>>> Dale Tronrud <[email protected]> 04/07/20 12:37 PM >>>
This topic has been discussing on the BB many times and a little
searching should give you some long-winded answers (some written by me).
The short version. If you refine a model with a common B factor for
all atoms, or keep a very narrow distribution of B's, you will end up
with an average B close to the Wilson B. If you don't something is
seriously wrong.
If you allow a distribution of B's in your model, that distribution
is skewed on the high side because B factors cannot go below zero but
there is no physical upper bound. The average of those B's will always
be larger than the Wilson B. How much larger depends on your refinement
method more than the properties of the crystal since it is determined by
how large a tail you allow your B distribution to have.
I think it is not just that the distribution is asymmetric and limited
to positive numbers-
it is due to the fact that "log" and "average" do
not commute (the average of the logs
is not the log of the averages),
and the logarithmic/exponential relation between intensity and B.
The wilson B is obtained from the slope of ln<Iobs> vs S^2.
from the relation <Iobs> ~ exp(-2BS^2).
We can take that slope as the rise over run in the range from S=0 to s=1/(3A),
even though
the real Wilson
plot will follow it only around 3A and beyond.
<Iobs> in a shell at resolution S will be down by a factor of
exp(-2BS^2) compared to <Iobs> at S=0
for S=(1/3A) this is a factor of exp(-0.222*B)
for B=10, this is a factor of 0.108
for B=100, this will be a factor of 2.3E-10
Now if for half the atoms B is 10 and for the other half B is 100
<Iobs> at 3 A will be something like (.108 + 2.3E-10)/2 = 0.054
(This is a little over my head because we are combining contribution of
the
two sets of atoms to each reflection in the shell, and they add
vectorially.
Maybe a factor of sqrt(2) is appropriate for random phases.
But for order of magnitude:)
The slope from S^2=0 to S^2=(1/3A)^2 =
-2B ={ ln(0.054) - ln(1) }/{.1111 - 0}
-2B = -26.3
B(wilson) = 13.2
Bave = (10+100)/2 = 55
The log of the average is larger than the average of the logs.
Another way of looking at it, the contribution of those atoms with B=100
at 3A is completely negligible compared to the contribution of the atoms
with B=10, so the slope around 3A reflects only the B-factor of the
well-ordered atoms, and the slope measured there should give B=10, not 13.
If atoms were randomly distributed and we could apply Wilson over the
entire resol range, we still wouldn't get a straight line if there is a range
of atomic B-factors.
It would be like "curve peeling" in analyzing two simultaneous
first-order reactions with different half-time: semilog plot of
dissociation of a mixture of fast and slow hemoglobin. Near zero time
the curve is steep as the fast molecules dissociate, then it flattens
out and becomes linear with a smaller slope as only slow molecules are still
dissociating. So
(in the absence of a curve-fitting program) you calculate the rate
constant for the slow molecules from the linear region at long time, and
extrapolate the line back to zero time to get the initial percent slow.
Then you can calculate the amount of slow at any time point and subtract
that from the total to get the amount of fast, and re-plot that on
semilog to get the fast rate constant.
By analyzing our Wilson plots at 3A (or more generally at the highest
resolution available) we are getting the B-factor for the
slowest-decaying (lowest B-factor) atoms, getting B=10 not 13 in this case.
You didn't say what your B factor model was when you achieved an
average value of 31 A^2. This value seems tiny to me since it implies
that your intensities are falling off in resolution so slowly that you
surely should have been able to measure data to a higher resolution. If
you decide to deposit this model you should look into why you have such
a low value.
On the other hand, the average B of 157 A^2 seems quite reasonable
for a 3 A model (using modern resolution cutoff criteria). It is higher
than your Wilson B, but that is expected. In addition, as you note, the
uncertainty of a Wilson B is quite large in the absence of high
resolution data.
Yes, this is the short version. ;-)
Dale Tronrud
On 4/7/2020 5:16 AM, Nicholas Keep wrote:
> I am at the point of depositing a low resolution (3.15 A) structure
> refined with REFMAC. The average B factors were 31 before I added the
> TLS contribution as required for deposition which raised them to 157-
> this is flagged as a problem with the deposition, although this did not
> stop submssion. The estimated Wilson B factor is 80.5 (although that
> will be quite uncertain) so somewhere between these two extremes.
>
> Is it only the relative B factors of the chains that is at all
> informative? Should I report the rather low values without TLS
> contribution or the rather high ones in any "Table 1"? Comments
> appreciated.
>
> Thanks
>
> Nick
>
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