Today’s Announcements 1.Next Tuesday: Diffusion (Why moving in a cell is like swimming in concrete.) 2. Homework assigned today Last graded Homework:

Today’s take-home lessons: FRET (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. You can measure ensemble FRET and single-molecule FRET An example of DNA and a major killer, Hepatitis C.

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

Example of FRET What is the end-to-end distance on a dsDNA? D= Fluorescein A = Tetramethylrhodamine

Donor quenching Acceptor “sensitized” emission The more Energy Transfer (donor and acceptor closer together), more donor decreases, more the acceptor increases. Example of FRET (on DNA)

Derivation of 1/R 6 Energy Transfer Donor Acceptor k ET E.T. = k ET /(k ET + k nd ) E.T. = 1/(1 + k nd /k ET ) Want k nd /k ET = (R/R o ) 6 =(R o /R) -6 k ET ~ R -6 and have some R- independent terms. k n.d. E.T. = 1/(1 + 1/k ET  D ) Energy Transfer = function (k ET, k nd ) k non-distance = k nd = k f + k heat = 1/  D = 1/ lifetime Can get from classical E.M. Do!

Classically: How is k ET dependent on R? So (classically) E.T. goes like R -6 and p e p a ~ R o 6 (a constant with right units) How does electric field go like? E ≈1/R 3 U ≈ E 2 ; E ≈ 1/R Dipole emitting: Energy = U = So light emission goes like p a E x p e E = p a p e E 2, p e p a /R 6 ≈ E.T. peEpeE Dipole absorption: Probability that absorbing molecule (dipole) absorbs the light paEpaE E.T. =1/(1 +k nd /k ET  = 1/(1 + (R 6 /R o 6 )) = 1/(1 + 1/k ET  D ) How does the energy go with distance? Think of a source emitting and look at R 1, then 2R 1, 3R 1. U ≈1/R 2 (Traveling: photons) In the Near-field (d << ) (must remember your E&M) This is in the “Far-field”: (d >> ) = p e /R 3

Energy Transfer goes like… or? Take limit…

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.

Terms in R o (homework details) in Angstroms where J is the normalized spectral overlap of the donor emission (f D ) and acceptor absorption (  A ). Does donor emit where acceptor absorbs? Spectral Overlap between Donor (CFP) & Acceptor (YFP) Emission R o ≈ 49-52Å.

Spectral Overlap leads to Energy Transfer: This leads to decrease in Donor Emission & Increase in Acceptor Emission

  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

Efficiency of energy transfer for Cy3, Cy5-labeled DNA duplexes as a function of duplex length. Iqbal A et al. PNAS 2008;105:  2, i.e. Orientation Effect observed, verified Two dyes, one the donor and one the acceptor are labeled on the ends of DNA. The dyes are fairly rigidly attached such that each one has a definite direction to them. They will therefore have an orientation effect to the energy transfer. That is, the amount of energy transfer will depend not only on the distance (R) (or the length of the DNA) between the donor and acceptor, but on their relative orientation, as well (as discussed on the previous slide). That is,  2 for each dye pair varies. If, on the other hand, the dyes were free to move, then  2 = 2/3 (see graph). Consequently, the amount of energy transfer at each position is more or less than if  2 =2/3.

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

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.