Indexing Dihedrals#

Dihedrals represent the torsion angle around a central chemical bond between four atoms in a molecule. There are four atoms, three bonds and two angles contained in a dihedral.

Dihedral is a molecular container that contains the four atoms, three bonds and two angles that make up the dihedral.

For example, let’s look at the aladip system again.

>>> import sire as sr
>>> mols = sr.load(sr.expand(sr.tutorial_url, ["ala.top", "ala.crd"]))
>>> mol = mols[0]
>>> print(mol)
Molecule( ACE:2   num_atoms=22 num_residues=3 )

We can get all of the dihedrals using the dihedrals() function.

>>> print(mol.dihedrals())
SelectorDihedral( size=41
0: Dihedral( HH31:1 <= CH3:2 = C:5 => N:7 )
1: Dihedral( HH31:1 <= CH3:2 = C:5 => O:6 )
2: Dihedral( CH3:2 <= C:5 = N:7 => CA:9 )
3: Dihedral( CH3:2 <= C:5 = N:7 => H:8 )
4: Dihedral( HH32:3 <= CH3:2 = C:5 => N:7 )
...
36: Dihedral( O:16 <= C:15 = N:17 => H:18 )
37: Dihedral( O:16 <= C:15 = N:17 => CH3:19 )
38: Dihedral( H:18 <= N:17 = CH3:19 => HH32:21 )
39: Dihedral( H:18 <= N:17 = CH3:19 => HH33:22 )
40: Dihedral( H:18 <= N:17 = CH3:19 => HH31:20 )
)

The result (a SelectorDihedral) is a molecular container for dihedrals. Like all molecular containers, it can be indexed,

>>> print(mol.dihedrals()[0])
Dihedral( HH31:1 <= CH3:2 = C:5 => N:7 )

sliced

>>> print(mol.dihedrals()[0:5])
SelectorDihedral( size=5
0: Dihedral( HH31:1 <= CH3:2 = C:5 => N:7 )
1: Dihedral( HH31:1 <= CH3:2 = C:5 => O:6 )
2: Dihedral( CH3:2 <= C:5 = N:7 => CA:9 )
3: Dihedral( CH3:2 <= C:5 = N:7 => H:8 )
4: Dihedral( HH32:3 <= CH3:2 = C:5 => N:7 )
)

or accessed by a list of indicies

>>> print(mol.dihedrals()[ [0, 2, 4, 6, 8] ])
SelectorDihedral( size=5
0: Dihedral( HH31:1 <= CH3:2 = C:5 => N:7 )
1: Dihedral( CH3:2 <= C:5 = N:7 => CA:9 )
2: Dihedral( HH32:3 <= CH3:2 = C:5 => N:7 )
3: Dihedral( HH33:4 <= CH3:2 = C:5 => N:7 )
4: Dihedral( C:5 <= N:7 = CA:9 => C:15 )
)

The Dihedral object is also a molecular container, so can be indexed, searched and sliced just like any other container.

>>> dihedral = mol.dihedrals()[0]
>>> print(dihedral[0])
Atom( HH31:1  [  18.45,    3.49,   12.44] )
>>> print(dihedral[1])
Atom( CH3:2   [  18.98,    3.45,   13.39] )
>>> print(dihedral.angles()[0])
Angle( HH31:1 <= CH3:2 => C:5 )
>>> print(dihedral.bonds()[0])
Bond( HH31:1 => CH3:2 )

Accessing dihedrals by atom, bond or angle#

You can also find dihedrals by looking for their constituent atoms. For example,

>>> print(mol.dihedrals("atomnum 1", "atomnum 2", "atomnum 5", "atomnum 7"))
SelectorDihedral( size=1
0: Dihedral( HH31:1 <= CH3:2 = C:5 => N:7 )
)

Returns the dihedrals between atoms with numbers 1, 2, 5 and 7. If there are no dihedrals that match, then an empty list is returned.

>>> print(mol.dihedrals("atomnum 1", "atomnum 2", "atomnum 3", "atomnum 4"))
SelectorDihedral::empty

If you are sure that there is only a single dihedral that matches, then you can use the dihedral() function

>>> print(mol.dihedral("atomnum 1", "atomnum 2", "atomnum 5", "atomnum 7"))
Dihedral( HH31:1 <= CH3:2 = C:5 => N:7 )

This will raise a KeyError if multiple dihedrals match, or if no dihedrals match.

You can use any valid atom identifier to identify the atoms. This includes search strings, e.g. here we can find all dihedrals where the carbon atoms are only in the middle two atoms.

>>> print(mol.dihedrals("not element C", "element C",
...                     "element C", "not element C"))
SelectorDihedral( size=16
0: Dihedral( HH31:1 <= CH3:2 = C:5 => O:6 )
1: Dihedral( HH31:1 <= CH3:2 = C:5 => N:7 )
2: Dihedral( HH32:3 <= CH3:2 = C:5 => O:6 )
3: Dihedral( HH32:3 <= CH3:2 = C:5 => N:7 )
4: Dihedral( HH33:4 <= CH3:2 = C:5 => O:6 )
...
11: Dihedral( HA:10 <= CA:9 = CB:11 => HB1:12 )
12: Dihedral( HA:10 <= CA:9 = CB:11 => HB2:13 )
13: Dihedral( HA:10 <= CA:9 = CB:11 => HB3:14 )
14: Dihedral( HA:10 <= CA:9 = C:15 => O:16 )
15: Dihedral( HA:10 <= CA:9 = C:15 => N:17 )
)

Passing in four atom identifiers, as above, will search for dihedrals by atom. Passing in three atom identifiers will search for dihedrals that contain the corresponding angle. For example

>>> print(mol.dihedrals("element H", "element C", "element C"))
SelectorDihedral( size=17
0: Dihedral( HH31:1 <= CH3:2 = C:5 => O:6 )
1: Dihedral( HH31:1 <= CH3:2 = C:5 => N:7 )
2: Dihedral( HH32:3 <= CH3:2 = C:5 => O:6 )
3: Dihedral( HH32:3 <= CH3:2 = C:5 => N:7 )
4: Dihedral( HH33:4 <= CH3:2 = C:5 => O:6 )
...
12: Dihedral( HA:10 <= CA:9 = C:15 => O:16 )
13: Dihedral( HA:10 <= CA:9 = C:15 => N:17 )
14: Dihedral( HB1:12 <= CB:11 = CA:9 => C:15 )
15: Dihedral( HB2:13 <= CB:11 = CA:9 => C:15 )
16: Dihedral( HB3:14 <= CB:11 = CA:9 => C:15 )
)

searches for dihedrals that contain angles between hydrogen-carbon-carbon atoms.

Passing in two atom identifiers will search for dihedrals that contain the corresponding bond.

For example, here we can find all of the dihedrals involving bonds between carbon and hydrogen atoms,

>>> print(mol.dihedrals("element C", "element H"))
SelectorDihedral( size=25
0: Dihedral( HH31:1 <= CH3:2 = C:5 => O:6 )
1: Dihedral( HH31:1 <= CH3:2 = C:5 => N:7 )
2: Dihedral( HH32:3 <= CH3:2 = C:5 => O:6 )
3: Dihedral( HH32:3 <= CH3:2 = C:5 => N:7 )
4: Dihedral( HH33:4 <= CH3:2 = C:5 => O:6 )
...
20: Dihedral( C:15 <= N:17 = CH3:19 => HH32:21 )
21: Dihedral( C:15 <= N:17 = CH3:19 => HH33:22 )
22: Dihedral( H:18 <= N:17 = CH3:19 => HH31:20 )
23: Dihedral( H:18 <= N:17 = CH3:19 => HH32:21 )
24: Dihedral( H:18 <= N:17 = CH3:19 => HH33:22 )
)

This would also work using atom identifying types, e.g. looking for dihedrals that contains the bond between atoms HH31:1 and CH3:2.

>>> print(mol.dihedrals(sr.atomid("HH31", 1), sr.atomid("CH3", 2)))
SelectorDihedral( size=2
0: Dihedral( HH31:1 <= CH3:2 = C:5 => O:6 )
1: Dihedral( HH31:1 <= CH3:2 = C:5 => N:7 )
)

You can even use complex search strings, here finding the dihedrals involving the bonds between atoms connecting two residues

>>> print(mol.dihedrals("atoms in residx 0", "atoms in residx 1"))
SelectorDihedral( size=10
0: Dihedral( HH31:1 <= CH3:2 = C:5 => N:7 )
1: Dihedral( CH3:2 <= C:5 = N:7 => H:8 )
2: Dihedral( CH3:2 <= C:5 = N:7 => CA:9 )
3: Dihedral( HH32:3 <= CH3:2 = C:5 => N:7 )
4: Dihedral( HH33:4 <= CH3:2 = C:5 => N:7 )
5: Dihedral( C:5 <= N:7 = CA:9 => HA:10 )
6: Dihedral( C:5 <= N:7 = CA:9 => CB:11 )
7: Dihedral( C:5 <= N:7 = CA:9 => C:15 )
8: Dihedral( O:6 <= C:5 = N:7 => H:8 )
9: Dihedral( O:6 <= C:5 = N:7 => CA:9 )
)

or mixing and matching searches

>>> print(mol.dihedrals(sr.atomid("C", 5), "element N"))
SelectorDihedral( size=10
0: Dihedral( HH31:1 <= CH3:2 = C:5 => N:7 )
1: Dihedral( CH3:2 <= C:5 = N:7 => H:8 )
2: Dihedral( CH3:2 <= C:5 = N:7 => CA:9 )
3: Dihedral( HH32:3 <= CH3:2 = C:5 => N:7 )
4: Dihedral( HH33:4 <= CH3:2 = C:5 => N:7 )
5: Dihedral( C:5 <= N:7 = CA:9 => HA:10 )
6: Dihedral( C:5 <= N:7 = CA:9 => CB:11 )
7: Dihedral( C:5 <= N:7 = CA:9 => C:15 )
8: Dihedral( O:6 <= C:5 = N:7 => H:8 )
9: Dihedral( O:6 <= C:5 = N:7 => CA:9 )
)

Passing in a single atom identifier will return all of the dihedrals that involve that atom (or atoms).

>>> print(mol.dihedrals("atomnum 2"))
SelectorDihedral( size=8
0: Dihedral( HH31:1 <= CH3:2 = C:5 => O:6 )
1: Dihedral( HH31:1 <= CH3:2 = C:5 => N:7 )
2: Dihedral( CH3:2 <= C:5 = N:7 => H:8 )
3: Dihedral( CH3:2 <= C:5 = N:7 => CA:9 )
4: Dihedral( HH32:3 <= CH3:2 = C:5 => O:6 )
5: Dihedral( HH32:3 <= CH3:2 = C:5 => N:7 )
6: Dihedral( HH33:4 <= CH3:2 = C:5 => O:6 )
7: Dihedral( HH33:4 <= CH3:2 = C:5 => N:7 )
)

This has returned all of the dihedrals that involve atom number 2, while

>>> print(mol.dihedrals("element C"))
SelectorDihedral( size=41
0: Dihedral( HH31:1 <= CH3:2 = C:5 => O:6 )
1: Dihedral( HH31:1 <= CH3:2 = C:5 => N:7 )
2: Dihedral( CH3:2 <= C:5 = N:7 => H:8 )
3: Dihedral( CH3:2 <= C:5 = N:7 => CA:9 )
4: Dihedral( HH32:3 <= CH3:2 = C:5 => O:6 )
...
36: Dihedral( O:16 <= C:15 = N:17 => H:18 )
37: Dihedral( O:16 <= C:15 = N:17 => CH3:19 )
38: Dihedral( H:18 <= N:17 = CH3:19 => HH31:20 )
39: Dihedral( H:18 <= N:17 = CH3:19 => HH32:21 )
40: Dihedral( H:18 <= N:17 = CH3:19 => HH33:22 )
)

gets all of the dihedrals that involve carbon.

Note that you can also use "*" to match anything, so

>>> print(mol.dihedrals("*", "element C", "element N", "*"))
SelectorDihedral( size=20
0: Dihedral( CH3:2 <= C:5 = N:7 => H:8 )
1: Dihedral( CH3:2 <= C:5 = N:7 => CA:9 )
2: Dihedral( C:5 <= N:7 = CA:9 => HA:10 )
3: Dihedral( C:5 <= N:7 = CA:9 => CB:11 )
4: Dihedral( C:5 <= N:7 = CA:9 => C:15 )
...
15: Dihedral( O:16 <= C:15 = N:17 => H:18 )
16: Dihedral( O:16 <= C:15 = N:17 => CH3:19 )
17: Dihedral( H:18 <= N:17 = CH3:19 => HH31:20 )
18: Dihedral( H:18 <= N:17 = CH3:19 => HH32:21 )
19: Dihedral( H:18 <= N:17 = CH3:19 => HH33:22 )
)

returns all of the dihedrals that are around carbon-nitrogen bonds.

Accessing dihedrals by residue#

You can also access dihedrals by residue, by passing in residue identifiers. Passing in two residue identifiers, such as here

>>> print(mol.dihedrals("residx 0", "residx 1"))
SelectorDihedral( size=10
0: Dihedral( HH31:1 <= CH3:2 = C:5 => N:7 )
1: Dihedral( CH3:2 <= C:5 = N:7 => H:8 )
2: Dihedral( CH3:2 <= C:5 = N:7 => CA:9 )
3: Dihedral( HH32:3 <= CH3:2 = C:5 => N:7 )
4: Dihedral( HH33:4 <= CH3:2 = C:5 => N:7 )
5: Dihedral( C:5 <= N:7 = CA:9 => HA:10 )
6: Dihedral( C:5 <= N:7 = CA:9 => CB:11 )
7: Dihedral( C:5 <= N:7 = CA:9 => C:15 )
8: Dihedral( O:6 <= C:5 = N:7 => H:8 )
9: Dihedral( O:6 <= C:5 = N:7 => CA:9 )
)

gives all of the dihedrals that involve bonds that are between those two residues.

While passing in a single residue identifier

>>> print(mol.dihedrals("residx 0"))
SelectorDihedral( size=13
0: Dihedral( HH31:1 <= CH3:2 = C:5 => O:6 )
1: Dihedral( HH31:1 <= CH3:2 = C:5 => N:7 )
2: Dihedral( CH3:2 <= C:5 = N:7 => H:8 )
3: Dihedral( CH3:2 <= C:5 = N:7 => CA:9 )
4: Dihedral( HH32:3 <= CH3:2 = C:5 => O:6 )
...
8: Dihedral( C:5 <= N:7 = CA:9 => HA:10 )
9: Dihedral( C:5 <= N:7 = CA:9 => CB:11 )
10: Dihedral( C:5 <= N:7 = CA:9 => C:15 )
11: Dihedral( O:6 <= C:5 = N:7 => H:8 )
12: Dihedral( O:6 <= C:5 = N:7 => CA:9 )
)

gives all of the dihedrals that involve atoms in this residue (including the dihedrals to other residues).

If you want the dihedrals that are contained only within the residue, then use the dihedrals function on that residue,

>>> print(mol["residx 0"].dihedrals())
SelectorDihedral( size=3
0: Dihedral( HH31:1 <= CH3:2 = C:5 => O:6 )
1: Dihedral( HH32:3 <= CH3:2 = C:5 => O:6 )
2: Dihedral( HH33:4 <= CH3:2 = C:5 => O:6 )
)

Calling the dihedrals function on any molecular container will return the dihedrals that involve only the atoms that are fully contained in that container.

Note

We have shown searching for dihedrals by residue. You can also search for dihedrals by chain or segment if your molecule has chains or segments. So print(mol.dihedrals("chainidx 0", "chainidx 1")) would print the dihedrals between the first two chains.

Uniquely identifying a dihedral#

Dihedrals are identified by their DihedralID. This is a quad of AtomID identifiers, one for each of the four atoms to be identified. While the atom identifier can be any type, it is best to use atom indexes, as these uniquely identify atoms in a molecule. A DihedralID comprised of four AtomIdx identifiers will uniquely identify a single dihedral.

You can easily construct a DihedralID using the sire.dihedralid() function, e.g.

>>> print(sr.dihedralid(0, 1, 2, 3))
Dihedral( AtomIdx(0), AtomIdx(1), AtomIdx(2), AtomIdx(3) )

constructs a DihedralID from atom indexes,

>>> print(sr.dihedralid("HH31", "CH3", "C", "O"))
Dihedral( AtomName('HH31'), AtomName('CH3'), AtomName('C'), AtomName('O') )

constructs one from atom names, and

>>> print(sr.dihedralid(sr.atomid(1), sr.atomid(2),
...                     sr.atomid(3), sr.atomid(4)))
Dihedral( AtomNum(1), AtomNum(2), AtomNum(3), AtomNum(4) )

constructs one from atom numbers.

You can mix and match the IDs if you want.

You can then use the DihedralID to index, just like any other identifier class.

>>> print(mols[sr.dihedralid("HH31", "CH3", "C", "O")])
Dihedral( HH31:1 <= CH3:2 = C:5 => O:6 )

gives the dihedral between the atoms called HH31, CH3, C and O in all molecules, while

>>> print(mols[sr.dihedralid(0, 1, 4, 5)])
Dihedral( HH31:1 <= CH3:2 = C:5 => O:6 )

gives the dihedral between the first, second, fifth and sixth atoms in each molecule.