Transcript Unfolding Proteins by Atomic Force Microscopy

How do proteins withstand force?
Examining the effect of force on a protein
folding landscape by combining atomic
force microscopy, protein engineering and
simulation
Jane Clarke
May 2004
Cambridge University Dept of Chemistry
MRC Centre for Protein Engineering
Robert Best
Susan Fowler
Annette Steward
Emanuele Paci (Zurich)
Martin Karplus
(Strasbourg/ Harvard)
Kathryn Scott
José Toca Herrera
Phil Williams
(U. Nottingham)
Wellcome Trust & MRC
The Protein Folding Problem
Proteins fold co-operatively into a unique 3-dimensional
structure that is the most stable conformation
Gene
sequence A
Unfolded
protein
Why? How?
How?
Folded protein
sequence B
Why not?
Misfolded protein
function
But proteins don’t just fold one time and
that’s it.
Mechanical unfolding of proteins may be
important in translocation and degradation
and in mechanically active proteins
For some proteins resisting unfolding may
be important
Protein folding pathways - and landscapes
‡2
‡1
D
I
N
We can explore the unfolding landscape
by folding and unfolding experiments
Karplus, Dobson
How does force modify the unfolding landscape?
TS
∆∆GTS-N
∆GTS-N
D
When you add FORCE (F):
Relative to the native state, N,
the barrier to unfolding (∆GTSN) is lowered by:
Fxu
and the free energy of
unfolding (∆GD-N) is lowered
by:
F(xu + xf)
xf
∆∆GD-N
N
xu
The protein is less stable and
unfolds more rapidly - the
unfolding rate (ku) is a
measure of the height of the
barrier between N and TS
What does AFM offer?
• Can investigate the way the energy landscape is
perturbed by force
• Known reaction co-ordinate (N-C length) making
it easier to do direct comparison with simulation
• Single molecule experiments offer the possibility
to observe rare events
The AFM Experiment
Asylum Research MFP
2
1
3
4
∆L
F
1. Non-specific adhesion 2. Unfolding of one domain
3. Unfolded protein stretching 4. Protein detaches
Analysis of AFM data
•
Unfolding proteins by AFM is a
kinetic measurement: mean
unfolding force depends on
pulling speed.
Unfolding rate constant
(extrapolated to 0 force) (ku0)
and unfolding distance (xu)
can be estimated by Monte
Carlo simulation or analytical
techniques.
Gaub, Fernandez, Evans
Slope gives
xu
Force (N)
•
Intercept gives
ku0
Interpreting the traces:
Which traces to choose?
The basic reminders about
kinetics and thermodynamics
Force measurements of protein unfolding are kinetic
measurements not thermodynamic measurements
So…
Beware of the word “stability” - what does it mean in the context of force
measurements?
“In a protein made up of a number of domains the least stable domains will
unfold first and the most stable domains will unfold last”
Titin I27 is significantly more stable than I28 (7.6 vs. 3.2 kcal/mol) but
I28 unfolds at significantly higher forces
The basic reminders about
kinetics and thermodynamics (2)
ku 
G  RTln 
kf 

But…
It is not possible to determine the stability of a protein using AFM
folding and unfolding data
It is possible to measure refolding rates using AFM
Carrion Vasquez
BUT - the unfolding and refolding pathway are not necessarily
(are unlikely to be?) the reverse of each other.
Titin - an elastic protein
1 µm
Titin - effect of force
Very low force
“working” forces
Very high force
Protein domains straighten out
Unstructured region unfolds
One or two domains unfold to
prevent the protein breaking
First experiments - using whole proteins with
heterogeneous domains
Gaub,
Bustamante,
Symmonds,
Fernandez
How do titin domains resist force?
Can we characterise the titin I27 forced
unfolding pathway in detail?
Using molecular biology
To make multiple repeats of one titin domain
A tag to allow
easy purification
A tag to allow
attachment to
AFM
In simulations the first step is to form an
intermediate by detachment of the A-strand
Fernandez, Schulten
Humps?
V4A
When we pull a protein with a
destabilising mutation in the
A-strand (V4A) it does not affect
the unfolding forces at all
Fowler et al. JMB 2002 322, 841
15N
This intermediate is stable and has
essentially the same structure as the
native state
1H
I is populated above 100pN
ku is the unfolding rate of I to ‡
and xu is the distance between I & ‡
‡F
‡F
‡F
ku≈10-4
I
N
∆G ≈ 3
kcal mol-1
N
N
I
I
3Å
0 pN
≈100 pN
>100 pN
Increasing force
Titin forced unfolding pathway
Native state
N
Transition
State
?
Intermediate
I
N
G
G
G
G
C
A
A’
G
N
C
‡
N
A’
ku
Free energy profile under force
N
I
~3 Å
Unfolded
D
C
Using protein
engineering to
analyse forced
unfolding
pathways:
C
V86
G
V13
A´
L60
B
L58
A mechanical Fvalue analysis
G I23
D
F73
C
E
V4
A
N
Best et al. PNAS 2002 99, 12143
L41
F
Theory
Protein engineering analysis - F = 1
•The unfolding force reflects the
difference in free energy between I and ‡
‡
U
N
I
• If the mutation removes a side
chain that is fully folded in the
transition state it will not affect the
unfolding force at all.
Protein engineering analysis - F = 0
•The unfolding force reflects the
difference in free energy between I and ‡
‡
U
N
I
•If the mutation removes a side
chain that is fully unfolded in the
transition state it will reduce the
unfolding force by a significant
amount - that we can predict
NB only works if the barrier we are examining
is the same in WT & mutant
(xu must remain the same)
The A’ strand is partly detached in ‡
250
WTF
Force (pN)
200
V13A
150
F
100
50
0
1
10
100
1000
Pulling Speed nm -s
10
4
Most f-values are ≈ 1
Most of the protein is intact in the transition state
250
250
F73L
L41A
200
150
Force (pN)
Force (pN)
200
100
50
0
10
150
100
50
100
1000
Pulling Speed nm -s
10
4
0
1
10
100
1000
Pulling Speed nm -s
10
4
BUT
xu changes - can’t do a F-value analysis
but this mutation clearly lowers the force, ie F must be <1
250
V86A
Force (pN)
200
150
100
50
0
10
100
1000
Pulling speed (nm/s)
10
4
Results:
The only part of the protein completely detached
in ‡ is A strand
A’ and G are partly disrupted. How?
Can molecular dynamics simulations help?
MD simulations - the protein unfolds via an intermediate
QuickTime™ and a YUV420 codec decompressor are needed to see this picture.
0.8
0.6
0.4
0.2
0
1
0.8
0.6
0.4
0.2
0
1
0.8
0.6
0.4
0.2
0
1
0.8
0.6
0.4
0.2
0
145
0.8
0.6
0.4
0.2
0
1
0.8
0.6
0.4
4
Analysing structures from the simulations experimental F-values are reproduced
Proportion
of native
contacts
13
intermediate
transition state
constant pulling force=350 pN
4
V482
V13
13
86
50
55
rNC (Å)
60
65
82
86
Mechanical unfolding pathway:
N
I
‡
D
Native State
N
G- strand
C
Step 1:
A-strand
pulled off to
form I
A
A’
N
Transition State
Step 2:
G-strand pulled
off, breaking
main chain &
sidechain
contacts with A’
& sidechain
contacts with AB loop. A’ loses
contacts with G
C & E-F loop
Titin forced unfolding pathway
Native state
N
Transition
State
‡
Intermediate
I
N
G
G
G
G
C
A
A’
G
N
C
A’
N
‡
N
I
Free energy profile under force
Unfolded
D
C
Force induced unfolding pathway
‡F
‡F
‡F
ku≈10-4
I
N
∆G ≈ 3
kcal mol-1
N
N
I
I
3Å
0 pN
≈100 pN
>100 pN
Increasing force
Does force change the energy landscape?
Do these transition states have
the same structure?
‡F
‡D
‡F
‡F
ku≈10-4
ku≈10-4
I
N
N
N
I
N
Denaturant
(0 pN)
I
3Å
0 pN
≈100 pN
>100 pN
Increasing force
Force
The A strand
unfolds very
early
Chemical denaturant
The A strand
remains partly
folded in TS
The core is
partly unfolded
in TS
The core plays
no role in
withstanding
force and
remains fully
folded in TS
The A’ G region
remains partly
folded in TS
The A’ G region
is fully unfolded
in TS
Force changes the energy landscape
Transition states have
different structures
‡F
‡D
‡F
‡F
ku≈10-4
ku≈10-4
?
I
N
N
N
N
?
when?
Denaturant
(0 pN)
I
0 pN
I
3Å
≈100 pN
>100 pN
Increasing force
Experimental limitations
300
250
200
150
Cannot measure at
rates below ~100
nm/s
100
50
0
-6
10
10
-4
10
-2
10
0
10
2
Pulling speed (nm/s)
10
4
10
6
All these mutants destabilise the protein significantly
250
V86A
•These mutants have a
significantly longer xu (~6Å) than
wt (~3Å)
Force (pN)
200
150
• BUT These very destabilising
mutants have apparently a lower
ku than wild type
100
50
0
10
100
1000
Pulling speed (nm/s)
10
4
Simplest model: these mutants are unfolding directly from N
So xu = xN ->‡ & ku = kN-‡
Williams et al: Nature 2003
In the mutant V86A
ku is rate constant for
unfolding from N to ‡F
and xu is the distance between
N & ‡F
In wildtype
ku is rate constant for unfolding
from I to ‡F
and xu is the distance between
I & ‡F
‡F
‡F
‡F
This will happen if
the mutation allows
the protein to unfold
before I is populated
(by destabilising I
&/or lowering the
unfolding barrier ‡F)
I
N
N
N
I
6Å
0 pN
I
3Å
≈100 pN
>100 pN
Increasing force
Force changes the energy landscape
Transition states have
different structures
‡F
‡D
‡F
‡F
ku≈10-4
ku≈10-4
ku ≈ 10-7
I
N
N
N
N
6Å
When?
Denaturant
(0 pN)
I
I
0 pN
3Å
≈100 pN
>100 pN
Increasing force
At v. low forces the “physiological” barrier may be the important one
How do titin domains resist force?
Why are some proteins more
mechanically stable than others?
Titin I27
(muscle)
Tenascin fnIII
(intracellular matrix)
Spectrin
(cytoskeleton)
Barnase
(enzymes)
T4 Lysozyme