Nerves, the heart, and the brain are electrical. How do these things work? This course presents fundamental principles, described quantitatively.

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来自 Duke University 的课程

Bioelectricity: A Quantitative Approach

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Nerves, the heart, and the brain are electrical. How do these things work? This course presents fundamental principles, described quantitatively.

从本节课中

Axial and Membrane Current in the Core-Conductor Model

This week we will examine axial and transmembrane currents within and around the tissue structure: including how these currents are determined by transmembrane voltages from site to site within the tissue, at each moment. The learning objectives for this week are: (1) Select the characteristics that distinguish core-conductor from other models; (2) Identify the differences between axial and trans-membrane currents; (3) Given a list of trans-membrane potentials, decide where axial andtrans-menbrane currents can be found; (4) Compute axial currents in multiple fiber segments from trans-membrane potentials and fiber parameters; (5) Compute membrane currents at multiple sites from trans-mebrane potentials.

- Dr. Roger BarrAnderson-Rupp Professor of Biomedical Engineering and Associate Professor of Pediatrics

Biomedical Engineering, Pediatrics

Hello again, this is Roger Coke Barr for the Bioelectricity course.

We're in week number 5, where we're laying track, and

we are at lecture segment number 9.

Segment 9 is about determining the transmembrane current.

And you were thinking,

finally, finally this guy is going to work on the transmembrane current.

Because you remember, we started this week with the equation that said, IM

= IC + I ion, and we said

to use this equation we need to have Im.

And I promised you back in the beginning that if you wanted to know Im,

the way to find it was from the neighbors.

Well after all these multitude of preliminaries,

we're finally be talking about the trans-membrane current.

It's just wonderful.

This is a simple argument but let me take it step by step.

We showed in the last segment.

That the axial current Ii1 could be found from Vm1 and Vm2.

This equation was our result.

That's saying that we can find this current right here, which I'll denote with

the blue arrow II1.

Now, if we did that for that segment, we can do it again for another segment.

So I can have another expression for the current and the next segment, Ii2.

And Ii2 is all the same things just with Vm2 and

Vm3 instead of Vm1 and Vm2.

It's just shifted over about one second.

Here's the thing, we know that in electricity, current is conserved.

If it goes one place, it has to keep on going.

It's not created at any one place, nor is it destroyed.

So conservation of current, sometimes called Kershaw's Law,

requires that the membrane current, at site two.

This current right here, Im2.

It has to be the current is coming in.

This current in blue minus the current that's going out.

This current in green and the difference has to go across the membrane and

here it is, its membrane current as site number two.

If you want to think about it this way, suppose it a bit current,

suppose Ii1 is a bit current, suppose Ii2 is a Tani current.

So a lot of current is pouring in along the axis to site number 2, but it can't go

on down the axis because the intrasellar current that's continuing on is just tiny.

So it has to be forced out, and it is forced out.

And that particular current that is forced out from the inside to the outside

we call that Im2.

The same thing is true, of course, in the other direction.

If II1 is tiny, II2 is big, then Im2 has to also be the difference,

but this time it'll be a negative value because it'll be coming inside.

If we do this mathematically, this same steps mathematically,

then we get this equation that's on the left, Im2.

Now I'm just subtracting one and two.

Take equation one as subtract to equation two and this is what I get.

Simply by subtraction of two from one.

And that difference then is this three terms series (Vm1- 2*Vm2 + Vm3).

An amazingly complicated expression often inscrutable when you just see it

written out there all by itself but coming from such a simple origin.

So now we have the units question again, let's talk about it just for a minute.

As before, the current is in amperes, this is in volts,

this is in ohms.

And everything is okay, as before, often times this is in milli-volts.

This is kili-ohms, so this would be in nano-amperes.

That's okay too, but we do have a little problem here.

The problem is that in the equation that we really cared about,

the one that started us on this path, the units for the membrane current were

not amperes but instead they were amperes per meter squared.

Or some other scaled variant of that,

maybe nano amp years per centimeter square but in the even there were per unit area,

units that were used in our motivating equation.

We want to be able to use our results.

We need to be able to use this result and this equation.

So we're going to have to have some kind of a unit conversion.

You could say we should start by spelling conversion correctly but

I won't, I'll just go on to the next slide.

We can get to the units we want simply by doing a division of

the current that we found by the surface area through which it flows.

So the way we do that is take our expression, divide by the surface area,

here it is right there, is the surface area of an individual segment.

The surface area, not the cross sectional area,

so that will be equal to 2 pi a * delta X surface area.

We take our original expression equation 1,

divided by the surface area 2, then wind up with equation number 3.

It's a little bit confusing looking,

but notice this, in the limit,

as delta x grows small,

we have Im = 1 / 2 pi a.

D square V m, d x square, where all

of it is evaluated at the point of interest.

So this is Im2, it will be this derivative evaluated at site number two.

Versus Im3 or 4 or 7,

their derivative on the right will luckily be evaluated in the same way.

So if you discretized the fiber.

When you discretize the fiber, the result you get is this result.

This is usually the result that you want if you're doing a numerical calculation.

If on the other hand, you're thinking of the problem mathematically or

looking in a reference book that shows the result mathematically,

or lots of times just in discussions of the problem.

You get this result that says Im = 1

over 2 Pi a * 2 square, Vm 2 x square.

You're thinking,

what is this partial derivative sign that this guy snuck in on me?

It's just done that way to say that it's the second derivative.

In space, even if things are changing in time.

So we have it, we've done it, hooray.

We have gotten the transmembrane current, and we got it from the neighbors.

And it is the second derivative, that's thought provoking.

The transmembrane current is not proportional to the transmembrane voltage.

It's not proportional to the slope of the transmembrane voltage.

It's proportional to the second spatial derivative of the transmembrane voltage.

You're going down the track, and you're looking for

the bumps, because the bumps are where there is big second derivatives.

And where there's a bump in the track,

that's where there's a memorang current in the fiber.

In the figure here of the Duke University dorms,

they're interesting because of the bumps.

I mean yes, there is a wall, and yes, there are straight lines on the wall.

But it's the bumps here, the bumps here, the irregularities there.

That's what makes the architecture interesting.

As is the case for all Gothic architecture.

And in our track, it's the bumps that get your attention if you're on the train.

And in our fiber, it's the bumps to get your attention as far as membrane current.

Thank you for watching.

We'll come back and look at this another way in the next segment.