# How does a higher channel density increase speed of propagation of an Action Potential?

If there is a higher density of sodium channels then there will more Na$^+$ flux in the cell. So it's like trying to pass more material through a pipe and therefore the speed of action potential (AP) propagation will be more. This is what comes to me intuitively, but I am not sure about my answer.

Can anyone explain in better words, how the speed of an AP increases with increasing channel density?

• Its actually quite opposite: action potential speed is inversely proportional to channel density. That's why myelinated neurons conduct impulse way faster than non-myelinated ones. – another 'Homo sapien' May 12 '16 at 16:35
• Can you give a reference from where you concluded these lines (first paragraph)? – another 'Homo sapien' May 12 '16 at 16:37
• I have edited your post to replace AP by action potential. Such abbreviations will not be familiar to many people and should not be used in titles. – David May 12 '16 at 18:55

As @another'Homosapien' mentioned in the comments, the reality is actually the opposite: action potential speed is inversely proportional to channel density.

• In other words: lower channel density increases speed of propagation of an Action Potential.

If you decrease the density of ion channels, you increase membrane resistance. This results in less "leakage" of cations, which allows voltages to spread farther and thus reduces the number of action potentials that need to be generated. This results in increased conduction velocity.

It's important to note two things about action potential propagation:

1. Each action potential takes time to occur.
2. The charge (i.e., voltage) that is created dissipates with $\uparrow$ distance.

In fact, we have equations to calculate both the time a voltage change takes to occur and how current flow decreases with distance.

• You can read more about the mathematics behind this and passive membrane properties in general here and here.

Importantly, these equations rely on two constants: length and time.

The time constant, $\tau$, characterizes how rapidly current flow changes the membrane potential. $\tau$ is calculated as:

$$\tau = r_mc_m$$

where r$_m$ and c$_m$ are the resistance and capacitance, respectively, of the plasma membrane. (See this previous answer for an explanation of resistance and capacitance. )

Importantly, these variables partially rely on membrane structure.

• c$_m$ (the capacitance of the membrane) decreases as you separate the positive and negative charges.

• r$_m$ (the resistance of the membrane potential) is the inverse of the permeability of the membrane.

• The higher the permeability, the lower the resistance.

• Lower membrane resistance means you lose ions quicker and therefore signals travel less far

But why? This is where that length constant becomes important. The length constant, $\lambda$, can be simplified to:

$$\lambda = \sqrt {\frac {r_m}{r_e + r_i} }$$

where, again r$_m$ represents the resistance of the membrane and r$_e$ and r$_i$ are the extracellular and intracellular resistances, respectively. (Note: r$_e$ and r$_i$ are typically very small).

Basically, if the membrane resistance r$_m$ is increased (perhaps due to lower average "leakage" of current across the membrane) $\lambda$ becomes larger (i.e., the distance ions travel before "leaking" out of the cell increases), and the distance a voltage travels gets longer.

### Why am I telling you all of this??

How are the time constant and the space constant related to propagation velocity of action potentials?

The propagation velocity is directly proportional to the space constant and inversely proportional to the time constant. In summary:

• The smaller the time constant, the more rapidly a depolarization will affect the adjacent region. If a depolarization more rapidly affects an adjacent region, it will bring the adjacent region to threshold sooner.

• Therefore, the smaller the time constant, the more rapid will be the propagation velocity.
• If the space constant is large, a potential change at one point would spread a greater distance along the axon and bring distance regions to threshold sooner.

• Therefore, the greater the space constant, the more rapidly distant regions will be brought to threshold and the more rapid will be the propagation velocity.

Sooo....

If you decrease the permeability of the membrane (i.e., if you prevent ion pumps from moving ions in/out of the axon), you increase the resistance of the axon membrane, which allows for the voltage created in the action potential to travel farther before dissipating.

• By allowing the voltage to spread farther before necessitating the generation of another action potential, you reduce the time it takes for signal propagation.

In other words, if you decrease the number of ion pumps, you increase membrane resistance (r$_m$). This causes voltage to spread farther and thus reduces the number of action potentials that need to be generated. The result? increased conduction velocity.

• Interesting answer, I'll think about it...certainly higher resistance means better propagation of electrical signals, but I am not as convinced that this is necessarily the correct answer to the OP's question. Also, note that ion pumps and channels aren't the same thing. – Bryan Krause Apr 3 '17 at 22:18