Individual Projects due today. Outline of rest of course April 9: FRET April 11: Diffusion

1. Today’s Announcements • Individual Projects due today. • Outline of rest of course • April 9: FRET • April 11: Diffusion • April 16: Diffusion II • April 18: Ion Channels, Nerves • April 23: Ion Channels II • April 26: Vision • April 30: Student Presentation (15 min. max; 5 min questions) • May 2: Student Presentation (15 min. max; 5 min questions) • May 4th (Friday) Papers due by Midnight • May 8, 8-11am: Final Exam

2. Today’s take-home lessons(i.e. what you should be able to answer at end of lecture) FRET – Fluorescence Resonance Energy Transfer (Invented in 1967) why its useful, R-6 dependence; R0 (2-8 nm), very convenient.

3. FRET: measuring conformational changes of (single) biomolecules FRET FRET useful for 20-80Å Distance dependent interactions between green and red light bulbs can be used to deduce the shape of the scissors during the function.

4. FRET is so useful because Ro (2-8 nm) is often ideal Bigger Ro (>8 nm) can use FIONA-type techniques (where you use two-different colored-labels)

5. E Energy Ro  50 Å Transfer Acceptor Donor R (Å) Dipole-dipole Distant-dependent Energy transfer Time Time Fluorescence Resonance Energy Transfer (FRET) Spectroscopic Ruler for measuring nm-scale distances, binding Look at relative amounts ofgreen& red

6. Energy Transfer Acceptor Donor Derivation of 1/R6 Energy Transfer. = function (kET, knd) E.T. = kET/(kET + knd) E.T. = 1/(1+ knd/kET) E.T. = 1/(1+ 1/kETtD) Kn.d. How is kET dependent on R? kET knon-distance = knd = kf + kheat= 1/tD= 1/ lifetime

7. Classically: How is kET dependent on R? How does electric field go like? Far-field: Near-field: E≈ 1/R U ≈1/R2(Traveling: photons) E ≈1/R3 (d << l) peE = pe/R3 Dipole emitting: Energy = U = Dipole absorption: paE Probability that absorbing molecule (dipole) absorbs the light So light absorbed goes like paE x peE = papeE2, pepa/R6 ≈ E.T. E.T. = 1/(1+ 1/kETtD) E.T. = 1/(1+ (R6/Ro6)) Classically: E.T. goes like R-6 , Depends on Ro

8. or ? Energy Transfer goes like… Take limit…

9. Terms in Ro in Angstroms • J is the normalized spectral overlap of the donor emission (fD) and acceptor absorption (eA) • qD is the quantum efficiency (or quantum yield) for donor emission in the absence of acceptor (qD = number of photons emitted divided by number of photons absorbed). • n is the index of refraction (1.33 for water; 1.29 for many organic molecules). • k2 is a geometric factor related to the relative orientation of the transition dipoles of the donor and acceptor and their relative orientation in space.

10. Terms in Ro in Angstroms where J is the normalized spectral overlap of the donor emission (fD) and acceptor absorption (eA) (Draw out ea(l)and fd(l) and show how you calculate J.)

11. Donor Emission Donor Emission http://mekentosj.com/science/fret/

12. Acceptor Emission Acceptor Emission http://mekentosj.com/science/fret/

13. Spectral Overlap terms Spectral Overlap between Donor & Acceptor Emission (J) For CFP and YFP, Ro, or Förster radius, is 49-52Å. http://mekentosj.com/science/fret/

14. With a measureable E.T. signal E.T. leads to decrease in Donor Emission & Increase in Acceptor Emission http://mekentosj.com/science/fret/

15. E.T. by changes in donor. Need to compare two samples, d-only, and D-A. E.T. by increase in acceptor fluorescence and compare it to residual donor emission. Need to compare one sample at two l and also measure their quantum yields. Where are the donor’s intensity, and excited state lifetime in the presence of acceptor, and ________ are the same but without the acceptor. Time Time How to measure Energy TransferDonor intensity decrease, donor lifetime decrease, acceptor increase.

16. y qD D qDA qA R • k2 is usually not known and is assumed to have a value of 2/3 (Randomized distribution) A z x • This assumption assumes D and A probes exhibit a high degree of rotational motion k2 : Orientation Factor The spatial relationship between the DONOR emission dipole moment and the ACCEPTOR absorption dipole moment (0< k2 >4) k2 often= 2/3 where qDA is the angle between the donor and acceptor transition dipole moments, qD (qA) is the angle between the donor (acceptor) transition dipole moment and the R vector joining the two dyes. k2 ranges from 0 if all angles are 90°, to 4 if all angles are 0°, and equals 2/3 if the donor and acceptor rapidly and completely rotate during the donor excited state lifetime.

17. Donor Linker Acceptor Example of Energy TransferStryer & Haugland, PNAS, 1967 Spectral Overlap

18. Energy Transfer Stryer & Haugland, PNAS, 1967

19. Orientation of transition moments of cyanine fluorophores terminally attached to double-stranded DNA. Iqbal A et al. PNAS 2008;105:11176-11181

20. Simulation of the dependence of calculated efficiency of energy transfer between Cy3 and Cy5 terminally attached to duplex DNA as a function of the length of the helix. Iqbal A et al. PNAS 2008;105:11176-11181

21. Efficiency of energy transfer for Cy3, Cy5-labeled DNA duplexes as a function of duplex length. Orientation Effect observed, verified! Iqbal A et al. PNAS 2008;105:11176-11181

22. Stairway to Unwinding Sua Myong* Michael Brunoϯ Anna M. Pyleϯ Taekjip Ha* * University of Illinois Ϯ Yale University

23. NS3 unwinds DNA in discrete steps of about three base pairs (bp). Dwell time analysis indicated that about three hidden steps are required before a 3-bp step is taken.

24. About HCV NS3 helicase • 1. Hepatitis C Virus (HCV) is a deadly virus affecting 170 million people in the world, but no cure or vaccine. • 2. Non-Structural protein 3 (NS3) is essential for viral replication. • 3.NS3 iscomposed of serine protease and a helicase domain • 4. NS3 unwinds both RNA and DNA duplexes with 3’ overhang • 5. In vivo, NS3 may assist polymerase by resolving RNA secondary structures or displacing other proteins.

25. Steps (~11 bp) and substeps (2-5 bp) in NS3 catalyzed RNA unwinding under assisting force • Dumont, Cheng, Serebrov, Beran,Tinoco Jr., Pyle, Bustamante. • Nature (2006)

26. Watching Helicase in Action!

27. SIX steps in 18 bp unwinding [NS3]=25nM [ATP]=4mM Temp = 32oC ATP ATP

28. More traces with Six steps [NS3]=25nM [ATP]=4mM Temp = 32oC tr1 tr2 tr3 1 2 3 4 5 6 tr4 tr5 tr6

29. Is each step due to one rate limiting step i.e. one ATP hydrolysis? -> Build dwell time histograms

30. k k k n irreversible steps: Here n = 3. 3 hidden steps involved with each 3 bp step Non-exponential dwell time histograms If the three base-pair steps were the smallest steps i.e. there were no hidden substeps within, the dwell time distribution will follow an exponential decay. Smallest substep = Single basepair ! dwell time (sec)

31. Independent measurements yield n value close to 3

32. A model for how NS3 moves (next lecture) Why are threonine’s involved? What’s the evidence for this? Read original article! Myong et al, Science,2007

33. A movie for how NS3 moves We made a little movie out of our results and a bit of imagination. The two domains over the DNA move in inchworm manner, one base at a time per ATP while the domain below the DNA stays anchored to the DNA through its interaction, possibly the tryptophan residue. Eventually, enough tension builds and DNA is unzipped in a three base pairs burst.

34. Class evaluation • What was the most interesting thing you learned in class today? • 2. What are you confused about? • 3. Related to today’s subject, what would you like to know more about? • 4. Any helpful comments. Answer, and turn in at the end of class.