February 8, 2005 :
• Homework
• Read Ch. 4,
5; Appendix
• Biophysics
Review
•
Cell Types;
Mechanical
Testing
• Percolation
• Kinetics
Prokaryotes
• Most have elevated osmotic pressure, I.e a few
tens of atmospheres.
• Challenger Deep sea dropped to 10,896 m in last 8
million years (Deepest place in ocean)
• Deficient in CaCO3 at that depth
• Foraminifera (shelled protists) quickly evolved
soft, non-calcareous shells
• Likely are highly pressurized.
Biomembrane as an isotropic
material
Bilayer compression resistance, KA = 4 g
g= 0.04 J/M2
(g
= surface tension)
Homogeneous lipid sheet:
Biomembrane
A
A
Tm = K A 
 =
•10 atm = 106
J/m3
Stretching membrane thins it
exposing hydrophobic core to
Water. Rupture occurs at 210% area expansion, so say
lysis tension ~ 0.016 J/M2. For
a 1 mm cell :
P= 64,000 J/M3 ~ 0.6 atm. at
rupture.
Life @ 1,200 atmoshperes
• How thick does the membrane need to be?
• How thick does the wall need to be?
• Compressibility properties:
KV d p
KA =
4 / 9  KV / m
KV  3m = E
K A  KV d p
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 when wall tension exceeds 5% of KA.
We can approximate KA by KVd, and for isotropic wall
material, Kv ~ E, so, assume a material E= 3 x 109 J/m3
tfailure= 0.05 KA= RP/2=0.05 E d
So to not fail,
d> RP/E
0.5 x10 6  106
d
So for R = 0.5 mM, P= 10 atm,
3x109
5
= nM
3
Thick wall sphere
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
Alternate method (not too thick
wall)
Rp i Rp o
 =

2h
2h
 fail = E max
Laplace Law for Cylinder under pressure
Homework:
1. Find wall stress for cylinder.
2. Calculate stress for a foraminefora not
Assuming a thin wall.
Compression of a network
•
•
A cartesian lattice of fibers (a) subject
to compressive strain (b). The
boundary conditions are that the
angles between the segments arising
from each junction are fixed at 90
and the deformation consists in
movement of the junctions toward
each other along the three orthogonal
axes of the lattice, with no shear
deformation. The junctions, however,
are allowed to rotate as
compression progresses. This is
equivalent to the boundary conditions
in Feynman et al.36 in which fiber
ends are fixed but direction changes
under compression.
Solution to Donnan Problem

•
A.



[ Na ]o [ K ]o [Cl ]i [ HCO3 ]i 150
=
=
=
=



[ Na ]i [ K ]i [Cl ]o [ HCO3 ]o 144
B.
Cout
58
E = log(
) = 1.03mV
z
Cin
C.
[ solutes ]i = [ solutes ]i
144  4 = 114  29  A ; A = 5mM
D.
1.03mV
Power from electrochemical
gradients (I.e batteries)
Distributed Model
Driving Force determined by Nernst
Electrical Model of Cell
Membrane
Molecular model
Outside
Cm
gK
EK
Inside
gNa
E Na
I x = g x  px  N x ( Ex  Vm )
Ca Wave in Oocyte
Mechanotransduction
What opens channels?
Types of mechanical analysis
•
•
•
•
•
Kinematics - just the connections
Statics- forces without motion
Dynamics- forces with motion
Rigid versus deformable body
FBDs
FEL
FBL
FBR
FER
Loading Types
•
•
•
•
•
•
•
Tension- compression
Shear
Reaction
Traction
Friction
Bending
Uniaxial/bi-axial
Cytomechanical forces:
•
•
•
•
•
•
Gravitational:
Muscle contraction:
Contact:
Buoyant:
Hydraulic: (Static or dynamic)
Pneumatic
Cell Deformation and Stiffness
• Most cells are constantly deformed in vivo
by both internal and external forces.
• Experimental deformations can be done by
poking, squishing, osmotic swelling,
electrical/magnetic fields, drugs, etc.
• Cells have both area and shear stiffness,
mostly due to the cytoskeleton, although
lipids contribute some.
Material Parameters
•
•
•
•
•
•
•
•
Moduli: Young’s, area, shear, bending (flexural)
Stiff versus compliant
Strength versus weakness
Brittle versus ductile
Incompressible/Compressible
Failure
Ultimate tensile strength
Hardness: Moh’s scale
Comparative Mechanical
Properties
Steel
Wood
Bone
Steel Wood Bone
Cells
Strain 
Cellular
‘pre-stress’
Cells
Comparative Stiffness
10000
1200
210
21
14
1
0.007
0.01
0.0002
di
am
on
d
l
st
ee
ne
bo
d
wo
o
er
ru
bb
su
e
0.0001
tis
Modulus (GPa)
100
Material
Elasticity
•
•
•
•
•
“ut tensio sic vis”
Young’s Modulus: Stress over strain
Shear Modulus: Related to Poisson
Comparative Strains
Comparative Stiffnesses
Poisson’s Effect
For most engineering materials,  < 0.3
Materials with  = 0.5 are "Incompressible."
Some materials have  > 1
Cauchy Strain
l  lo
x
lo
v =-(.7-1)/1 = 0.3
Y
ly  lyo
y
1
lyo
0.7
Poisson's Ratio
0
X
Incompressible
Means no volume change
2

swelling

y
x
Elastic Behaviours
Unixaxial stress
Pressure
<1
E = /
< 0
KA = P/A/A
1
2
Applying forces (testing types)
Uniaxial
Shear
Pressure
Biaxial
Tension or
Compression
Bending
Twisting
Testing methods
Q: What are the relative resolutions?
AFM
Q? Why doesn’t the AFM needle poke right through?
Micropipet methods
A. Whole Patched Cell
Micropipette
Pp
Qpp
Micropipette
Stretch
Tension
Pi
C
Mesangial
Cell
i
Q m (K w)
C o
 C i =constant
B. Isolated Cell
Stretch
Solutes
Pi
Solutes
C i (t)
Qm (Kw)
Mesangial
Cell
C o
Magnetic tweezers
Wang et al, Science
Pulling on CSK
Shear and compression
Example: Blood flow forces
Optical Tweezers
•
•
•
•
High resolution
Refractivity of bead
Trapping in the beam
Limited force
Optical Tweezer
Swelling
RBCs
•
•
Necturus erythrocytes loaded with fluo-4 (10 µM) and exposed to UV
light emitted from a mercury vapor bulb and filtered through a FITC cube
(400x). (A) Cells display little fluorescence under isosmotic
conditions (n=6). (B) Addition of A23187 (0.5 µM) to the
extracellular medium increased fluorescence under isosmotic
conditions (n=6). (C) Exposure to a hypotonic (0.5x) Ringer solution
increased fluorescence compared to basal conditions (n=6). (D) A
low Ca2+ hypotonic Ringer solution (5 mM EGTA) did not display the
level of fluorescence normally observed following hypotonic
swelling (n=6).
Light et al.
Stimulation Protocols
Impulse
Step
Sinusoid
Ramp
Magnitude
TIME
Figure 4.2 Modes (top) and timing protocols
(lower) of force application
Bone Loading Waveforms
Sickle Cell: A gel problem
• Single point defect causes Hbs- a polymerizing
tendency in deoxygenated state
• The stiff and deformed cells damage vessels
• Main approaches:
– 1. Controlling kinetics of polymerization
– 2. Regulating stiffness (rheology) of sickle cells.
Thermal shape variations
Stiff
Flexible
(a)-(c) Serial images of a 23 mm long
relatively stiff fiber. (b) and (c) are,
respectively, 21.9 and 41.4
seconds after (a). There is little visible
bending (see also Figure 2(a)),
consistent with a long persistence
length, lp . 12.0 mm. (d)-(f) Serial
images of a 20 mm long ¯flexible
fiber. (e) and (f) are, respectively,
51.8 and 60.8 seconds after (d).
There is marked bending and a short
persistence length, lp.0.28 mm (see
also Figure 2(b)). The fibers undergo
diffusional motion and hence are not
adhering to a glass surface, rather
are free in solution, a necessary
condition for using statistical
mechanics to obtain persistence
lengths. The width of each frame is
Statistics of fluctuations in 1
dimension
Statistical Mechanics
p =
f
kT
R=
p
R
p
= 2
2
p =
L3
L 2
48 (u
)
x 2
Dilute
Concentrated
Semi-concentrated
Floppy
Chains
Isotropic
Rods
Nema
Harmonic motion (undamped)
Gel motion follows simple rules
Model will predict dynamic and
Static equilibrium.

m x = 2 PAu(t )  k ( x  x0 )

mx =  k ( x )

x  2 x = 0
Natural Frequency
Damped Spring
Viscosity & Elasticity
• A complex material can be modeled as a purely
viscous material combined with a purely elastic
material, thus mathematically separating the
viscosity of a material from its elasticity. A purely
viscous component is a Newtonian fluid- it has no
memory and no elasticity; it cannot deform as a
solid. Cells generally behave as solid-liquid
composites. V-E tools can quantify their
behaviour, since the models separate viscosity
from elasticity in a kind of finite element model.
Maxwell Model: Differential
method
 T =  E  

1/E

 T =  E = 


 = 
E

d  1 d 
=

dt E dt 

1  
 = (  )
E

Maxwell model: Laplace Method
1 1
Z ( s) = 
E s

1/E

V
=
R
=
C
Viscosity: Pascal-sec

Compliance
+
Slipperiness
Z
o
s
o
=
Mechanical
Impedance.
For a step input
s
1 / E  1 / s
t=/E
Gel Model
• Make a complete model and label all
parameters
• Describe the output, relating what happens
and why.
• What is the time constant?
• State the assumptions and simplifications
Mechanical Terms Review
•
•
•
•
•
Statics and dynamics
Kinematics and kinetics
Vector and scalars
Forces, resultants
Deformation
Classwork
• Add damping to your model of cytogel
• Describe how you can model thermal
fluctuations in cell diameter, and list all the
elements. List assumptions.
• Write the model equation for the above.
• Complete a simulink model of the above,
and do all labelling, including all parameter
values.