1. Individual Projects due today.2. Outline of rest of course

a. April 9: FRETb. April 11: Diffusion

c. April 16: Diffusion IId. 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 4th (Friday) Papers due by Midnightj. 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; R0 (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 Ro (2-8 nm) is often ideal

Bigger Ro (>8 nm) can use FIONA-type techniques(where you use two-different colored-labels)

Fluorescence Resonance Energy Transfer (FRET)

EnergyTransfer

Donor Acceptor

Dipole-dipole Distant-dependentEnergy transfer

AD

R0AD

D A

R (Å)0 25 50 75 100

0.0

0.2

0.4

0.6

0.8

1.0

E

Ro 50 Å

60 )/(1

1

RRE

Spectroscopic Ruler for measuring nm-scale distances, binding

Time

Time

Look at relative amounts of green & red

Derivation of 1/R6

60 )/(1

1

RRE

knon-distance = knd = kf + kheat = 1/D = 1/ lifetime

EnergyTransfer

Donor Acceptor

kET

E.T. = kET/(kET + knd)

E.T. = 1/(1 + knd/kET)

How is kET dependent on R?Kn.d.

E.T. = 1/(1 + 1/kETD)

Energy Transfer. = function (kET, knd)

Classically: How is kET dependent on R?

Classically: E.T. goes like R-6

How does electric field go like?

Far-field:

Near-field: E ≈1/R3 (d << )

E≈ 1/R U ≈1/R2 (Traveling: photons)

Dipole emitting: Energy = U =

So light absorbed goes like paE x peE = papeE2, pepa/R6 ≈ E.T.

peE = pe/R3

Dipole absorption:Probability that absorbing molecule (dipole) absorbs the light paE

E.T. = 1/(1 + 1/kETD)

E.T. = 1/(1 + (R6/Ro6))

, Depends on Ro

Energy Transfer goes like…

60 )/(1

1

RRE

6

0 )/(1

1

RRE

or ?

Take limit…

60 )/(1

1

RRE

Terms in Ro

in Angstroms

• J is the normalized spectral overlap of the donor emission (fD) and acceptor absorption (A)

• 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).• is a geometric factor related to the relative orientation of the transition dipoles of the donor and acceptor and their relative orientation in space.

60 )/(1

1

RRE

Terms in Ro

61

2421.0 nJqRo D in Angstroms

where J is the normalized spectral overlap of the donor emission (fD) and acceptor absorption (A)

(Draw out a() and fd() and show how you calculate J.)

Donor Emission

http://mekentosj.com/science/fret/

Donor Emission

http://mekentosj.com/science/fret/

Acceptor Emission

Acceptor Emission

Spectral Overlap terms

http://mekentosj.com/science/fret/

Spectral Overlap between Donor & Acceptor Emission

For CFP and YFP, Ro, or Förster radius, is 49-52Å.

(J)

With a measureable E.T. signal

http://mekentosj.com/science/fret/

E.T. leads to decrease in Donor Emission & Increase in Acceptor Emission

How to measure Energy TransferDonor 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.

AD

R0AD

D A

Time

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

AR A

DDA

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< 2 >4)2 often = 2/3

Example of Energy TransferStryer & Haugland, PNAS, 1967

Donor Linker AcceptorSpectral 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.

Iqbal A et al. PNAS 2008;105:11176-11181Orientation Effect observed, verified!

Sua Myong*Michael Brunoϯ

Anna M. Pyleϯ

Taekjip Ha*

* University of IllinoisϮ 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 helicase1. 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!

0 5 10 15 20 25 300.0

0.2

0.4

0.6

0.8

1.0

FR

ET

time (sec)

0 5 10 15 20 25 30 35

0

500

1000

1500

2000 B C

Inte

nsi

ty (a

u)

ATP

5 10 15 20 25 30 35 400.0

0.2

0.4

0.6

0.8

1.0

FR

ET

time (sec)

5 10 15 20 25 30 35 40

0

500

1000

1500 B C

Inte

nsi

ty (a

u)

ATP

SIX steps in 18 bp unwinding [NS3]=25nM[ATP]=4mMTemp = 32oC

-2 0 2 4 6 8 10 12 14 160.0

0.2

0.4

0.6

0.8

1.0 D 5 point AA Smoothing of hel15tr244_D

-2 0 2 4 6 8 10 12 14 16 180.0

0.2

0.4

0.6

0.8

1.0 D 5 point AA Smoothing of hel15tr80_D

0 2 4 6 8 100.0

0.2

0.4

0.6

0.8

1.0 D 5 point AA Smoothing of hel15tr269_D

-2 0 2 4 6 8 10 12 14 16 18 200.0

0.2

0.4

0.6

0.8

1.0 D 5 point AA Smoothing of hel15tr289_D

-5 0 5 10 15 20 25 30 35 400.0

0.2

0.4

0.6

0.8

1.0 D 9 point AA Smoothing of hel15tr363_D

tr1 tr2 tr3

tr4 tr5 tr6

[NS3]=25nM[ATP]=4mMTemp = 32oC

12

34

5

6

More traces with Six steps

Is each step due to one rate limiting stepi.e. one ATP hydrolysis?

-> Build dwell time histograms

Non-exponential dwell time histograms

Smallest substep = Single basepair !

dwell time (sec)

k k k

n irreversible steps: Here n = 3. 3 hidden steps involved with each 3 bp step

ktn et 1

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.

0 2 4 6 8 100

10

20

30

40

50

n 3.06573k 1.54921

0 2 4 6 8 10 12 140

5

10

15

20

25

30

35

n 2.71815k 0.89507

0 2 4 6 8 10 12 140

10

20

30

40

50

60

70

n 2.976k 0.816

0 5 10 15 20 25 30 350

40

80

120

160

n 3.13406k 0.47806

0 2 4 6 8 100

10

20

30

40

50

n 3.00507k 1.01296

0 2 4 6 8 10 12 14 16 18 200

10

20

30

40

50

60

70

n 3.27347k 0.75546n

um

be

r o

f m

ole

cule

s

time (sec)

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.