neurotransmitter storage (all or one?)

Richard Norman rsnorman at mediaone.net
Fri Sep 8 20:03:28 EST 2000


At the risk of prolonging an already long thread, yes you are a bit late
and possibly missing some of the main points.  There are large numbers
of nerve cells that do not make action potentials at all, yet participate in
neuronal integration or "computation".  And there are large numbers of
"microcircuits" where portions of dendrites and axons from a variety
of cells all interact and where many of the participants do not produce
action potentials at all.  And that almost all of the true "computation" in
the nerve cell, that is, what is termed synaptic integration, occurs in a
part of the cell that does not produce an action potential.

The digital action potential is only one of the tools in the toolkit of the
neuron.  It is rather an oversimplification to say that a neuron "works" by
having a state of AP or not.

"TonyJeffs2" <tonyjeffs2 at aol.comTonyJ> wrote in message
news:20000908112330.20405.00000662 at nso-fp.aol.com...
> Hi,
> I'm a latecomer to this discussion and am probably missing the point, but
here
> is my view. Here Goes.
>
> A computer memory address works by having a state of 0v or 5v .  To get to
5
> volts, that memory address requires the movement of a specific large
number of
> electrons, specific to that particular memory address, and similar to
other
> similar memory addresses.  So the movement of a large number of electrons
is
> necessary to generate a binary change
>
> A neuron works by having a state of no AP or AP.  To generate an AP, it
> requires a specific large number of nerutransmitter molecules, specific to
that
> particular neuron, and similar to other similar neurons.  So the movement
of a
> specific large number of nt molecules is necessary to generate a binary
change.
>
> Thus, the two situations are the same.
>
> I've ignored the fact that there are inhibitory and excitatory nts. That
makes
> for a slightly more complicated equation, but the end argument is the
same.
>
> And then ther is the matter of potentiation, at which point the two
systems are
> not the same.
>
> But they are both digital; not analog.
>
> Tony
>
>







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