From: Arthur Rowe <Arthur.Rowe@nottingham.ac.uk>
  To  : j.matthews@biochem.usyd.edu.au, RASMB@bbri.org
  Date: Thu, 4 Nov 1999 15:12:41 GMT0BST

Re: cavities

Dear Jacqui -

you wrote:

" I have been performing sediemtation equilibrium experiments on a protein 
which has a large cavity at the centre, which apparantly not accesible to the solvent.  
Based on the crystal structure, this cavity contains a number of well-ordered waters.  
If I calculate vbar based on amino acid composition and cofactors, the apparant 
molecular weight I obtain is consistently and significantly lower than expected 
(having tried out a number of different conditions including different salt 
concentrations). I am about to see if I can account for this difference by a) including 
the ordered waters in my vbar estimations and b) assuming a completely 
water-filled cavity. 

 Does anybody have any experience with thsi type of phenonemon? 

 I should add that it is impractical to determine the vbar for this protein using 
densiometric techniques. "

Sounds a fascinating protein !  Dealing with the second (b) case first, if the cavity is 
entirely filled with water, packed at a density which is equivalent to the external 
(aqueous) solvent, then such water has no more effect on the vbar (and hence the 
mass estimated) than does externally bound 'water of solvation'.  There will be a 
small discrepancy for the case of salt solutions, due to the difference in density 
between pure water and the said salt solution, but for practical purposes any errors 
here (for DILUTE salt solutions) are likely to be minimal relative to other errors in 
AUC estimations of M.

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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