Greater the magnitude of receptor potential, greater is the rate of discharge of action potentials in the nerve fibre.1

Textbook of physiology, by Indu Khurana, pgno796

Now consider a case where stimulus ( strength ) is large , so there is more accumulation of positive charges near the spike generator region, this would then form action potential , this action potential should then travel in both directions just like at initial segment , where SD spike clears the existing EPSPs, so if I apply same logic here then antidromic Action potential should clear those generator potentials. If I am right then how is more stimulus causing more frequent action potentials?

Case2: If we take the scenario where there is no antidromic conduction of action potential ( for some unknown reasons) then more and more generator potentials are coming at spike generator region(1st node of ranvier) then also how it is causing more frequent action potential generation , if we consider that fact refractory period is constant for all action potentials( in a particular neuron)?


1 Answer 1


There are several important points to answering your question, each somewhat independent of the others. First, lets think about this problem from the perspective of the axon hillock, where action potentials are thought to be generated.

Conduction of action potentials requires voltage-gated sodium channels

When you talk about antidromic action potentials, you mean when they start at the "end" of an axon and return towards the cell body. You can also get backpropagating action potentials into the cell body and dendrites, but these are impaired by two things: 1) fewer voltage-gated sodium channels, so the action potential is weaker or not really an action potential at all, and 2) impedance mismatch. The axon is very narrow; the soma is very big in comparison (this is less of a factor in the context of peripheral sensory receptors where the soma is located far from the site of action potential initiation, but it is still true for the neurites there). A few sodium ions coming in around the axon hillock is enough to depolarize that membrane enough to start an action potential, but when those ions diffuse passively into the rest of the soma, they have a lot more membrane area to cover, and they don't cause as much depolarization.

What all of this means is that the "strength" of a backpropagating action potential isn't less than that of an action potential in the axon.

Frequency is relative

When people talk about frequency coding of intensity, they are talking about a gradual increase in frequency, not going immediately to refractory period. For example, a cell may fire at 1 Hz, then fire at 4 Hz, then fire at 16 Hz, then fire at 64 Hz. If the cell has a refractory period of 5 ms, even at 64 Hz it is nowhere near it's theoretical maximum firing rate.

Related to that point...moving ions takes time and cells are not isopotential

Threshold isn't reached immediately in the axon hillock when a "refractory period" ends: that's the difference between an absolute and a relative refractory period. Ion concentrations and ion permeabilities set an equilibrium potential, but, it takes time for the potential to actually reach that equilibrium, and both the present voltage and equilibrium potential can be different in different parts of the cell: this leads to current flow, which takes time.

After an action potential, the axon hillock typically hyperpolarizes for a bit, sometimes followed by a brief depolarization. During that time, if there are other parts of the cell (such as dendrites) that are still relatively depolarized from a receptor potential, ions will be flowing from those areas into the axon hillock. This depolarizes the axon hillock, but again, this takes time (I'm purposely repeating that to convey a feeling of this all being a dynamic, moving process, with ions moving through each step). The amount of time it takes will depend on the voltage difference, so a bigger depolarization in the dendrites will bring the axon hillock back to threshold sooner.

Not that many ions flow during an action potential

This has been a recurring theme here, see this answer: Why is it possible to calculate the equilibrium potential of an ion using the Nernst equation from empirical measurements in the cell at rest?

The change in membrane potential isn't just because ions flow: it's because permeabilities change, briefly creating a new equilibrium potential. If you have in your mind massive quantities of sodium and potassium ions flowing, completely upsetting the ionic balance in the cell and drowning out all other electrical activity, you have it wrong.

Stimuli are often long-lasting

Especially if you are talking about a mechanical stimulus, most will last a lot longer than an individual spike, which is only ~1ms long. So although one transient stimulus can cause several action potentials, often what actually happens is that those receptor potentials are quite long lasting. Effectively, they set a new "resting potential" for the cell which is above the cells' firing threshold.

In summary:

Receptor potentials depolarize the cell, bringing them to or beyond firing threshold. An action potential starts in the axon hillock and propagates down the axon, but only has a minor impact on the rest of the cell. Importantly, the action potential is really brief, not many ions move, and there is current flow in both directions, so the depolarized parts of the cell are still depolarized somewhat even after a spike. As the initial axon segment recovers from post-action potential hyperpolarization and sodium channels leave their inactivated state, current from the receptor potential is flowing in, depolarizing the cell to threshold and causing another spike. Repeat.

  • $\begingroup$ "So although one transient stimulus can cause several action potentials, often what actually happens is that those receptor potentials are quite long lasting. Effectively, they set a new "resting potential" for the cell which is above the cells' firing threshold." --> Would this mean that it then takes less for cell to fire next (as the new resting potential is above the initial resting potential)? I am asking because that would be not consistent with adaptation (which I assume would increase the threshold) $\endgroup$
    – Pugl
    Jul 26, 2019 at 22:33
  • 1
    $\begingroup$ @Pugl Both are possible, on different time scales. When held at a depolarized potentials, cells can somewhat paradoxically become less excitable: this is because of voltage-gated sodium channel inactivation. The more VG Na channels are inactivated, the fewer are available to open, so more depolarization is required to reach the positive feedback underlying an AP (it's possible to depolarize a cell enough that it can never spike, called "depolarization block"). Adaptation can also occur at multiple stages. I can't say much more about a general case. $\endgroup$
    – Bryan Krause
    Jul 26, 2019 at 22:41

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