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AnnouncementsAssignment due on Wednesday.
I strongly suggest you come and talk to me.(but don’t leave it to the last minute!)
Pick: 1) an article, 2) a review article, and 3) some general textbook(s)
You will then give a 10 min. talk on (plus 2-3 min. for questions).This Wednesday, hand in your articles (1 and 2),
plus a ½ - 1 page write up.1) Why you are interested in it,
2) What you think major technical innovation is, 3) What is main biological point.
March 10th : Mid-term Exam!
March 27th & 28th (evenings) : Your presentations (?)
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
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
Energy Transfer goes like…
60 )/(1
1
RRE
6
0 )/(1
1
RRE
or ?
Take limit…
FRET is so useful becauseRo (2-8 nm) is often ideal
Bigger Ro (>8 nm) can use FIONA-type techniques
Terms in Ro
61
2421.0 nJqRo Din 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
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
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.)
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
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
Example of Energy TransferStryer & Haugland, PNAS, 1967
Donor Linker AcceptorSpectral Overlap
Energy Transfer
Stryer & Haugland, PNAS, 1967
Sua Myong*Michael Brunoϯ
Anna M. Pyleϯ
Taekjip Ha*
* University of IllinoisϮ Yale University
Unzipping mystery of helicases
Ha et al, Nature (2002)
FR
ET
time
Active form of helicase
Helicases separate (unzip) two DNA strands. We can measure very transient intermediates during unzipping using single molecule techniques.
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 16 18 200.0
0.2
0.4
0.6
0.8
1.0 D 5 point AA Smoothing of hel15tr210_D
-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
8 12 16 20 24
0
400
800
1200
1600
emis
sion
(au
)
8 12 16 20 240.0
0.2
0.4
0.6
0.8
1.0
FR
ET
time (sec)
ATP
5 10 15 20 25 30 350.0
0.2
0.4
0.6
0.8
1.0
time (sec)
5 10 15 20 25 30 35
0
400
800
1200
1600
2000
ATP
THREE steps in 9 bp unwinding [NS3]=25nM[ATP]=4mMTemp = 32oC
0.0 0.2 0.4 0.6 0.8 1.00.0
0.2
0.4
0.6
0.8
1.0
FRET enter
FR
ET
exi
t00.66671.3332.0002.6673.3334.0004.6675.3336.0006.6677.3338.0008.6679.33310.0010.6711.3312.0012.6713.3314.0014.6715.3316.0016.6717.3318.0018.6719.3320.00
Transition Density Plot (6 steps)
0.0 0.2 0.4 0.6 0.8 1.00.0
0.2
0.4
0.6
0.8
1.0
FRET enter
FR
ET
exi
t
00.76001.5202.2803.0403.8004.5605.3206.0806.8407.6008.3609.1209.88010.6411.4012.1612.9213.6814.4415.2015.9616.7217.4818.2419.00
0.0 0.2 0.4 0.6 0.8 1.00.0
0.2
0.4
0.6
0.8
1.0
FRET enter
FR
ET
exi
t
00.36330.72671.0901.4531.8172.1802.5432.9073.2703.6333.9974.3604.7235.0875.4505.8136.1776.5406.9037.2677.6307.9938.3578.7209.0839.4479.81010.1710.5410.90
0.0 0.2 0.4 0.6 0.8 1.00.0
0.2
0.4
0.6
0.8
1.0
FRET enter
FR
ET
exi
t
01.0802.1603.2404.3205.4006.4807.5608.6409.72010.8011.8812.9614.0415.1216.2017.2818.3619.4420.5221.6022.6823.7624.8425.9227.00
70 molecules 60 molecules
167 molecules50 molecules
Transition Density Plot (3 steps)
67 molecules 65 molecules
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
ktn et 1
0 2 4 6 8 10
0
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 (next lecture)
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.