Lecture 10 - Rutgers School of Engineering

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Transcript Lecture 10 - Rutgers School of Engineering

Lecture 10
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Background for cell propulsion
Fluid dynamics
Enzyme kinetics
How do animals swim?:
1. pushing fluid backward by limb action;
2. pushing fluid forward by resistance of
body.
• I.e fish starting from release will accelerate
until the backward & forward momentum (of
the fluid) balance. Viscosity is only
significant at the boundary layer.
Cell Propulsion
• Small scale phenomenon: slow velocities
driven by surface forces: pressure and
viscous stress. Fluid resistance is
significant, and balances propulsive force.
• Motion of a body depends on the ratio of
viscous and inertial effects: Reynold’s
number: Small for cells, large for almost all
animals. Cellular world is ruled by friction.
• Reynold’s number quantifies the relative
magnitudes of frictional and inertial forces
  vR / 
 UL / 
 L  / 
2
Cellular Motors
• Molecular motors must move (swim) in
fluids, where most of the work is dissipated
• What forces must they overcome?
• Where do the motors get their fuel?
• How do they exhaust spent fuel?
• What is the efficiency?
Creature
R
Bacteria
10-4
Spermatozoa
10-2
Flying Insects
Birds
104
105
Oscillatory muscles
Stretch activation
Synchronous
Asynchronous
Stretch- activated currents
Sliding filamentds
Myosin
• 5.3 pN for each
myosin molecule
• 100 molecules per
filament.
• Each filament has
c.s.a. of 1.8 X 10 –15
m2 in the relaxed
muscle.
Strain in solids and fluids
A
f
z
 ( )G
A
d
d
f  v

A
d
f
Sample fluid properties
Fluid
Air
Water
Olive Oil
Glycerine
Corn Syrup
m (kg m ) Pa-S fcrit (N)
-3
1
1000
900
1300
1000
When f > fcrit- inertial forces dominate
-5
2 X 10
0.0009
0.08
1
5
-10
4 X 10
-10
8 X 10
-6
7 X 10
0.0008
0.03
Swimming: is it worth it?
• Cilium with velocity, v, length, d, time
scale:
ts  d / v
• Diffusion time scale :
tD  d / D
• Swimming time, ts should be < tD
D
v
d
2
Viscous flow
• Newtonian fluids are isotropic
• What is a viscous fluid?
Shear
• When f< fcrit
f  vo A / d
f crit   / 
2
A
vo
d
Planar geometry
f
• I.e., 1 mm cilium, D =
10-5 cm2/sec,
• so
v> 103 mm
/sec:
• stirring and swimming
is not energetically
favorable for nutrition.
Comparative motors
ATP SYNTHASE — A MARVELLOUS ROTARY ENGINE OF THE CELL
< previous next >
Rotary Cellular Motors
• The rotary mechanism of ATP synthase , Stock D, Gibbons C,
Arechaga I, Leslie AGW, Walker JE
CURRENT OPINION IN STRUCTURAL BIOLOGY ,10 (6): 672679 DEC 2000
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• 2. ATP synthase - A marvellous rotary engine of the cell, Yoshida
M, Muneyuki E, Hisabori T
NATURE REVIEWS MOLECULAR CELL BIOLOGY 2 (9):
669-677 SEP 2001
•
• 3. The gamma subunit in chloroplast F-1-ATPase can rotate in a
unidirectional and counter-clockwise manner Hisabori T, Kondoh
A, Yoshida M FEBS LETTERS 463 (1-2): 35-38 DEC 10 1999
•
• 4. Constructing nanomechanical devices powered by biomolecular
motors.C. Montemagno, G Bachand, Nanotechnology 10: 225-2312,
1999.
ATP SYNTHASE — A MARVELLOUS ROTARY ENGINE OF THE CELL
< previous next >
F1 ATPase: A rotary motor
• Can either make or break ATP, hence is
reversible
• Torque of 40 pN-nM; work in 1/3 rev. is 80
pn-nM (40 * 2p/3) equivalent to free energy
from ATP hydrolysis
• Can see rotation by attaching an actin
filament
For rotary motion:
2
d
I
 M
d t2
 w L
2
M 
4
1  2
I
mL
3
Nature Reviews Molecular Cell Biology 2; 669-677 (2001)
ATP SYNTHASE — A MARVELLOUS ROTARY ENGINE OF THE CELL
< previous next >
Current is coulombs per second. How many charges in a coulomb?
For this y ou need Faraday 's constant 96,500 Coulombs per mole of
charged m olecules, in this case potassium ions.
0.24  12  18 moles
Q K Kflux
10 2.510
96 500
sec
I f work , W , is done on the partic le during dif f us ion, t hen the t ime is
increased as:
W
t w t d e
kT
So s ay W = 10 KT,
then t w = 20 ms
So how f ast c an t he mot or go? Ass uming a bac k -and-f ort h motion
it would t ake at leas t 40 ms , s o the max f requenc y = 250 H z or
10 nM X 250 per sec ond = 2. 5 mic rons per second. (linear m ot ion).
Elasticity
Nano versus macro elasticity
Behaviour relative to kT: Stretch a rubber band and a
string of paper clips.
Significant for The nanometer-scale monomers of a
macromolecule, but not for a string of paper clips.
The retracting force exerted by a stretched rubber
band is entropic. It increases disorder.
Do most polymers have persistence lengths longer
than their total (contour) length?
• When L>> xthe chain has many bends and
is always crumpled in solution – the FJC
model applies, with each link approximated
as 2 xand perfectly flexible joints.
• To count all possible curved states in a
smooth-bending rod in solution- it’s a
WLC- supercoiling is possible.
• Promoters have different abilities to uncoil
• Twisting DNA
torsional buckling
instability
• Unwinding and causes local denaturation
• Many motors are needed: RNA plymerase,
DNA polymerase: 100 nucleotides/sec.
• Forces (pN) can stop transcription
Mechano - regulation
• Growth, proliferation, protein synthesis,
gene expression, homeostasis.
• Transduction process- how?
• Single cells do not provide enough material.
• MTC can perturb ~ 30,000 cells and is
limited.
• MTS is more versatile- more cells, longer
periods, varied waveforms..
Markov Chains
• A dynamic model describing random
movement over time of some activity
• Future state can be predicted based on
current probability and the transition matrix
Transition Probabilities
Tomorrow’s
Game Outcome
Today’s Game Outcome
Win
Lose
Win
3/4
1/2
Lose
1/4
1/2
Sum
1
1
Need a P for
Today’s game
Grades Transition Matrix
This Semester
Grade
Tendencies
Good
Bad
Good
3/4
1/2
Bad
1/4
1/2
Sum
1
1
To predict future:
Start with now:
What are the grade
probabilities for this
semester?
Markov Chain
1/4
Win
Lose
3/4
1/2
Pi 1  APi
1/2
 a11 a12   3 / 4 1 / 2 
  

A  
 a21 a22   1 / 4 1 / 2 
3 / 4

Pi  
1/ 4 
Pwin,i 1  3 / 4  3 / 4  1 / 2  1 / 4  11/ 16
Plose ,i 1  1 / 4  3 / 4  1 / 2  3 / 4  5 / 16
Intial Probability
Set independently
Computing Markov Chains
% A is the transition probability
A= [.75 .5
.25 .5]
% P is starting Probability
P=[.1
.9]
for i = 1:20
P(:,i+1)=A*P(:,i)
end