# Are ion channels/resistancies in Rall's model voltage-independent or not?

In the description of the electrical circuit that models a piece of membrane

in Rall's Scholarpedia article on Ralls model, I read:

"[...] this conductance [$G_e$] is zero under resting conditions, but is turned on by an excitatory synaptic input. [...] this conductance is [$G_i$] zero under resting conditions, but is turned on by inhibitory synaptic input."

Does this mean, Rall is assuming voltage-dependent ion channels? Or why does he stress, that the conductances can (or even must) be turned on and off?

From an electrical point of you, I would guess, that the three resistancy-battery pairs can be replaced by one such pair that would not be voltage-dependent, and maybe even the battery is not needed, as in cable theory:

Or would Rall's model not work anymore, then? This doesn't get clear to me, and would not have been stressed by the author. (Later in his article, Rall doesn't mention $G_r$, $G_e$, $G_i$ anymore, but talks only about membrane resistance $r_m$.)

• Is it clear for you that the switch for those conductances turning on or off is synaptic transmission? Channels at the synapse will only conduct ions when a neurotransmitter binds on them. $G_e$ and $G_i$ could be replaced by a single element, except we know that excitatory and inhibitory synapses exist and may be active at different times and in different places of the cell. So you need them there if you try to describe further attributes in more complicated models (timing, localization). $G_r$ I guess cannot be fused with the other conductances as it is always on. Sep 15, 2017 at 11:56
• Yes, this is clear for me, but I didn't want to think of the initial creation of the potentials but only how the travel and spread (driven and controlled by membrane capitance and resistance). The conductances then will not be turned on and off by neurotransmitters but by the traveling potentials themselves. This means: I am only interested in voltage-gated but not in ligand-gated ion channels. (The initial potentials at time t=0 may just be "given".) Sep 18, 2017 at 7:28
• Then Rall's model is not appropriate for your purposes. Once $G_e$ and $G_i$ are turned on there is no additional voltage modulation - synaptic transmission is their only purpose. His is a 'passive' model, as we call it, and you seem to be interested in 'active' models. Sep 18, 2017 at 8:01
• @vkehayas: In this question I actually was interested in passive transport. Thanks to your explanations I now understand: that travel and spread of EPSPs - once initiated - is controlled solely by $C_m$, $G_r$, and $E_r$ - the other conductances are just off. And they are not voltage- but ligand-gated channels, so won't be influenced by passing-by potentials, right? Sep 22, 2017 at 7:28
• Mostly correct. The passive diffusion is described by the cable equation $$\frac{1}{r_l}\frac{\partial^2 V}{\partial X^2} = c_m\frac{\partial V}{\partial T}+\frac{V}{r_m}$$ Each compartment like the one you showed is assumed to be connected to its neighbours with $r_l$. Sep 22, 2017 at 7:57

If you are interested in passive diffusion and not just membrane potential generation in a dendritic compartment, then you need to look at the cable equation:

$$\frac{1}{r_l}\frac{\partial^2 V}{\partial X^2} = c_m\frac{\partial V}{\partial T}+\frac{V}{r_m}$$

Here, every compartment is connected in series with resistances $r_l$.

We define the length constant as $$\lambda = \sqrt{\frac{r_m}{r_l}}$$

and the time constant as $$\tau = r_mc_m$$

Then the equation can be re-written as:

$$\lambda^2\frac{\partial^2 V}{\partial X^2} = \tau\frac{\partial V}{\partial T}+V$$

The reason $G_e, G_i$ are not mentioned when talking about diffusion along the dendrite is that they are only involved in the generation of voltage $V$ in the compartment.

• Final remark: the ion channels that are responsible for $G_r$, resp. $r_m$ are the leak channels, right? Sep 22, 2017 at 9:40
• OK, $G_r$, resp. $R_m$ depend solely on physical properties of the leak channels, but $r_m = R_m/2\pi a$ also depends on the diameter $a$. Now I got the picture! Sep 22, 2017 at 9:49