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Protein Geometry Motivation

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Protein Geometry Motivation Protein Geometry Measuring Geometric Properties of Proteins Protein Geometry 1 Motivation One of the goals of structural bioinformatics is to aid the biochemist in modeling molecular functionality on a computer. Notice the change in experimental setting: in vivo in vitro in silico Although reaction mechanisms are essentially the final statement in characterizing molecular interactions, there is often a need to track conformational changes and other geometric aspects of the molecules. Protein Geometry 2 1 Introduction The geometry of molecules deals with computations related to bond length and inter-atomic distances in general, bond angles, and dihedral angles. More complicated calculations deal with the construction of molecular surfaces and quantities such as charge densities. These are calculations related to a static molecule. In more dynamic setting we may attempt to evaluate these quantities as they change with time due to flexibility of the molecule. Modeling the flexibility is itself a big challenge. Protein Geometry 3 Given Atomic Coordinates We start with the assumption that we know the positions of atoms in 3D space. We then develop the formulae that give us various other measurements. bond length, bond angles, dihedral angles. In some applications you are given the inter- atomic distances and need to derive the coordinates. Protein Geometry 4 2 Distance Between Atoms If the position of atoms in 3-space is given by (x, y, z) coordinates then we can use the standard Pythagorean calculation of distance. T If atom a has coordinates (ax, ay, az) and atom b T has coordinates (bx, by, bz) the distance between a and b is given by: 222 d a, b ax b x a y b y a z b z This is the same as the norm calculation: a b a bT a b. Protein Geometry 5 Bond Angle Consider two atoms (vectors a and b) bonded to a third atom represented by vector c. We calculate the bond angle at c by first calculating cos . cos u , v u v Recall that an inner product of normalized vectors can be viewed as the cosine of the a angle between these vectors. In this case: u a c v b c. c So: b a cT b c cos . a c b c Protein Geometry 6 3 Dihedral Angles (1) Consider atoms in the protein backbone: N1 Ca 1 C 1 N 2 C a 2 C 2 N 3 C a 3 C 3 N 4 Going from residue to residue: Corresponding bond lengths do not change much. For example: Ni – Cai has approximately the same bond length as Nj – Caj. Similarly, corresponding bond angles tend to be the same. You cannot make this same statement for dihedral angles. Protein Geometry 7 Dihedral Angles (2) ? Dihedral angles are due to a “swivel” action Cai-1 around a single bond. C O For example, we could i-1 i-1 keep all bond angles Hi+1 constant while moving N Hi i Ni relative to Ni+1 swiveling Ni+1 around the Ci - Cai bond. Ca So, while a bond angle is i Ci determined by 3 atoms a dihedral angle is defined O Cbi Cb carbon in i by 4 atoms. Residue i. Protein Geometry 8 4 Dihedral Angles (3) Ca The alpha carbon PHI i-1 dihedral angle is the C angle between two i-1 Oi-1 planes: Unit normal of Hi+1 Ci-1, Ni, Cai plane. Hi Ni Ni+1 Unit normal of Cai Ni, Cai, Ci plane. Cai Ci Ci-1 Ci O dihedral Cbi i Ni Protein Geometry 9 Dihedral Angles (4) The alpha carbon PSI Cai-1 dihedral angle is the angle between two C i-1 Oi-1 planes: Unit normal of N , Ca , C plane. i i i Hi+1 Hi Ni NNi+1 Unit normal of i+1 Ci Cai, Ci, Ni+1 plane. Cai C Ni i Ni+1 OO dihedral Cbi i i Cai Protein Geometry 10 5 Dihedral Angles (5) So, given three points in a plane, we have to know how a unit normal to that plane is calculated. This is done by calculating a cross product. Suppose we have two vectors u and v. The cross product is a vector that is perpendicular to both u and v and it has a magnitude that is equal to the area of the parallelogram spanned by the vectors. It can be shown that this is: i j k uv det uu u uvuvi uvuvj uvuvk . 123 2332 3113 1221 v1 v 2 v 3 The next slides show this. Protein Geometry 11 Dihedral Angles (6) If the magnitude of the normal is the area of the parallelogram defined by u and v then its magnitude is given by the formula: u h u v sin where is the angle v h between u and v. u Then we can write: u v2 u 2 v 2sin22 u 2 v 2 1 cos 2 u22 v u v cos 2 u22 v uT v Protein Geometry 12 6 Dihedral Angles (7) Continuing: 3 3 3 2 2 2 2 2 u v u v uT v u 2 v 2 u v k k k k k1 k 1 k 1 We can now apply Lagrange’s Identity to see that the last expression is actually: 23 2 To prove this you can simply ui v j u j v i . expand both expressions i11 j i (tedious). (A proof of the general case can be found at: Now note that: http://en.wikipedia.org/wiki/Lagrange's_identity) 2 i j k 2 det uu u uvuvi uvuvj uvuvk 123 2332 3113 1221 v v v 1 2 3 is the same as this last sum. Protein Geometry 13 Dihedral Angles (8) Recall: The cross product is a vector that is perpendicular to both u and v and it has a magnitude that is equal to the area of the parallelogram spanned by the vectors. So far we have shown that working with this definition we get a vector that has a magnitude given by i j k detu u u . 1 2 3 Is the vector defined by this v1 v 2 v 3 determinant perpendicular to both u and v? Yes. If you calculate the inner product of this vector with u or v you get 0. For example: T u2 v 3 u 3 v 2 u 1 u v u v u 0. 3 1 1 3 2 u1 v 2 u 2 v 1 u 3 Protein Geometry 14 7 Dihedral Angles (9) Recall: The cross product is a vector that is perpendicular to both u and v and it has a magnitude that is equal to the area of the parallelogram spanned by the vectors. So far our determinant based calculation for the cross product has given the correct magnitude and we have the “perpendicular to both u and v” requirement. BUT which of these is correct? (They both show a vector perpendicular to both u and v). v v u z y u x For uv we use the “right-hand rule” that is also seen in the usual 3D Euclidean coordinate system: If the fingers of the right hand curl around the uv normal going in the direction from u to v then the thumb points in the direction assigned to . So the first diagram is for while the next diagram is for vu . Protein Geometry 15 Dihedral Angles (10) One last vector algebra issue: The cross product evaluated using the determinant does not necessarily have a unit length. Recall that our dihedral angle calculation requires vectors that are unit normal. In such a situation, we will have to normalize: Given we calculate: uv . uv Back to proteins Protein Geometry 16 8 Dihedral Angles (11) Suppose we are calculating the PHI dihedral. For simplicity, we will Start with the calculation of just use the atom names to label vectors Unit normal of that are their coordinates. Ci-1, Ni, Cai plane. We let u = Ni – Ci-1 and v = Cai – Ni. Unit normal of i j k Cai Ni, Cai, Ci plane. u v det u u u 1 2 3 Ci-1 C i v1 v 2 v 3 uv23 uvi 32 uv 31 uv 13 j uv 12 uvk 21 . Dividing this by its norm will give the Ni PHI dihedral unit normal (call it n(Ci-1, Ni, Cai)). Protein Geometry 17 Dihedral Angles (12) Calculating the PHI dihedral (continued): Then calculate the other unit normal: Unit normal of We let v = Cai – Ni Ci-1, Ni, Cai plane. and w = Ci – Cai. Then use: Unit normal of i j k Cai Ni, Cai, Ci plane. v w det v v v 1 2 3 Ci-1 w w w Ci 1 2 3 vw2332 vwi vw 3113 vw j vw 1221 vwk . Dividing this by its norm will give the Ni PHI dihedral unit normal (call it n( Ni, Cai,Ci,)). Protein Geometry 18 9 Dihedral Angles (13) Finally with both unit normals computed we can get the value of PHI for this alpha carbon using the arccos() function: CNCCi11, i , a i , i arccos nCNC i , i , a i nNCC i , a i , i . Sign of the dihedral angle: By convention, a dihedral angle is assumed to be in the range ,. Since the calculation of arccos() may lead to an angle in the range 0, we typically have to adjust the sign. (See next slide) Protein Geometry 19 Dihedral Angles (14) To derive the sign of a dihedral angle: We look along the bond lying in the intersection of the two planes (be sure to look in the increasing i direction).
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