Chapter 4: Molecular statistics: random walks, friction, and diffusion and how to relate this to the cellular world

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Chapter 4Molecular statistics random walks,
friction, and diffusionand how to relate this
to the cellular world
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Topics

• Dissipative processes (friction and diffusion)
• Friction in a process is the cause of heat
  production and the loss of order in a system
• Diffusion can be understood from the random
  motion of molecules
• These two processes are heavily coupled, and
  their relation enables us to determine Avogadros
  number (and kB) (Einstein)

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Brownian motion

• Colloidal suspension small particles (about 1
  ?m) in water
• Brown (1828) noticed that these particles
  performed a perpetual dance, no matter how long
  one waited also lifeless particles.
• Explanation by others this motion is caused by
  the thermal motion of the (much, much smaller)
  water molecules.
• Paradox 1 How is it possible that we can see the
  motion of the much much bigger particles? The
  step size of the grain must be many times larger
  than the size of the water molecules.
• Paradox 2 an elementary estimate (gas law)
  yields a collision frequency of about 1012 Hz
  (103 m/s)/(10-9 m).
• How can we observe the effect of such a high
  frequency with the naked eye?

Answer (Einstein) large (visible)
displacements are extremely rare!
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The one-dimensional random walk

• Start in x0
• At each time step take a random step ?xkL with
• p(k1) 0.5 and p(k-1) 0.5
• What is the average distance after N steps?
• ltxNgt N p() ?x N p(-) ?x
• N 0.5 (L-L)
• 0
• What is the mean quadratic displacement?
• ltx2Ngt lt(xN-1kNL)2 gt
• ltxN-12gt 2LltxN-1kNgt L2ltkN2gt
• ltxN-12gt L2ltkN2gt
• We obtain
• ltx2Ngt N L2

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The diffusion equation

• The total time in the previous process is found
  by Nt/?t
• The 1D diffusion equation
• ltx2Ngt 2Dt
• in which D?L2/(2?t) is the diffusion
  constant (in m2/s)
• Generalisation to three dimensions L? L

See Assignments Computer exercise
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Diffusion is a random process

• The paradox of small displacements and high
  frequencies is resolved because we can only
  observe the effect of very large displacements.
  These have a very low probability, and therefore
  occur at a very low frequency!
• The diffusion equation does not depend on the
  distribution of step sizes (You are going to show
  this in the computer exercise!)
• Diffusion is a collective random walk of many
  particles
• The position-probability distribution becomes
  observable as a particle density profile

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Friction

• Friction is quantitatively related to diffusion
• Simple model of a particle suspended in a liquid
  subjected to a force f in direction x
• Between two collisions the force acts freely on
  the particle, so that by Newtons law dvx/dt
  f/m
• After each collision the velocity is entirely
  random
• Between two collisions, the displacement is

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• Determine the mean displacement
• Starting velocity is zero on average
• The mean drift velocity, due to the net force is
• with the viscous
  coefficient of friction kg/s
• There is a simple relation between this
  coefficient and the liquids viscosity, ?
• In Stokes formula, R is the particles radius,
  and the viscosity depends on the liquid and the
  temperature.

For water, ?0.001 kg/ms
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The Einstein-relation

• We now have expressions for the constants
• These can be measured in a simple macroscopic
  diffusion experiment, and measurement of the
  velocity of a macroscopic particle in water,
  subjected to the gravitational and viscous
  forces.
• We also know that
• and
• From which follows Einsteins famous equation

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Finally, Avogadros number NA

• Take a thermometer and measure T
• Measure the diffusion constant, D
• Determine the friction coefficient, ?
• Compute kB
• Apply the ideal gas law and find NA

Note Einsteins equation is universal! It does
not depend on the type of particle or
liquid (although ? and D do in a very
complicated way!)
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Biological application 1 Polymers

• Polymer long chain-like molecule that is
  constructed from many (identical) small building
  blocks.
• Q What is the typical head-tail length of a
  polymer?
• A Consider the polymer
• as a random walk. It then
• follows that the average
• length is given by

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• The head-tail length of a polymer increases with
  the square-root of the mass.
• This model allows to
• distinguish different
• spatial configurations
• for polymers, e.g. caused
• by electrical charges.
• Incorporating that each
• corner point can be used
• only once, yields an
• exponent of 0.58

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Diffusion density profileswhat is the
distribution of the molecules over time and space?

• Again, we set up a simple physical model

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• Suppose that the particle density only varies
  along
• the x-dimension
• In time step ?t each particle will jump to a
  compartment either left or right
• The net number of particles passing the wall at
• x-L/2 from left to right then equals
• Step size L is chosen such that the derivative of
  the number of particles does not change at that
  scale, then
• The density c(x) is given by c(x)N(x)/(LYZ)

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Ficks law

• Dividing the number of particles, N(x)c(x)LYZ,
  by the time interval ?t and the wall surface YZ,
  gives the particle current density ( flux)
• And because we defined (1 dim) L2/(2?t) D
• we obtain Ficks law

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The diffusion equation

• We now look at the change in the number of
  particles in a compartment per time interval
• When the derivative of j doesnt change over the
  width of a compartment (small L) we obtain the
  continuity eqn
• Combining this result with Ficks law gives the
• diffusion equation

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Remarks

• The diffusion equation describes Brownian motion
  for a large number of particles.
• For a large number of particles, the local
  probability to encounter a particle is
  represented by the density.
• The density evolves in a deterministic way
  according to the diffusion equation.
• When the number of particles is not large, the
  statistical approach is not accurate enough,
  because of the large influence of statistical
  fluctuations.

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Diffusion through a thin tube

• A thin tube, length L, connects two basins with
  ink solution at different concentrations, c1 en
  c2
• The tube does not affect c1 and c2
• A constant concentration profile develops
• Equilibrium dc/dt0, so d2c/dx20
• Solve diffusion eqn. c(x)c1(c2-c1)x/L
• Current density j -D(?c)/L

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Membrane permeability

• The thin tube is a model for a pore in a cellular
  membrane. We then postulate that
• Ps is the membrane permeability, which depends on
  the fractional surface of the pores ?, the
  diffusion constant D, and
• the pore length.

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Cell in alcohol solution

• Concentration outside cout, initial concentration
  inside cin(0). Cell radius R10?m. Find cin(t)
• Solution
• cin(0) 3N(0)/(4?R3)
• cout is constant
• dcin/dt 3j 4?R2 /(4?R3) -3Ps (cin-cout)/R
• d(?c)/dt -3Ps ?c /R
• This yields

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Remarks

• This simple membrane model only accounts for
  diffusion through a channel
• Some molecules dissolve in the membrane, diffuse
  to the other side and then leave the membrane
• This process is described by a slightly different
  diffusion model PsBD/L, with B the partition
  coefficient

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Bacterial metabolism ( your turn.)

• What is the largest possible oxygen consumption
  of a bacterium (sphere, radius R) in water with
  oxygen concentration c0?
• Assume stationary concentration profile
• with c(?) c0 en c(R) 0
• Inward flux jD(dc/dr), and also jI/(4?r2)
• This yields c(r) A-(1/r)(I/4?D)
• Apply boundary conditions Ac0 en I4?DRc0
• c(r) c0(1-R/r)
• The maximal influx is proportional to R, but the
  consumption to R3. This limits the maximum volume
  of a(ny!) bacterium! see YT 4F

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The Nernst-Planck equation

• Consider ions in a solution, across which an
• electric field E is applied
• The ion velocity then is v f/?qE/ ?
• ions passing surface dA in time dt is cdAvdt.
  Flux j cqE/?
• In the presence of concentration differences, we
  add the diffusive flux from Ficks law

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• Use Einsteins relation to eliminate the viscous
  friction-coefficient
• Nernst-Planck equation
• In equilibrium, no net flux
• Integration over the distance l between the two
  plates with electric field EdV(x)/dx gives the
  Nernst relation

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• Suppose a concentration difference cin/cout10
  for positively charged sodium (Na) ions accross
  the cell membrane, then kBTr/e1/40 Volt gives a
  potential difference of ?V58mV across the cell
• Even though the real cell potential is not equal
  to the Nernst potential, the magnitude is quite
  right!
• The Nernst relation also yields

Hey, thats Boltzmann!
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Electric resistance

• Another consequence of Nernst-Planck
• If the plates are in the ionic solution, so that
  there is no concentration difference, an electric
  current will run. With c uniform
• so from IqAj we
  obtain

Ohms law!
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Diffusion Gaussian profiles

• A concentration pulse will diffuse as a Gaussian
  profile. The variance will increase with time
  according to 2Dt (one dimension), the peak
  decrases with time

YT 4G (also show that the inflexion points
move away from each other)
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Assignments

• Your Turn 4G (show also that the inflextion
  points move away from each other)
• Exercises 4.2, 4.4, 4.5 and 4.8 (lt Nov 23)
• PLUS Computer exercise (counts triple, and you
  get more time lt Xmas)