Talk:Isoelectric point
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[edit] This article is or was the subject of a Wiki Education Foundation-supported course assignment. Further details are available on the course page. Student editor(s): Kj0367. Peer reviewers: Kj0367.
Above undated message substituted from Template:Dashboard.wikiedu.org assignment by PrimeBOT (talk) 00:55, 17 January 2022 (UTC)
Untitled
[edit]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)
Lead paragraph
[edit]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)
Translation into Chinese Wikipedia
[edit]The 20:00, 12 March 2009 130.91.112.68 version is translated into Chinese Wikipedia.--Wing (talk) 13:44, 28 March 2009 (UTC)
derivation of formula needed
[edit]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)
Equilibrium of neutral and zwitterionic glycine
[edit]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
- H2NCH2CO2H + H3O+ ↔ H3N+CH2CO2H + H2O
- step 2, deprotonation of the carboxylic acid:
- H3N+CH2CO2H + H2O ↔ H3N+CH2CO2- + H3O+
- K = 4.6 x 10-3
- H3N+CH2CO2H + H2O ↔ H3N+CH2CO2- + H3O+
- step 1, protonation of the amino group:
- 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
- H2NCH2CO2H ↔ H3N+CH2CO2-
- 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)
Wrong charge states for adenosine monophosphate.
[edit]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 (pKa < 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 pKa 6-6.5
HOPO2-OR + H2O ↔ OPO22-OR + H3O+
The quoted pKas 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- ↔ AMP2-
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)
What I find confusing
[edit]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)
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