# Re: AMBER: RESP charges of RNA nucleotides with 3' and 5' ends

From: Ilyas Yildirim <yildirim.pas.rochester.edu>
Date: Wed, 6 Jul 2005 18:24:45 -0400 (EDT)

Dear AMBER users, Qing Zhang and Francois,

I have been trying to understand how Cieplak et al. 95 paper calculated
the charges for RNA residues, and have some questions on this. I have
read Qing Zhang's "Procedure of Computing Partial Charges of
Carcinogen-Modified B-Deoxynucleotides" paper, in order to understand
the RESP procedure. I will appreciate if you could help me to get some
answer on my questions. My main concern is how to get the RESP charges
for 3' and 5' end of a strand. I would like to write what I have
understood from that paper, and have your ideas whether I understood it
right or not.

Let's say that we want to get the RESP charges for cytosine residue,
and call it "RC" (As it is called in xleap this way). We want to calculate
the charges for RC, RC3 and RC5. RC3 and RC5 are the same residues when
they are at 3' or 5' end of a strand. (As it is used in xleap this way)

So, in order to find the RESP charges for these 3 structures we would
follow the following procedure (Thats what I have extracted from Qing
Zhang and Cieplak et al. 95 papers):

1. Start with the nucleoside structure (here it is cytidine). Use an ab
initio program to calculate the ESP grid points.

2. Start another structure; dimethylphosphate (DMP). Use an ab initio
program and get the ESP grid points.

3. Use the program 'resp' to calculate the charges.

This is what I have extracted from these both papers. But the 3rd step
looks pretty confusing. In Qing Zhang's paper, at step 5, there are 2
stages defined.

On the first stage, it says that we have to define the charge
constraints. As far as I understood from that, it means that if we want to
calculate the RESP charges for RC5, this constraint is I + II = 0 ( I, II,
III, and IV are defined below). If we want to calculate the charges for
RC3, this constraint is III + IV = 0. If we want to calculate the RESP
charges for RC, the constraints are I + II = III + IV = 0. After defining
the charge constraint, on the first fit, we equivalence the O1P and O2P
atoms, and the H atoms in NH2 group. And a hyperbolic restraint with a
force constant a=0.0005 is used. I hope I am right till now.

Here, a couple of questions arises:

1. What is the meaning of 'equivalencing the atoms'? I know that at the
end, they will have the same charges, but this term 'equivalencing' is
used in Cieplak et al. 95. page 1366, such that all sugar atoms are
equivalenced. How can someone equivalence the sugar atoms of 4 different
residues? Does that mean that they do a multiple molecule fit to these 4
different residues and equivalence the sugar atoms?

2. And how can we include the charge constraints in the resp input file? I
have read the AMBER manual's RESP part, but without an example and the
meanings of each lines in the resp input file, it is very hard to
understand what each line means. (Especially I am confused after the 7th
area of resp input file definitions)

On the second stage, the CH2 and CH3 are refit while the rest of the
charges are kept. And a hyperbolic restraint with a force constant of
a=0.001 is used. And the H atoms of CH2 and CH3 are equivalenced.

Then, I have tried to analyze the sample case in Qing Zhang's paper. In
the first fit, in the 2nd section of the resp input file, it is defined
that we are doing a multiple molecule fit (here it is 2). On the 6th area,
the atomic numbers and ivaries are defined. There are 3 different
definitions for the ivary: 0, n and -99. When it is -99, this means that
"Dont change the charge anymore". But when it is defined as '0' or 'n',
the charges are varied. Here another term is used; 'center'. What does
this mean? Is this just used when we equivalence some of the atoms?

After the 6th area, I could not understand the rest of the lines. Here is
the copy of those lines used in this sample:

6 0.000
1 6 1 14 2 2 2 3 2 4 2 5
6 0.000
1 1 1 9 2 10 2 11 2 12 2 13

The 7th area is for charge constraints and the 7.1th area is for the list
of the atoms that will be constrained according to this charge
constraints. Now, here, I would assume that "6 0.0000" is the charge
constraint and the next line is the atoms that will be constrained on the
first molecule, while the 3rd line is again the charge constraint and the
4th line is the atoms in the second molecule (DMP) that will be
constrained.

As I said in the beginning of my email, my main concern is to find the
RESP charges of a residue when it is on 3' or 5' end. If there is any
automated procedure to find the RESP charges for a 3' and 5' end, that
will be very helpful. I have been using RED to calculate the RESP charges
for an arbitrary residue, but when it is 3' or 5' end, I am not sure how
to use RED to calculate the charges (because the total charge of the
molecule is not anymore 0 or -1). I will greatly appreciate any

Best,

PS:

The structure of DMP is as follows:

O1P
|
CH3 - O3' - P - O5' - CH3
|
--- O2P ---
II IV

The structure for the cytidine is as follows:

| HO
III | \
| O5'
|
C5' O4' BASE
\ / \ /
C4' C1'
\ /
C3' - C2'
|
O3' |
\ | I
H |

PS: I skipped the relevant hydrogens.

On Wed, 22 Jun 2005, FyD wrote:

> > I was checking out the RESP charges of some of the RNA residues: RC, RC5
> > and RC3, for example. The net charge of RC is -1, while the net charge of
> > RC5 and RC3 are -0.3081 and -0.6919, respectively.
>
> & RC5 + RC3 = -1
>
> > When I compare RC with
> > RC3, only O3' and H3T charges are different. When I compare RC with RC5,
> > only O5' and H5T are different. I was wondering how the charges are
> > defined in 5' and 3' ends.
>
> See Cieplak et al. J. Comput. Chem 1995, 16, 1357-1377
>
> > Are they defined in a way such that the 3'
> > residues are going to have -0.6919 net charge while 5' residues are going
> > to have -0.3081 net charge? I will use this information to create my own
> > 3' and 5' end for some non-standard residues. Thanks in advance.
>
>
> Regards, Francois
>
>

```--
Ilyas Yildirim
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Received on Wed Jul 06 2005 - 23:53:00 PDT
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