Tuesday, March 01, 2005

• Geometric and Mechanical Properties

• Mechanical Statics

• Review-

Thick walled sphere

• Equilibrium• Pressure inside• Average stress in wall

• Pressure from outside

• Pressurized both sidesh

pr

h

pr

rrh

rp

rrh

rp

rr

rp

prrr

oi

PP

io

ooP

io

ii

io

iiP

iiPio

outin

out

in

in

22

)(

)()(

)(

2

2

22

2

222

Charged polymers: Electromechanical Chemistry

I.e. Alanine charge+H3N-CH-COO-

CH3

pKa =9.9 pKa =2.7

pH pKalog A

HA( )

- = fractional charge = A

A

AH 1

11

alog pH pKa( )

- =

Aqueous charge

X Y Z

Meridions

Latitudes

Losing volume, not gainingarea;

21

11

RRC

Curvature

Shape : Oblate sphere

Slow cell squishing

b 0FRAME

5

a 3FRAME

10

Curvature

Membrane Tension

T m T m

T m

T m

P

P

R

T m dyT m dy

P

1. Hemisphere

2. Patch

3. Patch in x-z plane

R

dx

d

Tdy

Tdy

Tdy d

4. Vertical Resultant

dxdy

T

T

Tension on membrane patch

Ri

Rc

T

FT = 2 Ri T sin()

Tension force pulling down:

Fappl + FT = P Ri 2

Force Balance

Fappl

Tangent-Curvature

2

2

;)(

s

rC

s

rst

Cs

t

R(s)= position

t1

t2

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.

4

)(4

)(2

2

2/122

0

2

2/122

2

R

dyyRyI

dyyRdA

dAyI

R

For hollow cylinder,subtract the hollow portion.

Step (2) Bending a rod

y

s

s+s

dA

)1.......(..........

1

........

0....0

dAR

yEdF

dA

dF

EER

y

s

s

ctrcommonhavearcsR

yR

s

ss

ywhens

l

R (at neutral surface) is assumed constant onthe small segment.

Step (2) reiteration(Landau & Lifschitz, 1986 , Theory of Elasticity)

R

Ey

dA

dFR

y

s

sstrain

strainE

yatdA

dF

}{

}{

)2(..........)(

)(

2

sec

xR

EIxM

R

EI

dAyR

EydFM

ydFdMnowfinddAR

EydF

tioncross

Step (2) continued: Integrate

dAyI 2

Step (3) Moment due to appl F

)2.......()(

..........

)3)......(()(

xR

EI

xPhxM

P

h(x)

x

P

P P

)sin()(

......

)4.........()(

/1

max

22

2

2

2

2

2

2

2

cL

xhxh

xdt

xdasformsame

EI

xPh

dx

hdso

dx

hd

ds

rdcurvatureR

Note similarity to harmonicMotion :

Minus sign because Curvature is negative.

From before:

(5)

Step (4)

.......... curvesgentlefor

)()()(

xR

EIxPhxM

Hmax occurs at Lc/2 and h(0) = h(Lc)= 0.

22

2

2

2max

22

2

)()(

.....

)4.......()(

.............................

)5)...(()()sin()(

cc

cc

LEI

Lf

fPSoEI

xPh

dx

hdand

xhLL

xh

Ldx

hd

fbuckle

buckle

c

Step (5) Differentiate h twice

• 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?

22

2

2

2

2

max2

2

2

)/()/(

)/(

)(

)()/(

)/sin()/(

LcLcEIP

LcEI

P

xhEI

P

dx

hd

xhLc

LcxhLcdx

hd

f

2

2

1 L

EIP

2

2

2 4L

EIP

2

2

3 4L

EIP

Buckling ofRods withDifferentFixations

Buckling of cell without reinforcement

))1(3

(2

2

2

2

h

L

Ecrit

Tissue Mechanical Environment Normal Range Cell Types

Bone,Cartilage

Weight bearing forces Continuous: 1X -4X Body weight

Osteocytes,Osteoblasts,Chondrocytes

ArterialEndothelium

Fluid pressure and shear Pulsatile: 60-140mm Hg;

Endothelial

Tendon Tension Up to 560 +- 9Kg/cm2

Nerve

Skin Compression and shear NerveOrgan ofCorti

Fluid shear Hair

Muscle(Intrafusal)

Tension Nerve/specializedmuscle

Muscle(Extrafusal)

Tension; active contraction Smooth, cardiac,and skeletalmyocytes

Mesangium Fluid pressure and shear Pulsatile: 60-140mm Hg

Mesangial

• 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 efficiencyStrength/weight

RodSquareBar

Fiber orientation for strength

A: Actin fibers in two C2C12 cells. B,C: C2C12 cell with a schematic representation of the actin cytoskeleton, whichis predominantly orientated along the first principal axis of the cell. As a result of the actin fibers, deformation of the cell and itsnucleus is restricted in this direction.

Cell Walls for strengthHow 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,failure= 0.5 KA= RP= 0.5 E d. So to not fail,d> 2RP/E . So for R = 0.5 M, P= 1 atm,

nM

x

xd

3

10103

10105.027

56

Homogeneous rigid sheet: Biomembrane

Bilayer compression resistance, KA = 4 J/M2

Stretching membrane thins itexposing hydrophobic core toWater. Rupture at 2-10% areaExpansion, so say lysis tension~ 0.2 J/M2. For a 5 m cell ,

P= ~ 8000 J/M3 ~ 0.08 atm. at rupture.

Comparative Forces

• To pull a 5 m cell at a speed of 1 /sec:

• F= 6Rv = 0.1 pN

• Compare this with force to bend or buckle hair, 10 cm length, R = 0.05 mm:

• 5 x 10 4 pN

• or to move it 1 cm:

• F = 3 f z/L3 = 1.5 x 10 6 pN

bucklefF

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

y

y

y

y

y

Uni- and Bi-axial Stress and Strain 

Take the case of unconstrained isotropic object compressed in the y direction:

 

Before strain After strain

 

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:

•

Ey

y

• Because the transverse direction is unconstrained:

•

• and,

•

xyy E 2'

yxy 2'

Thus the new stress in the y direction is the original unconstrained stress plus the stress caused by transverse constraint:

xyy E

Now, Consider the case where the x direction is constrained from movement. I.e. transverse

movement is resisted, making y

y

xx

Solving for y we have the biaxial strain equation:  

)(1

xyy E

Z

Y

X

xx

yx

zx

xy

yy

zy

xz

yz

zz

3-Dimensional stresses (stress tensor)

Stress components @ Equilibrium

0

0

0

3

33

2

32

1

31

3

13

2

22

1

21

3

13

2

12

1

11

xxx

xxx

xxx

Blood Forces

Y.C. Fung

Analyze a Small element of upper EC membrane : (Also a mult-part solution)

x

y

z

Cell 1 cell 2 cell 3

Flow

Analysis of EC upper membrane

0

,,

zyyzxzzx

xzzxzyyzyxxy

xx

yx

zx

xy

yy

zy

xz

yz

zz

Symmetrical

(Fluid Mosaic)

x

y

z

0

,

yxxy

yxxy

On surface facingblood

On surface facingcytosol

x

y

z

h

xx

xzxyxx

dyT

zyx

0

0

x

y

z

h

xxx

yxxy

yxxy

dyT0

0

,

On surface facing

blood

We need membrane tension as f()x

y

z

Define

LTdxx

T

x

T

dyT

ce

dyy

dyx

egrateanddymultbyzyx

x

Lxx

h

xxx

hxy

hxx

xzxyxx

0

0

00

0

sin

0

int0

(if Tx= 0 @ x=0)

x

y

z

Stress on cell from flow

h

L

h

T

so

hT

xxx

xxx

@ x = -L

For = 1 N/m2 , L= 10 m, h = 10 nm

2310m

Nxx m

NTx

610

Shear stress from flow in a pipe

rL

P

raL

PrU

dl

dU

2

)(4

)( 22

P1 P2

Fluid Pressure is omnidirectional

>

A

dZ

dx

dy

P1

P2

P3

P4q

P5

Rotate by 90, and see also:

P4 = p5

P1 dy dz = P2 sin(q) dz dy/sin(q)

P1= P2

Fx = 0

P3 dx dz = P2 cos(q) dz dx/cos(q)

P2=P3

Fy = 0

Fz = 0

Hence P1=P2=P3=P4=P5 =P

Two State Transitions

2121212

2121211

XKXKdt

dX

and

XKXKdt

dX

2

1

2

12221

1211

X

X

X

Xaa

aa

XAX

1221

1221

kk

kkA

Entropic springs

Large reeFew Configurations

Small reeMany Config-urations

4-segment chain configurations RNA

24

Applying a tension to the zero ree statereduces possible configurations to 10.S drops from ln(16) to ln (10). Hence tension translates to loss of entropy.

tension

Sudden extensions of 22 nM (unfolding) when forces above 14 pN are applied

22 nM

Rate Constants

Kopen = 7 sec-1

Kfold = 1.5 sec-1

Kopen = 0.9 sec-1

Kfold = 8.5 sec-1

RNA unfolding

Sudden extensions of 22 nM (unfolding) when forces above 14 pN are applied

22 nM

Coding of Probability

1

))(1(i

i

t

t

dttnKT

Integral pulse frequency modulation

Probability Pulse frequency and width Modulation

Pulse Width Modulator

2

1

)(t

t

dttu

Leaky integrator

Thresholder

Pulsesout

Inputs

Reset

Mechanical Models

Voigt solution1

Z

1

Rs C Z

1

C

s1

Laplace domain V s( ) I s( ) Z s( )I o

s

1

C 1

s 0( ) s1

taat eeaa

21

12

1 = bi-exponential decay

Time domainV t( ) I o R 1 e

t

E o

E1 e

t

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

Sensor Plant-

--

-

Output

Error

Perturbation

Feedback

Set Point

Temperature Control

BAsIC

su

sysG

CXY

BUAXX

T 1)(

)(

)()(

0

1

1

0

03

10

C

B

A

G(s)Y(s)U(s)

1/s 1/s+

-1

3

X2 X1

0

1

1

0

03

10

C

B

A

)()(

)()()(

tCXty

tBUtAXtX

• 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

Balance

)(12

12

32 RR

RR

R

FT

TCP

Plates coated with poly-HEMA

Compression of cells reduces theload measured by the balanceby an equivalent amount

Liquid behaviour: Surface tensions of embryonic tissue

0

5

10

15

20

25

Neu

ral

reti

na

hear

t

mes

oder

mdynes/cm

NRLHEpM

120/12/13/14/15/1 P

Liquid Properties