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

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Del curso dictado por Duke University

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.

De la lección

Passive and Active Resonses, Channels

This week we'll be discussing channels and the remarkable experimental findings on how membranes allow ions to pass through specialized pores in the membrane wall. The learning objectives for this week are: (1) Describe the passive as compared to active responses to stimulation; (2) Describe the opening and closing of a channel in terms of probabilities; (3) Given the rate constants alpha and beta at a fixed Vm, determine the channel probabilities; (4) Compute how the channel probabilities change when voltage Vm changes.

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

Biomedical Engineering, Pediatrics

Hello, again. This is Roger Coke Barr once again for the

Bioelectricity course. We're in Week three and this is segment

eleven. I thought, once again, it would be useful

to have a short problem session so as to talk about the idea of channel

probabilities. Let's suppose we consider this question.

There are thousand channels in our patch. Our initial measurement shows 100 are

open, 900 are closed. So, most of them are closed.

100 are open, 900 are closed. The trans-member involves it's changes to

a new value. That means, there are new values of alpha

and new values for beta. The new values for alpha and beta, alpha

is 0.8 per millisecond, beta 0.2 per millisecond.

You notice that alpha and beta are right constants so they are per some unit of

ten. So, now the question is, how many channels

are expected to be open one millisecond after the initial measurement?

And how many channels are expected to be open ten milliseconds after the initial

measurement? Part big.

Let's do each part in turn. How many channels are expected to be open

one millisecond after the initial measurement?

Well, you recall we had this equation that said that dN0 / dt was equal to alpha

times the number closed minus beta times the number open.

I think we have all of the quantities that we need here that are given in the

problem. So, alpha, we know that's 0.8.

Beta, we know that's 0.2. 0.8, 0.2.

The number closed at the beginning of the segment was 900.

The number open at the beginning of the voltage transition, so right at the end of

the voltage transition, the beginning of the time with the new voltage is 100.

So, if we put all those numbers in there, we can compute all the quantities on the

right and if that we then, if we then rewrite our equation and say, well, look

then, delta N zero will be equal to delta t times what we had before, alpha times

the number of closed minus beta times the number open.

We know this part and we'll find the change in the number of channels that are

open. So, the number that are open at the end of

one millisecond will be 100, because we started with 100 plus delta N zero is

computed, as computed by the calculation we have just completed.

So, if you would work that out please and find out what the actual number is at the

end of one millisecond. Now, let's go on and do the question for

ten milliseconds. Now, we can't just do this like we did

part A because there is too much of an extrapolation.

So, I'd say there's just too much, too long, ten milliseconds.

Too long to do in one step. I mean, for that matter, one millisecond

might have been too long to do it in one step.

We did it that way, but if we had done it in say, ten short steps, each 100 micro

seconds, our answer might have been a little bit different and it would have

been better. Ten milliseconds is definitely too long,

so we have to do it some other way. Well, there are a couple of alternatives.

One alternative is to write a computer program and to do it as a whole bunch of

short steps. But a better alternative, a better

alternative is to look at our equation, dN0 / dt equals whatever it is, and then

to solve that equation, solve. And when we solve the equation, we'll get

N0 as a function of time equal to whatever it is over here on the right hand side.

Maybe you can solve it. And if you do, then, you'll be able to

look at the result and to find the result after one millisecond for part A, and

after ten milliseconds for part B. So now, let me make a comment about this

part. When you solve the equation, solve dN0 /

dt, for N0 as a function of t, it should be, it should be that N0 after ten

milliseconds. Is getting close, it's approaching alpha

over alpha plus beta. Now, I say that because I've done this

problem before. Why don't you do it and see if that's the

case. See if, if N0 after ten milliseconds is

pretty close to alpha over alpha plus beta.

In fact, is it close enough that alpha over alpha plus beta can be used as a good

estimate of the answer? Or is it still some distance away?

Please check that out, and get those details straight.

So, thank you for watching this problem segment.

And we'll go on soon to the week in review.

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