James,
Thank you for your help.
I appreciate the very thorough explanation. I have never heard of "noise"
although I have produced a 'kicked' map in phenix and refined the
structure using those maps..though I guess this is different. I will try
it. Thanks.
-Yarrow
>
> 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.
>>
>>
>
>
--
Yarrow Madrona
Graduate Student
Molecular Biology and Biochemistry Dept.
University of California, Irvine
Natural Sciences I, Rm 2403
Irvine, CA 92697
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