如何计算蛋白质db文件中的键角? [英] How to calculate bond angle in protein db file?

问题描述

如果我对蛋白质骨架(N-Ca-C-N-Ca-C ....)具有三个x，y，z坐标，例如:

If I have three x,y,z coordinates for protein backbone (N-Ca-C-N-Ca-C....) as such:

N -14.152 0.961 4.712
CA -13.296 0.028 3.924
C -11.822 0.338 4.193
N -11.121 -0.642 4.703
CA -9.669 -0.447 4.998
C -8.861 -1.586 4.373

如何计算键角"(Ni-Cai-Ci，Cai-Ci-Ni + 1，Ci-Ni + 1-CAi + 1)?

how can I calculate the "bond angles" (Ni-Cai-Ci, Cai-Ci-Ni+1, Ci-Ni+1-CAi+1)?

推荐答案

基本矢量几何.两个归一化向量的点积是它们之间夹角的余弦.

Basic vector geometry. The dot-product of two normalized vectors is the cosine of the angle between them.

弗林斯:

N -14.152 0.961 4.712
CA -13.296 0.028 3.924
C -11.822 0.338 4.193

(N-Ca)=(-14.152 0.961 4.712)-(-13.296 0.028 3.924)=(-0.856，0.933，0.778)
归一化:(-0.576，0.628，0.524)

(N-Ca) = (-14.152 0.961 4.712) - (-13.296 0.028 3.924) = (-0.856, 0.933, 0.778)
normalized: (-0.576, 0.628, 0.524)

(C-Ca)=(-11.822 0.338 4.193)-(-13.296 0.028 3.924)=(1.474，0.310，0.269)
归一化:(0.963，0.203，0.176)

(C-Ca) = (-11.822 0.338 4.193) - (-13.296 0.028 3.924) = ( 1.474, 0.310, 0.269)
normalized: (0.963, 0.203, 0.176)

点积:(-0.576、0.628、0.524)x(0.963、0.203、0.176)= -0.335
角度= acos(-0.335)= 109.57度

dot-product: (-0.576, 0.628, 0.524) x (0.963, 0.203, 0.176) = -0.335
angle = acos(-0.335) = 109.57 degrees

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