Perhaps I was a bit gruff, but what I was trying to do was throw out a
voice of caution.  In crystallography, it is very tempting to
"normalize" everything by setting the mean to zero and the RMS
deviation to one, but this is a "linear transformation", and linear
transformations only work when the things you are comparing are
"linear".  True, everything is linear to a first-order approximation,
and you may well be "okay" comparing B factors between two crystals
that are "not all that different" in average B, but the assumption of
"linearity" with something fundamentally non-linear like B factors
will break down at some point.  I'm not really sure where, but I would
suspect that "normalizing" B factors from a 5 A structure to those of
a 2.0 A structure might be "ill advised".

My opinion is that multiplying and dividing B factors makes no
physical sense.  If you must "normalize" something, I'd suggest
normalizing the "half-occupancy volume" (proportional to sqrt(B)^3),
or the width of the half-occupancy volume (proportional to sqrt(B)).
If you want to try to make a leap to "entropy", then if you assume the
half-occupancy volume contains an ideal gas, then the entropy should
be proportional to log(B).  But B itself is proportional to the
surface area of the half-occupancy volume, and I can't think of too
many physical-chemical properties that would be "proportional" to
that.

Nevertheless, purely on theory, I predict you will much more easily
hit "pitfalls" (or quicksand) by trying to normalize "B" directly
instead of first converting it into something more tangible (like a
volume).

That's all I'm sayin'

-James Holton
MAD Scientist

On Mon, Mar 4, 2013 at 12:31 PM, John Fisher <[log in to unmask]> wrote:
> Seriously?
> I believe this specific forum has become quicksand rather than a useful
> tool. Normalization based simply on ratios of one struct to the other should
> allow normalization. Correct?, or have I just simply lost my mind here?
> J
>
> John Fisher, M.D./PhD
> St. Jude Children's Research Hospital
> Department of Oncology
> Department of Structural Biology
> W: 901-595-6193
> C: 901-409-5699
>
> On Mar 4, 2013, at 1:26 PM, James Holton <[log in to unmask]> wrote:
>
>
> No, you can only add and subtract B values because that is mathematically
> equivalent to multiplication in reciprocal space (which is equivalent to
> convolution in real space):
>
> exp(-B1*s^2) * exp(-B2*s^2) = exp(-(B1+B2)*s^2)
>
> Multiplying and dividing B values is mathematically equivalent to applying
> fractional power-law or fractional root functions in reciprocal space (and I
> don't even want to think about what that does in real space).
>
> exp(-B1*B2*s^2) = ???
>
> -James Holton
> MAD Scientist
>
>
> On 3/4/2013 11:19 AM, Bosch, Juergen wrote:
>
> 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 <[log in to unmask]> 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: [log in to unmask]
> *******************************************
>
>
> ......................
> 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
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>
>
>
>
>