Transcript No Slide Title

Tuesday, March 01, 2005
• Geometric and Mechanical Properties
• Mechanical Statics
• Review-
Thick walled sphere
• Equilibrium
P
• Pressure inside
2
2
ri
pi ri
 pi 2

• Average stress in wall
2
h
(
r

r
)
(ro  ri )
o
i
 (ro 2  ri 2 ) 



Pin
 ri pi
2
in
2
Pout
po ro

h(ro  ri )
Pin  Pout

r pi
2h

r po
2h
• Pressure from outside
• Pressurized both sides
Charged polymers:
Electromechanical Chemistry
pKa =9.9
Aqueous charge
I.e. Alanine charge
+H N-CH-COO3
pKa =2.7
CH3
pH
pKa
- = fractional
 

log A

( HA )
charge =
1
- =
1
1
alog ( pH  pKa)
A 
A

 AH
Shape : Oblate sphere
Meridions
Curvature
1
1
C

R1 R2
Latitudes
X Y Z
Losing volume, not gaining
area;
Slow cell squishing
b 0
FRAME
5
a 3
FRAME
10
Membrane Tension
R
3. Patch in x-z plane
dx
P
1. Hemisphere
Tm dy
Tm dy
P
Tm
Tm
R

T
Tm
Tm
dy
dx
P
Tdy
d
2. Patch
T
Tdy d 
Tdy
4. Vertical Resultant
Tension on membrane patch
Fappl
Ri
f
T
f
Rc
Tension force pulling down:
FT = 2  Ri T sin(f)
Force Balance
Fappl + FT = P Ri 2
Tangent-Curvature
t1
R(s)= position
t2
t
C
s
r
t (s)  ;
s
2
 r
C  2
s
Forces on Rods
• Does compressive force play a role?
• Failure mode is buckling-To analyze must
consider geometry when it buckles• (1) get m.o.I;
• (2) general formula for moment in the rod.
(3) moment as a fxn of applied F.
• (4) relation between R of curvature and x,
(5) simplify eqn.
Step (1) Moment of Inertia of c.s.
I   y dA
2
dA  2( R  y ) dy
2
2 1/ 2
R
I  4 y ( R  y ) dy
2
2
2 1/ 2
0

R
4
2
For hollow cylinder,
subtract the hollow
portion.
Step (2) Bending a rod
s+Ds
s
dA
Ds  0..when.. y  0
s  Ds R  y

...arcs..have..common.ctr
s
R
Ds y
 1 dF
  l  
s
R
E E dA
yE
dF 
dA.................(1)
R
y
R (at neutral surface) is
assumed constant on
the small segment.
Step (2) reiteration
(Landau & Lifschitz, 1986 , Theory of Elasticity)
dF

at  y
dA
 E  {strain}
Ds y
{strain} 

s
R
dF Ey


dA R
Step (2) continued: Integrate
Ey
dF 
dA  nowfind  dM  ydF
R
E 2
M
ydF

y
dA


R
crosssec tion
EI

R
EI
M ( x) 
..........(2)
R( x)
I   y dA
2
Step (3) Moment due to appl F
P
P
M ( x)  Ph( x)......(3)
EI
.......... 
.......(2)
R( x)
h(x)
x
P
P
Step
(4)2
2
d r
d h
1 / R  curvature  2   2
ds
dx .. for...gentle..curves...
2
Minus sign because
d h  Ph( x)
so  2 
.........(4)
Curvature is negative.
dx
EI
2
From before:
d x
2
same.. form..as.. 2   x
EI
M ( x)  Ph( x) 
dt
R( x)
x
h( x)  hmax sin( )
Note similarity to harmonic
Lc
Motion :
(5)
Hmax occurs at Lc/2 and h(0) = h(Lc)= 0.
Step (5) Differentiate h twice
d h
 2
x
 2
 ( ) hmax sin( )  ( ) h( x)...(5)
2
dx
Lc
Lc
Lc
2
d h  Ph( x)
....and ......................... 2 
.......(4)
dx
EI
So...P..  f buckle
2
 f buckle  (

Lc
) EI   f (
2

Lc
)2
• Use spring equation. Hmax occurs at Lc/2.
h(0) = h(Lc)= 0. We can relate F to Lc by
double differentiating h, and then
comparing it to the previous formula for the
moment.
• Buckle force is independent of hmax . Rod
will buckle when P> Pbuckle
• Can a microtubule withstand typical forces
in a cell?
2
d h
2
 ( / Lc) hmax sin( x / Lc)
2
dx
2
 ( / Lc) h( x)
2
d h
P

h( x )
2
dx
EI
P
2
 ( / Lc)
EI
2
2
P  EI ( / Lc)   f ( / Lc)
Buckling of
Rods with
Different
Fixations
P1 
 EI
2
2
L
P2  4
 EI
2
2
L
P3 
 EI
2
2
4L
Buckling of cell without
reinforcement
 crit
 E
2
2
h
 2 (
)
2
L 3(1  )
Tissue
Mechanical Environment
Normal Range
Cell Types
Bone,
Cartilage
Weight bearing forces
Continuous: 1X 4X Body weight
Arterial
Endothelium
Tendon
Fluid pressure and shear
Pulsatile: 60-140
mm Hg;
Up to 560 +- 9
Kg/cm2
Osteocytes,
Osteoblasts,
Chondrocytes
Endothelial
Skin
Organ of
Corti
Muscle
(Intrafusal)
Muscle
(Extrafusal)
Compression and shear
Fluid shear
Nerve
Hair
Tension
Mesangium
Fluid pressure and shear
Nerve/specialized
muscle
Smooth, cardiac,
and skeletal
myocytes
Mesangial
Tension
Tension; active contraction
Pulsatile: 60-140
mm Hg
Nerve
• Living cells are both affected by and
dependent upon mechanical forces in their
environment. Cells are specialized for life
in their own particular environments, whose
physical stress patterns become necessary
for normal functioning of the cells. If the
forces go outside the normal range, then the
cells are likely to malfunction, possibly
manifesting as a disease or disability.
Material efficiency
Strength/weight
Square
Bar
Rod
Fiber orientation for strength
A: Actin fibers in two C2C12 cells. B,C: C2C12 cell with a
schematic representation of the actin cytoskeleton, which
is predominantly orientated along the first principal axis of
the cell. As a result of the actin fibers, deformation of the cell a
nucleus is restricted in this direction.
Cell Walls for strength
How thick does wall need to be to withstand normal
pressures inside a bacterium, I.e. 30-60 atm. ?
Lets say lysis occurs @ 50% strain. We can approximate KA
By KVd, and for isotropic wall material, Kv ~ E, so,
tfailure= 0.5 KA= RP= 0.5 E d. So to not fail,
d> 2RP/E . So for R = 0.5 mM, P= 1 atm,
2  0.5 x10 6  105
d
3 x107
10
 nM
3
Homogeneous rigid sheet:
Biomembrane
Stretching membrane thins it
exposing hydrophobic core to
Water. Rupture at 2-10% area
Expansion, so say lysis tension
~ 0.2 J/M2. For a 5 mm cell ,
P= ~ 8000 J/M3 ~ 0.08 atm. at
rupture.
Bilayer compression resistance, KA = 4 g
g 0.04 J/M2
Comparative Forces
• To pull a 5 mm cell at a speed of 1 m/sec:
• F= 6hRv = 0.1 pN
• Compare this with force to bend or buckle
hair, 10 cm length, R = 0.05 mm:
• f buckle  5 x 10 4 pN
F
• or to move it 1 cm:
• F = 3 f z/L3 = 1.5 x 10 6 pN
Comparative Forces
• Adhesion force between proteins on cell
and on matrix: tens of pN.
• Spectrin spring constant = 1-2 x 10 –5 J/m2
so to stretch by 0.1 um takes 1 pN.
Properties of the CSK
• A dynamic structure that changes
both its properties and composition
in response to mechanical
perturbations.
Pulling on CSK
Uni- and Bi-axial Stress and Strain
Take the case of unconstrained isotropic object compressed in the y
direction:
Before strain
After strain
y
y
y
y
y
x
• Note that for an elastic material the strain occurs almost
instantaneously upon application of the stress. Also note
that to maintain constant stress, y , the applied force must
be reduced if the face area increases, but this would be a
negligible change for all practical situations.
• The strain in the y direction is:
•
y 
y
E
• Because the transverse direction is
unconstrained:
•
• and,
•
Now, Consider the case where the x direction is
constrained from movement. I.e. transverse
movement is resisted, making

'
y
  x    y
2
 y  E  y   x
'
y
x
x
2
y
Thus the new stress in the y direction is the original unconstrained stress plus the stress caused by transverse constraint:
 y  E y   x
Solving for y we have the biaxial strain equation:
1
 y  ( y   x )
E
3-Dimensional stresses
Y
(stress tensor)
X
t xx t xy t xz
Z
tyx t yy tyz
t zx t zy tzz
Stress components @ Equilibrium
t 11 t 12 t 13


0
x1 x2 x3
t 21 t 22 t 13


0
x1
x2
x3
t 31 t 32 t 33


0
x1
x2
x3
Blood Forces
Y.C. Fung
Analyze a Small element of upper
EC membrane : (Also a mult-part
solution)
y
z
x
Flow
Cell 1 cell 2 cell 3
Analysis of EC upper membrane
t xy  t yx ,t yz  t zy ,t zx  t xz
Symmetrical
t zx  t xz  t yz  t zy  0 (Fluid Mosaic)
y
z
x
 txx txy txz 
 tyx t tyz 
yy


 tzx tzy tzz 


t xy  t yx  t ,
t xy  t yx  0
y
z
x
On surface facing
blood
On surface facing
cytosol
 xx t xy t xz


0
x
y
z
h
Tx    x dy
y
0
z
x
t xy  t yx  t ,
On surface facing
blood
t xy  t yx  0
h
Tx    xx dy
y
0
z
x
We need membrane tension as f(t)
Define
 xx t xy t xz


 0  multby  dy  and  int egrate
x
y
z
t xy
 xx
0 x dy  0 y dy  0
h
h
y
z
x
sin ce
h
Tx    xx dy 
0
Tx
Tx
t  0  
dx  Tx  tL
x
x
0
L
(if Tx= 0 @ x=0)
Stress on cell from flow
Tx   xx h
so
 xx
Tx
L
 t
h
h
@ x = -L
For t = 1 N/m2 , L= 10 mm, h = 10 nm
 xx
N
 10 2
m
3
Tx  10
6
N
m
Shear stress from flow in a pipe
dU
t h
dl
DP 2 2
U (r ) 
(a  r )
4 Lh
DP
t 
r
2L
P1
P2
Fluid Pressure is omnidirectional
dZ
P2

Fx = 0
P1 dy dz = P2 sin(q) dz dy/sin(q)
dy
P5
P1= P2
>
P1
q
P4
dx
A

Fy = 0
P3 dx dz = P2 cos(q) dz dx/cos(q)
P2=P3
P3
 Fz = 0
Rotate by 90, and see also:
P4 = p5
Hence P1=P2=P3=P4=P5 =P
Two State Transitions
  
 X 1   a11
    a
 X   21
 2


a12   X 1 

a 22  

X
 2

X  AX
 k21 k12 
A 

k

k
12 
 21
dX 1
  K 21 X 1  K 12 X 2
dt
and
dX 2
 K 21 X 1  K 12 X 2
dt
Entropic springs
4-segment chain configurations
RNA
24
22 nM
Sudden extensions of 22 nM
(unfolding) when forces
above 14 pN are applied
Large ree
Few Configurations
tension
Small ree
Many Configurations
Applying a tension to the zero ree state
reduces possible configurations to 10.
S drops from ln(16) to ln (10). Hence tension
translates to loss of entropy.
Rate Constants
Kopen = 7 sec-1
Kopen = 0.9 sec-1
Kfold = 1.5 sec-1
Kfold = 8.5 sec-1
RNA unfolding
22 nM
Sudden extensions of 22 nM (unfolding)
when forces above 14 pN are applied
Coding of Probability
Integral pulse frequency modulation
ti 1
T  K  (1  n(t )) dt
ti
Probability
Pulse frequency and width Modulation
Pulse Width Modulator
Inputs
Leaky integrator
t2
 u(t )dt
t1
Reset
Thresholder
Pulses
out
Mechanical Models




Voigt solution
1
1
1
Z R
s .C
C
Z
s
t
Io 1
1
.
.
.
V ( s ) I ( s ) Z( s )
s C ( s 0) . s
Laplace domain

1

1
e at1  e a2t = bi-exponential decay
a 2  a1
t
Time domain
V ( t ) I o .R . 1 e
t

1
t
o
E
t
. 1
e
t
t
h
E
Classwork
• Make a simulink model of the RNA unfolding kinetics. Your model
should be well documented, according to the following guidelines:
• All parameter boxes should be labeled
• Document boxes should be included to describe operations
• Internal parameters, such as initial conditions, should be specified
• Sub-systems should be used so that the entire model can be fit onto 1
page and each sub-system can be printed separately, with
documentation.
• A separate description of the system and all formulae should be made.
• Outputs should be the predicted, as well as measured probabilities
• A reasonable noise level should be placed in the model
Control System, I.e. climate
control
Set Point
Perturbation
Output
Sensor
-
-
Plant
-
Error
Feedback
Temperature Control
 0 1
A

 3 0
0 
B 
1 
1 
C 
0 

X  AX  BU
Y  CX
y( s)
G ( s) 
u ( s)
1
 C ( sI  A) B
T
G(s)
U(s)
Y(s)
1/s
+
1
X2
1/s
X1
3

X (t )  AX (t )  BU (t )
y (t )  CX (t )
 0 1
A

 3 0
0 
B 
1 
1 
C 
0 
• Patterns on silicon
with fibronectin.
• Cells grown on small
pattern : Apoptosis
• On a line they
differentiate
• On a large surface,
they grow.
Mechanical Terms Review
•
•
•
•
•
Statics and dynamics
Kinematics and kinetics
Vector and scalars
Forces, resultants
Deformation
Homework
Using the data shown in Figure previous, and
the ground free energy, Fo = 79 kT, graph
the unfolding and folding probabilities,
using Excel or other program. Put actual
data points for the selected forces on your
theoretical curve.
Tensiometry
Plates coated with poly-HEMA
Balance
P  C T
Compression of cells reduces the
load measured by the balance
by an equivalent amount
F
R2 R1
T
(
)
2
  R 3 R2  R1
Liquid behaviour: Surface
tensions of embryonic tissue
25
Liquid Properties
20
15
10
dynes/cm
mesoderm
heart
0
Neural
retina
5
P  1/ 5 1/ 4 1/ 3 1/ 2  1/ 120
N
R
L
H
Ep
M