Introductory slide
Free energy transduction
Force
Conjugate flux
Chemical
potential
Chemical flux
Force
Chemical
potential
Torque
Motor: eg. myosin
Displacement
Chemical flux
eg. F1 ATP-synthase
rotation
A molecular motor: acto-myosin
“food”
electrons
e.g. glucose
reproduction
photons
pmf
ATP
growth
transport
movement
R.D. Vale & R. Milligan
The mechanism of muscle contraction
organization of proteins within a sarcomere
insect flight
muscle
hierarchical
organization
Actin-myosin in vitro motility assay
10μm
J.E. Molloy
Actin-myosin crossbridge cycle
ATP binding completes the cycle
Once bound, Phosphate release
allows myosin head to relax.
This “powerstroke” pulls the thick
filament even against an external
force, doing Work = Force x distance
ATP binding makes
myosin release actin
ATP hydrolysis transfers free energy to
“strained” straight form of myosin head.
This form also binds actin.
Conformational states of the myosin motor
revealed by X-ray crystallography
4
structural model of
power stroke
5
Houdusse, A. & Sweeney, H.L. Curr. Opin. Struc. Biol. 11, 182 (2001)
P
r
Reaction coordinate r
Free energy landscape: phosphate release with power stroke
5
4
Mechanical coordinate x
x
Free energy landscape: phosphate release with power stroke
5
5
Free energy
Reaction coordinate r
4
4
Mechanical coordinate x
Mechanical coordinate x
Barrier-crossing rates



kYX pY
pX, pY are probabilities that
states A, B are occupied;
kXY , kYX are rate constants
Y
Free energy
Reaction coordinate r
probability
fluxes
kXY pX
G*
X
Y
Reaction coordinate r
Mechanical coordinate x
Rate constants often obey Arrhenius equation:
ΔG
X
 G* 

k XY  A exp  

k
T
B



 and

 G *  G
k YX  A exp  
k BT





G* is an activation barrier; reaction rate depends on probability that system has sufficient energy to cross it.
Consistent with detailed balance:
k XY
k YX
  G XY 
  G XY 
p
 and, in equilibrium, Y  exp  

 exp  


k BT 
pX
k B T 



k XY p X  k YX p Y

Free energy landscape: position-gated transition
Free energy
Reaction coordinate r
lowest barrier
a
b
c
a bc
Mechanical coordinate x
Reaction coordinate r
transition is most probable at b, where
the activation barrier is lowest
Rates and free energies in crossbridge cycle
1
4
2
3
5
6
1’
Key
AM actin-myosin
M free myosin
T ATP
DP ADP+Pi
D ADP
- empty
Crossbridge model for muscle acto-myosin
AMDP (4)
3 → 4 actin binding
AMD (5)
4 → 5 phosphate release
2→3 ATP hydrolysis
0
AM (6)
AMT (1)
MT (2)
MDP (3)
-10 kT
ΔGATP  -25kBT
5→6 ADP release
Free
energy
6→1 ATP binding
-20 kT
1→2´ unbinding from actin
MT (2’)
Rate
constant
position
power-stroke
-5nm
0
x0
0
binding
un-binding
position
x0+Δx
Each cycle:
• hydrolyzes one molecule of ATP,
releasing ~25 kT of free-energy,
• produces one ~5 nm power stroke
(about half of the available free
energy can be converted to work)
Reaction-Diffusion Equation
Let P(x)dx be the probability of finding a motor in the range x→x+dx.
Consider diffusion in the presence of an external force:
“Probability flux”:
J x, t    D
P x, t 
x

F x 

P  x , t  where F  x   
dU  x 
& D,  are diffusion, drag coefficients
dx
Continuity equation:
P  x, t 
t

J  x, t 
x
 P x, t 
2
 D
x
2


  F x 


P
x
,
t


x  

Fokker-Planck Equation
Including the possibility of transitions between states:
 Pi  x , t 
t
 Pi  x , t 
2
 D
x

2
 k
ji


  F x 


P
x
,
t
i


x  

( x ) P j ( x , t )  k ij ( x ) P ( x , t ) i
j
Reaction-Diffusion Equation
mechanical:
motor movement while in state i

chemical:
transitions into and out of
state i at position x
Minimal model for muscle acto-myosin
actin binding and
phosphate release
power stroke
unbinding
-5nm
o
0
o
x
If we assume that the muscle contracts at a constant
speed, so x = vt, we can use the minimal model to predict
the relationship between force and speed…
Special case: constant velocity, two states
Reaction-Diffusion Equation:
 Pi  x , t 
t
-5nm
 Pi  x , t 
2
 D
x

0
2
 k
ji


  F x 


P
x
,
t
i


x  

( x ) P j ( x , t )  k ij ( x ) P ( x , t ) i

j
o
o
2 states:
P ≡ Pbound = 1-Punbound
rates:
kon, xo < x < xo + Δx
koff, x > 0
constant velocity:
F/γ = v = constant
x = vt
&D=0
(motion determined by ensemble of motors)
steady state:
P/ t = 0
P x, t 
t
 k on  x 1  P  x , t   k off  x P  x , t   v
P x, t 
x
 0
Solve in 3 pieces:
1.
Binding zone around x = xo (= -5 nm)
(assume initial condition P(xo) =0;
binding zones are widely separated)
2.
power stroke zone xo < x < 0
3.
drag-stroke zone
x>0
If stiffness of crossbridge is κ, average force
exerted per crossbridge is:
2
2
v
 k on  x    x o
F 
 2

1  exp  

d 
v
2
k off

 
 
N.B. please use d, not D as in
problem set, for distance
between binding sites




Model vs data
Monte Carlo: chemical transitions
At time t, motor is in state i at position x.
State i has chemical transitions to states j ( j´ )
The probability of making a transition from
state i to state j during a time step Δt is
Pi  j  k ij  t
j,x
For sufficiently small Δt:
kij
i,x
kij
P( any
transition )

k
ij
t
 1
j
j´,x
´
Rate constants for these transitions ensure
detailed balance:
k ij  x 
U i x   U
 exp 
k ji  x 
k BT

j
A random number between 0 and 1 decides
whether any transition should occur in time
t and, if so, which:
x 
Pj


0
(Non-zero rate constants define positions
where the motor can switch states.)
P(any)
Random number in this
interval, transition to j
1
Random number
here, no transition
Monte Carlo: mechanical motion
At time t, motor is in state i at position x, moving with
instantaneous velocity v.
“Thermal force” due to collisions with
water molecules
2
m
d x
dt
2
 
dU
dx
 Ft  
i
dx
“Langevin Equation”
dt
Inertial forces are negligible (Re = 10-8 ~10-6):
dx
 
dt
1 dU i

 dx
1

Ft
The distance moved in time t is…
x  
1 dU
t
 dt
i
 dx
 
1 dU

dx
motion due to potential
(deterministic)
0
i

1

t
F
t
dt
The second term is the distance the particle would
have travelled by free diffusion in time t :
substitute a random distance drawn from a Gaussian
distribution with mean square 2Dt.
0
 t  Ν 0 , 2 D  t 
Brownian motion (stochastic)
normally distributed: mean 0, variance 2Dt