Oliver,<div><br></div><div>peak_cutoff only changes the tolerance on the fTF correlation coefficient; however the program is probably still not finding a solution with a reasonable weighted phase-difference score Phi(g).  The tolerance for this (default=0.25) can be increased with the score_tol= command line argument.  You can send me your output file offlist for a closer look.</div>

<div><br></div><div>Nick</div><div><br><div class="gmail_quote">On Thu, Jul 12, 2012 at 8:44 AM, &quot;Weiergräber, Oliver H.&quot; <span dir="ltr">&lt;<a href="mailto:o.h.weiergraeber@fz-juelich.de" target="_blank">o.h.weiergraeber@fz-juelich.de</a>&gt;</span> wrote:<br>

<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">Hello,<br>
<br>
I am using labelit.check_pdb_symmetry in order to evaluate a crystal structure with sg P21 (2 copies/asu) and all angles indistinguishable from 90 degs for potential higher (i.e. orthorhombic) symmetry. Comparison of refinement statistics clearly indicates that the most likely scenario is perfect pseudomerohedral twinning of a true P21 crystal (Rfree=0.21). Accordingly, labelit.check_pdb_symmetry reports a striking discrepancy between Rsymop(Iobs) and Rsymop(Icalc): 8 % vs 31 %, i.e. there is an additional (apparent) 2-fold symmetry in the data which cannot be explained by the model.<br>


In an attempt to force a solution for comparison and visualization purposes, I have increased the r_symop_tol parameter to 0.35; the program starts to evaluate possible space groups with fTF, yielding scores (these are correlation coefficients, right?) way above 90 %. Still, in the end it reports there are no higher symmetry solutions.<br>


<br>
Why is this? Does the program check for the contrast between the best solution and the others? Changing peak_cutoff to a &quot;discriminative&quot; value did not help...<br>
<br>
Best regards,<br>
Oliver<br>
<br>
<br>
<br>
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