Doug RaifordLesson 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
04/21/23 2Protein Conformation Prediction (Part I)
Isolate protein and crystalize Time consuming process Slowly evaporate Many experiments in parallel Different conditions
X-ray crystallographyGet XYZ spatial coordinates
04/21/23 Protein Conformation Prediction (Part I) 3
Store these XYZ coordinates in text files
PDB website
04/21/23 Protein Conformation Prediction (Part I) 4
X Y Z Occu Temp ElementATOM 1 N THR A 5 23.200 72.500 13.648 1.00 51.07 N ATOM 2 CA THR A 5 23.930 72.550 12.350 1.00 51.27 C ATOM 3 C THR A 5 23.034 72.048 11.220 1.00 50.34 C ATOM 4 O THR A 5 22.819 72.747 10.228 1.00 51.19 O ATOM 5 CB THR A 5 25.221 71.703 12.416 1.00 51.94 C ATOM 6 OG1 THR A 5 26.159 72.326 13.305 1.00 53.51 O ATOM 7 CG2 THR A 5 25.849 71.583 11.046 1.00 53.33 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
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Recall that proteins are able to fold because of the torsional rotation of the aa bonds
04/21/23 Protein 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 groupsWhat places in space are occupied?
Bump checking
04/21/23 Protein Conformation Prediction (Part I) 7
Tetrahedron
04/21/23 Protein Conformation Prediction (Part I) 8
04/21/23 9Protein Conformation Prediction (Part I)
almost always 180
Know distancesEach angle is 109.5
04/21/23 10Protein 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
(180-109.5; in the plane of Y)
Rotate by torsion angle04/21/23 11Protein Conformation Prediction (Part I)
To transform a vector space…
04/21/23 Protein Conformation Prediction (Part I) 12
X
Y
Z
A
B
C
To transform a vector space…
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X
Y
Z
A
B
C
New X axisNew Y axis
New Z axis
It’s all about projections If target vector is a unit vector then
simple dot product
04/21/23 Protein 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
04/21/23 Protein Conformation Prediction (Part I) 15
X
Y
Z
A
B
C
New XNew Y
New Z
The new X is BC (as a unit vector)
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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 [XC,YC,ZC]-[XB,YB,ZB]
04/21/23 Protein Conformation Prediction (Part I) 17
B
C
Origin
Magnitude of BC
04/21/23 Protein Conformation Prediction (Part I) 18
X’
Y’
Z’
A
B
C
First row of transformation matrix
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X
Y
Z
A
B
C
New X
AB in plane of new Y so Z component is zero
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X
Y
Z
A
B
C
Important piece: Y component
Second row of transformation matrix
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X
Y
Z
A
B
C
New Y
Third row of transformation matrix easy once have first two: Cross Product
04/21/23 Protein Conformation Prediction (Part I) 22
X
Y
Z
A
B
C
New Y
Know distance to next atomKnow angle is 70.5° (180-109.5)
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
04/21/23 Protein 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
04/21/23 Protein Conformation Prediction (Part I) 24
Z
Y
C
Dnew in plane of xy
YC Dnew in
plane of xy
X
Dfinal
Θ (torsional angle)
70.5°
Transform next xyz into new vector space coordinates (same as before
Determine ||CD||
04/21/23 Protein 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.
04/21/23 Protein Conformation Prediction (Part I) 26
Atom X Y ZN 2.863 -15.219 -0.703C 3.920 -14.209 -0.705C 5.265 -14.836 -1.065
BC?
04/21/23 Protein Conformation Prediction (Part I) 27
Atom X Y Z A N 2.863 -15.219 -0.703 B C 3.920 -14.209 -0.705 C C 5.265 -14.836 -1.065
X
Y
Z
A
B
C
[XC,YC,ZC]-[XB,YB,ZB]
[5.265 -14.836 -1.065]-[3.920 -14.209 -0.705]
[1.345 -0.627 -0.36]
Magnitude of BC?
distance B to C: 1.527
New X axis:[0.880 -0.410 -0.236]
Calculator makes life easier:
[2.863,-15.219,-0.703] sto A[3.920,-14.209,-0.705] sto B[5.265,-14.836,-1.065] sto CunitV (C-B)
unitV under “VECTR / MATH”
Calculator makes life easier:
[2.863,-15.219,-0.703] sto A[3.920,-14.209,-0.705] sto B[5.265,-14.836,-1.065] sto CunitV (C-B)
unitV under “VECTR / MATH”
Actually forgot a stepNeed to translate all three pointsMove in direction of negative CWill place C and origin and
keep A and B relative to C
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X
Y
Z
A
B
C
No change to X
CalculatorA-C sto AB-C sto BC-C sto CB-A sto ABC-B sto BC
unitV BC (same answer)
unitV under “VECTR / MATH”
CalculatorA-C sto AB-C sto BC-C sto CB-A sto ABC-B sto BC
unitV BC (same answer)
unitV under “VECTR / MATH”
New Y?
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X
Y
Z
A
B
C
New Y axis:[0.440 0.894 0.088]
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?
04/21/23 Protein Conformation Prediction (Part I) 30
X
Y
Z
A
B
C
New Z axis:[0.174 -0.181 0.968]
CalculatorunitV BC enter sto XunitV(AB-(dot(AB,BC)/(norm BC)2 * BC)) enter sto Ycross(X,Y)
Cross under “VECTR / MATH”
CalculatorunitV BC enter sto XunitV(AB-(dot(AB,BC)/(norm BC)2 * BC)) enter sto Ycross(X,Y)
Cross under “VECTR / MATH”
De novo From first principles
Comparative/Homology Based Sequence similarity
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04/21/23 32Protein Conformation Prediction (Part I)