Questions regarding fragmentation using Robetta

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    • #2945
      Anonymous

        First I want to make sure that I understand this correctly. Fragmentation creates L-K+1 fragments, blasts them against some database and finds the homologous secondary structures that are shown below (small portion as an example).  Correct?

        I actually mean to ask about the many information below. The first column shows the pdb ID, then chain, residue number, residue, then what is H, L and E? Then coordinates, then I’m not sure what the rest of the columns are. I also want to ask how I can identify the fragments that correspond to a certain fragment in my sequence, because there seems to be no information about that in the fragmentation file. Finally, Robetta always gives 9mers and 3mers, why is that, and can I generate fragments of other lengths as well?


        1n40 A 165 S H -66.309 -17.535 177.170 3.855 9.042 22.902 7 0.000 P433 F197
        1n40 A 166 I H -104.711 -44.261 -172.345 3.855 9.042 22.902 7 0.000 P433 F197
        1n40 A 167 A H -62.841 -30.837 -178.413 3.855 9.042 22.902 7 0.000 P433 F197
        1n40 A 168 F L -105.367 22.303 167.891 3.855 9.042 22.902 7 0.000 P433 F197
        1n40 A 169 M E -65.784 151.772 172.783 3.855 9.042 22.902 7 0.000 P433 F197
        1n40 A 170 S L -119.403 91.178 -165.517 3.855 9.042 22.902 7 0.000 P433 F197
        1n40 A 171 S L -114.587 153.248 172.822 3.855 9.042 22.902 7 0.000 P433 F197
        1n40 A 172 A L -68.005 -24.031 -172.183 3.855 9.042 22.902 7 0.000 P433 F197
        1n40 A 173 D L -128.425 155.523 173.971 3.855 9.042 22.902 7 0.000 P433 F197

        1yht A 71 A H -60.918 -32.578 179.256 3.865 9.717 30.230 7 0.000 P433 F198
        1yht A 72 E H -65.272 -27.117 -178.195 3.865 9.717 30.230 7 0.000 P433 F198
        1yht A 73 N H -87.412 -14.097 -169.362 3.865 9.717 30.230 7 0.000 P433 F198
        1yht A 74 A L -93.084 161.236 164.697 3.865 9.717 30.230 7 0.000 P433 F198
        1yht A 75 V E -90.702 124.401 -178.577 3.865 9.717 30.230 7 0.000 P433 F198
        1yht A 76 Q E -90.968 130.726 179.279 3.865 9.717 30.230 7 0.000 P433 F198
        1yht A 77 G L -81.024 171.512 177.497 3.865 9.717 30.230 7 0.000 P433 F198
        1yht A 78 K L -65.057 -24.081 176.723 3.865 9.717 30.230 7 0.000 P433 F198
        1yht A 79 D L -81.104 0.318 177.046 3.865 9.717 30.230 7 0.000 P433 F198

        1m6i A 88 P H -53.890 -23.727 -179.641 3.869 9.371 49.107 7 0.000 P433 F199
        1m6i A 89 S H -76.655 -9.239 179.912 3.869 9.371 49.107 7 0.000 P433 F199
        1m6i A 90 F H -81.070 -15.028 179.415 3.869 9.371 49.107 7 0.000 P433 F199
        1m6i A 91 Y L -92.365 154.958 177.727 3.869 9.371 49.107 7 0.000 P433 F199
        1m6i A 92 V E -106.044 162.942 -179.911 3.869 9.371 49.107 7 0.000 P433 F199
        1m6i A 93 S L -73.799 144.915 180.031 3.869 9.371 49.107 7 0.000 P433 F199
        1m6i A 94 A L -58.576 -34.034 -179.515 3.869 9.371 49.107 7 0.000 P433 F199
        1m6i A 95 Q L -65.783 -35.614 -179.785 3.869 9.371 49.107 7 0.000 P433 F199
        1m6i A 96 D L -88.457 -22.462 -177.905 3.869 9.371 49.107 7 0.000 P433 F199

         

      • #14314
        Anonymous

          H/L/E are secondary structure assignments – these are DSSP letters. helix loop sheet.

          The “coordinates” are backbone torsions – phi, psi, omega probably – the third is definitely omega.  Only those three are stored because fragments assume ideal coordinates (so bond lengths and angles are assumed, and sidechains are irrelevant because we repack anyway).  I don’t know what the rest of the columns do without reading the code.

          I think 3 and 9 was chosen empirically: 15 years ago they tried several lengths and found that ab initio performed best at those lengths, so we standardized on it.

          (I am aware this is an incomplete answer, but this is what I know off the top of my head)

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