Firstly lets assume that the protein concentration bound is less than the
immobilised ligand concentration due to its size (and 7mM of a protein is
Secondly lets assume a Kd for the free ligand bound to protein, of 1mM.
The same ligand immobilised, will have a Kd which is much higher since the
off-rate will be the same, but the on-rate will be lower (due to
Kd = k(off)/k(on)
So lets assume the immobilised ligand now has a Kd of 7.3mM.
This means that we require an effective concentration of 7.3mM immobilised
ligand to bind half of the protein to the resin, but we can compete it off
the column with only >1mM free ligand.
As for why you need 100mM I can only guess. This may mean you have some
other modes of binding to the resin/gel, eg. aggregation on the resin once
bound. You then need higher concentration of ligand to disrupt the
immobilised complex. Another possibility is obviously degraded ligand
giving a lower effective concentration.
> I have a transmembrane protein which is hypothesised to bind a ligand which
> has been shown to bind to other (unrelated) transmembrane proteins. The
> is commercially available in an immobilised form, covalently attached to
> agarose @ 7.3umole per ml of packed gel.
>> Now, am I being simplistic in assuming that to compete with immobilised
> bound to protein binding sites, the free ligand must be applied to the column
> @ > 7.3mM (i.e 7.3umole/ml gel = 7.3mM) ?
>> My reason for asking is that, when reading through the literature, I came
> across a paper where the authors used this immobilised ligand, from the same
> suppliers, in a one-step purification method of an unrelated protein and
> (assuming they used the same amount of immobilised ligand / ml gel) were able
> to elute bound protein using 2mM free ligand, whereas I can only elute bound
> protein using 100mM free ligand.
>> Any suggestions gratefully received,
>> Rob Leach,
> School of Biochemistry & Molecular Biology,
> University of Leeds
brothers gonna work it out