I have had several people now ask me where the heck my "Crick-Magdoff"
equation came from.  So it looks like brevity has once again conquered
clarity.  So, those of you you enjoy my long emails read on.  Those of
you who do not, I direct you to the "Delete" button that should have
come with your email client.


My little formula is a rough approximation, and it begins with the
Crick-Magdoff equation, which you can find in Blundell and Johnson
section 6.5 (page 161), paraphrased here:

rms(deltaI)/I_P = sqrt(2*N_H/N_P)*F_H/F_P

where:
deltaI   change in spot intensity due to heavy atom contribution
I_P      spot intensity with no heavy atom contribution
N_H   number of heavy atoms in cell
N_P   number of protein atoms in cell
F_H   structure factor of the heavy atom
F_P   average structure factor of the protein atom

I have always found it more useful to rearrange equations like this to
include the values you might have on hand, such as the molecular
weight.  So, if we assume that the average protein atom weighs 14 amu
and has 7 electrons in it, and that the average B-factor of protein and
heavy atoms are similar, we can replace F_P with 7 and replace N_P with
MW/14 where MW is the molecular weight of the protein in Daltons and
call N_H the number of heavy atoms per protein.  It is also not
completely wrong to replace F_H with fpp.

If we then require that the rms change in intensity be greater than the
average noise, then we can write down the requirement:
sigma(I_P) < rms(deltaI)

So, after doing these substitutions and rearranging, we get:

sigma(I_P)/I_P < sqrt(2*N_H/(MW/14))*fpp/7
sigma(I_P)/I_P < 0.756*sqrt(N_H/MW)*fpp

I_P/sigma(I_P) > 1.3*sqrt(MW/N_H)/fpp

There are some obvious approximations here.  Probably the biggest is
assuming that fpp = F_H.  In actual fact, anomalous differences "count
double" since fpp contributes both to F+ and F-.  I think Peter Zwart
pointed this out earlier.  There is also another sqrt(2) in the opposite
direction because sigma(delta-I) is the quadrature sum of two
sigma(I_P).  It also matters if you are interested in the rms anomalous
difference or the mean absolute anomalous difference, as these are not
the same thing.  Nonetheless, I think this last formula should be
accurate to at worst a factor of two.

In general, it is a good idea to have your signal be more than equal to
noise, so I consider this formula a limit to be avoided rather than a
goal to be met.  The skill and expertise required to solve the structure
increases quite sharply as your I/sigma(I) approaches this limit, but
you can always double I/sigma(I) by merging data from four crystals.
The latter is a better strategy.

-James Holton
MAD Scientist