# Why do big neurons have lower threshold potentials than small neurons?

1. ### JS-Student

7
Why is that large neurons have smaller threshold potentials than small neurons during external stimulation?

My confusion is because the time constant should be larger, right?

tau = (r_m) * (c_m). where r_m is membrane resistance and c_m is membrane capacitance

Capacitance increases proportionally to surface area (proportional to radius squared) and membrane resistance decreases only in proportion to diameter (proportional to radius). The length constant only decreases in proportion to the square root of the radius while the time constant increases in proportion to radius. A higher time constant with only a slightly smaller length constant seems to indicate that larger neurons should be stimulated after smaller neurons. My understanding is definitely flawed somewhere, but I do not know what I am getting wrong. Any help is greatly appreciated.

2. ### atyy

9,955
If I understand your simple model, your reasoning is incomplete, because when you make the neuron bigger, the membrane resistance will also be less. This is because resistance ~ resistivity/area, where resistivity is a property of the material. In a larger neuron, the membrane area is greater, so resistance will drop.

That being said, could you please give a reference for larger neurons having smaller threshold potentials?

3. ### JS-Student

7
Hi and thanks for the response,

Reference: http://www.kumc.edu/AMA-MSS/Study/neuro3.htm This website states that for externally stimulated neurons, larger neurons have lower threshold potentials and are thus stimulated first. Internally (Physiologically) stimulated neurons result in smaller neurons being stimulated first. Ctrl-F "threshold potential" will take you to the spots I got this information from. I also heard this in my neuroscience class.

Also resistance being proportional to resistivity / area only applies to axial resistance (across the length of the neuron) which is not used for the time constant, but rather the length constant. Length constant = square root of (membrane resistance/axial resistance). Thus the length constant increases less than the time constant. For some reason, everywhere I look, it is stated that resistance across the membrane is proportional to the diameter (or radius) only and not the area of the membrane. I believe this is because membrane resistance is determined primarily by the number of open channels. Increasing the diameter of the axon by 2 will only double the amount of open channels not quadruple it.

4. ### JS-Student

7
I think i am confusing dimensions. The axons are constant length tubes, then increasing the radius only increases the surface area in proportion to the radius, not the radius squared. The capacitance thus only increases the same as the membrane resistance and the time constant shouldn't change. The length constant is thus the only thing that matters?

5. ### atyy

9,955
The time constant is (membrane resistance X membrane capacitance). Let's take a small patch of membrane for simplicity with a certain area and thickness. The resistance is ((resistivity X thickness) / area). The capacitance is ((constant X area) / thickness). So keeping the thickness the same but increasing the area affects resistance and capacitance in opposite ways, so that the time constant is unchanged.

Looking up http://en.wikipedia.org/wiki/Length_constant, the length constant is approximately sqrt(membrane resistance/axial resistance). The membrane resistance will decrease as the radius (since the length is constant as you said), but the axial resistance will decrease as the radius squared (since the axial resistance depends on the cross sectional area of the axon). So the length constant will increase as the axon diameter gets larger, and there will be more spatial summation. [STRIKE]So if the stimulation is distributed spatially, then because spatial summation is larger for the larger axon, the threshold will be smaller.[/STRIKE]

Edit: I'm not sure the final sentence above is right. Here is an explanation by John Koester and Steven A. Siegelbaum that seems more convincing. http://www.ib.cnea.gov.ar/~redneu/2...dnjhkoalgmeho00dbookimagebookdb_7c_2fc~11.htm
"In examination of a neurological patient for diseases of peripheral nerves the nerve often is stimulated by passing current between a pair of extracellular electrodes placed over the nerve, and the population of resulting action potentials (the compound action potential) is recorded farther along the nerve by a second pair of voltage-recording electrodes. In this situation the total number of axons that generate action potentials varies with the amplitude of the current pulse.

To drive a cell to threshold, the current must pass through the cell membrane. In the vicinity of the positive electrode, current flows across the membrane into the axon. It then flows along the axoplasmic core, eventually flowing out through more distant regions of axonal membrane to the second (negative) electrode in the extracellular fluid. For any given axon, most of the stimulating current bypasses the fiber, moving instead through other axons or through the low-resistance pathway provided by the extracellular fluid. The axons into which current can enter most easily are the most excitable.

In general, axons with the largest diameter have the lowest threshold for extracellular current. The larger the diameter of the axon, the lower the axial resistance to the flow of longitudinal current because of the greater number of intracellular charge carriers (ions) per unit length of the axon. Therefore a greater fraction of total current enters the larger axon, so it is depolarized more efficiently than a smaller axon. For these reasons, larger axons are recruited at low values of current; smaller-diameter axons are recruited only at relatively greater current strengths."

Last edited: Jul 18, 2014