Ok Thank you Justin for this clarification<br><br><div class="gmail_quote"><br><blockquote class="gmail_quote" style="border-left: 1px solid rgb(204, 204, 204); margin: 0pt 0pt 0pt 0.8ex; padding-left: 1ex;"><br>
<br>
<br>
<br>
sa wrote:<br>
> Dear All,<br>
><br>
> I have simulated 6 peptides (with 7 AA each capped in N and C termini)<br>
> in water and trehalose. During all the simulation time, the six peptides<br>
> have b-sheet conformations. I would like to calculate the average % of<br>
> secondary structure for the 6 peptides over the course of run. So I have<br>
> read the subject reported in the following link<br>
> <a href="http://redmine.gromacs.org/issues/683" target="_blank">http://redmine.gromacs.org/issues/683</a> and used the following command for<br>
> the two first frames<br>
><br>
><br>
> /work/sa001/gmx-post4.5.3/bin/do_dssp_mpi -f *-Center_All.xtc -s<br>
> run_1.tpr -tu ps -dt 1 -b 1 -e 5 -o<br>
> 6_Peptide_53A6_Trehal_Pref_SS.xpm -sss<br>
> 6_Peptide_53A6_Trehal_Pref_HEBT.dat -ssdump<br>
> 6_Peptide_53A6_Trehal_Dump_SS.dat -sc test.xvg<br>
><br>
> I obtained the following output for my six peptides<br>
><br>
> @TYPE xy<br>
> @ subtitle "Structure = + + + + + + + + + + + + + + + +<br>
> + + + + + + + + + + + + B-Sheet + + + + + + "<br>
> @ view 0.15, 0.15, 0.75, 0.85<br>
> @ legend on<br>
> @ legend box on<br>
> @ legend loctype view<br>
> @ legend 0.78, 0.8<br>
> @ legend length 2<br>
> @ s0 legend "Structure"<br>
> @ s1 legend "Coil"<br>
> @ s2 legend "B-Sheet"<br>
> @ s3 legend "Chain_Separator"<br>
> 2 30 12 30 5<br>
> 4 30 12 30 5<br>
> # Totals 60 24 60 10<br>
> # SS % 0.64 0.26 0.64 0.11<br>
><br>
><br>
> I can understand how the %SS values are obtained in the example given in<br>
> <a href="http://redmine.gromacs.org/issues/683" target="_blank">http://redmine.gromacs.org/issues/683</a>, but not in my case. Could you<br>
> tell me how the %SS is obtained the output above.<br>
><br>
<br>
Like any other average. From the code:<br>
<br>
/* now print percentages */<br>
fprintf(fp, "%-8s %5.2f", "# SS %", total_count / (real) (mat->nx * mat->ny));<br>
for(s=0; s<mat->nmap; s++)<br>
{<br>
fprintf(fp," %5.2f",total[s] / (real) (mat->nx * mat->ny));<br>
}<br>
fprintf(fp,"\n");<br>
<br>
So the total number of secondary structure elements is divided by the product of<br>
(number of frames * number of total residues).<br>
<br>
Your results are affected by the problem I mentioned in the issue report you<br>
quote. You have 42 residues, but since chain separators count as residues, the<br>
calculations are all done out of 47 residues instead. You'll have to either<br>
modify the code to account for this problem or simply re-calculate the averages<br>
yourself.<br>
<br>
-Justin<br>
<br>
> Thank you in advance for your help<br>
><br>
> SA<br>
><br>
><br>
><br>
<br>
--<br>
========================================<br>
<br>
Justin A. Lemkul<br>
Ph.D. Candidate<br>
ICTAS Doctoral Scholar<br>
MILES-IGERT Trainee<br>
Department of Biochemistry<br>
Virginia Tech<br>
Blacksburg, VA<br>
jalemkul[at]<a href="http://vt.edu" target="_blank">vt.edu</a> | (540) 231-9080<br>
<a href="http://www.bevanlab.biochem.vt.edu/Pages/Personal/justin" target="_blank">http://www.bevanlab.biochem.vt.edu/Pages/Personal/justin</a><br>
<br>
========================================<br>
</blockquote></div><br>