r norman rsnorman_ at _comcast.net
Tue Jun 10 09:31:51 EST 2003

```On 10 Jun 2003 06:07:56 -0700, dcapelle at gmx.de (David Capelle) wrote:

>hi,
>
>i have got a question about so called activated membrane channels.
>having read in "from neuron to brain" that membrane channels can be
>activated by extracellular application of ACh I was wondering what
>exactly this means. i suppose it refers to the fact that activated
>channels are constantly switching between an open and a closed state
>whereas unactivated channels are most likely to be closed at any time.
>however in the mentioned book it is stated (in box 1 in chapter 2)
>that if all channels were activated there would not be any noise in
>the overall membrane current to be expected.
>therefore it is argued the formula relating variance, mean current and
>the current  through an individual channel ought to be
>
>c=Var/I(1-p) rather than c=Var/I; where c is the individual current, I
>the mean current and p the fraction of activated channels.
>i would appreciate if anyone could explain this to me as it does not
>make sense at the moment (possibly becos i am not getting the meaning
>of "activated"). besides i fear the book which is supposedly so
>brilliantly elucidating just does not suit the level of medical
>
>david

You do have activation/inactivation a little incorrect.  The
inactivated channel is closed, the activated channel is open.
As you apply ACh to these channels they "tend" to activate.  That is,
the probability of opening, p, increases.  Yes, p is the fraction of
open channels but it is also the probability of any one channel beign
open.  If p = 0, all the channels are closed.  There is no current and
there is no fluctuation or variance.  If p=1, all the channels are
open.  The current is maxium but there is again no fluctuation or
variance.  If p has some intermediate value, the average number of
open channels is pN where N is the total number of channels and the
variance is p(1-p)N.

More specifically,  let
c = current through one open channel
p = probability of channel being open
N = number of channels

Then the average current will be
I = c * p  * N
and the variance will be
Var = c-squared p*(1-p)*N = c * I * (1-p).

As a result,
c = Var/I * (1-p)

The text really is "brilliantly elucidating" just as you say.  And it
is, in fact, an excellent book for pre-med undergraduates.  That does
not mean it is easy!

Two major ideas.  One, as you increase the ACh concentration, you
increase p, the probability of any one channel being open.  That deals
with the mechanism of ACh activation.  Two, you can experimentally
measure the amplitude of a single channel event, c, from a
"fluctuation" or "membrane noise" analysis of the statistical
aggregate of many channels acting simultaneously.  You can even do
more. By measuring the frequency spectrum of the 'noise', you can even
get data on the mean duration of the open-time of each individual
channel.  Before the days of patch clamping, this was the only way to
"study" individual channel events.

```