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

Energy into Voltage

This week we will examine energy, by which pumps and channels allow membranes to "charge their batteries" and thereby have a non-zero voltage across their membranes at rest. The learning objectives for this week are: (1) Describe the function of the sodium-potassium pump; (2) State from memory an approximate value for RT/F; (3) Be able to find the equilibrium potential from ionic concentrations and relative permeabilities; (4) Explain the mechanism by which membranes use salt water to create negative or positive trans-membrane voltages.

- 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 two, Segment eleven.

And this segment is simply a problem session, involving membrane resistance and

membrane capacitance. I thought I would start off the problem

and then, maybe you could finish it up. Suppose, the question that I ask is, we

have a cylindrical fiber, ten microns in diameter, that has segments that are 100

microns long, you know, micron as in micrometer ten^-6 meters.

It's often shorthand written as a, a [inaudible] followed by an m.

And when people write it, with on the screen, sometimes they write it as and

that's what's happened here in this slide. So, we have a fiber that has, that a

segment of the fiber has those dimensions. We know, for this particular fiber, the cm

is one microfarad per centimeter square, or MS 1500 ohms per centimeter square.

So, for one segment, what is the resistance and the capacitance.

So, the thing you have to know here, just because, you know how to do problems like

this is you have to know that you, you don't attack cm or rm or r or c directly.

You first find the surface area because the surface area comes into both

calculations. So, we would say here that the surface

area for this simple, cylindrical structure, we have this arrangement, this

is 100 micrometers. This diameter is given as ten micrometers

so this will be called that D and I'll call this length L, just to be able to

represent them symbolically. So, I'll have the surface area will be the

circumference. So, that will be pi times the diameter.

And then, to get the area, we'll multiply it times the length.

So, surface area's pi times the diameter, times the length, and if one substitutes

in those numbers, then one gets a numerical answer.

And you have to be careful because our other values are in per centimeter square

to convert A as to centimeter square. Probably, the easiest way to do the

conversion is by putting in value, a value of d and l in centimeter square rather

than putting them in those micrometers. So, if you figure out the relation between

those, I think it's ten^-4. If you put in the relation between those,

then you should get the surface area in centimeter square.

I'll leave it to you to complete that calculation if you would and now, we need

to find the actual resistance and the capacitance for this segment.

As it was explained in earlier, in earlier segments, the calculation itself is quite

easy, resistance will be the membrane resistance.

1500 divided by whatever you found for As, capacitance will be won and result will be

in ohms. Capacitance will be one, microfarad per

centimeter square times As centimeter square.

So, that will be, whatever you find, and that will be in microfarads.

So, as calculations like this are normally done, you could put these numbers together

and you will have the results. I'll ask you to complete these

calculations so that you have the actual number values.

One needs number values like this in doing calculations involving membranes which

always require that you have the actual resistance and the actual capacitance for

the dimensions that are present in a particular problem.

I'd like to just mention a few notes even though these are simple, to come up in

this problem. So, in our notation, what has happened

here is that in some places, such as right here, is, we have cm2, but what's meant by

that is centimeter square. In other words, cm2 means sum of majors

squared. Similar thing happens with Cm, which

really means C subscript m to capacitance of a segment of membrane.

In the olden days, meaning, before computers, people generally got the right

thought, and set these as subscripts and superscripts.

Nowadays, with the advent of computers and computer programs, they often write them

in text, at, on the line, so that, that means that you have to do a mental

translation back and forth between what's written on the line, superscripts, and

subscripts, so as to have everything in this proper notation, that's a second

note. It's helpful to bring to mind the fact

that the units of membrane resistivity are ohm centimeters square.

They're not the same as those for the bulk resistivity.

That is to say the resistivity of the material inside the cell or the material

outside the cell. We used bulk resistivity in Week One and

there, and the units we used there were ohm centimeters.

So, the units are different and then the process one goes through to do the

conversion is different. But then again, the structure is

different. Bulk resistivity exists in three

dimensions where as a membrane is from the viewpoint of resistance, essentially a

flat plane structure, a two-dimensional structure.

And then finally, it's just helpful once again to bring to mind, that as one moves

to get a resistance or a capacitance, one divides, divides Rm by the area to get R

and multiplies Cm by the area to get c. Oh, I wish I had a dollar for every time I

had made this mistake, either on a calculation on paper or on the board or on

a, in a computer program. You just don't think about it.

So, I'm, I'll talk about it here just to give it a little extra impetus.

You'll make the mistake of doing it wrong and dividing both or multiplying both, you

just have to be aware that it's they are different.

And when you come back through to try to find the error, this is one of the places

that the error may exist. So, with this picture of the Duke

University Medical Center, we'll conclude this segment.

We'll comeback in a short while and discuss in the next segment, some of the

conclusions from the second week of work.