www.neuron.yale.edu

Hi again!

Lets say I am trying to investigate the effect of the different holding voltages in a neuron having an active soma and passive dendrites.
I control the holding voltage by adjusting the background currents in the model.

Can I subject the model to a current clamp, and the look at the voltage change (Vh-V@Iclamp) to determine the effective membrane resistance??

Rm effective = (Vh - V@Iclamp)/current

Also, lets say I am applying the current clamp for 10ms, since the soma is active, the Vm changes with time, ie the Vm rises very quickly within the 1st second and then sort of goes down to a new lower level after that. After 10 seconds the Vm goes back to its holding potential.

Should I read the voltage@Iclamp towards 9.5ms to ensure the voltage is steady in order to estimate the effective resistance ??

Please do advice, thank you so much for your time.

Meena

Meena wrote:Can I subject the model to a current clamp, and the look at the voltage change (Vh-V@Iclamp) to determine the effective membrane resistance??

Try it with a simple model consisting of a cylinder that has these properties:
Length 1000 um
Diameter 1 um
Discretize according to the d_lambda rule with the default value of d_lambda
Insert pas mechanism with default g_pas.
Make sure that cm and Ra have their default values (1 uf/cm2 and 35.4 ohm cm,
respectively).

Don't forget to set v_init to -70 mV, so your simulation starts with a steady baseline at -70.
Make sure that the current pulse lasts long enough for charging to stabilize before
measuring membrane potential. Compare your result with the actual value (1/g_pas).
Can you explain your result?

Try again with Ra = 100 ohm cm. Now what do you get, and can you explain why?

Restore Ra to its default value, and divide g_pas by 100, and try again. What value did
you get, and why?

Hi Ted,

I tried this and I am going to take a stab at explanaining it...

This cable has surface area of 3143.57 um^2=3.143e-05 cm^2

Setting a g_pas=0.001 S/cm^2 for this structures, implies that the Rm would be 1k ohm.cm^2, or RM = 31.82 Mohm.

Applying a current clamp of 1nA for 50ms resulting the Vm rising to -7.5mV.

Thus, (-7.5-(-70))/1nA= 62.5Mohm (double) , and hence this somewhat suggests that the g_pas is half of what it was set to.

Is it possible that I am seeing this because I am looking at the cable as whole with all the parameters included ?
Thus this 62.5Mohm reflects the the capacitor and Rm in parallel with the Ra in series for all the nsegs ?

And thus the "Ra" portion is drawing a little current, making the effective resistance seem higher?

When Ra=100, Vm becomes 29.9mV
and this makes the effective resistance ~100 Mohms.
This probably has to do with increasing the Ra, increasing the resistance value in series, that increases the impedance of the circuit ?

Or has to do with the fact that when Ra=100 ohm.cm, the parallel RC portion draw less current from the Istim resulting in a smaller current and thus a greater effective membrane resistance is seen ?


When Ra is returned to its default value and g_pas is divided by 100, Voltage charges and then discharges like an RC circuit with a tau of about 100ms.

I could not really determine what the effective resistance here would be as the Vm did not stabilize , but its much much greater than the previous values. (much-much actually)

I think when the g_pas here is reduced, the current across the capacitor rises rapidly charges it and starts to discharge. The Vm discharges with tau ~100ms.

Is this still a legit method to determine effective membrane resistance ?
I am a little confused....I hope that I am thinking about this in the right way....
Please advice

Thank you so much for all your time.
Meena

Meena wrote:This cable has surface area of 3143.57 um^2=3.143e-05 cm^2

PI*diam*L = 3141.59 um2. Perhaps you were adding the area of the circular faces at
each end? Don't.

Setting a g_pas=0.001 S/cm^2 for this structures, implies that the Rm would be 1k ohm.cm^2, or RM = 31.82 Mohm.

Close enough. Omitting surface area of the ends gives 31.83 Mohom.
Instead of RM, I'm going to use RN (input resistance of a neuron as measured from the
location of the electrode).

Applying a current clamp of 1nA for 50ms resulting the Vm rising to -7.5mV.
Thus, (-7.5-(-70))/1nA= 62.5Mohm (double) , and hence this somewhat suggests that the g_pas is half of what it was set to.

True if the current is applied at the middle of the section. What would you expect to see
if you moved the point of current injection and voltage measurement to one end, e.g. the
0 end? Test your prediction.

Is it possible that I am seeing this because I am looking at the cable as whole with all the parameters included ?
Thus this 62.5Mohm reflects the the capacitor and Rm in parallel with the Ra in series for all the nsegs ?

These measurements are made in the steady state, so cm has nothing to do with it.

And thus the "Ra" portion is drawing a little current, making the effective resistance seem higher?

Here's a good way to think about what is going on:
The current delivered by your electrode must escape the cell by flowing through the
membrane. If the cytoplasm was a perfect conductor, then the injected current would
spread uniformly over the entire surface area of the cell--i.e. the current density would
be uniform, and equal to Iinj / totalarea. This is the assumption you made in your initial
query. If current density is uniform, the method you proposed would work nicely.

But cytoplasm is not a perfect conductor, and real cells have long skinny branches.
For locations that are anatomically close to the electrode, the intervening cytoplasm
introduces only a very small net series resistance, so the membrane potential at close
locations is very close to the value Vo that you measure at the injection site, and
membrane current density is also very close to the value at the injection site. With
increasing distance x from the injection site, the net resistance of the intervening
cytoplasm also increases, and the membrane potential Vx must decrease as well.
Lower membrane potential means smaller membrane current density--the larger the
distance x, the smaller the current density for the membrane at that distance.
Therefore the total current delivered by the electrode cannot spread equally over the
entire membrane. Instead, most of it is concentrated in membrane that is near the
electrode. This means that the local current density is higher than it would be if
cytoplasm were a perfect conductor, so the local membrane potential is also higher,
and your measurement of RN must give you an overestimate of Rm.

So your intuition about this result

When Ra=100, Vm becomes 29.9mV
and this makes the effective resistance ~100 Mohms.

as stated here

This probably has to do with increasing the Ra, increasing the resistance value in series, that increases the impedance of the circuit ?

is correct.

Or has to do with the fact that when Ra=100 ohm.cm, the parallel RC portion draw less current from the Istim resulting in a smaller current and thus a greater effective membrane resistance is seen ?

I suppose that sentence could be rewritten in a way that says "most of the injected
current is restricted to the vicinity of the electrode" (ignoring the mention of capacitance,
which isn't relevant to steady state measurements).

When Ra is returned to its default value and g_pas is divided by 100, Voltage charges and then discharges like an RC circuit with a tau of about 100ms.

I could not really determine what the effective resistance here would be as the Vm did not stabilize , but its much much greater than the previous values. (much-much actually)

Don't quit so easily. Let the simulation run until it reaches steady state. It only takes
1000-2000 ms. Be sure to let the injected current persist throughout (set dur to 1e9).
There is something yet to be learned.

I think when the g_pas here is reduced, the current across the capacitor rises rapidly charges it and starts to discharge. The Vm discharges with tau ~100ms.

If the injected current is still on, why would cm start to discharge?

As to the question of how to estimate Rm, more later.

Hi again Ted,

Ok, so first thing I did now is to move the point of current injection and voltage measure to the "0"
The Vm measured was 50.39mV.
The RN is (50.39mV+70mV)/1nA=120.39 Mohm.

I guess I have much higher value here than when i injected the current and measured the voltage at the center of the cable because the current injected at any end would be more concentrated/higher in density at the ends. This leads to a higher Vm reading and thus higher RN ?
I am right on this ?

Your right, the next lesson was very valuable.
Dividing the g_pas by 100, and injecting 1na at the center at the cable, the Vm measured was 3150mV.
The RN = (3150+70)mV/1nA ~ 3220 Mohms.

Repeating this, but changing the point of current injection and voltage measure to the "0" end, the Vm measured was 3261.93 mV
The resulting RN = (3261.93+70)mV/1nA ~ 3331.93 M ohms.

I guess since the g_pas was divided by 100, thus the Rm increased by 100. I can also divide the RN values above by 100.

For the "0.5" , RN ~ 32.2 Mohms
For the "1" , RN ~ 33.1 M ohms

Obviously this estimation is a lot better to estimate the effective membrane resistance?

Is the reduced conductance of the membrane somehow causing more even distribution of the injected current into the cable as now the I_pas rises more smoothly vs abruptly in the earlier situations ? A little unsure of how to think of this?

And another side note, I may have mis-expressed what I wanted to say, when I said this :

"I think when the g_pas here is reduced, the current across the capacitor rises rapidly charges it and starts to discharge. The Vm discharges with tau ~100ms."

Your right, the cm does not discharge when the injected current is still on.
Its just that here I only had the injected current for 50ms, and Vm started to discharge after that.

Meena wrote:first thing I did now is to move the point of current injection and voltage measure to the "0"
The Vm measured was 50.39mV.
The RN is (50.39mV+70mV)/1nA=120.39 Mohm.

I guess I have much higher value here than when i injected the current and measured the voltage at the center of the cable because the current injected at any end would be more concentrated/higher in density at the ends. This leads to a higher Vm reading and thus higher RN ?

Here's how to think about this:
Imagine a cable running from left to right that stretches out infinitely in both directions.
You poke it with an electrode and inject a constant current I, wait for membrane potential
to reach its new steady state, and find that the voltage change is dV1. So the input resistance
of the cable is R1 = dV1/I.

By symmetry you know that half of the current you inject flows into the right half of the
cable, and the other half flows into the left half. So the input resistance of the right half
of the cable is equal to the input resistance of the left half, and both are given by
R2 = dV/(I/2) = 2*R1.

The cable in your computational model isn't infinitely long, but it's long enough that when
you inject current in the middle, the resulting voltage change at the sealed ends is a tiny
fraction of the voltage change at the location of your electrode. As the saying goes,
"that's good enough for government work."

Dividing the g_pas by 100, and injecting 1na at the center at the cable, the Vm measured was 3150mV.
The RN = (3150+70)mV/1nA ~ 3220 Mohms.

Repeating this, but changing the point of current injection and voltage measure to the "0" end, the Vm measured was 3261.93 mV
The resulting RN = (3261.93+70)mV/1nA ~ 3331.93 M ohms.

So far, so good. The input resistance is now almost independent of the location at which
it is measured.

I guess since the g_pas was divided by 100, thus the Rm increased by 100. I can also divide the RN values above by 100.

I understand what you're saying, and it is true, but it's not the take home message. The
object of this exercise is to demonstrate that when membrane resistance is low, the
input resistance of an anatomically extensive structure will depend on where you make
the measurement, because the current you inject cannot spead uniformly over the entire
surface of the cell. When membrane resistance is high, current can spread much more
evenly and the potential change that you measure is a much better indication of the
potential change over the whole cell. As a result, input resistance is also much more
uniform (less dependent on location), and, if you knew the total surface area of the cell,
you would be able to use the input resistance to find a better estimate of membrane
resistance.

The problem in the real world is that, unless your cell is an oocyte or has some other
compact geometry that is easy to measure, you can't determine its actual surface area.
Even with compact geometries, there can be surface irregularities (especially infoldings)
that make the apparent surface area a poor underestimate of the actual area. And you
can almost forget about real neurons--most of the surface area of "interesting" cells is
in their fine dendritic brances, whose diameters are 1 um or less. The practical limits of
optical imaging, combined with artifacts introduced by staining and fixation, mean that
branches with diameters < 1um are subject to errors of about 0.2 um. Then there's the
problem of noncylindrical profiles--if you examine electron micrographs, you'll see that
almost no neurites have a cylindrical profile (or elliptical, if cut obliquely). There are all
kinds of irregularities, including infoldings and profiles that are convex on one side and
concave on the other. So the relationship between apparent diameter and true surface
area is not a constant proportion, nor is the apparent diameter a reliable indicator of the
transverse area of the lumen of a neurite (very important, since transverse area is a
critical determinant of the longitudinal resistance that impedes the flow of current through
a cell).

Pretty depressing, eh?


A quick and dirty trick that some people use to estimate Rm:
Apply a current pulse or step and try to determine the slowest time constant in the
resulting charging curve. This is an estimate of the mebrane time constant tau_m
Assume that Cm is 1 uf/cm2. Since Rm * Cm = tau_m, it is easy to calculate the Rm
estimate.
Too quick and dirty by far in some cases.

Let me suggest that you read at least a few of the following papers, which present
a few of the strategies used by real neuroscientists to study the basic electrical
signaling properties of neurons. Most if not all of them should be available at no cost
from the journals' web sites, or via links from PubMed--

Spruston, N. and Johnston, D.
Perforated patch-clamp analysis of the passive membrane properties of
three classes of hippocampal neurons.
Journal of Neurophysiology 67:508-529, 1992.

Chitwood, R.A., Hubbard, A., and Jaffe, D.B.
Passive electrotonic properties of rat hippocampal CA3 interneurones.
Journal of Physiology 515:743-756, 1999.

Clements JD, Redman SJ.
Cable properties of cat spinal motoneurones measured by combining
voltage clamp, current clamp and intracellular staining.
J Physiol. 1989 Feb;409:63–87.

L J Gentet, G J Stuart, and J D Clements
Direct measurement of specific membrane capacitance in neurons.
Biophys J. 2000 July; 79(1): 314–320.

This next one has what some may regard as an unfavorable "insight/effort" ratio, but it
is included as an example of (excessive ?) rigor:

Major G, Larkman AU, Jonas P, Sakmann B, Jack JJ.
Detailed passive cable models of whole-cell recorded CA3 pyramidal neurons
in rat hippocampal slices.
J Neurosci. 1994 Aug;14(8):4613–4638.

Hi,

I like to know, the follwoing code, is a right methods to calclulate input resistance or, should we consider other points?
thanks in advance,

See https://www.neuron.yale.edu/phpBB2/viewtopic.php?p=2498
The input resistance at any point is the input impedance at 0 Hz at that point.

Hi,

Ted has written

Omitting surface area of the ends gives 31.83 Mohom.

But, when for a section has L=1000 um^2, diam=1 and g_pas=0.001 s/cm^2 (then Rm=1 kohm.cm^2 )
should be our input resistance, which is equal to 31.43 Mohm not, 31.83 or 31.82!

and

the code written by Karlo, is the way to calculate input resistance of the cell?

[/quote]

chinhou wrote:

Code: Select all

Rn=a/Rm *10^4  [Mohm]

should be our input resistance, which is equal to 31.43 Mohm not, 31.83 or 31.82!

Think about this for a moment. If what you write is true, then the following statements
are also true:
1. Increasing Rm will make input resistance smaller.
2. Increasing surface area will make input resistance larger.

the code written by Karlo, is the way to calculate input resistance of the cell?

It will compute the input impedance at the midpoint of A

1. Increasing Rm will make input resistance smaller.
2. Increasing surface area will make input resistance larger.

Yes, i think these are true.


how did you calculate Rn, for given area and Rm, where is my fault?

chinhou wrote:

1. Increasing Rm will make input resistance smaller.
2. Increasing surface area will make input resistance larger.

Yes, i think these are true.

Both are false.

how did you calculate Rn, for given area and Rm

Rm is specific membrane resistance--resistance per unit area of membrane.
Rn is input resistance of a neuron, typically measured by injecting a current at some location
and measuring the voltage change at that location.

where is my fault?

1. Imagine a spherical cell to which a patch clamp electrode is attached.
You inject 1 nA current, and membrane potential depolarizes by 100 mV.
What is Rn?
Surface area of the cell is 1000 um^2.
What is Rm?

2. Imagine another spherical cell that has surface area of 10,000 um^2
and Rm = 10,000 ohm cm^2.
What is Rn?

3. A third cell has surface area 20,000 um^2 and Rm = 10,000 ohm cm^2. What is its Rn?

Hello Ted,
First number is input resistance from Impedance method,second number is (V2-V1)/I where V2,V1 are membrane potential before and after injected current, respectively.Third number is calculated from Rn=area*g_pas*10^2
then,

I said stupid thing about Rn=a*g_pas*10^2!!!,

and
input resistance decreasing by increasing area,
input resistance decreasing by decreasing Rm



but i didn't understand, how error in area effect input resistance calculation?

Ted:......PI*diam*L = 3141.59 um2. Perhaps you were adding the area of the circular faces at each end?