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

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; R 0 (2-8 nm), very convenient.

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

Derivation of 1/R 6 k non-distance = k nd = k f + k heat = 1/  D = 1/ lifetime Energy Transfer Donor Acceptor k ET E.T. = k ET /(k ET + k nd ) E.T. = 1/(1 + k nd /k ET ) How is k ET dependent on R? K n.d. E.T. = 1/(1 + 1/k ET  D ) Energy Transfer. = function (k ET, k nd )

Classically: How is k ET dependent on R? Classically: E.T. goes like R -6 How does electric field go like? Far-field: Near-field: E ≈1/R 3 (d << ) E≈ 1/R U ≈1/R 2 (Traveling: photons) Dipole emitting: Energy = U = So light absorbed goes like p a E x p e E = p a p e E 2, p e p a /R 6 ≈ E.T. p e E = p e /R 3 Dipole absorption: Probability that absorbing molecule (dipole) absorbs the light paEpaE E.T. = 1/(1 + 1/k ET  D ) E.T. = 1/(1 + (R 6 /R o 6 )), Depends on R o

Energy Transfer goes like… or? Take limit…

Terms in R o 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). n is the index of refraction (1.33 for water; 1.29 for many organic molecules).   is a geometric factor related to the relative orientation of the transition dipoles of the donor and acceptor and their relative orientation in space.

Terms in R o in Angstroms where J is the normalized spectral overlap of the donor emission (f D ) and acceptor absorption (  A ) (Draw out  a ( ) and f d ( ) and show how you calculate J.)

Donor Emission Donor Emission

Acceptor Emission

Spectral Overlap terms Spectral Overlap between Donor & Acceptor Emission For CFP and YFP, R o, or Förster radius, is 49-52Å. (J)

With a measureable E.T. signal E.T. leads to decrease in Donor Emission & Increase in Acceptor Emission

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. E.T. by changes in donor. 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. Time

  Orientation Factor 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

Example of Energy Transfer Stryer & Haugland, PNAS, 1967 DonorLinkerAcceptor Spectral Overlap

Energy Transfer Stryer & Haugland, PNAS, 1967

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

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:

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!

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

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.

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.

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

Watching Helicase in Action!

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

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

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.

Independent measurements yield n value close to 3

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

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. A movie for how NS3 moves

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.