:Talk:Isoelectric point

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Should add an example of pI calculation for lysine.

I added a simple method, someone might put a blurb about Henderson-Hasselbach...

Should there be mention of ZPC? Lizz612 21:47, 29 January 2006 (UTC)

tried to introduce the topic more. Expanded on information and elaborated on some topics. It's not perfect but at least I tried.140.203.16.53 19:54, 11 October 2007 (UTC)

[http://en.wikipedia.org/w/index.php?title=Isoelectric_point&oldid=276814139 The 20:00, 12 March 2009 130.91.112.68 version] is translated into Chinese Wikipedia.--Wing (talk) 13:44, 28 March 2009 (UTC)

The article pulls the average formula out of nowhere. Horrible, horrible way to introduce the concept. I don't have time to fix it right now, but someone should. John Riemann Soong (talk) 06:49, 13 April 2010 (UTC)

:A full derivation can be found in the pKa article. 96.54.32.44 (talk) 20:05, 20 February 2011 (UTC)

From the article:

"Glycine may exist as a zwitterion at the isoelectric point, but the equilibrium constant for the isomerization reaction in solution is not known"

H2NCH2CO2H is in equilibrium with H3N+CH2CO2-"

:The reason that one can't find a published equilibrium constant for this reaction is simply because it is a two step process. The equilibrium constants for each step are easily obtained from the published pKa1 = 2.34 and pKa2 = 9.58 for glycine (see table in proteinogenic amino acid):

::step 1, protonation of the amino group:

:::H₂NCH₂CO₂H + H₃O⁺ ↔ H₃N⁺CH₂CO₂H + H₂O

::::: K = 4.0 x 10⁹

::step 2, deprotonation of the carboxylic acid:

:::H₃N⁺CH₂CO₂H + H₂O ↔ H₃N⁺CH₂CO₂^- + H₃O⁺

:::::K = 4.6 x 10^-3

:There is no particular order for the two steps, and deprotonation of the carboxylic acid can just as easily precede protonation of the amino group.

:The overall equilibrium constant for the two step process is the product of the two K values shown.

:::H₂NCH₂CO₂H ↔ H₃N⁺CH₂CO₂^-

:::::K = 1.9 10⁷

:So there is one molecule of H₂NCH₂CO₂H for every 19 million zwitterion H₃N⁺CH₂CO₂^-

:96.54.32.44 (talk) 18:31, 20 February 2011 (UTC)

There is a problem with the pKa for phosphate esters, in that the first deprotonation of the phosphoric acid group is almost a strong acid (pK_a < 1) and very difficult to measure. For this reason, phosphate esters are often quoted with a single pKa, but the ionization is for the second deprotonation, typical pK_a 6-6.5

HOPO₂^-OR + H₂O ↔ OPO₂^2-OR + H₃O⁺

The quoted pK_as for 5'-AMP are 0.9, 3.8, 6.1, and should be attributed as

:pKa₁ = 0.9 - first phosphoric ester ionization

:pKa₂ = 3.8 - deprotonation from adenine-H⁺

:pKa₃ = 6.1 second deprotonation of hydrogen phosphate ester

That would make the charge states

AMP⁺ ↔ AMP ↔ AMP^- ↔ AMP^2-

so pI = ½(0.9 + 3.8) = 2.4

See Data for Biochemical Research by RMC Dawson et al. Oxford. 96.54.32.44 (talk) 05:07, 25 February 2011 (UTC)

::The exact page reference for the above is Data for Biochemical Research by RMC Dawson et al 2nd Ed. Oxford (1969) pp. 103-114. The same information is in the 3rd Ed., but I don't have the page numbers. 96.54.32.44 (talk) 00:31, 21 April 2011 (UTC)

Although free adenine can be protonated twice, formation of the N-ribofuranoside in adenosine deactivates the stronger of the two basic N atoms leaving only the pKa 3.8. 96.54.32.44 (talk) 18:51, 1 March 2011 (UTC)

In the first paragraphs, a surface is mentioned and a solid, but later, the substance going in and out of solution.

Should I be imagining a solution with various ion species, or a surface equilibrium process -- or is it all happening at once? ? If so, where is the solid in the concentration pictures?

Is the charge on the solid or is it some kind of aggregate charge in the liquid?

I would appreciate it if the main components of the system were laid out and described before the ancillary stuff. 84.227.252.109 (talk) 18:33, 24 September 2014 (UTC)