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
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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?