How are sections connected?

The basics of how to develop, test, and use models.
Post Reply
cluff
Posts: 4
Joined: Fri May 14, 2021 11:47 am

How are sections connected?

Post by cluff »

Hi there,

I would like to know how sections in the models are connected to each other, e.g. what the equivalent circuit would look like? Is it the same way that the segments are?

I have looked in the NEURON book and across the forum but can't seem to find a definitive answer.

In the following question about total membrane capacitance, you mention that this can be calculated by summing the capacitance across all segments of all sections:
viewtopic.php?t=2481

this suggests to me that across the sections, the capacitances are in parallel?

If I wanted to compute total input resistance for example, could I compute the input resistances of the different sections and then calculate the total using the equation

1/Rtot = 1/Rsection1 + 1/Rsection2...

Many thanks in advance for your help!
ted
Site Admin
Posts: 6100
Joined: Wed May 18, 2005 4:50 pm
Location: Yale University School of Medicine
Contact:

Re: How are sections connected?

Post by ted »

I would like to know how sections in the models are connected to each other, e.g. what the equivalent circuit would look like?
Fig. 5.6D in The NEURON Book shows the equivalent circuit for a section that has nseg = 1. Its 0 and 1 ends are labeled clearly.
Suppose you had two sections, both with nseg = 1, and you connected the 0 end of one of them to the 1 end of the other. What do you think the equivalent circuit would be?
If you don't have a copy of The NEURON Book, you can download a preprint of chapter 5 https://www.neuron.yale.edu/ftp/ted/boo ... xedref.pdf
In the following question about total membrane capacitance, you mention that this can be calculated by summing the capacitance across all segments of all sections:
viewtopic.php?t=2481

this suggests to me that across the sections, the capacitances are in parallel?
"Total membrane capacitance" is the integral of the membrane capacitance over the entire surface of the cell. The existence of the term "total membrane capacitance" should not be taken to suggest that that all membrane capacitance is in parallel. Since cytoplasm has finite resistivity, and neurites have finite diameters, there is always a finite axial resistance that separates the internal surface of one patch of membrane from another. This resistance is negligible only in the case of a cell whose diameter and electrical properties of cytoplasm and membrane are such that its anatomical length is small compared to the length constant of a cylindrical cable with the same diameter and electrical properties of cytoplasm and membrane. Likewise, extracellular resistance is nonzero, and can be quite large if satellite cells are closely apposed to the external surface of a cell (e.g. myelinated regions), so the external surface of one patch of membrane is in general not in parallel with the external surface of any other patch of membrane. However, except for myelinated regions, the extracellular resistances are much smaller than the intracellular axial resistance.

That's why the answer to your next question
If I wanted to compute total input resistance . . .
is no.
cluff
Posts: 4
Joined: Fri May 14, 2021 11:47 am

Re: How are sections connected?

Post by cluff »

Thank you for your reply.

From the book figure, it looks as if the sections are indeed connected in the same way that segments are, with a resistor that represents axial resistance.

I understand the point that the elements are not in parallel, however I am still unclear as to how the total membrane capacitance can thus be calculated by the summation across segments/sections?

Regarding input resistance (Rin), from my understanding this can be calculated using the formula:

Rin = rm/A, where rm is specific membrane resistance and A is the area.

However, it seems that it is then not trivial to calculate this across the whole multicompartment model due to your explanation?

If I was interested in the Rin of a specific compartment/segment of the multicompartment model and I had blocked all conductances other than g_pas, could I calculate the Rin of this compartment only using the formula rm/A, where rm = 1/g_pas?

Is the recommended way to calculate total Rin then to apply a current, record the steady state voltage and apply Ohms law?

Thanks in anticipation!
ted
Site Admin
Posts: 6100
Joined: Wed May 18, 2005 4:50 pm
Location: Yale University School of Medicine
Contact:

Re: How are sections connected?

Post by ted »

I understand the point that the elements are not in parallel, however I am still unclear as to how the total membrane capacitance can thus be calculated by the summation across segments/sections?
Total membrane capacitance is total membrane capacitance, i.e. the integral over the surface of the cell of membrane capacitance. It is not the "input capacitance of the cell as seen by a patch electrode attached at some point."
Regarding input resistance (Rin), from my understanding this can be calculated using the formula: Rin = rm/A, where rm is specific membrane resistance and A is the area.
Only if the cell is electrotonically compact--which means that its anatomical length must be a small fraction of the length constant of a cable with the same diameter. Consider this simple thought experiment. Cell A is spherical with surface area a, and its input resistance is RA. The membrane of cell B has the same specific resistance (in ohm cm2) as A, but B is dumbbell-shaped, consisting of two spheres, each with surface area a/2, that are connected by a cylindrical neck. Let the length and diameter of the neck be such that (1) the neck's surface area is 1e6 (one million times) smaller than a, and (2) the axial (end-to-end) resistance of the neck is 1e6 (one million) times larger than the input resistance of cell A. You attach a patch electrode to one of the dumbbells of cell B. What is the input resistance that you observe?
If I was interested in the Rin of a specific compartment/segment of the multicompartment model and I had blocked all conductances other than g_pas, could I calculate the Rin of this compartment only using the formula rm/A, where rm = 1/g_pas?
A meaningful term for the result of that calcuation would be "membrane resistance of the compartment after all conductances other than g_pas have been blocked." If the other conductances weren't blocked, the term "membrane resistance of the compartment" would apply. The term "input resistance" is typically used to refer to measurements made at a specific location of a more or less intact cell, not to refer to a measurement or calculation performed on an isolated compartment.
Is the recommended way to calculate total Rin then to apply a current, record the steady state voltage and apply Ohms law?
What you describe is a way to measure input resistance Rin. "Total Rin" is an abuse of language--the adjective "total" is not merely irrelevant but invites confusion. There is no "total Rin" any more than there is a "partial Rin."
cluff
Posts: 4
Joined: Fri May 14, 2021 11:47 am

Re: How are sections connected?

Post by cluff »

Consider this simple thought experiment. Cell A is spherical with surface area a, and its input resistance is RA. The membrane of cell B has the same specific resistance (in ohm cm2) as A, but B is dumbbell-shaped, consisting of two spheres, each with surface area a/2, that are connected by a cylindrical neck. Let the length and diameter of the neck be such that (1) the neck's surface area is 1e6 (one million times) smaller than a, and (2) the axial (end-to-end) resistance of the neck is 1e6 (one million) times larger than the input resistance of cell A. You attach a patch electrode to one of the dumbbells of cell B. What is the input resistance that you observe?
Would the input resistance be RA*2? As the measurement is now restricted to this new "compartment" which has half the surface area?
A meaningful term for the result of that calculation would be "membrane resistance of the compartment after all conductances other than g_pas have been blocked." If the other conductances weren't blocked, the term "membrane resistance of the compartment" would apply. The term "input resistance" is typically used to refer to measurements made at a specific location of a more or less intact cell, not to refer to a measurement or calculation performed on an isolated compartment.
Ah right, sorry I think I was confusing input resistance with membrane resistance due to some inconsistencies in the literature and online resources.

I am interested in the time constant of the neuron. A final question would be then...

In an electrotonically compact single compartment neuron model, can the time constant be defined as:

tau = rmcm where rm is the specific membrane resistance of the compartment (in ohm cm2) and cm is the membrane capacitance of the compartment (in F)?

So if you increased surface area you would then increase capacitance and thus time constant? And rm is independent of area due to it being the reciprocal of conductance which has the unit S/cm2?

(In case I have said something wrong here, the reference I used was: https://link.springer.com/content/pdf/1 ... 6_32-1.pdf)

If you did this calculation on a single compartment of a multicompartment model, would you be correctly calculating the time constant of that compartment?

I see many papers report a time constant of an entire real neuron. I wonder, given that many neurons are not electrotonically compact, if this impacts the recorded measures? And how can you do the same on a multicompartment model e.g. calculate the time constant of the whole neuron model?

Sorry for all the questions!
ted
Site Admin
Posts: 6100
Joined: Wed May 18, 2005 4:50 pm
Location: Yale University School of Medicine
Contact:

Re: How are sections connected?

Post by ted »

Would the input resistance be RA*2
Yes.
can the time constant be defined as:

tau = rmcm where rm is the specific membrane resistance of the compartment (in ohm cm2) and cm is the membrane capacitance of the compartment (in F)?
No, since the resulting units are s/cm2. cm needs to be specific membrane capacitance, with units capacitance/area. Most published literature reports specific membrane capacitance in uF/cm2 and specific membrane resistance in ohm cm2, the product of which is us (microseconds)--but membrane time constant is almost always stated in ms (milliseconds).
if you increased surface area you would then increase capacitance and thus time constant?
No, because if you have twice the area you get not just twice the capacitance but also half the resistance (because doubling the area doubles the number of ion channels). Result: time constant is independent of surface area.
(In case I have said something wrong here, the reference I used was:
If that chapter by Golowasch and Nadim led you astray, perhaps the section in question is written in an ambiguous way or an inadvertent error slipped by proofreading. You might want to re-read the section in question and if you still think that it leads to the conclusion that membrane time constant depends on surface area, contact the authors.
I see many papers report a time constant of an entire real neuron. I wonder, given that many neurons are not electrotonically compact, if this impacts the recorded measures?
Before the development of the patch clamp method, electrical properties of neurons were studied by using "sharp electrodes", i.e. microelectrodes that poked a hole in the cell membrane, to inject current and record membrane potential. Injection of a brief current pulse or a prolonged current step typically produced voltage transients that could not be adequately fit by a single exponential function. This was in agreement with the predictions of cable theory, which had shown that the time course of v was the sum of an infinite series of decaying exponentials. With the advent of patch electrodes, which made a tight seal with the cell membrane instead of ripping a hole that allowed solute and charge to leak into/out of the cell at the point of impalement, it was soon observed that the charging properties of many cells could be well fit by a single exponential, or the sum of two exponentials. This doesn't invalidate cable theory; it merely indicated that many kinds of cells were electrically far more compact than had been thought. Given a cell with a real morphology, the higher order terms would still be present in a mathematical solution, but their amplitudes were far too small to be resolved experimentally.
And how can you . . . calculate the time constant of the whole neuron model?
You don't. The software generates a simulation by numerical methods that approximately solve the equations that describe the model's properties. You can analyze that simulation to see how well it can be fit by a sum of exponential functions.
Post Reply