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使用此 PDB 文件和以下PyMOL代码：

cd /Users/foo/Desktop/
reinitialize
load pdp_4gg6CD1_I.pdb
as cartoon 
select chainI, chain I
select chainC, chain C 
select chainD, chain D 
show sticks, chainI
spectrum count, cyan_red, chainI
color yellow, chain C

我可以制作这个图像：

在此处输入图像描述

我想要做的是显示chainC（黄色）中所选残基与chainI（棒）之间的距离。

我想要的chainC的选定残基是这样的：

[9, 23, 25, 44, 53, 54, 55, 59, 62, 63, 66]

Y    Y   H   W  R   R   F    F   T  N    V

我怎样才能做到这一点？

4

1 回答 1

2
  • 让我们cmd.iterate用来获取选择中的所有原子（链 C 和链 I），并将坐标和原子名称写入字典以供以后使用。
  • 然后我们可以计算所有原子之间的所有距离，并将最近的残基和原子名称写入两个列表（min_c和min_i）
  • 在最后一步中，我们只需要绘制最近的原子之间的距离对象，我们就完成了:)
from math import sqrt

def closestAtoms(list1=[9, 23, 25, 44, 53, 54, 55, 59, 62, 63, 66], list2=[5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17]):
    atoms = {}
    atoms2 = {}

    for r in list1:
        command = "chain C and resi %s" % (r)
        coordinates = {'atoms': []}
        cmd.iterate_state(-1, command, 'atoms.append([x, y, z, name])', space=coordinates)
        atoms[r] = coordinates['atoms']

    for i in list2:
        command = "chain I and resi %s" % (i)
        coordinates = {'atoms': []}
        cmd.iterate_state(-1, command, 'atoms.append([x, y, z, name])', space=coordinates)
        atoms2[i] = coordinates['atoms']

    for i in list2:
        min_dist = 10**3
        for c in list1:
            for cc in atoms[c]:
                for ii in atoms2[i]:
                    dist = sqrt((cc[0] - ii[0])**2 + (cc[1] - ii[1])**2 + (cc[2] - ii[2])**2)
                    if dist < min_dist:
                        min_dist = dist
                        min_c = [c, cc[3]]
                        min_i = [i, ii[3]]
        cmd.distance('dist_%s_%s' % (min_c[0], min_i[0]), 'chain C and resi %s and name %s' % (min_c[0], min_c[1]), 'chain I and resi %s and name %s' % (min_i[0], min_i[1]))

cmd.extend("closestAtoms", closestAtoms)
于 2016-06-29T22:14:48.803 回答