hey guys.,x1=[6.630]; y1=[12.294]; z1=[-1.457]; x2=[7.613]; y2=[12.686]; z2=[-0.404]; x3=[8.324]; y3=[11.427]; z3=[0.102]; x4=[9.372]; y4=[11.018]; z4=[-0.564]; these are coordinates to calculate phi and psi angles, please tell me how i should do.?

hey guys.,

x1=[6.630]; y1=[12.294]; z1=[-1.457];  x2=[7.613]; y2=[12.686]; z2=[-0.404]; x3=[8.324]; y3=[11.427]; z3=[0.102]; x4=[9.372]; y4=[11.018]; z4=[-0.564];

these are coordinates to calculate phi and psi angles, please tell me how i should do.?

[KSSV]

How these coordinates are related to phi and psi? Do you have any formula?

[Stephen23]

Having lots of variables with names x1, x2, x3, etc, is not a good use of MATLAB: MATLAB works best on arrays, which means keeping your data together as much as possible makes code faster and easier to write.

Using lots of variables also makes accessing them slow and buggy, as is explained here:

https://www.mathworks.com/matlabcentral/answers/304528-tutorial-why-variables-should-not-be-named-dynamically-eval

Answer 1

[Roger Stafford]

(Corrected) I am assuming that the four sets of coordinates you describe correspond to four points P1 = [x1;y1;z1], P2 = [x2;y2;z2], P3 = [x3;y3;z3], P4 = [x4;y4;z4], and that these four points are positions of four atoms in a polypeptide chain for which you wish to determine their torsion angle, as defined in the website:

http://www.proteinstructures.com/Structure/Structure/Ramachandran-plot.html

for the four corresponding points A, B, C, and D in the “standard IUPAC definition” given there. This torsion angle can vary from -180 to +180 degrees as the rotation varies from extreme counterclockwise to extreme clockwise, respectively, while moving forward from point B (P2) to point C (P3). This is a signed dihedral angle between the plane defined by points A, B, and C, and the plane defined by points B, C, and D.

That computation can be performed in Matlab as follows:

Q12 = cross(P3-P2,P1-P2);
Q43 = cross(P3-P2,P4-P3);
Q1243 = cross(Q12,Q43);
ang = atan2d(sign(dot(P3-P2,Q1243))*norm(Q1243),dot(Q12,Q43));

[Note that the function ‘atan2d’ is used here to obtain angles in degrees, and the quantity “sign(dot(P3-P2,Q1243))” is used to distinguish between clockwise and counterclockwise rotations.]

[Andrei Bobrov]

+1, yes!

[Ram kumar Masilamani]

@Roger., I followed your answer. finally I got This is for psi angle p1=[6.630, 12.294, -1.457];(N atom) p2=[7.613, 12.686, -0.404];(c alpha) p3=[8.324, 11.427, 0.102];(c atom) p4=[9.372, 11.018, -0.564];(n atom)+1 q1 = cross(p1-p2,p3-p2)

q1 =

   -1.5241   -0.2513    1.5163

q2=cross(p2-p3,p4-p3)

q2 =

   -1.0454   -1.0038   -1.0286

ang = atan2(norm(cross(q1,q2)),dot(q1,q2))

ang =

    1.4964

is this right value for psi. please do me a reply

[Ram kumar Masilamani]

@Roger, And also I have a very basic doubt. I saw some phi and psi angles value like -58 degree, 123 degree. but I carried out different method to find out values of dihedral angles .. but I obtain only like single digit answers.. Am i going in right way..? please tell me

[Steven Lord]

The atan2 function returns angles in radians. If you want degrees, use atan2d.

[Roger Stafford]

@Ram. I have modified the answer I gave previously so as both to allow for counterclockwise rotations with negative values and to produce angles in degrees rather than radians. (For your example I got the torsion angle, psi, equal to 85.73846504695527 degrees.)