如何使用BioPython识别受体原子和施主原子？

Question

我知道如何使用BioPython (which is very easy)遍历蛋白质链中的结构、模型、链、残基和原子。

我如何通过遍历来识别链中的给体和受体原子？

Answer 1

gues Matteo很忙，所以我在这里尝试使用

供体-受体距离和供体-受体-受体前行角( ^H)

来自https://www.sciencedirect.com/science/article/pii/S2665928X20300246：肽键平面性约束氢键几何形状并影响二级结构构象

不确定代码是正确的，特别是H角计算，这可能也不是最快的方法，但尽管如此：

使用https://www.rcsb.org/structure/16PK中的16pk.pdb作为输入.pdb文件

我的代码是从How can I get a list of neighboring Hydrogen atoms of an alpha-carbon?复制的

import numpy as np

from Bio.PDB import Chain, Atom, NeighborSearch, PDBParser, Selection

def Hangle(ac: "C atom.coord" ,bc : " O atom.coord" ,cc: "N atom.coord") -> "H_angle" :
    """
    https://www.sciencedirect.com/science/article/pii/S2665928X20300246
    
    Peptide bond planarity constrains hydrogen bond geometry and influences secondary structure conformations
    
    and https://stackoverflow.com/questions/35176451/python-code-to-calculate-angle-between-three-point-using-their-3d-coordinates

    Parameters
    ----------
    a : "C atom.coord"
        DESCRIPTION.
    b : " O atom.coord"
        DESCRIPTION.
    c : "N atom.coord"
        DESCRIPTION.

    Returns
    -------
    TYPE
        DESCRIPTION.

    """
    a = np.array(ac)
    b = np.array(bc)
    c = np.array(cc)
    
    ba = a - b
    bc = c - b
    
    cosine_angle = np.dot(ba, bc) / (np.linalg.norm(ba) * np.linalg.norm(bc))
    angle = np.arccos(cosine_angle)
    
    return np.degrees(angle)




def get_putative_acceptors_list(n_donor_atom: Atom.Atom, po):
    ns = NeighborSearch(po)
    return [h for h in ns.search( n_donor_atom.get_coord(), 3.3, 'A') if h.element == 'O']



p = PDBParser()
structure = p.get_structure('16pk', '16pk.pdb')

n_donor_list = [atom for atom in structure.get_atoms() if atom.name == "N"]
    
    
po = Selection.unfold_entities(structure, 'A')

acceptor_list = []

water = 0
  
for n in range(len(n_donor_list)):
    
    o = get_putative_acceptors_list(n_donor_list[n], po)
    
    for i in o:
        if  i.get_parent().resname == 'HOH':
            water +=1
            # print('water', water)
            continue
        
        if i.name != 'O':
            # print(type(i))
            # # print(dir(i.get_parent()))
            # print(i, i.get_parent().resname, i.get_parent().id[1])
            continue
        else:
            if i.get_parent().id[1] != n_donor_list[n].get_parent().id[1]-1:
                 if i.get_parent().id[1] != n_donor_list[n].get_parent().id[1]:
                     
                     # if i.get_parent().id[1] == n_donor_list[n].get_parent().id[1]-3: #catches alpha elics ???
                
                         dist = np.linalg.norm(n_donor_list[n].coord - i.coord)
                         
                         h_angle = Hangle(i.get_parent()['C'].coord, i.coord,  n_donor_list[n].coord  )
                         
                         if 99 < h_angle < 156:
                
                             acceptor_list.append((n_donor_list[n], n_donor_list[n].get_parent().resname,n_donor_list[n].get_parent().id[1],
                                  i, i.get_parent().resname, i.get_parent().id[1], dist,
                                  i.serial_number, i.get_parent()['C'], i.get_parent()['C'].serial_number, h_angle))
                                              
    
for i in acceptor_list:
    print(i)
print('water', water)        

记住https://proteopedia.org/wiki/index.php/Hydrogen_bonds的精细指纹，我得到了主链O(来自C=O)和N(假定的施主)之间假定的H键的这种输出。

.....
(<Atom N>, 'GLU', 403, <Atom O>, 'ALA', 400, 3.247074, 3024, <Atom C>, 3023, 106.052284)
(<Atom N>, 'LEU', 404, <Atom O>, 'SER', 401, 3.2268648, 3029, <Atom C>, 3028, 109.48323)
(<Atom N>, 'LEU', 404, <Atom O>, 'ALA', 400, 3.1475434, 3024, <Atom C>, 3023, 154.70815)
(<Atom N>, 'LEU', 405, <Atom O>, 'LEU', 402, 3.0635853, 3035, <Atom C>, 3034, 112.344406)
(<Atom N>, 'GLU', 406, <Atom O>, 'GLU', 403, 3.075921, 3043, <Atom C>, 3042, 105.97161)
(<Atom N>, 'GLY', 407, <Atom O>, 'LEU', 404, 2.9155347, 3052, <Atom C>, 3051, 115.01412)
(<Atom N>, 'LYS', 408, <Atom O>, 'GLU', 403, 3.044416, 3043, <Atom C>, 3042, 152.44513)
....