`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.