Apologies for my previous email appearing to put words in Dale's mouth- I'm
using my school's
webmail and it apparently doesn't indicate the quoted text.
The following is what I added:
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.
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