0:00
So now we come, in segment 5 of week 2, to the most complex topic that we'll discuss in the present week.
How does the membrane make use of these ionic differences
that have been established by the sodium-potassium pumps
in order to create a membrane voltage?
It does so in the following series of steps;
First, it adjusts itself, so that it is
permeable to potassium only; in other words, not sodium.
Now of course, it can't do that perfectly,
But what we mean by that is it adjusts itself
so that the permeability of potassium is much greater than the permeability of sodium
or, in other words, ions of potassium can pass through the membrane relatively
easily as compared to ions of sodium.
It's what it does first. It's not an easy trick to do,
because what we're asking the membrane to do--here is the membrane
, and here's a
potassium channel...right here...
that's the channel, here comes a potassium!
HEre comes a potassium, and we're asking the membrance to let that potassium go right on through.
The potassium is coming fast. Fast, fast, fast!
It's not just drifting over there; think in terms of a car traveling very fast.
It's coming at high speed; the membrane recognizes it, and lets it
though the channel.
On the other hand, what we're asking the membrane to do
is if we have a sodium coming, a sodium ion coming, up to that same channel
well, we wanted to X out the sodium and bounce it back
so that teh sodium ion does not get through.
It's not an easy trick to do,
because these are both small ions
Both a positive charge--just has to recognize it and act quickly.
But it is able to do it, so it's a
great accomplishment.
So the membrane begins by being permeable to potassium
by a much higher degree than it is to sodium.
Then, the memrbane lets potassium ions <i>leak</i> back
out.
through these channels where it is permeable to ]
potassium only.
So, in a, uh, conceptual or pictorial sense, you'd say,
"What's happening here
is that the membrane
The membrane is letting potassium leak, from the inside to the outside.
And when it lets it leak from the inside to the outside,
it carries a positive charge with it.
So the result of the potassium ion moving out
is that there is excess positive outside, excess
negative inside.
The corollary is that an electric field is established,
pointing inward, from higher potential to lower potential.
So I've drawn an arrow here representing the direction of the
electric field.
Every time a potassium ion goes out,
The electric field gets stronger
If one describes this with a mathematical equation,
one has the equation that is, uh, given here.
The movement of potassium outward because of the diffusion gradient
is the term here on the left.
The electric field, on teh other hand,
is pushing potassium back in
That's the term over here on the right.
This is inward.
Here.
This is outward. There.
Both processes are occurring at the same time,
to the degree that they are not in balance--
there is a flux of potassium ions, indicated by the J<u>p symbol</u>
that is moving across the membrane.
6:23
The electric field pushes potassium back in.
(!)
At some point, enough potassium has moved out
that the electric field very strongly is pushing back in.
So an equilibrium results
when the flow from diffusion equals the flow because of the
electric field.
The result is equilibrium,
and in a one-ion system,
a system in which only one ion
can move across the membrane, this equilibrium
is called the Nernst V<u>m</u>
So the Nernst transmembrane voltage for potassium occurs
when the two forces, diffusion and the electric field,
are in balance.
7:21
What is the voltage at which that equilibrium
will occur for potassium?
So, if we had more weeks in our course :),
I would derive this result mathematically.
Since our time is limited, I will just show teh result.
There will be an equilibrium when the transmembrance
voltage Vm is equal to RT/F.
You remember that from last week.
times the logarithm of the concentrations:
the extracellular concentration over the intracellular concentration.
In the particular case of squid,
that means that the logarithm 20/397
and you recall that R<u>TO/F from last week is about 26 millivolts.</u>
9:42
If you write it out step by step like that,
then it's just one step following another.
You think of the process as a whole and you say,
This is just amazing!
This membrane takes advantage of the fact
that there is a potassium concentation
difference.
It's <i>letting</i> potassium ions move outside the membrane!
That's creating an electric field.
Diffusion, the force of diffusion,
the process of diffusion is being used to create an electric field!
It all comes to equilibrium, and in the case
of squid axon that's at
-77mV
That means the inside is negative by 77 mV
with respect to the outside.
Now the fact that it has a special symbol is to be expected.
What's extraordinary is that it happens at all!
The thing is just sitting there,
there's no electricity, and it charges itself up.
If we talk about sodium, it's all the same things,
but the sodium concentration starts outside instead of inside
but same thing, same equation--
the result is different because the concentration inside and outside is at a different ratio,
But the result now is that the equilibrium value is <i>positive<i>.</i></i>
(plus 57mv)
Wow.
Potassium equilibrium was negative, and the sodium equilibrium is positive.
Quite a voltage shift!
Of course, the membrane to accomplish this, has to become
permeable to sodium,
no longer permeable to potassium,
so that the system is now a sodium-ion system
instead of a potassium-ion system.
Just as before, the system depends on the
concentration gradient
being established first
so that it can be exploited to create the voltage difference.
We have not discussed chlorine ions,
but there's a similar process.
for chlorine.
One might ask, can you do the mathematics
so that all the different ions appear in the same system?
And in fact you can, and that's this
multi-ionic expression, sometimes called
"Goldman's expression."
And it appears on the slide.
The thing that you
notice is that the ratios <i>ahem</i>
the expression is written in terms of
these permeabilities, and if you work with the equation
somewhat, you'll find that what's important is the permeability <i>ratios</i>
more than the permeabilities.
You can do that, for example, by taking the expression as it's given
and dividing each term by, say, P<u>k.</u>
And if you do divide each term by P<u>k,</u>
you'll get a result that will be in terms of the ratios.
Once again, the mystery factor, RT/F
Dr. Roger BarrAnderson-Rupp Professor of Biomedical Engineering and Associate Professor of Pediatrics
