Hi,
>
> 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)
>
.
.
.
> I_P/sigma(I_P) > 1.3*sqrt(MW/N_H)/fpp
one can actually require that
abs(delta I) > 3 sigma(delta I)
Using eq. 13 from Acta Cryst. D61, 1437–1448, and the same (sometimes
(very) questionable assumptions) James used, the magic factor 1.3
inflates to 2.0
I once wrote a jiffy that allows one to simulate FOM's for various
phasing + errors scenarios. I still have the code around, if someone
is interested.
Note that sometimes you cannot find sites, but with knowledge of the
sites itself, phasing + density modification would result in
interpretable maps (Acta Cryst. D60, 1085–1093).
Also, the distribution of errors is rather important. I suspect that
what matters in the end is that you have enough well-measured Bijvoet
pairs to get the substructure. An empirical analysis possibly could
shift the magic number back to 1.3 ;-)
Cheers,
P
SAD Scientist ;-)
> 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
>
|