Action potential with SEclamp

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nikkip
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Action potential with SEclamp

Post by nikkip »

Hello,

I'm using an SEclamp to obtain Vm and Im data. However, if the clamp's rs is too big or if my cell is too big, I'm getting an action potential at sufficiently depolarized clamp values. How/why is the clamp allowing the cell to fire an action potential?

Thanks!

Nikki
ted
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Re: Action potential with SEclamp

Post by ted »

The phenomenon is called "escape" and it has a long, inglorious history. A voltage clamp is supposed to act like a perfect voltage source, i.e. have a source impedance of 0. Of course this is impossible to achieve with a physical device, but performance is "good enough" if the clamp's source impedance is 2 or more orders of magnitude smaller than that of the cell that is to be clamped. Also, if a cell has significant electrotonic extent, one may achieve "excellent" clamping of membrane in the near vicinity of the clamp electrode, but distal parts of the cell may escape regardless of how low the clamp's source impedance may be . . . because of the electrical separation between the clamp's electrode and the membrane that "escapes."
nikkip
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Re: Action potential with SEclamp

Post by nikkip »

Thanks for the quick reply!

So I'm trying to understand the two issues from a circuits perspective.

Image

My model is a single compartment (i.e. space-clamped). That being said, I realize that NEURON computes voltages at 0 and 1, even though there are only membrane properties at 0.5. I'm not sure how to properly represent this fact in my circuit.

Issue #1: rs > Rm/100
Under this condition, most of Vclamp is dropped across rs, rather than achieving the goal of Vm ~= Vclamp. But even if this is the case, why can this cause AP's? Won't I simply obtain a small deltaVm?

Issue #2: Cell too large ("significant electrotonic extent")

So this relates to my earlier question of how to represent the fact that NEURON computes Vm at the ends of the cell. Or even more basically, what equation NEURON uses to compute these end voltages. Assuming I'm following the correct line of thought, would these 'escaped' action potentials initiate at the cell ends, then propagate to Vm(0.5) which I record?

Thank you!

Nikki
ted
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Re: Action potential with SEclamp

Post by ted »

First realize that NEURON uses the central difference approximation for the second derivative of v with respect to distance along a cable. Then realize that this means each compartment is properly represented by an equivalent T circuit where each compartment's axial resistance is divided into two parts (left and right arms of the T) and the compartment's membrane properties are "lumped" into the T's vertical limb--see Figs. 5.6 and 5.7 in The NEURON Book, or read
Hines, M.L. and Carnevale, N.T.
The NEURON simulation environment.
Neural Computation 9:1179-1209, 1997
which is available at http://www.neuron.yale.edu/neuron/nrnpubs

What this means: attaching a signal source to the 0 or 1 end of a section that has nseg=1 puts some axial resistance between the signal source and the lumped membrane properties. Attaching the signal source to any internal location actually connects the source directly to the ungrounded end of the lumped membrane properties.
nikkip wrote:Issue #1: rs > Rm/100
Under this condition, most of Vclamp is dropped across rs
False and a moment's reflection will tell you why. A correct statement is "if rs < Rm/100 then in the steady state difference between Vclamp and the voltage on the ungrounded side of membrane capacitance will be < 1% of Vclamp."
even if this is the case, why can this cause AP's?
In brief, that's what happens when an excitable cell is subjected to a depolarizing current that is large enough to activate ion channels that produce inward current, and the phenomenological negative conductance produced by activating those channels is not swamped out by the positive source conductance of the depolarizing signal source.
Issue #2: Cell too large ("significant electrotonic extent")
So this relates to my earlier question of how to represent the fact that NEURON computes Vm at the ends of the cell.
If nseg is 1, v at the 0 and 1 locations is calculated algebraically from the boundary conditions at those locations. If there are no other sections, v at the 0 and 1 locations is identical to v(0.5). If other sections are attached to either end, the potential at that point is the weighted average of the potentials at the adjacent internal nodes of those sections that area attached at that point--follows from the notion of conservation of charge (there's no membrane at a 0 or 1 end, so any current that enters along one path to that node must exit along one more other paths from that node).

In a model that has significant electrotonic extent, there must be more than one compartment. The signal source is attached directly to one compartment, but the membrane associated with other compartments is separated from it by intervening axial resistance and the low pass filtering effect of membrane capacitance.
Assuming I'm following the correct line of thought, would these 'escaped' action potentials initiate at the cell ends, then propagate to Vm(0.5) which I record?
That's a good guess. You can test your intuition very quickly--use a CellBuilder to make a model axon 1um diam x 1000 um long, give it hh properties, attach an SEClamp with rs = 1 megohm to its 0 end (or even its first internal node--won't affect the qualitative results) that starts with amp1=-65 dur1=1. Change dur2 to 1e9 and start with amp2 = -60 mV and run a simulation (let tstop be 10 ms just to make sure you give the axon time enough to respond fully) and see if you get a spike. Then

Code: Select all

REPEAT
  add +1 to amp2
  run a simulation
UNTIL you get a spike or it seems clear that the axon won't spike
Where did the spike start, if you got one? What happens to the spike trigger zone as you make amp2 more depolarized?

Try again with rs = 1000 megohms. You'll want to start with amp2 = -20 mV.

What happens if the axon is only 200 um long?
nikkip
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Re: Action potential with SEclamp

Post by nikkip »

First realize that NEURON uses the central difference approximation for the second derivative of v with respect to distance along a cable. Then realize that this means each compartment is properly represented by an equivalent T circuit where each compartment's axial resistance is divided into two parts (left and right arms of the T) and the compartment's membrane properties are "lumped" into the T's vertical limb. What this means: attaching a signal source to the 0 or 1 end of a section that has nseg=1 puts some axial resistance between the signal source and the lumped membrane properties. Attaching the signal source to any internal location actually connects the source directly to the ungrounded end of the lumped membrane properties.
So, I have a fundamental question: At what point does NEURON compute the second difference of v? I realize that the second difference figures in the cable equation, but I didn't think it was explicitly computed in NEURON.

Nevertheless, I believe that understand your explanation wrt the circuit. I am simulating a single compartment with the SEclamp at 0.5. This means that the arms of the T are simply hanging branches and there's no axial resistance between the source (Vclamp and rs) and the lumped membrane properties.
Issue #1: rs > Rm/100
Under this condition, most of Vclamp is dropped across rs
False and a moment's reflection will tell you why. A correct statement is "if rs < Rm/100 then in the steady state difference between Vclamp and the voltage on the ungrounded side of membrane capacitance will be < 1% of Vclamp."
Oops. My mistake. I meant if rs > 100Rm, then most of Vclamp is dropped across rs (or phrased with your more accurate wording).
In brief, that's what happens when an excitable cell is subjected to a depolarizing current that is large enough to activate ion channels that produce inward current, and the phenomenological negative conductance produced by activating those channels is not swamped out by the positive source conductance of the depolarizing signal source.
So is this another way of saying that the SEclamp cannot provide enough (negative) current to compensate for the inward current from the activated ion channels? If yes, I understand how this may be an experimental issue, but why is the simulation current-limited?
In a model that has significant electrotonic extent, there must be more than one compartment. The signal source is attached directly to one compartment, but the membrane associated with other compartments is separated from it by intervening axial resistance and the low pass filtering effect of membrane capacitance.
Ok that makes sense. However, in my case, I only have one compartment. So why do I think I'm seeing escaped AP's with a larger cell? Well, a larger cell means smaller Rm, so the escaped action potentials are actually due to issue #1 (need Rm>100*rs), not due to electrotonic extent, right?

Thank you!
ted
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Re: Action potential with SEclamp

Post by ted »

nikkip wrote:At what point does NEURON compute the second difference of v?
The central difference approximation to the second derivative of potential with regard to distance is inherent in the way NEURON sets up the system matrix for the equations that are integrated numerically. One could say it happens as soon as the total number of segments in a model is > 1.
In brief, that's what happens when an excitable cell is subjected to a depolarizing current that is large enough to activate ion channels that produce inward current, and the phenomenological negative conductance produced by activating those channels is not swamped out by the positive source conductance of the depolarizing signal source.
So is this another way of saying that the SEclamp cannot provide enough (negative) current to compensate for the inward current from the activated ion channels?
No, the issue is the stability or instability of equilibria in a system in which a two terminal signal source with positive source resistance drives another two terminal device that has an I-V relationship that contains a region of negative slope. A classical example would be an NE-2 neon lamp in parallel with a 1 uf capacitor, driven by a 100 V battery through a resistor. If the resistor is small enough that 1/R is equal to or larger than the most negative slope conductance of the lamp's I-V curve, the lamp stays lit; otherwise, the lamp repeatedly flahes on and off. You'd probably find an introductory course on nonlinear dynamics quite interesting and useful.
I only have one compartment. So why do I think I'm seeing escaped AP's with a larger cell? Well, a larger cell means smaller Rm, so the escaped action potentials are actually due to issue #1 (need Rm>100*rs), not due to electrotonic extent, right?
Correct!
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