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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)
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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
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.
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.
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)
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
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
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
or ? Energy Transfer goes like… Take limit…
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.
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.)
Donor Emission Donor Emission http://mekentosj.com/science/fret/
Acceptor Emission Acceptor Emission http://mekentosj.com/science/fret/
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/
With a measureable E.T. signal E.T. leads to decrease in Donor Emission & Increase in Acceptor Emission http://mekentosj.com/science/fret/
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.
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.
Donor Linker Acceptor Example of Energy TransferStryer & Haugland, PNAS, 1967 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;105:11176-11181
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
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
Stairway to Unwinding Sua Myong* Michael Brunoϯ Anna M. Pyleϯ 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 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.
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)
SIX steps in 18 bp unwinding [NS3]=25nM [ATP]=4mM Temp = 32oC ATP ATP
More traces with Six steps [NS3]=25nM [ATP]=4mM Temp = 32oC tr1 tr2 tr3 1 2 3 4 5 6 tr4 tr5 tr6
Is each step due to one rate limiting step i.e. one ATP hydrolysis? -> Build dwell time histograms
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)
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
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.
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.