<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML xmlns="http://www.w3.org/TR/REC-html40" xmlns:v =
"urn:schemas-microsoft-com:vml" xmlns:o =
"urn:schemas-microsoft-com:office:office" xmlns:w =
"urn:schemas-microsoft-com:office:word" xmlns:m =
"http://schemas.microsoft.com/office/2004/12/omml"><HEAD>
<META content="text/html; charset=us-ascii" http-equiv=Content-Type>
<META name=GENERATOR content="MSHTML 8.00.6001.19019">
<STYLE><!--
/* Font Definitions */
@font-face
{font-family:Calibri;
panose-1:2 15 5 2 2 2 4 3 2 4;}
@font-face
{font-family:Tahoma;
panose-1:2 11 6 4 3 5 4 4 2 4;}
@font-face
{font-family:Consolas;
panose-1:2 11 6 9 2 2 4 3 2 4;}
/* Style Definitions */
p.MsoNormal, li.MsoNormal, div.MsoNormal
{margin:0in;
margin-bottom:.0001pt;
font-size:12.0pt;
font-family:"Times New Roman","serif";
color:black;}
a:link, span.MsoHyperlink
{mso-style-priority:99;
color:blue;
text-decoration:underline;}
a:visited, span.MsoHyperlinkFollowed
{mso-style-priority:99;
color:purple;
text-decoration:underline;}
pre
{mso-style-priority:99;
mso-style-link:"HTML Preformatted Char";
margin:0in;
margin-bottom:.0001pt;
font-size:10.0pt;
font-family:"Courier New";
color:black;}
span.HTMLPreformattedChar
{mso-style-name:"HTML Preformatted Char";
mso-style-priority:99;
mso-style-link:"HTML Preformatted";
font-family:Consolas;
color:black;}
span.EmailStyle20
{mso-style-type:personal-reply;
font-family:"Calibri","sans-serif";
color:#1F497D;}
.MsoChpDefault
{mso-style-type:export-only;
font-size:10.0pt;}
@page WordSection1
{size:8.5in 11.0in;
margin:1.0in 1.0in 1.0in 1.0in;}
div.WordSection1
{page:WordSection1;}
--></STYLE>
</HEAD>
<BODY style="MARGIN: 4px 4px 1px; FONT: 10pt Tahoma" lang=EN-US link=blue
bgColor=white vLink=purple>
<DIV dir=ltr align=left><SPAN class=371431315-18032011>Mark, if you write out
the equation for a monomer-tetramer association you will then realize that the
units of the molar association constant are M^^-3. Thus 1/Ka cannot give you a
dissociation constant in molar. </SPAN></DIV>
<DIV dir=ltr align=left><SPAN class=371431315-18032011></SPAN> </DIV>
<DIV dir=ltr align=left><SPAN class=371431315-18032011>I don't actually think
there is a universally-accepted definition of "dissociation constant" for
oligomers above dimer. You can define a concentration where monomer and tetramer
have equal molar concentrations (and I believe that is what is being reported in
the thermo_results file), or you could define one where the weight
concentration of monomer and tetramer is equal, and it would be reasonable to
call either one a "dissociation constant", but you need to define your
terms. However either definition will involve taking the cube root of the
association constant.</SPAN></DIV>
<DIV dir=ltr align=left><SPAN class=371431315-18032011></SPAN> </DIV>
<DIV dir=ltr align=left><SPAN class=371431315-18032011>John</SPAN></DIV><BR>
<DIV dir=ltr lang=en-us class=OutlookMessageHeader align=left>
<HR tabIndex=-1>
<B>From:</B> rasmb-bounces@rasmb.bbri.org [mailto:rasmb-bounces@rasmb.bbri.org]
<B>On Behalf Of </B>Mark Agacan<BR><B>Sent:</B> Friday, March 18, 2011 6:25
AM<BR><B>To:</B> Mark Agacan; rasmb@server1.bbri.org<BR><B>Subject:</B> [RASMB]
Sedphat tetramerization constant<BR><BR></DIV>
<DIV></DIV><BR>
<DIV>
<DIV>Hi,</DIV>
<DIV> </DIV>
<DIV>I'm confused about which value to use in the calculation of
dimerization and tetramerization constants in Sedphat.</DIV>
<DIV> </DIV>
<DIV>e.g. monomer-tetramer-octamer self-association model:</DIV>
<DIV> </DIV>
<DIV>logKa14 = 3.127, then Ka14 = 1339.677 M3, so 1/Ka14 gives 7.465 x 10-4, or
746.5 uM (which should be the equilibrium dissociation constant for
the monomer : tetramer interaction).</DIV>
<DIV> </DIV>
<DIV>But the thermo_results file says c1 = c4 at 9.069 x 10-2 M = 90.69
mM. Is this then the tetramerization constant?</DIV>
<DIV> </DIV>
<DIV> </DIV>
<DIV>I also processed the same data using monomer-n-mer self association model
with n = 4:</DIV>
<DIV> </DIV>
<DIV>logKa1n = 3.047, then Ka1n = 1.114 x 10+3 M3, so 1/Ka1n = 4.487 x 10-4, or
448.7 uM.</DIV>
<DIV> </DIV>
<DIV>Again the thermo_results file says c1 = c4 at 9.642 x 10-2 M = 96.42
mM.</DIV>
<DIV> </DIV>
<DIV>Can someone explain please?</DIV>
<DIV> </DIV>
<DIV>Many Thanks,<BR><BR>Mark</DIV>
<DIV> </DIV>
<DIV> </DIV>
<DIV>~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~<BR>Dr Mark
Agacan, Scientific Officer for the Division of Biological Chemistry and Drug
Discovery,<BR>Wellcome Trust Biocentre, College of Life Sciences, University of
Dundee, Dundee, DD1 5EH <BR>Tel: +44 1382 386095 Fax: +44 1382
345764 Mobile: 07525 451
117<BR>~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~</DIV>
<DIV> </DIV>
<DIV>
<DIV>************************************************************ </DIV>
<DIV>
<DIV><FONT color=#666666>Please consider the environment. Do you really need to
print this email?</FONT> </DIV></DIV></DIV></DIV><BR>
<P>The University of Dundee is a registered Scottish charity, No: SC015096
</P></BODY></HTML>