In article <1993Jan26.123642.11726 at ringer.cs.utsa.edu> senseman at lucy.brainlab.utsa.edu (David M. Senseman) writes:
>From: senseman at lucy.brainlab.utsa.edu (David M. Senseman)
>Subject: Re: primary afferent depolarization
>Date: Tue, 26 Jan 1993 12:36:42 GMT
>In article <9301251254.AA21883 at xray1.cshl.org> anderson at CSHL.ORG (John Anderson) writes:
>>Could someone please explain this statement? How does presynaptic
>>depolarization reduce the presynaptic spike amplitude? Seems like it
>>should enhance it.
>>>>PAD (primary afferent depolarization) isn't all that mysterious.
>Why would expect depolarization to ENHANCE transmitter release?
>Assuming that the axonal spike completely invades all the terminals,
>(probably a reasonable assumption for most systems Eva was
>referring to), then depolarization would DECREASE syanptic release
>simply because the membrane potential of the presynaptic terminal
>would be closer to V which is going to be somewhere above +100 mv.
> Ca
>That means the electrical force driving Ca inward would be reduced
>so that calcium entry during the terminal spike would be reduced.
>Since transmitter release is dependent on entry of external Ca,
>less transmitter release would occur.
>
I don't agree. The driving force on Ca++ is determined by the
voltage *during* the spike when the voltage-sensitive Ca channels
are opened. Besides, the driving force on Ca++ is so large that
a few millivolts depolarization from rest wouldn't make much
difference.
>If you get enough PAD so that the terminal membrane is really
>depolarized, then additional factors could come into to play
>such as channel inactivation.
This is probably more important--inactivation can occur
near resting membrane potential.
>>>--
>David M. Senseman, Ph.D. | Imagine the Creator as a low
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