Molecular Biophysics Diffraction theory (1)

1. Molecular Biophysics Diffraction theory (1)

2. How do we know what we know about protein structure? Wave phenomena: X-ray diffraction Electron diffraction Electron microscopy Resonance phenomena: Nuclear magnetic resonance

3. Why not use light and many lenses?

4. X-rays electrons Neutrons nuclei Electrons charge

5. Unfortunately, no lenses exist for X-rays

6. so we have to use diffraction

7. Interaction of X-rays with electrons

9. A wave has an AMPLITUDE and a PHASE A.exp{i(2px/l-a)} A.sin(2px/l-a) The anatomy of a wave

10. e2 s0 r e1 s |S|=2sinq/l Scattering from two electrons |s0|=|s|=1/l phase difference = 2pr.(s-s0) = 2pr.S

11. (e2) 2pr.S (e1) Scattering from two electrons The two waves have equal amplitudes (because e1 and e2 are identical), but a phase difference of2pr.S |S|=|s-s0|=2sinq/l Total scatter = (e1) + (e2) = 1 + exp (2pir.S)

12. e3 r3 e4 e2 s0 r4 e1 r2 s r1 Scattering from many electrons Total scatter = (e1) + (e2) + ... = S exp (2pirj.S) Total scatter = (e1) + (e2) = 1 + exp (2pir.S)

13. Electron density r(x,y,z) Scattering from an atom Heisenberg – we can‘t actually position an electron No. of electrons in box: r(x,y,z)DxDyDz f(S) = Sr(r)D3r exp (2pir.S) f(S) = r(r)d3r exp (2pir.S) atomic scattering factor

14. Scattering from an atom Schrödinger – we can calculate where an electron is likely to be f(S) = r(r)d3r exp (2pir.S) atomic scattering factor The electron distribution is centrosymmetric: r(r) = r(-r)

15. 2pr.S 2p(-r).S Scattering from an atom f(S) = r(r)d3r exp (2pir.S) atomic scattering factor The electron distribution is centrosymmetric: r(r) = r(-r) f(S) = r(r){exp (2pir.S) + exp (2pi-r.S)} d3r = r(r)cos (2pr.S) d3r If the electron distribution is also spherically symmetric [r(r) = r(|r|)], then f(S) is independent of the direction of S, only on it‘s magnitude |S|=2sinq/l

16. Scattering from an atom f(S) = r(r)cos (2pr.S) d3r If the electron distribution is also spherically symmetric [r(r) = r(|r|)], then f(S) is independent of the direction of S, only on it‘s magnitude |S|=2sinq/l

17. Scattering from an atom f(S) = r(r)cos (2pr.S) d3r f(0) = r(r)cos (2pr.0) d3r = r(r)1 d3r = Z If the electron distribution is also spherically symmetric [r(r) = r(|r|)], then f(S) is independent of the direction of S, only on it‘s magnitude |S|=2sinq/l

18. r2 f2 r1 f3 r2 f1 r3 r1 F r3 Diffraction from a molecule f1(S) = {r1(r)cos (2pir.S) d3r } exp (2pir1.S) = f1(S) exp (2pir1.S) F(S) = f1(S) + f2(S) + f3(S) = Sfj(S) exp (2pirj.S)

20. Diffraction from a molecule F(S) = f1(S) + f2(S) + f3(S) = Sfj(S) exp (2pirj.S)

21. Diffraction from a molecule F(S) = f1(S) + f2(S) + f3(S) = Sfj(S) exp (2pirj.S)

22. Diffraction from a molecule F(S) = f1(S) + f2(S) + f3(S) = Sfj(S) exp (2pirj.S)

23. Diffraction from a molecule F(S) = f1(S) + f2(S) + f3(S) = Sfj(S) exp (2pirj.S)

24. Diffraction from two molecules

25. Crystal symmetries • The twofold axis can be used to generate the crystal • but now a=g=90o • The unit cell is divided into two • asymmetric units

26. c c b b b a a g g b a a Diffraction from a crystal The unit cell Fcell (S) = Sfj(S) exp (2pirj.S)

27. vc 2c ub c 2b b 0a 2a 3a 4a ta F0(S) = Sfj(S) exp (2pirj.S) Diffraction from a crystal

28. Diffraction from a crystal vc ta + ub + vc 2c ub c 2b b 0a 2a 3a 4a ta F0(S) = Sfj(S) exp (2pirj.S) Ftuv(S) = F0(S) exp {2pi [ta + ub + vc].S}

29. Diffraction from a crystal Ftuv(S) = F0(S) exp {2pi [ta + ub + vc].S} Ftuv(S) = F0(S) exp {2pi ta.S} exp {2pi ub.S} exp {2pi vc.S} Fcryst(S) = SSSFtuv(S) Fcryst(S) = F0(S) Sexp {2pi ta.S} Sexp {2pi ub.S} S exp {2pi vc.S} F0(S) = Sfj(S) exp (2pirj.S)

30. t=3 t=2 t=4 2pi ta.S t=1 2pi ta.S 2pi ta.S 2pi ta.S 2pi ta.S t=5 t=0 t=5 t=6 t=0 t=3 t=2 t=4 t=1 2pi ta.S t=6 Diffraction from a crystal Fcryst(S) = F0(S) Sexp {2pi ta.S} Sexp {2pi ub.S} S exp {2pi vc.S} Sexp {2pi ta.S} but if2pi a.S = 0 or 2p or 4p etc.

31. Fcryst(S) = nanbncF0(S) if and only if2pa.S = 2p b.S = 2p c.S = 0 or 2p or 4p etc. } h, k,l are integers, called the Miller indices of the scattered beam Diffraction from a crystal Fcryst(S) = F0(S) Sexp {2pi ta.S} Sexp {2pi ub.S} S exp {2pi vc.S} a.S = h b.S = k c.S = l These conditions are called the Laue conditions

32. Diffraction from a crystal