Yep, I agree calculate the average B per structure and divide each B by this value, then multiply it by any value that is reasonable so you can visualize color differences :-) Jürgen
On Mar 4, 2013, at 2:16 PM, Jacob Keller wrote: 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<mailto: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<mailto:j-kell...@northwestern.edu> ******************************************* ...................... Jürgen Bosch Johns Hopkins University Bloomberg School of Public Health Department of Biochemistry & Molecular Biology Johns Hopkins Malaria Research Institute 615 North Wolfe Street, W8708 Baltimore, MD 21205 Office: +1-410-614-4742 Lab: +1-410-614-4894 Fax: +1-410-955-2926 http://lupo.jhsph.edu
