myelin reduces capacitance, one first has to understand a bit about how a neuron cell membrane is constructed and the structure of the myelin sheath. In the neuron cell membrane, capacitance occurs because of the hydrophobic center of the plasma membrane, which is an insulator, and the hydrophilic surfaces (both extracellular and intracellular) that are good conductors.

Myelin, like neurons, also is composed of cells with cell membranes constructed by a lipid bilayer with a hydrophobic center and hydrophilic surfaces (this is common to all cells). Myelin wraps around neurons MANY times. Each turn of myelin around the neuron works like another capacitor connected in SERIES. This is the key concept. If you have capacitors in series, the total capacitance remains constant while the capacitance of each cell membrane is reduced by a factor proportional to the number of capacitors (wraps of cell membrane). So, if there were 25 wraps of myelin around a section of neuron, the capacitance in that location would be 1/25th of the capacitance in an area that did not have myelin.

The reason that myelin speeds the action potential is that an action potential is NOT generated along the areas covered by the myelin sheath. The membrane depolarization, divided among all those layers of myelin, isn't adequate to reach the threshold potential required to produce an action potential. So, a full action potential only occurs at the nodes of Ranvier, which are the gaps between the myelinated sheaths.

 

[granpa]

capacitors in series. that makes it so simple. why didnt I think of that? the formula though is:
1/net capacitance=1/c1+1/c2+...+1cn
so the net capacitance decreases.

but on the ether hand I didnt ask about nonmyelinated axons. I was asking about de-myelinated myelinated axons.

 

[Moonbear]

capacitors in series. that makes it so simple. why didnt I think of that? the formula though is:
1/net capacitance=1/c1+1/c2+...+1cn
so the net capacitance decreases.

No reason you'd think of that without it being explained. Yes, that is the formula for capacitors in series.

but on the ether hand I didnt ask about nonmyelinated axons. I was asking about de-myelinated myelinated axons.

Non-myelinated would be the same as de-myelinated, except they're actually alive and viable so you could test conductance. De-myelination would only occur in a pathological condition, and I'm not sure that other problems wouldn't be associated with that to affect any results.

Edit: Though, if you need to see a reference to believe it (which I understand), here is one that's freely available where they experimentally demyelinated axons.
http://www.jgp.org/cgi/reprint/95/5/867.pdf

That should take you directly to a PDF file of the article. Let me know if it doesn't work.

 

[Dale]

De-myelination would only occur in a pathological condition, and I'm not sure that other problems wouldn't be associated with that to affect any results.

I'm not 100% sure of this (say 85% sure), but I believe that the voltage gated channels are not normally present to any significant degree underneath the mylein sheath, only at the nodes. So after de-mylienation the action potential can no longer jump to the next node, nor are there enough channels to carry it without jumping, so the AP dies out entirely and does not propagate.

 

[atyy]

Non-myelinated would be the same as de-myelinated, except they're actually alive and viable so you could test conductance. De-myelination would only occur in a pathological condition, and I'm not sure that other problems wouldn't be associated with that to affect any results.

I'm not 100% sure of this (say 85% sure), but I believe that the voltage gated channels are not normally present to any significant degree underneath the mylein sheath, only at the nodes. So after de-mylienation the action potential can no longer jump to the next node, nor are there enough channels to carry it without jumping, so the AP dies out entirely and does not propagate.

I agree with DaleSpam's very helpful comment.

Just to expand a little, comparing myelinated and unmyelinated axons requires some care. Koch (Biophysics of Computation, OUP 1999) gives u ~ d for myelinated axons, u ~ sqrt(d) for unmyelinated axons, v ~ sqrt(d) for the pseudovelocity of a linear passive cable, and also states that the velocity for a myelinated axon is limited by the time constant of a linear passive cable.

 

[atyy]

The reason that myelin speeds the action potential is that an action potential is NOT generated along the areas covered by the myelin sheath. The membrane depolarization, divided among all those layers of myelin, isn't adequate to reach the threshold potential required to produce an action potential. So, a full action potential only occurs at the nodes of Ranvier, which are the gaps between the myelinated sheaths.

"...divided among all those layers of myelin..."

What an interesting concept! that makes it so simple. why didnt I think of that?

So the internode depolarization would actually be greater in a demyelinated axon compared to a myelinated axon?

And the initial depolarization at a node (where it is no longer divided among all the layers of myelin) would actually be larger than the initial depolarization at a point in the preceding internode (where it is divided among all the layers of myelin)?

 

[atyy]

Passive propagation is very fast, but decays over time for two reasons. First, the current leaks out of the membrane. To prevent this we should increase the membrane resistance R. This is the first job of myelin. Second, the membrane capacitance C filters the signal, so a big narrow signal will become a short broad signal. To prevent this we should decrease membrane capacitance C. This is the second job of myelin.

Obviously, myelin can only do so much, and the action potential will eventually decay, even in the presence of myelin, to a point where it has to be regenerated. Thus the presence of the nodes.

The nodes cannot be myelinated, otherwise the voltage-dependent channels that generate the action potential will not have access to the extracellular sodium whose inward flow generates the action potential. You cannot eat your cake and have it.

So it seems that the game should be to rely on fast passive propagation for speed, and place the nodes as far apart as possible, only where you need to regenerate the action potential. Of course you want to build in some safety factor, and it is indeed found that blocking a single node does not impede action potential propagation in a myelinated axon.

Edit: But there is something very fishy with this explanation. Time constant Tm~RmCm. What's the point of decreasing Cm, but increasing Rm by the same amount?

Edit: Another part of the puzzle. Space constant Lm~Rm/Ra

 

[somasimple]

I'm not 100% sure of this (say 85% sure), but I believe that the voltage gated channels are not normally present to any significant degree underneath the mylein sheath, only at the nodes. So after de-mylienation the action potential can no longer jump to the next node, nor are there enough channels to carry it without jumping, so the AP dies out entirely and does not propagate.

It is propagated =>
http://www.pubmedcentral.nih.gov/articlerender.fcgi?tool=pubmed&pubmedid=4647244

 

[somasimple]

http://www.ncbi.nlm.nih.gov/pubmed/...nel.Pubmed_DefaultReportPanel.Pubmed_RVDocSum

Interestingly, the overall morphology of Schwann cells lacking claudin-19 expression appeared to be normal not only in the internodal region but also at the node of Ranvier,

These CAPs showed a characteristic double-peak waveform: a peak around a normal conduction velocity and an additional delayed peak (Fig. 5).

 

[atyy]

Edit: But there is something very fishy with this explanation. Time constant Tm~RmCm. What's the point of decreasing Cm, but increasing Rm by the same amount?

Edit: Another part of the puzzle. Space constant Lm~Rm/Ra

I am baffled as to how the argument is quantitatively made. I will just quote from the authority I best know, but don't understand (Koch, Biophysics of Computation, OUP 1999; bolding added by me):

The function of ... myelin ... is to reduce the huge capacitive load ..., as well as to reduce the amount of longitudinal current that leaks across the membrane. The effective membrane capacitance ... is Cm/250 ... while the effective resistance is 250 times higher than the Rm of one layer of myelin.

Measurements (Huxley and Stampfli, 1949) and computations indicate that the time it takes for the currents at one node to charge up the membrane potential at the next node is limited by the time it takes to charge up the intervening internodal membrane. This is determined by the time constant of the membrane tau, which is independent of the geometry of the axon. In this time the spike will have moved across the internodal distance, making the propagation velocity proportional to this distance divided by tau. Since anatomically the internodal distance is linearly related to the diameter of the axon, the velocity of propagation will be proportional to the fiber diameter, u ~ d, rather than the square-root dependency found for unmyelinated fibers.

 

[somasimple]

I'm not 100% sure of this (say 85% sure), but I believe that the voltage gated channels are not normally present to any significant degree underneath the mylein sheath, only at the nodes. So after de-mylienation the action potential can no longer jump to the next node, nor are there enough channels to carry it without jumping, so the AP dies out entirely and does not propagate.

http://www.ncbi.nlm.nih.gov/pubmed/...nel.Pubmed_DefaultReportPanel.Pubmed_RVDocSum

Although K+ channels were not yet segregated

As K+ channels were increasingly sequestered in juxtaparanodes, conduction became progressively insensitive to K+ channel block.

 

[atyy]

Edit: But there is something very fishy with this explanation. Time constant Tm~RmCm. What's the point of decreasing Cm, but increasing Rm by the same amount?

Edit: Another part of the puzzle. Space constant Lm~Rm/Ra

I could understand an argument along these lines:

Lm ~ distance after which the action potential has decayed too much.

Tm ~ time taken to charge the membrane

What we'd like to do is increase Lm and decrease Tm. However, if we increase Lm by increasing Rm, then we also increase Tm. So to offset the increase in Tm by Rm, we also decrease Cm, so that Tm at least remains constant.

However, that is not identical with Koch's argument cited above.

 

[somasimple]

So to offset the increase in Tm by Rm, we also decrease Cm, so that Tm at least remains constant.

However, that is not identical with Koch's argument cited above.

=> Divergence of opinion with Prof C Koch?
That is a dangerous personal opinion.

 

[atyy]

=> Divergence of opinion with Prof C Koch?
That is a dangerous personal opinion.

Yes, it is. But Prof C Koch is a nice guy, so he won't hit me, although he might punish me by discussing consciousness.

 

[Dale]

It is propagated =>
http://www.pubmedcentral.nih.gov/articlerender.fcgi?tool=pubmed&pubmedid=4647244

Thanks for the info. They do appear to propagate. In number 5 they do mention varying degrees of demyleination, so I am not sure if it still propagates with complete demyleination, but there is at least some "extra" mylein than is strictly needed.

 

[Moonbear]

Thanks for the info. They do appear to propagate. In number 5 they do mention varying degrees of demyleination, so I am not sure if it still propagates with complete demyleination, but there is at least some "extra" mylein than is strictly needed.

The reference I cited earlier also measured propagation in demyelinated neurons. Nonetheless, this is why I advised caution in comparing the concept of DEmyelinated to NONmyelinated, because I'm not sure that all else is equal if you remove the myelin, either via pathological or experimental mechanisms.

 

[somasimple]

Yes, it is. But Prof C Koch is a nice guy, so he won't hit me, although he might punish me by discussing consciousness.

That's fine but there is another problem!
If the capacitance is effectively 50 time less with the wrapping of myelin, you forgot a crucial parameter in your computation: length => area.
if a node is 2.2 pF for 0.5 µm then an internode (1 mm) is (2.2/50)*2000= 88 pF.
Did I said Divergence?

 

[Dale]

I advised caution in comparing the concept of DEmyelinated to NONmyelinated, because I'm not sure that all else is equal if you remove the myelin, either via pathological or experimental mechanisms.

I agree.

 

[somasimple]

Thanks for the info. They do appear to propagate. In number 5 they do mention varying degrees of demyleination, so I am not sure if it still propagates with complete demyleination, but there is at least some "extra" mylein than is strictly needed.

I agree.
But what about this?
https://www.physicsforums.com/showpost.php?p=1886218&postcount=19

 

[Dale]

I agree.
But what about this?
https://www.physicsforums.com/showpost.php?p=1886218&postcount=19

It doesn't really matter as long as you are consistent. If you measure resistance in ohms*cm² and capacitance in farads/cm² then you have the same time constant as if you measure resistance in ohms and capacitance in farads. You just have to be consistent.

 

[somasimple]

I have some difficulty with your reply:
A cylindrical resistance is proportional to volume => proportional to length.
http://en.wikipedia.org/wiki/Electrical_resistance
A plane capacitor is proportional to area => proportional to length.
http://en.wikipedia.org/wiki/Capacitor#Capacitance
See computations =>
http://butler.cc.tut.fi/~malmivuo/bem/bembook/21/21.htm
I'm consistent.

Where did you find a unit of ohm*cm²?

 

[Dale]

A cylindrical resistance is proportional to volume

No, resistance is proportional to length/area. Volume is length*area. So, resistance is not proportional to volume.

Where did you find a unit of ohm*cm²?

The "area specific resistance" is in ohm*cm². It is the appropriate "normalized" resistance for current through a membrane. To the best of my knowledge it is used primarily for characterizing fuel cell membranes and neuron membranes.

Note Gm listed in your last reference (http://butler.cc.tut.fi/~malmivuo/bem/bembook/21/21.htm ). Conductance/unit area is simply the inverse of area specific resistance and is, IMO, more convenient.

 

[atyy]

That's fine but there is another problem!
If the capacitance is effectively 50 time less with the wrapping of myelin, you forgot a crucial parameter in your computation: length => area.
if a node is 2.2 pF for 0.5 µm then an internode (1 mm) is (2.2/50)*2000= 88 pF.
Did I said Divergence?

Yes, my solution is problematic. I am completely baffled as to how the "standard answer" would be justified quantitatively with all holes in the argument filled in. Koch mentions numerical computations and a "precise" argument called dimensional scaling, which he does not describe. In short, I do not think his text contains the details which would enable one to reach his conclusions firmly.

 

[somasimple]

Yes, my solution is problematic. I am completely baffled as to how the "standard answer" would be justified quantitatively with all holes in the argument filled in. Koch mentions numerical computations and a "precise" argument called dimensional scaling, which he does not describe. In short, I do not think his text contains the details which would enable one to reach his conclusions firmly.

Atyy,
I bought the book of Koch. I'll see.
I'll ask Ted Carnevale...
Edit:
I'm unable to draw the equivalent circuit since the capacitor must be oriented to the external milieu but connected to the internal one. The resistance must be longitudinal.
It gives a worst solution than before.

 

[somasimple]

No, resistance is proportional to length/area. Volume is length*area. So, resistance is not proportional to volume.

You're right. So resistance is still proportional to length. So is Capacitance

 

[somasimple]

Yes, my solution is problematic. I am completely baffled as to how the "standard answer" would be justified quantitatively with all holes in the argument filled in. Koch mentions numerical computations and a "precise" argument called dimensional scaling, which he does not describe. In short, I do not think his text contains the details which would enable one to reach his conclusions firmly.

See table 2
http://www.pubmedcentral.nih.gov/picrender.fcgi?artid=1260098&blobtype=pdf
http://www.ncbi.nlm.nih.gov/pubmed/...nel.Pubmed_DefaultReportPanel.Pubmed_RVDocSum
The time constant is already 10 fold too high.

 

[Dale]

You're right. So resistance is still proportional to length. So is Capacitance

Yes for resistance.

Capacitance is proportional to area/distance. So capacitance is inversely proportional to length if you wish to say it that way. (Although "distance" is a better description of the separation between plates than "length" since "length" connotes the largest dimension of an object and the separation between the plates is the smallest dimension)

 

[somasimple]

we are speaking about the area of the plates, here.
the distance between the plates, d was already implied.
two plates of an area of A and separated by a distance d.
C is proportional to A and inversely proportional to d since C =e*A/d

Thus if A is augmented (internode), even if d is augmented, C is augmented
since a node has a length of L1= 0.5 µ => area = 2*pi*R*L1 => 2.2 pf

the same plates where d is *50 => C/50
an internode is 2000 time longer
2*pi*R*L1*2000 => (2.2/50)*2000 = 88 pf
and the computation is simplified since the perimeter augments with each wrap => C>88pf

 

[Dale]

C is proportional to A and inversely proportional to d since C =e*A/d

Yes.

Thus if A is augmented (internode), even if d is augmented, C is augmented

That depends entirely on which is augmented more. If they are both doubled then C is unchanged. If A is doubled and d is tripled then C is reduced to 2/3 of its original value. On the other hand if A is tripled and d is doubled then C is augmented to 3/2 of its original value.

 

[somasimple]

DaleSpam,
Give the results in our example.
50 turns of myelin (50*d) and a length that is 2000 time longer (2000*A). :zzz:

 

[Dale]

50 turns of myelin (50*d) and a length that is 2000 time longer (2000*A). :zzz:

This is not correct. The presence or absence of the mylein doesn't change the length at all. It may slightly change A by a small increase in the circumference.

 

[somasimple]

This is not correct. t may slightly change A by a small increase in the circumference.

Thats is not a result at all!
Please give us your result (and computation)?

The presence or absence of the mylein doesn't change the length at all.

Where did I said the length was modified?

 

[Dale]

Thats is not a result at all!
Please give us your result (and computation)?

What result and computation are you talking about?

Where did I said the length was modified?

Your previous post where you said:

50 turns of myelin (50*d) and a length that is 2000 time longer (2000*A). :zzz:

It is very difficult for me to communicate with you. I know that a large part of that is a language barrier, so I am trying to be patient.