Today’s Announcements 1. Today: finish FRET 2. Diffusion (Why moving in a cell is like swimming in concrete.) 2.Your written Paper, due Nov. 21 st . Today’s take-home lessons: FRET Quick review of FRET – Fluorescence Resonance Energy Transfer. An example of DNA and a major killer, Hepatitis C. Mechanism—hidden sub-steps detected by FRET
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Today’s Announcements 1.Today: finish FRET 2.Diffusion (Why moving in a cell is like swimming in concrete.) 2.Your written Paper, due Nov. 21 st. Today’s.
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Today’s Announcements1. Today: finish FRET
2. Diffusion(Why moving in a cell is like swimming in concrete.)
2.Your written Paper, due Nov. 21st.
Today’s take-home lessons: FRET
Quick review of FRET – Fluorescence Resonance Energy Transfer.
An example of DNA and a major killer, Hepatitis C. Mechanism—hidden sub-steps detected by FRET
Can follow basic Reaction:e.g. Melting Temperature
Clegg at al, PNAS, 1993
As [NaCl] goes up/down, what happens to melting temperature?
Terms in Ro (homework details)
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).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.
60 )/(1
1
RRE
Compare to known standard.
Varies from 0 to 4; usually = 2/3.
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.
AD
R0AD
D A
Time
Time
E.T. by decreases in donor emission.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.
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
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
Orientation of transition moments of cyanine fluorophores terminally attached to double-stranded DNA.
Iqbal A et al. PNAS 2008;105:11176-11181
Cy3 Cy5
(Ensemble) FRET on DNA
Fluorescein to Rhodamine
Clegg at al, PNAS, 1993
A separate series of data withreversed strand labeling gives a similar
curve, and the independentfits agree within +2 Å for all the values
L, a, and d, and within ±15° for .
Notice A, D, separated by length L,
but not on-axis!
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!
Why is there a bump around 14 bp and 19 bp?
Sua Myong*Michael Brunoϯ
Anna M. Pyleϯ
Taekjip Ha*
* University of IllinoisϮ Yale University
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.
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.
Conclusion of FRET studies on NS3
Watching Helicase in Action!
Helicase
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
Myong, Ha, Science, 2007
(vary [ATP], °T, dsDNA = 9 bp)
Three steps in 9 bp unwinding
1 step per 3 base pairs (again)
Is each step due to one rate limiting stepi.e. one ATP hydrolysis?
-> Build dwell time histograms (just like with molecular motors, i.e. myosin, kinesin, F1Fo-ATPase
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.