Affinity or dissociation constant?
jimhu at tamu.edu
Wed Jun 21 02:15:53 EST 2000
Nathan Bays wrote:
> Dissociation constant: When two molecules are in solution together, what is
> the ratio of the two molecules NOT stuck together vs. the two molecules
> stuck together? (quantitative)
> Please anyone feel free to clarify or correct.
Almost, but not exactly -
For A+B --> Complex
Kd = [A][B]/[Complex] :products over reactants in the REVERSE reaction,
hence the name - "dissociation" constant
Thus, the numerator is the product of the concentrations of the unbound
molecules. The vernacular reading of your definition would be the sum.
It's important to remember that [A] and [B] are the concentrations of unbound
molecules, not the total that is in the tube. Thus,
Atotal = [A]+[Complex] and Btotal = [B] + [Complex].
Consider the situation where Atotal >> Btotal and find out what fraction of the
total B is in a complex at any given Atotal
Using the definition of Kd and express [B] in terms of Btotal
Kd = [A](Btotal - [Complex])/[Complex]
Kd/[A] = Btotal/[Complex] - 1
Kd/[A] + 1 = Btotal/[Complex]
invert and rearrange:
_________ = ___________
Use Atotal ~ [A] since the contribution of [Complex] to Atotal is negligible.
After all we defined the conditions so that you put in a lot more A than B, so
you can't get more Complex than the amount of B you put into the reaction.
_________ = ________________
The ratio on the left is what you want - the fraction of the B you put in your
tube that is in the complex. Try programming this into an Excel spreadsheet
and substitute in some different values for Kd - you will see why lower Kd
means more binding. For a reaction with just 1 mole of each kind per mole of
complex, half of Btotal will be bound when Kd/Atotal = 1. Thus, the amount of
Atotal where half-maximal binding occurs approximates Kd. Note that this rule
of thumb fails if Btotal > Atotal, and that it fails when the complex is A2B
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