From: Karen Fleming <kgf@mail.csb.yale.edu>
  To  : arthur.rowe@nottingham.ac.uk
  Date: Sat, 06 Nov 1999 09:46:59 -0500

Re: cavities

Thanks to Arthur for a very thoughtful response. However, I should add that I look forward with great interest to the results of calculations that consider the
cavity to be entirely empty. I myself would be quite astonished if this was actually, physically the case. Nature does after all abhor a vacuum. And in fact, I
am not sure that I could be convinced of the existence of an entirely empty cavity even if the crystallographers did not include, for whatever reason, any or
all cavity waters in their model building and refinement. On the contrary, a lack of evidence of ordered water in the cavity would only suggest to me that the
electron density was not sufficiently precise enough to place ordered waters. I am not a crystallographer, but it is my impression that depending on the method
and the tricks used in the crystal structure refinement, the presence of waters in the cavity with an equivalent electron density to that of bulk solvent would
result in a minimization of their contribution to the model of the well ordered crystallized protein.

Just my humble opinion.....

Cheers,

Karen F.

> At the other extreme, if the cavity is entirely empty, then vbar which you calculate
> will be too small by the ratio (volume of protein incl cavity)/(volume of protein
> residues alone).  Hence the M which you calculate will be too low, roughly by 3%
> for every 1% by which numerator in the above exceeds the denominator.
> Potentially quite a big effect.  Presumably if you have a crystal structure you can
> make a fair estimate to the above ratio, and see what a modified vbar does to your
> M value.  If the explanation I have given is correct (I nearly said 'holds water' !)
> then the new M value will be either equal to or larger than the true M value.
>
> For any intermediate percentage of the cavity being filled, which you might again
> base on your crystallographic structure, you can simply estimate a vbar by
> proportionality.  A bit of a potential problem however might be that as you say the
> water is 'ordered', its packing density may differ from free water (as in solvent).
> That however is a second order refinement - I suggest to try the above calculations
> first.
>
> Best wishes
>
> Arthur Rowe
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