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[modeller_usage] RE:



Prasad,

 

The error message tells you exactly what the problem is. The number of residues in the alignment and PDB files are different. There are more or less residues in the alignment than what is read in from the PDB file.

 

Eswar.

 

---

Eswar Narayanan, Ph.D

Mission Bay Genentech Hall
600 16th Street, Suite N474Q
University of California, San Francisco
San Francisco, CA 94143-2240 (CA 94158 for courier)

Tel +1 (415) 514-4233; Fax +1 (415) 514-4231

http://www.salilab.org/~eashwar

-----Original Message-----
From: SLN Prasad Reddy [mailto:]
Sent:
Saturday, January 31, 2004 5:59 AM
To:
Subject:

 


  Sir,
        I am struggling in modeling a protein with ligand (ATP) and metal atom (Mg). I am requesting you to help me in this regard.

        My  problem : I want to model a protein with ligand and metal ion. The template I have chosen  is a crystal structure having both lignad and metal atom. I tried with procedure given  in the FAQ's question #16 of Modeller6v2 manual . I found symbols for ATP and Mg form restyp.lib  of Modeller 6v2 and incorporated at the end of the alignment like  following

>P1;templ1 structureX:templ1:
1::10::
FAYVI/.$@*
>P1;targ1 sequence:targ1:1::8::
-GWIV/.$@*


The top file is :

INCLUDE
SET OUTPUT_CONTROL = 1 1 1 1 1
SET ALNFILE = 'fi.ali'
SET KNOWNS = 'temp11'
SET SEQUENCE = 'targ1'
SET HETATM_IO = on
CALL ROUTINE = 'model'


But I am getting a log file as fallows
 
TOP_________>  105  705 SET ALNFILE = 'fi.ali'

TOP_________>  106  706 SET KNOWNS = 'temp11'

TOP_________>  107  707 SET SEQUENCE = 'targ1'

TOP_________>  108  708 SET HETATM_IO = ON

TOP_________>  109  709 CALL ROUTINE = 'model'

TOP_________>  110  399 CALL ROUTINE = 'getnames'

TOP_________>  111  509 STRING_IF STRING_ARGUMENTS = MODEL 'undefined', OPERATION;
                      = 'EQ', THEN =    'STRING_OPERATE OPERATION = CONCATENA;
                      TE, STRING_ARGUMENTS = SEQUENCE .ini, RESULT = MODEL'

TOP_________>  112  510 STRING_IF STRING_ARGUMENTS = CSRFILE 'undefined', OPERATI;
                      ON = 'EQ', THEN =    'STRING_OPERATE OPERATION = CONCATE;
                      NATE, STRING_ARGUMENTS = SEQUENCE .rsr,  RESULT = CSRFILE;
                      '

TOP_________>  113  511 STRING_OPERATE OPERATION = 'CONCATENATE',                ;
                        STRING_ARGUMENTS = SEQUENCE '.sch',  RESULT = SCHFILE

TOP_________>  114  512 STRING_OPERATE OPERATION = 'CONCATENATE',                ;
                        STRING_ARGUMENTS = SEQUENCE '.mat',  RESULT = MATRIX_FI;
                      LE

TOP_________>  115  513 SET ROOT_NAME = SEQUENCE

TOP_________>  116  514 RETURN

TOP_________>  117  400 CALL ROUTINE = 'homcsr'

TOP_________>  118  106 READ_ALIGNMENT FILE = ALNFILE, ALIGN_CODES = KNOWNS SEQUE;
                      NCE


Dynamically allocated memory at        amaxseq [B,kB,MB]:      2205269    2153.583    2.103
openf5__224_> Open      11  OLD  SEQUENTIAL  fi.ali

Dynamically allocated memory at        amaxbnd [B,kB,MB]:      5865109    5727.646    5.593
openf5__224_> Open      11  OLD  SEQUENTIAL  fi.ali
read_al_374_> Non-standard residue type,position,sequence:  $              1
read_al_374_> Non-standard residue type,position,sequence:  @              1
read_al_374_> Non-standard residue type,position,sequence:  $            2
read_al_374_> Non-standard residue type,position,sequence:  @            2

Read the alignment from file      : fi.ali
Total number of alignment positions: 

  #  Code        #_Res #_Segm PDB_code    Name
-------------------------------------------------------------------------------
  1      temp11        1        temp11
  2      targ1          1        targ1
TOP_________>  119  107 CHECK_ALIGNMENT

check_a_343_> >> BEGINNING OF COMMAND
openf5__224_> Open      11  OLD  SEQUENTIAL  ./temp11.pdb
rdabrk__290E> Number of residues in the alignment and  pdb files are different:     
              For alignment entry:        1
recover____E> ERROR_STATUS >= STOP_ON_ERROR:        1      1

Dynamically allocated memory at          finish [B,kB,MB]:      5865109    5727.646    5.593
Starting time                                            : 2000/08/03  19:06:47.190
Closing time                                            : 2000/08/03  19:06:52.888
Total CPU time [seconds]                                :      0.00


                              Kindly suggest me a solution to my problem.


Thanking you


With regards

Prasad .