This is a very good point. I never was very happy with calculating the average of all B-factors. E.g. if one adds a lot of high-B water molecules, the average B goes up, but with the structure itself nothing changes. Instead of calculating the average B, it would probably better to calculate the “Wilson B-factor” of the calculated intensities and compare this to the observed Wilson B-factor.
My 2 cts, Herman Von: CCP4 bulletin board <CCP4BB@JISCMAIL.AC.UK> Im Auftrag von Edward Berry Gesendet: Mittwoch, 8. April 2020 03:52 An: CCP4BB@JISCMAIL.AC.UK Betreff: [EXTERNAL] Re: [ccp4bb] Average B factors with TLS EXTERNAL : Real sender is owner-ccp...@jiscmail.ac.uk<mailto:owner-ccp...@jiscmail.ac.uk> 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. ________________________________ To unsubscribe from the CCP4BB list, click the following link: https://www.jiscmail.ac.uk/cgi-bin/webadmin?SUBED1=CCP4BB&A=1<https://urldefense.proofpoint.com/v2/url?u=https-3A__www.jiscmail.ac.uk_cgi-2Dbin_webadmin-3FSUBED1-3DCCP4BB-26A-3D1&d=DwMFaQ&c=Dbf9zoswcQ-CRvvI7VX5j3HvibIuT3ZiarcKl5qtMPo&r=HK-CY_tL8CLLA93vdywyu3qI70R4H8oHzZyRHMQu1AQ&m=9Nm8tYLQdnLGaXZn9fIHeHI9UC6HUdDn7fCqWuCzghA&s=ecSRP0i1CUvJt3-jIzdvJwSj62hNWLjqzMccZnHzkpQ&e=> ######################################################################## To unsubscribe from the CCP4BB list, click the following link: https://www.jiscmail.ac.uk/cgi-bin/webadmin?SUBED1=CCP4BB&A=1
