Today’s Announcements 1.Today: finish FRET 2.Diffusion (Why moving in a cell is like swimming in concrete.) 2.Your written Paper, due Nov. 21 st. Today’s take-home lessons: FRET Quick review of FRET – Fluorescence Resonance Energy Transfer. An example of DNA and a major killer, Hepatitis C. Mechanism—hidden sub-steps detected by FRET

Cool video (remember GFP?) =sXM_WmwT2v4

Green Fluorescent Protein: Genetically-encoded dye Fluorescent protein from jelly fish

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

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

Fluorescence Resonance Energy Transfer (FRET) Energy Transfer Donor Acceptor Dipole-Dipole distant-dependent energy transfer R (Å) E R o  50 Å Spectroscopic Ruler for measuring nm-scale distances, binding Time Look at relative amounts of green & red

cWa7WA&list=PLEGwTYt9TZjkRBwbZxyw mg536QRtRbh4V FRET between two proteins, one labeled with the donor (CFP) and a one with the acceptor (YFP). A third protein gets in between the donor and acceptor.  10 nm– no FRET: donor emission, very little acceptor emission  Protein is displaced; < 10nm FRET A video of FRET Start at 2:40pm

Can follow basic Reaction: e.g. Melting Temperature Clegg at al, PNAS, 1993 As [NaCl] goes up/down, what happens to melting temperature?

Terms in R o (homework details) in Angstroms J is the normalized spectral overlap of the donor emission (f D ) and acceptor absorption (  A ) q D is the quantum efficiency (or quantum yield) for donor emission in the absence of acceptor (q D = number of photons emitted divided by number of photons absorbed). How do you measure this? n is the index of refraction (1.33 for water).   is a geometric factor related to the relative orientation of the transition dipoles of the donor and acceptor and their relative orientation in space. Compare to known standard. Varies from 0 to 4; usually = 2/3.

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

  Orientation Factor (homework details) where  DA is the angle between the donor and acceptor transition dipole moments,  D (  A ) is the angle between the donor (acceptor) transition dipole moment and the R vector joining the two dyes.  2 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. x y z D A R AA DD  DA This assumption assumes D and A probes exhibit a high degree of rotational motion  2 is usually not known and is assumed to have a value of 2/3 (Randomized distribution) The spatial relationship between the DONOR emission dipole moment and the ACCEPTOR absorption dipole moment (0 4)  2 often = 2/3

Orientation of transition moments of cyanine fluorophores terminally attached to double- stranded DNA. Iqbal A et al. PNAS 2008;105: Cy3  Cy5

(Ensemble) FRET on DNA Fluorescein to Rhodamine Clegg at al, PNAS, 1993 A separate series of data with reversed strand labeling gives a similar curve, and the independent fits agree within +2 Å for all the values L, a, and d, and within ±15° for . Notice A, D, separated by length L, but not on-axis!

Efficiency of energy transfer for Cy3, Cy5-labeled DNA duplexes as a function of duplex length. Iqbal A et al. PNAS 2008;105: Orientation Effect observed, verified! Why is there a bump around 14 bp and 19 bp?

Sua Myong* Michael Bruno ϯ Anna M. Pyle ϯ Taekjip Ha* * University of Illinois Ϯ Yale University

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 is composed 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.

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. Conclusion of FRET studies on NS3

Watching Helicase in Action! Helicase

ATP SIX steps in 18 bp unwinding [NS3]=25nM [ATP]=4mM Temp = 32 o C

tr1 tr2tr3 tr4 tr5tr6 [NS3]=25nM [ATP]=4mM Temp = 32 o C More traces with Six steps

Myong, Ha, Science, 2007 (vary [ATP], °T, dsDNA = 9 bp) Three steps in 9 bp unwinding 1 step per 3 base pairs (again)

Is each step due to one rate limiting step i.e. one ATP hydrolysis? -> Build dwell time histograms (just like with molecular motors, i.e. myosin, kinesin, F 1 F o -ATPase

Non-exponential dwell time histograms Smallest substep = Single basepair ! dwell time (sec) kkk n irreversible steps: Here n = 3. 3 hidden steps involved with each 3 bp step 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. 18 bp DNA9 bp DNA

See (from Myong Science, 2007: 7/26/ DC1/ s1.mov (Summarized on next page.)

A model for how NS3 moves Myong et al, Science,2007

Class evaluation 1.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.