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This is a biology+physics question. Since there is no biology section, I am asking here.

In brain, when a neuron reaches to threshold value, it fires an action potential. In most graphics, that is shown as 40 mV. An example is here

Neuron's this potential reaches to all synapses where this neuron is connected to, and chemically transferred to other neuron and transformed into electric charge again.

My question is, because in artificial models not shown, if that whole 40 mV is distributed to those synapses according to synapse strengths, or do all synapses take a total 40 mV which is like creating a potential while it is not there?

Another small question is if the collected potential is stored in neuron if it cannot reach to threshold value like a capacitor in circuit?

enter image description here

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I'm migrating this because you probably would've posted on Biology first if you'd known about it, and it doesn't look like you're going to get the answer you're looking for here. –  Shog9 Nov 27 '12 at 1:10
    
Can you revise your question? I am not sure what you mean when you say the "whole 40 mV is distributed" versus "a total 40 mV" which is "not there." –  yamad Nov 28 '12 at 17:38
    
Well, it is about electron movements actually. That's the reason I have written it is biology+physics question. Normally, in an electric circuit, if you have 40 mV in circuit, this voltage is divided according to resistance in parallel wires. I mean electrons are divided, but with resistance it can be defined as voltage. Anyway, neurons send their signals to other neurons by both electric charge on axon and chemical at synapses. My question was if that electric charge is shared when the active potential reaches to where axon starts making branches to axon terminals. –  tcak Nov 28 '12 at 18:05
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3 Answers

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I think you are really wondering how axonal morphology affects the voltage along the axon. The short answer to your question is that each synaptic terminal sees the same voltage because that voltage is generated "fresh" at each site. This has nothing to do with synaptic strength, which determines the amount of voltage deflection in the postsynaptic terminal but not in the presynaptic terminal.

The axon has two major properties that allow for this even activation across its branches. First, as @Armatus said, the axon is an active structure, meaning that the action potential (AP) regenerates as it travels along the axon. This means that there is essentially no decay of the voltage as the AP propagates.

Second, the axonal geometry appears to be carefully regulated. As you seem to have guessed, there has to be sufficient current flow to activate an action potential in the neighbouring patches of membrane or else there is a propagation failure and the AP stops. This is a problem at axonal branch points (see Box 1 in Debanne, 2004), where the diameters of daughter branches have to be correctly sized so that there is little or no extra electrical load. (For instance, if one branch stays the same size as the mother, the other branch is an extra load that will "steal" current and leave less current travelling down the mother branch). Recent work using voltage-sensitive dyes have shown that in mammalian neurons, in practice, propagation failures are rare (Foust et al, 2010). This suggests that there is fine regulation to ensure that the geometry is "just right." (Those "just right" rules are basically governed by cable theory, which I've previously described here.)

To your last question, yes, there is time-dependent decay of subthreshold potentials because the membrane is a capacitor (~1 uF / cm^2). I wouldn't call this "storage" exactly, but it does work just like a capacitor in a circuit. Many people think of the membrane as a capacitor and a resistor in parallel.


Debanne, D., 2004. Information processing in the axon. Nature Reviews Neuroscience 5, 304–316.

Foust, A., Popovic, M., Zecevic, D., McCormick, D.A., 2010. Action Potentials Initiate in the Axon Initial Segment and Propagate through Axon Collaterals Reliably in Cerebellar Purkinje Neurons. J. Neurosci. 30, 6891–6902.

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Action potentials are governed by the ion concentrations on different sides of the membranes. When an AP is triggered locally, it spreads because the ions which travel across the membrane during the AP spike diffuse to neighbouring regions and trigger the same response in the ion channels there.

Hence, it is irrelevant where an AP is, it is always the same voltage. The 40mV are not "distributed" at all - they diffuse and cause new depolarisations around them, triggering new 40mV potentials until they reach the neuron terminals. So in a way, yes, the AP propagates by "creating a potential while it is not there". All terminals will have the same 40mV potential (the exact voltage of course may be different between cell types).

Potentials which do not reach the threshold value are stored to some extent. This is in the sense that potentials around neurons are always established by ion gradients - a potential in your sense means a change in concentrations from the resting state. Until the concentrations are normalised again by the various transport mechanisms for ions across the membrane, the potential across the membrane will remain changed and it may be closer or further away from the AP threshold.

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I need to get my essay in by thursday so I can't spare much more time right now, but if someone bumps this on thursday night I can gather some further reading :) –  Armatus Nov 27 '12 at 19:45
    
I liked your answer. As far as I understand, there is no connection between strengths of different synapses of same axon also. This has collapsed all the theory I was telling myself. Anyway, I want to ask one last question. Is it discovered that what kind of events change the value of threshold, or maybe is it fixed for all neurons? –  tcak Nov 28 '12 at 15:50
    
I would make the guess that the voltages (resting potential, AP, threshold) are characteristic for each neuron type, but can be changed for individual cells according to circumstances, like most things about neurons. –  Armatus Nov 28 '12 at 22:45
    
Threshold is not fixed, even within a single neuron over short periods of time. Fundamentally, threshold is reached when enough voltage-gated sodium channels open so that a rapid depolarization occurs--that is, when inward current is much greater than outward current. When this occurs is a complicated function of membrane resistance, leak currents, the recent history of the membrane, other inward currents present, the kinetics and voltage-dependence of outward currents, etc., etc. –  yamad Nov 28 '12 at 23:10
    
@Armatus, the ions that flow in through ion channels during the action potential are almost certainly not the same ions that carry current axially to depolarize adjacent sites. The charge travels (that's what current is) but the ion fluxes are local. –  yamad Nov 28 '12 at 23:12
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Here is a pretty good animation of neurotransmission: google:Neurotransmitter action potential

What they don't show there very well is that the action potential signal does not travel like a voltage on a wire, but as a propagating chemical condition.

At the synapse, how strongly the connection is made to the next neuron is addressed in the 'long term potentiation' module.

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I have read these documents before. But no matter how the potential is carried, I think the potential must be divided in its circuitry. But unfortunately, I have never seen any information about this yet. Either it is not discovered or no body cared. Maybe it was better I asked this question to biology section. –  tcak Nov 26 '12 at 8:34
    
Each discrete Na/K pump contributes to the action potential, but it isn't until a threshold is met that the action propagates. It's closer to fluid dynamics than electronics: When a pump opens, the ions exchange and equalize, then recover. It's not circuitry, it's concentration gradients that maintain the Voltage and propagate the action potential. In circuits, electrons move freely on a metal surface, in neurons Ions move towards equilibrium. –  MCM Nov 27 '12 at 1:36
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@MCM: don't have the time to reply now, but just wanted to say that Na/K pump does not contribute to the generation of AP, but rather to the mainteinance of the resting membrane potential –  nico Nov 27 '12 at 7:56
    
@nico - Yeah, I worded that oddly. Thanks for clearing that up. –  MCM Nov 27 '12 at 13:43
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