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I saw this article http://www.ncbi.nlm.nih.gov/pubmed/16710314 and it mentioned pA(2) values and I had no idea what they were. What are they? What do they mean?

If possible it'd be just dandy if you could cite some journal articles/books/other reliable sources for me to dig up.

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    $\begingroup$ I don't know enough about this to write a good answer, but you can find an explanation here. A p(A2) value describes the affinity of an antagonist for its receptor. $\endgroup$
    – Alan Boyd
    Commented Jun 27, 2013 at 16:21

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Digging around a bit, I found a few resources:

Design and models for estimating antagonist potency (pA2, Kd and IC50) following the detection of antagonism observed in the presence of intrinsic activity.

and PHARMACODINAMIC on page 30, describes pA2 as:

indicates affinity of antagonist for receptors

(as Alan Boyd just commented on)

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Below are responses generated using the isolated guinea-pig ileum preparation to determine a pA2 value for a sample of the cholinergic antagonist homatropine. Responses are expressed in millimetres +/- SDM (n =4). Plot the data and then determine by Schild analysis the pA2 value and Schild slope. What conclusions can you draw; does the experiment/data fulfil the criteria for a valid Schild analysis?

Acetylcholine ACH Alone ACH +Homotropine ACH +Homotropine ACH +Homotrop Concentration 3x10^-9 M 1x10^8M 3x10^-8M [Molar]

3 x 10^-9 3+/-1
6 x 10^-9 8+/-3 Not Response 3 x 10^-8 28+/-6 10+/-4 No response 6 x 10^-8 38+/-8 21+/-9 5+/-2 No response 3 x 10^-7 55+/-5 43+/-7 27+/-9 10+/-4 6 x 10^-7 56+/-5 52+/-3 38+/-7 20+/-7 3 x 10^-6 - 55+/-4 53+/-6 43+/-6 6 x 10^-6 - - 56+/-2 54+/-5

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    $\begingroup$ This is a well-written answer; however, it does not answer the question being asked. If you could state what exactly it is as opposed to how it behaves when tested on a guinea pig that would be helpful. Also, it is advisable that you put in references. Aside from this, you did well for a first answer and I welcome you to Biology S.E.! If you have additional questions, please visit The Help Center. $\endgroup$
    – L.B.
    Commented Mar 20, 2015 at 2:23

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