Doug Raiford Lesson 17
Framework model Secondary structure first Assemble secondary structure segments Hydrophobic collapse Molten: compact but denatured Formation of secondary structure after: settles in van der Waals forces and hydrogen bonds require close proximity 11/8/20152Protein Conformation Prediction (Part I)
Isolate protein and crystalize Time consuming process Slowly evaporate Many experiments in parallel Different conditions X-ray crystallography Get XYZ spatial coordinates 11/8/2015Protein Conformation Prediction (Part I)3
Store these XYZ coordinates in text files PDB website 11/8/2015Protein Conformation Prediction (Part I)4 X Y Z Occu Temp Element ATOM 1 N THR A N ATOM 2 CA THR A C ATOM 3 C THR A C ATOM 4 O THR A O ATOM 5 CB THR A C ATOM 6 OG1 THR A O ATOM 7 CG2 THR A C
To fully model the folding action of a polypeptide chain Must know all the forces acting on each aa Must be able to predict the motion of the aa’s given the forces 11/8/2015Protein Conformation Prediction (Part I)5
Recall that proteins are able to fold because of the torsional rotation of the aa bonds 11/8/2015Protein Conformation Prediction (Part I)6 almost always 180
Must be able to take phi and psi angles and transform into xyz coordinates of various atoms Don’t forget about R groups What places in space are occupied? Bump checking 11/8/2015Protein Conformation Prediction (Part I)7
Tetrahedron 11/8/2015Protein Conformation Prediction (Part I)8
11/8/20159Protein Conformation Prediction (Part I) almost always 180 Know distances Each angle is 109.5
11/8/201510Protein Conformation Prediction (Part I) 4 atoms on same plane , , and ω all relative to R group (O in case of ω)
One approach Given xyz of last three, and next torsion angle… Transform so that C is at origin, BC on new X, AB on plane of new Y Then apply torsion Start D on X Swing out 70.5 ( ; in the plane of Y) Rotate by torsion angle 11/8/201511Protein Conformation Prediction (Part I)
To transform a vector space… 11/8/2015Protein Conformation Prediction (Part I)12 X Y Z A B C
To transform a vector space… 11/8/2015Protein Conformation Prediction (Part I)13 X Y Z A B C New X axis New Y axis New Z axis
It’s all about projections If target vector is a unit vector then simple dot product 11/8/2015Protein Conformation Prediction (Part I)14 A B
Dot product of a row with vector yields the projection of the vector onto the vector represented by the row All three dot products yields all three components 11/8/2015Protein Conformation Prediction (Part I)15 X Y Z A B C New X New Y New Z
The new X is BC (as a unit vector) 11/8/2015Protein Conformation Prediction (Part I)16 X’ Y’ Z’ A B C
Remember, all we have is the last xyz coordinates All vectors are assumed to originate at the origin So BC is actually [X C,Y C,Z C ]-[X B,Y B,Z B ] 11/8/2015Protein Conformation Prediction (Part I)17 B C Origin
Magnitude of BC 11/8/2015Protein Conformation Prediction (Part I)18 X’ Y’ Z’ A B C
First row of transformation matrix 11/8/2015Protein Conformation Prediction (Part I)19 X Y Z A B C New X
AB in plane of new Y so Z component is zero 11/8/2015Protein Conformation Prediction (Part I)20 X Y Z A B C Important piece: Y component
Second row of transformation matrix 11/8/2015Protein Conformation Prediction (Part I)21 X Y Z A B C New Y
Third row of transformation matrix easy once have first two: Cross Product 11/8/2015Protein Conformation Prediction (Part I)22 X Y Z A B C New Y
Know distance to next atom Know angle is 70.5° ( ) X component = ||CD|| cos(70.5°) Y component starts out at ||CD|| sin(70.5°) This is the distance from X to the new D 11/8/2015Protein Conformation Prediction (Part I)23 X Y Z A B C D
Z component is that distance times sinθ (torsion angle) Y = ||CD|| sin(70.5°)*cos θ Z = ||CD|| sin(70.5°)*sin θ 11/8/2015Protein Conformation Prediction (Part I)24 Z Y C D new in plane of xy Y C X D final Θ (torsional angle) 70.5°
Transform next xyz into new vector space coordinates (same as before Determine ||CD|| 11/8/2015Protein Conformation Prediction (Part I)25 X Y Z A B C D
XYZ coordinates for an amino acid Build the linear transform matrix used to transform the original vector space into the space defined by the three atoms above. 11/8/2015Protein Conformation Prediction (Part I)26 AtomXYZ N CC C
BC? 11/8/2015Protein Conformation Prediction (Part I)27 AtomXYZ A N B C C C X Y Z A B C [X C,Y C,Z C ]-[X B,Y B,Z B ] [ ]-[ ] [ ] Magnitude of BC? distance B to C: New X axis: [ ] Calculator makes life easier: [2.863, ,-0.703] sto A [3.920, ,-0.705] sto B [5.265, ,-1.065] sto C unitV (C-B) unitV under “VECTR / MATH” Calculator makes life easier: [2.863, ,-0.703] sto A [3.920, ,-0.705] sto B [5.265, ,-1.065] sto C unitV (C-B) unitV under “VECTR / MATH”
Actually forgot a step Need to translate all three points Move in direction of negative C Will place C and origin and keep A and B relative to C 11/8/2015Protein Conformation Prediction (Part I)28 X Y Z A B C No change to X Calculator A-C sto A B-C sto B C-C sto C B-A sto AB C-B sto BC unitV BC (same answer) unitV under “VECTR / MATH” Calculator A-C sto A B-C sto B C-C sto C B-A sto AB C-B sto BC unitV BC (same answer) unitV under “VECTR / MATH”
New Y? 11/8/2015Protein Conformation Prediction (Part I)29 X Y Z A B C New Y axis: [ ] Calculator unitV(AB-(dot(AB,BC)/(norm BC) 2 * BC)) Norm under “VECTR / MATH” Calculator unitV(AB-(dot(AB,BC)/(norm BC) 2 * BC)) Norm under “VECTR / MATH”
New Z? 11/8/2015Protein Conformation Prediction (Part I)30 X Y Z A B C New Z axis: [ ] Calculator unitV BC enter sto X unitV(AB-(dot(AB,BC)/(norm BC) 2 * BC)) enter sto Y cross(X,Y) Cross under “VECTR / MATH” Calculator unitV BC enter sto X unitV(AB-(dot(AB,BC)/(norm BC) 2 * BC)) enter sto Y cross(X,Y) Cross under “VECTR / MATH”
De novo From first principles Comparative/Homology Based Sequence similarity Structure prediction methods De novo Homology modeling 11/8/201531Protein Conformation Prediction (Part I)
11/8/201532Protein Conformation Prediction (Part I)