Today’s take-home lessons (i.e. what you should be able to answer at end of lecture) FRET – why it’s useful, R -6 dependence; R 0 (3-7 nm), very convenient. 1.Homework assigned #6 due next Monday, 4/12 in class. 2.Next Monday 4/12: Klaus Schulten birds and magnetotaxis. 3.Next Wednesday 4/14: VMD (computer analysis) Today’s Announcements

1. Two models of DNA: A ________________________model assumes that DNA chains are completely straight, unstretchable. No thermal fluctuations away from straight line are allowed. The polymer can only disorder at the joints between segments. A ______________________model assumes that DNA chains have a correlation length called the persistence length. 3. Resolution of a microscope is (classically) limited by Heisenberg uncertainty principle and is equal to __________________. Quiz 2. DNA has both twist and writhe. What is the twist and writhe number in the follow graph? Left:_______ ________, Center: ____ _________, Right: _____ _________ Freely jointed chain (FJC) Worm-like chain (WLC) T w = 0, W r = 0 T w = 2, W r = 0 T w = 0, W r = 2 λ/2 N.A 5. There are three families of motor proteins called: _______ ______ and _______. 6. List two advantages of two-photon microscopy: ______________________ and ______________________________. 4. If you excite a fluorescent molecule with light of a certain wavelength, the molecule will emit light at a ______________ wavelength. longer Kinesin, Myosin, Dynein Low light scattering Inherent z-resolution (or confocality)

FRET FRET: measuring conformational changes of single biomolecules The distance 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

Reminder: Fluorescence properties. Absorption Wavelength Fluorescence Stokes Shift ( nm) Excitation from lowest ground state to excited states (fsec). Relaxation to lowest excitation singlet state (psec). Transition to multiple ground states (nsec).

Lifetimes and Quantum Yields Lifetime (  d ): exp(-  /  d ): 1-100nsec  d = 1/k Quantum yield(  ) : k rad / (k rad +k non-rad ) [how much of excited-state energy goes into light vs. heat]  d, k

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 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 nd E.T. = 1/(1 + 1/k ET  D ) E.T. = function (k ET, k nd )

Classically: How is k ET dependent on R? Classically: E.T. goes like R -6, Depends on R o How does electric field go like? Far-field: Near-field: 1/R 3 (d << ) 1/R 2 Dipole emitting: Energy = U = So light absorbed goes like: p e E = p e /R 3 (E = electric field) 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 )) p a E x p e E= p a p e E 2 = p e p a /R 6 E.T. = 1/(1 + k nd /k ET )

Quantum Mechanically Determine Hamiltonian (Energy) of Interaction. By Fermi’s Golden Rule, rate goes like H 2. Dipoles interacting: Dipolar Interaction depends on R 3 : E.T. = H 2 = R 6

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

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!

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.