Error with pdb2gmx when peptide is capped with N-terminal acetyl group

GROMACS version: 2021.2
GROMACS modification: No

I am trying to produce a .gro file using the charmm36m ff. My peptide is 21 residues long and has an acetyl group at the N-terminus.

When I ran gmx pdb2gmx -f peptide_charmm.pdb -o peptide.gro -ignh -ter I got the following error:

Residue 2 named ACE of a molecule in the input file was mapped to an entry in the topology database, but the atom C used in that entry is not found in the input file. Perhaps your atom and/or residue naming needs to be fixed.

The atom names for ACE in the .rtp file are as follows:

[ ACE ] ; N-terminal acetyl patch
[ atoms ]
CAY CT3 -0.270 1
HY1 HA3 0.090 1
HY2 HA3 0.090 1
HY3 HA3 0.090 1
CY C 0.510 2
OY O -0.510 2

and in my pdb file the ACE and first residue SER are shown below:

ATOM 1 CAY ACE P 1 -2.160 -0.094 1.070 1.00 0.00 PROA
ATOM 2 HY1 ACE P 1 -2.594 -0.183 2.088 1.00 0.00 PROA
ATOM 3 HY2 ACE P 1 -2.503 -0.953 0.455 1.00 0.00 PROA
ATOM 4 HY3 ACE P 1 -2.503 0.860 0.614 1.00 0.00 PROA
ATOM 5 CY ACE P 1 -0.672 -0.102 1.160 1.00 0.00 PROA
ATOM 6 OY ACE P 1 -0.105 -0.197 2.247 1.00 0.00 PROA
ATOM 7 N SER P 2 0.000 0.000 0.000 1.00 0.00 PROA
ATOM 8 HN SER P 2 -0.453 0.078 -0.885 1.00 0.00 PROA
ATOM 9 CA SER P 2 1.461 0.000 0.000 1.00 0.00 PROA
ATOM 10 HA SER P 2 1.786 -0.916 -0.471 1.00 0.00 PROA
ATOM 11 CB SER P 2 2.051 1.206 -0.743 1.00 0.00 PROA
ATOM 12 HB1 SER P 2 1.854 2.133 -0.163 1.00 0.00 PROA
ATOM 13 HB2 SER P 2 1.581 1.291 -1.746 1.00 0.00 PROA
ATOM 14 OG SER P 2 3.466 1.063 -0.911 1.00 0.00 PROA
ATOM 15 HG1 SER P 2 3.758 1.849 -1.379 1.00 0.00 PROA
ATOM 16 C SER P 2 1.915 0.000 1.471 1.00 0.00 PROA
ATOM 17 O SER P 2 1.128 0.000 2.417 1.00 0.00 PROA

I would be really grateful if anyone could offer advice on how to overcome this error. Thank you in advance!

I had the same problem, and eventually I have changed the ACE in the merged.rtp using the older definition of the atom names in ACE:
[ atoms ]
CH3 CT3 -0.270 1
HH31 HA3 0.090 1
HH32 HA3 0.090 1
HH33 HA3 0.090 1
C C 0.510 2
O O -0.510 2
[ bonds ]
C CH3
C +N
CH3 HH31
CH3 HH32
CH3 HH33
O C
[ impropers ]
C CH3 +N O

Bests,
Sandor

Hi Sandor,

Thank you so much for the advice. It worked for me too :) I really appreciate your help!

Best wishes,

Alice