You only entertain addition+subtraction--why not use multiplication/division to normalize the b-factors?
JPK On Mon, Mar 4, 2013 at 2:04 PM, James Holton <jmhol...@lbl.gov> wrote: > 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. >> >> >> -- ******************************************* Jacob Pearson Keller, PhD Postdoctoral Associate HHMI Janelia Farms Research Campus email: j-kell...@northwestern.edu *******************************************
