Subject: Re: CCL:A little geometrical problem

From: Michael Jakusch <[email-masked]>
Date: Fri, 1 Mar 2002 09:45:57 +0100

On Thu, Feb 28, 2002 at 05:43:41PM +0100, Logean Antoine wrote:
> A little geometrical problem :
> 
> Taken a residue in a PDB file, how can I calculate the Cartesian
> coordinates of CB when I know the coordinates of N, CA and C, the angle
> N-CA-CB, C-CA-CB and the bond length CA-CB ? 
> 

I think the most general approach is as follows:

vn ... vector from CA to N  
vc ...           "       C
vx ...                   CB

an ... angle N-CA-CB
ac ...       C-CA-CB

ln ... length of vn
.                .
.                .
.                .

The scalar product

vn . vx = ln * lx * cos an   (1).

all terms on the right side are known (we know the length of the bond
CA-CB!), so (1) is essentially the equation of a plane (the one in
which the circle defined by the cone (from the angle) and the sphere
(of radius lx around CA) lies). Intersecting this plane with the other
one (vc, etc.) yields a line, the intersections of which with the
sphere of radius lx around CA give the position(s) of CB. (Of course
there are the three possible cases of 0, 1, or 2 solutions -
corresponding to the line not touching, just touching and intersecting
the sphere).

Regards,

Michael

-- 
Dr. Michael Jakusch

ETH - Swiss Federal Institute of Technology
Department of Applied Biosciences
Winterthurerstrasse 190
CH-8057 Zürich
Switzerland
Phone: +41.1.635 60 71
Fax:   +41.1.635 68 84
email: [email-masked]

---

From: Ivan Tubert <[email-masked]>
Date: Fri, 1 Mar 2002 11:21:36 -0500 (EST)

Of course there are up to two solutions!

It is possible to reach that conclusion without resorting to cones or
equations.

Think about the stereochemistry for a moment:

The three atoms that you know define a _plane_. The fourth atom can be off
the plane on either face, forming either _enantiomer_. If you know the
configuration of your aminoacid--usually (L)-- you can choose the correct
solution.

There is also the degenerate case in which the four atoms are on the same
plane, so you have only one solution; and you can have zero solutions if
the angles are not wide enough.

Ivan Tubert

On Fri, 1 Mar 2002, Ingo Brunberg wrote:

Obviously this problem is not as trivial as it seems. Though I can
think of a rather complicated solution, I just want to point out, that
there is most likely no single solution. The point you are searching
is located on the intersection of two cones with the angles N-CA-CB
and C-CA-CB originating from CA. So there are three different cases:

1) N-CA-CB + C-CA-CB < N-CA-C: no solution
2) N-CA-CB + C-CA-CB = N-CA-C: one solution
3) N-CA-CB + C-CA-CB > N-CA-C: two solutions

Seems like the problem leads to a quadratic equation.

Regards,

Ingo

> It seems to be trivial but I am unable to find a solution !
> 
> I am sure that somebody can help me.
> 
> In advance thank you
> 
> LOGEAN Antoine                              

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