1.individual projects due today. 2.outline of rest of course a.april 9: fret b.april 11: diffusion...
TRANSCRIPT
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.