How to extract the atomic coordinates of a specific amino acid from a pdb file

Hello, Is it possible in Matlab bioinformatics tool, to extract the coordinates of a specific amino acid like lysine and store it in a Matrix.. There will be many lysine residues in a protein. How to store the atomic coordinates of each and every lysine residue from a protein in matrices Lysine1_coordinates, Lysine 2_coordinates and so on from a pdb file using Matlab?

Answer 1

[Arthur Goldsipe]

What is your definition for the location of lysine amino acid? The reason I ask is that a PDB file contain information about the coordinates of specific atoms, not amino acids. As I'm sure you know from the other questions you've already posted, you can use getpdb to access this information in MATLAB. This information includes whether specific atoms are part of a lysine (from the resName field) and which amino acide in the seuqence (from the resSeq field), as well as the coordinates in the X, Y, and Z fields.

My assessment is that this question is a bit too general right now for you to get much help. I suggest you to attempt to sketch out what you want to do in more detail in a combination of MATLAB code and pseudocode. Then, perhaps we can offer more specific advice on whatever parts you're stuck on.

[Masha]

Thank you for your reply sir.

uxc = getpdb('1UXC'); 
isLYS = ({uxc.Model.Atom.resName} == "LYS") ;
uxcLYS = uxc.Model.Atom(isLYS)  % Extracts particulars of only LYS

uxcLYS = 1×110 struct array with fields:
    AtomSerNo
    AtomName
    altLoc
    resName
    chainID
    resSeq
    iCode
    X
    Y
    Z
    occupancy
    tempFactor
    segID
    element
    charge
    AtomNameStruct

LysXYZ = [uxcLYS.X; uxcLYS.Y; uxcLYS.Z]' %Coordinates of all lysine residues

LysXYZ = 110×3

-5.5370 8.9610 -2.1930 -5.2160 8.1380 -3.4020 -4.0490 7.1820 -3.0960 -3.4180 7.2650 -2.0570 -4.8430 9.0600 -4.5710 -6.0970 9.7390 -5.1370 -5.8120 11.2200 -5.4190 -6.3110 11.6180 -6.8150 -7.7190 12.1140 -6.7520 -5.1600 9.8650 -2.1270

%There are 22 atoms in a lysine residue. So after 22 atoms the coordinates
%of the next lysine residue come in. So based on that, I typed this. 
Lys1cd = LysXYZ(1:22,:) %Coordinates of first lysine residue

Lys1cd = 22×3

-5.5370 8.9610 -2.1930 -5.2160 8.1380 -3.4020 -4.0490 7.1820 -3.0960 -3.4180 7.2650 -2.0570 -4.8430 9.0600 -4.5710 -6.0970 9.7390 -5.1370 -5.8120 11.2200 -5.4190 -6.3110 11.6180 -6.8150 -7.7190 12.1140 -6.7520 -5.1600 9.8650 -2.1270

Lys2cd = LysXYZ(23:44,:) %Coordinates of second lysine residue

Lys2cd = 22×3

-1.5690 -7.8990 -0.0580 -0.7890 -8.8680 -0.8780 0.7290 -8.7220 -0.6950 1.4840 -9.2680 -1.4810 -1.1170 -8.5070 -2.3120 -2.3950 -9.1970 -2.7870 -2.4770 -9.0240 -4.3000 -3.7400 -9.6840 -4.8550 -3.9400 -9.2170 -6.2570 -1.4470 -6.9500 -0.2720

Lys3cd = LysXYZ(45:66,:)

Lys3cd = 22×3

3.1410 -10.0930 1.1860 3.8720 -11.3930 1.1360 3.8640 -11.9640 -0.2960 4.8350 -12.5720 -0.7040 3.2080 -12.3780 2.1090 4.0770 -12.5970 3.3570 3.3880 -11.9970 4.5880 4.1550 -12.3460 5.8710 3.6580 -13.6230 6.4660 2.3880 -10.0250 1.8000

For loop

Lys(1) = LysXYZ(1:22,:)

Unable to perform assignment because the indices on the left side are not compatible with the size of the right side.

for n=1:100
    h=n+1
Lys(h) = LysXYZ((22*n)+1:22*(h),:)
end
%How to write a for loop for this instead of manually typing in the range
%of rows required to extract the corresponing coordinates of lysine
%residues.

[Arthur Goldsipe]

The simplest way I can think of to fix your code is to do the following:

Lys = {};
for i = 1:5
    Lys{i} = LysXYZ((22*i-21):(22*i),:);
end

Note that you have to write Lys{i} instead of Lys(i). That's because I'm creating Lys as a cell array. You can read more about them here.

[Masha]

Thank you so much!