[RASMB] Sedphat tetramerization constant

John Philo jphilo at mailway.com
Fri Mar 18 08:34:24 PDT 2011


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. 
 
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.
 
John

  _____  

From: rasmb-bounces at rasmb.bbri.org [mailto:rasmb-bounces at rasmb.bbri.org] On
Behalf Of Mark Agacan
Sent: Friday, March 18, 2011 6:25 AM
To: Mark Agacan; rasmb at server1.bbri.org
Subject: [RASMB] Sedphat tetramerization constant



Hi,
 
I'm confused about which value to use in the calculation of dimerization and
tetramerization constants in Sedphat.
 
e.g. monomer-tetramer-octamer self-association model:
 
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).
 
But the thermo_results file says c1 = c4 at 9.069 x 10-2 M = 90.69 mM.  Is
this then the tetramerization constant?
 
 
I also processed the same data using monomer-n-mer self association model
with n = 4:
 
logKa1n = 3.047, then Ka1n = 1.114 x 10+3 M3, so 1/Ka1n = 4.487 x 10-4, or
448.7 uM.
 
Again the thermo_results file says c1 = c4 at 9.642 x 10-2 M = 96.42 mM.
 
Can someone explain please?
 
Many Thanks,

Mark
 
 
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