Modeling a neuron in a spatially smooth Ve

[Qroid_montreal]

Hello,

I'm a grad student trying to model the effects of spatially-distributed, time-varying charges on an action potential. To do this, I'm simulating the Mainen and Sejnowski layer V pyramidal neuron (http://senselab.med.yale.edu/modeldb/sh ... model=2488) and changing the conductance in each section, causing it to spike. As it spikes, I'm playing pre-computed extracellular potential time-series (calculated from the distribution of charges) to each section via the extracellular mechanism. The code runs fine and the results look good.

I just have a question about whether I'm getting the physics right. What I want to do is model the neuron being placed in a dynamic Ve. I want the potential to be spatially realistic, and so have a smooth 1/r spatial dependency. My physics is rusty, but I think this means the field should have an effect on Vi, even for a passive membrane. So do I need to worry about explicitly controlling Vi? Or does the extracellular mechanism take care of this, and it's simply sufficient to play with Ve via the extracellular mechanism?

Let me know if I'm being unclear and I'll clarify!

[ted]

Time to brush up your intuition with a bit of super fine steel wool and a dab of oil.

Think about a cell that is so small you can consider its membrane potential to be uniform. Its membrane potential is given by q/c where c is the capacitance of the cell membrane and q is the charge that has accumulated on the inside of the cell (which happens to be equal and opposite to the charge that has accumulated on the outside of the cell). The cell is sitting in a bath of Ringer's solution. The bath also contains an Ag-AgCl electrode (the "bath electrode") that is attached to ground, so the electrical potential in the bath (call it Vo) is 0 mV. Let's call the electrical potential inside the cell Vi. The transmembrane potential--the potential difference from one side of its membrane to the other--is Vm = Vi-Vo. The cell is at rest, and its resting membrane potential Vm is -70 mV. What is the value of Vi?

Now suppose you take the bath electrode and attach it to the positive end of a 1.5 volt battery; the negative end of the battery is attached to ground. What are the new values of Vo, Vi, and Vm?

[Qroid_montreal]

Embarrassed how rusty I am at this... Let's see what I can do.

In the first situation, Vo = 0 mV, Vm = -70 mV. So Vi = Vm+Vo = -70 mV.

In the second situation, the voltage difference between Vi and ground remains the same (Vi=-70 mV). Vo has been changed by the battery, so Vo = 1.5 V. The new membrane potential is now Vm = Vi-Vo = (-70 - 1500) mV = -1570 mV.

Is this right?

[ted]

In the first situation, Vo = 0 mV, Vm = -70 mV. So Vi = Vm+Vo = -70 mV.

Correct.

In the second situation, the voltage difference between Vi and ground remains the same (Vi=-70 mV). Vo has been changed by the battery, so Vo = 1.5 V. The new membrane potential is now Vm = Vi-Vo = (-70 - 1500) mV = -1570 mV.

You got Vo OK, but not Vm or Vi. Remember the cell's membrane capacitance? If membrane potential were to change from -70 to -1570 mV, the charge on membrane capacitance would have to increase by more than 20 fold. Specifically, the net negative charge inside the cell would have to increase by more than 20 times. In this Gedankenexperiment, the cell is a little sphere sitting in a bath, with no electrode attached to it, so there's no way to inject charge into the cell.* So the transmembrane potential Vm should do what?

*--You might want to consider the electrical diagram that represents the physical system.
Before the battery is inserted:

Code: Select all

interior of cell
   o
   |
+--+--+
|     |
Cm    Rm
|     |
|   Vrest
|     |
+--+--+
   |
   o exterior of cell
   |
 ground

After the battery has been inserted:

Code: Select all

interior of cell
   o
   |
+--+--+
|     |
Cm    Rm
|     |
|   Vrest
|     |
+--+--+
   |
   o exterior of cell
   |
  pos
1.5 V cell
  neg
   |
   o
   |
 ground

[Qroid_montreal]

Am I incorrect in thinking that Vi remains at -70 mV (relative to ground)?

Is Vi also changed by the addition of the battery, such that now Vi = (-70 + 1500) mV = 1430 mV? If that's the case then Vm = (1430 - 1500) mV = -70 mV. This would mean that the battery doesn't change Vm. Is that right?

Thanks.

[ted]

Am I incorrect in thinking that Vi remains at -70 mV (relative to ground)?

You were. Might still be, if you're still thinking that.

Is Vi also changed by the addition of the battery, such that now Vi = (-70 + 1500) mV = 1430 mV?

True.

then Vm = (1430 - 1500) mV = -70 mV. This would mean that the battery doesn't change Vm.

True. How could it? It can't do anything to the charge stored in membrane capacitance. Look at the equivalent circuit. Membrane capacitance is represented as a two terminal device. One terminal is on the inside of the cell, and the other is on the outside of the cell. The only way that charge stored on membrane capacitance can change is if current flows in one of those terminals and then flows out the other. The only way the 1.5 V battery can change the charge stored on membrane capacitance is if current flows out one of the battery's terminals, along a path that passes through membrane capacitance, and finally returns to enter the battery's other terminal. Does such a path exist? The only signal source that can deliver charge to membrane capacitance is the battery labeled Vrest, which represents the net effect of all the open ion channels in the cell membrane. After membrane capacitance accumulates sufficient charge q so that q/cm = Vrest, no more current will flow from the ion channels into the membrane capacitance.