# AMBER Archive (2006)Subject: AMBER: AMBER 8 - dihedral angles when clambda=1

From: Ilyas Yildirim (yildirim_at_pas.rochester.edu)
Date: Thu May 11 2006 - 23:57:37 CDT

Dear amber users/developers,

I am doing free energy calculations using the TI Approach. Here is a
structure I am dealing with:

cytidine isocytidine

C6 - C5 H41 C6 - C5 DH41
/ \ / / \ /
C1' - N1 C4-N4 ---- > C1' - N1 C4-O4
\ / \ \ / \
C2 - N3 H42 C2 - N3 DH42
| |
O2 N2
/ \ / \
DH21 DH22 H21 H22

I have set the dihedral parameters of the dummy atoms (the ones with DH..)
equal to zero.

DH-O -C -N* 1 0.00 180.0 2. [X -CA-N2-X]
DH-O -C -NC 1 0.00 180.0 2. [X -CA-N2-X]
DH-O -C -CM 1 0.00 180.0 2. [X -CA-N2-X]

These are the necessary dihedral parameters for the dummy atoms (I have
the other necessary parameters, too, but for my question they are not
relevant).

When I do a minimization using icfe=1, klambda=6, clambda=0; at the
end, I see that the C2-DH21-O2-DH22 is not planar to the base (which is
expected). The C4-H41-N4-H42 is planar (as expected).

When I do a minimization using icfe=1, klambda=6, clambda=1; at the end, I
see that the C2-H21-N2-H22 is planar (expected), but C4-DH41-O4-DH42 is
planar, too (unexpected).

When clambda=1, that means that the structure is 'isocytidine'. And having
defined the dummy atom dihedral angles as zero, C4-DH41-O4-DH42 should not
be planar at the end.

I would like to hear your ideas on what I might be doing wrong here.

Best,

```--
Ilyas Yildirim
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