Formally, the "best" way to compare B factors in two structures with 
different average B is to add a constant to all the B factors in the 
low-B structure until the average B factor is the same in both 
structures.  Then you can compare "apples to apples" as it were.  The 
"extra B" being added is equivalent to "blurring" the more well-ordered 
map to make it match the less-ordered one. Subtracting a B factor from 
the less-ordered structure is "sharpening", and the reason why you 
shouldn't do that here is because you'd be assuming that a sharpened map 
has just as much structural information as the better diffracting 
crystal, and that's obviously no true (not as many spots).   In reality, 
your comparison will always be limited by the worst-resolution data you 
have.

Another reason to add rather than subtract a B factor is because B 
factors are not really "linear" with anything sensible.  Yes, B=50 is 
"more disordered" than B=25, but is it "twice as disordered"? That 
depends on what you mean by "disorder", but no matter how you look at 
it, the answer is generally "no".

One way to define the "degree of disorder" is the volume swept out by 
the atom's nucleus as it "vibrates" (or otherwise varies from cell to 
cell).  This is NOT proportional to the B-factor, but rather the 3/2 
power of the B factor.   Yes, 3/2 power.  The value of "B", is 
proportional to the SQUARE of the width of the probability distribution 
of the nucleus, so to get the volume of space swept out by it you have 
to take the square root to get something proportional the the width and 
then you take the 3rd power to get something proportional to the volume.

An then, of course, if you want to talk about the electron cloud (which 
is what x-rays "see") and not the nuclear position (which you can only 
see if you are a neutron person), then you have to "add" a B factor of 
about 8 to every atom to account for the intrinsic width of the electron 
cloud.  Formally, the B factor is "convoluted" with the intrinsic atomic 
form factor, but a "native" B factor of 8 is pretty close for most atoms.

For those of you who are interested in something more exact than 
"proportional" the equation for the nuclear probability distribution 
generated by a given B factor is:
kernel_B(r) = (4*pi/B)^1.5*exp(-4*pi^2/B*r^2)
where "r" is the distance from the "average position" (aka the x-y-z 
coordinates in the PDB file).  Note that the width of this distribution 
of atomic positions is not really an "error bar", it is a "range".  
There's a difference between an atom actually being located in a variety 
of places vs not knowing the centroid of all these locations.  Remember, 
you're averaging over trillions of unit cells.  If you collect a 
different dataset from a similar crystal and re-refine the structure the 
final x-y-z coordinate assigned to the atom will not change all that much.

   The full-width at half-maximum (FWHM) of this kernel_B distribution is:
  fwhm = 0.1325*sqrt(B)
and the probability of finding the nucleus within this radius is 
actually only about 29%.  The radius that contains the nucleus half the 
time is about 1.3 times wider, or:
r_half = 0.1731*sqrt(B)

That is, for B=25, the atomic nucleus is within 0.87 A of its average 
position 50% of the time (a volume of 2.7 A^3).  Whereas for B=50, it is 
within 1.22 A 50% of the time (7.7 A^3).  Note that although B=50 is 
twice as big as B=25, the half-occupancy radius 0.87 A is not half as 
big as 1.22 A, nor are the volumes 2.7 and 7.7 A^3 related by a factor 
of two.

Why is this important for comparing two structures?   Since the B factor 
is non-linear with disorder, it is important to have a common reference 
point when comparing them.  If the low-B structure has two atoms with 
B=10 and B=15 with average overall B=12, that might seem to be 
"significant" (almost a factor of two in the half-occupancy volume) but 
if the other structure has an average B factor of 80, then suddenly 78 
vs 83 doesn't seem all that different (only a 10% change).  Basically, a 
difference that would be "significant" in a high-resolution structure is 
"washed out" by the overall crystallographic B factor of the 
low-resolution structure in this case.

Whether or not a 10% difference is "significant" depends on how accurate 
you think your B factors are.  If you "kick" your coordinates (aka using 
"noise" in PDBSET) and re-refine, how much do the final B factors change?

-James Holton
MAD Scientist

On 2/25/2013 12:08 PM, Yarrow Madrona wrote:
> Hello,
>
> Does anyone know a good method to compare B-factors between structures? I
> would like to compare mutants to a wild-type structure.
>
> For example, structure2 has a higher B-factor for residue X but how can I
> show that this is significant if the average B-factor is also higher?
> Thank you for your help.
>
>