Basic references:

H. Berg,"Random Walks in Biology", Ch. 1,2 (Princeton University Press, 1983)

R. Feynman, "Lectures on Physics", Vol. 1, Ch.43

J. Crank, "The Mathematics of Diffusion" (Oxford University Press, 1956; 2nd ed. 1976)

Parameters:

Diffusion coefficients for

CO

O

H

PO

ATP in cytoplasm: 0.15 x 10

O

CO

Concentrations

CO

O

PO

O

O

The basic diffusion equation, sometimes called Fick's law, states that the flux per unit area (flux density),

D is a diffusion coefficient, which we usually assume to be a constant.

Diffusion fluxes can cause changes in the concentration distribution of the diffusing molecules, and to deal with these we need the full differential equation for diffusion, which can be written

One dimensional diffusion occurs when we have two infinite parallel boundary planes and concentrations that are constant along any plane parallel to these boundaries. It can also be used exactly for diffusion in a straight tube of constant diameter, and it is often a useful approximation to real situations that do not exactly meet these requirements.

Many situations in which we are interested are steady-state situations, in which dC/dt = 0. In simple cases, steady-state solutions for the diffusion equation are usually easier to find than complete, time-dependent solutions. However, in more complex situations, the easiest way to find the steady-state solution may be by numerical integration (with a computer) of the complete equation until an equilibrium is nearly reached.

We can approach this as a one-dimensional steady-state problem. Therefore, the concentration gradient must be constant over the range we are considering. We assume that the concentration at some depth, say 1000 meters, is maintained at 3 M, and that the concentration falls to 0 at the surface.

The gradient is therefore 3 x 10

The flux is D * gradient, or 10

We can convert this to ~1 x 10

This would support the incorporation of 10

We can compare this with an estimate of oceanic net productivity of approximately 70 gm of carbon per year per square meter.

Obviously, diffusion does not solve this problem. But wait! If the productivity of 70 gm C per year per m

An estimate of the CO

Again, diffusion is inadequate, by a factor of 3 x 10

This is a steady-state problem involving spherical symmetry. The appropriate equation could be derived from the general vector equation, but let's instead use a simpler approach. In the steady state, we assume that all of the molecules reaching the surface of the cell at radius a are adsorbed, resulting in C=0 at r=a, and a constant rate, Q, of uptake of nutrient molecules. This means that the total flux through any spherical shell surrounding the cell must also equal Q. Therefore, for any radius, r;

Q = -4pr

Therefore

C = C

Since C=0 at r=a, we find that Q = 4pDaC

The somewhat suprising result is that the uptake is not proportional to the surface area (~a

Let's compare the effects on two hypothetical cells:

Volume(4pr

(assume 95% water, 5% dry weight, 2.3% carbon)

C content:: 8 x 10

P content:: 8 x 10

(assume C

P uptake:: 1.3 x 10

Doubling time:: 500 sec=~8 min _or_ 50,000 sec = ~14 hours

This comparison suggests that diffusion considerations really do make a difference in the lives of such cells, especially if they are abundant enough to decrease the phosphorus concentration below the level used in these calculations. However, we should note that some large phytoplankton like diatoms have a large vacuole, so that it is not correct to compare diatoms with small phytoplankton on the assumption that the C content is proportional to the volume.

Can a cell increase its uptake of a nutrient by stirring the surrounding medium with flagella, or by swimming rapidly through the medium so that it is always exposed to a high concentration?

A simple-minded (and incorrect) approach: Assume that a spherical cell swims through the medium at velocity V, and absorbs all of the molecules in a cylinder with a radius equal to the radius of the cell -- e.g. pr

For the large cell (100 microns radius) the velocity is 40 microns/sec, which is reasonable. However, this "barn door" approach is wrong, because it neglects the fact that a small object moving through a fluid will drag fluid along that is sticking to its surface.

The correct treatment is given in a classic paper by Berg & Purcell (Biophysical Journal 20: 193-219, 1977). The true relationship between uptake and swimming speed is non-linear. The uptake can be doubled by a swimming speed equal to about 2.5 D/a. (250 microns/sec for the small cell or 25 microns/sec for the large cell) However, to triple the uptake requires a swimming speed of about 20 D/a. (2000 microns/sec for the small cell or 200 microns/sec for the largecell). This method for increasing uptake is worthwhile for a dinoflagellate that can swim at 200 microns/sec, barely useful for a cell the size of Chlamydomonas that can swim at 100 microns/sec, and useless for a bacterium that might be able to swim at 20 microns/sec. It does mean that the disadvantage of being large is not quite so great as indicated by our initial calculation, if the large cell can swim at a reasonable velocity. Swimming efficiency may not be important, if the limiting factor is nutrient availability rather than energy input.

dC/dt = Dd

This can be used, for instance, for the diffusion of ATP along the length of a flagellum, if we assume that [ATP] is everywhere high enough that q is independent of C. The steady state solution is easily found to be:

C = Co -(q/D)(xS-x

where C

C(S) = C

and the solution is only valid if C(S)>>0. Also, if C(S)-->0, we have a situation where the rate of diffusion is not adequate to supply the chemical reaction. We can define a dimensionless criterion:

Ø = qS

If Ø << 2, we can be sure that diffusion is adequate to supply the chemical reaction.

We could do similar calculations for cylindrical or spherical geometries, and we find that the form of the expression for Ø is the same -- the only change is in the numerical factor. It is easy to show that for diffusion into a sphere, the critical value of Ø is 6. For the degree of accuracy needed in biology, we can generalize to volumes of arbitrary shape, and suggest that whenever Ø<<1, we can be quite sure that diffusion is adequate. This allows some leeway to cover cases where the rate of reaction depends on the concentration.

An example: flight muscles of insects and birds are probably the most actively metabolizing tissues known. A value given for the honeybee is 2.2 kcal/gm of muscle/hour (about 2.5 watts/gm). This metabolism will require 0.37 liters of O

S

S = 0.8 x 10

This gives us a rough estimate of the size of a cell or a mass of tissue, exposed to atmospheric oxygen concentrations, that can easily maintain this high metabolic rate.

Problem 1: If an insect flight muscle contains tracheal tubules which allow air to diffuse into all parts of the muscle, and the tracheal tubules make up 20% of the volume of the muscle, how large can the muscle be? Use the same metabolic rate as in the example just above.

Problem 2: Derive the expression for Ø for spherical symmetry and a zero-order chemical reaction (reaction rate independent of concentration).

Problem 3: ATP is generated in mitochondria at the base of a sea urchin sperm flagellum. Assume that a concentration of 10 mM ATP is maintained at the base of the flagellum. Assume that the flagellum is a cylinder of constant diameter = 0.2 microns. A flagellum beating at 30 cycles per second consumes 45,000 ATP molecules per second per micron of length. What is the maximum useful length of such a flagellum?