Affinity or dissociation constant?

Jim Hu jimhu at tamu.edu
Wed Jun 21 02:15:53 EST 2000


Nathan Bays wrote:
<snip>

> 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.

<snip>

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:

[Complex]        1
_________ = ___________
  Btotal     (Kd/[A])+1

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.


[Complex]        1
_________ = ________________
  Btotal     (Kd/Atotal)+1

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
etc.






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