MathBench- Australia
Diffusion Part 2
December 2015
page 1
Cell Processes:
Diffusion Part 2: Fick’s Second Law
URL: http://mathbench.umd.edu/modules-au/cell-processes_diffusion_2/page01.htm
Fick’s Second Law
Learning Outcomes
After completing this module you should be able to:
 Explain how the rate of movement of a substance is proportional to the distance squared
(Fick’s Second Law).
 Measure the time it takes for a substance to move from one area to another as a function of
distance and the permeability of the medium.
It’s about “time to diffuse”
Fick’s first law tells us HOW MUCH flux to expect. Fick’s second law gets into more detail, telling
us how fast the concentration is changing at any given point in space. This law involves both rates of
change with respect to space (position) and time and takes the form of something called a
“partial differential equation”, but we won’t need to go into that much detail here.
If this sounds complicated … it is. We won’t go into how this law works, except to say that when you
integrate over time, you get a very useful (and easy to understand) result: a formula for how much
time diffusion takes over a specified distance:
Assuming you know a diffusion coefficient, this formula allows you to determine how much time is
necessary (on average) for particles moving by diffusion to traverse any given distance.
Try it out:
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Diffusion Part 2
December 2015
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If D = 0.00001 cm2/s (for oxygen through water), how long would it take the oxygen to
diffuse 0.01 cm below the surface of a still pond?


Remember: T = Δx2 / 2D
Δx = 0.01 cm!
Answer: (0.01cm)2 / (2 * 0.00001cm2 / sec)) = 5 sec
How about 0.1 cm?
Answer: (0.1cm)2 / (2 * 0.00001cm2 / sec)) = 500 sec
What about 1 cm?
Answer: (1cm)2 / (2 * 0.00001cm2 / sec)) = 50,000 sec
Any fish hanging out at 0.1 cm below the water surface would die from lack of oxygen if they had to
rely on diffusion to bring them oxygen – – luckily there are also water currents.
A Very Useful Formula
On the last page, every time we multiplied distance by 10, time to diffuse increased 100 fold!
10 times as far --> 100 times as much time
What’s going on here? If time to diffuse were directly proportional to distance, then 10 times the
distance would require 10 times as long -- not 100. This overinflation is like if you went to
McDonald’s to buy 10 hamburgers and they charged you 100 times (not 10 times) the price of a single
hamburger. Not fair? Well, that means time to diffuse is NOT directly proportional to distance.
Time-to-diffuse increases dramatically
How does a graph of time to diffuse as a function of distance look? In other words,
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Diffusion Part 2
December 2015
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1. Use the equation T = Δx2/2D
2. Put distance (delta x) on the x-axis
3. Put time (T) on the y-axis
4. Assume D = 0.00001 cm2/sec
So, what does the graph look like?
Remember that T = Δx2 / 2D is a quadratic equation, analogous to y = ax2 and so takes the shape of
a parabola. In this case, the coefficient “a” is represented by 1/2D. Here is the graph, using D =
0.00001.
Notice that as the distance increases a little, time increases a lot. We say that:

“time to diffuse increases with the square of distance”, or equivalently,

“distance diffused increases with the square root of time”.
General Functions and Diffusion
Before we leave this section, we just want to stress one REALLY, REALLY important concept.
All of the graphs and functions that we have shown should already be familiar to you from your
general maths courses. Just because they are formulated here in reference to flux, diffusion and
gradients, doesn't mean they have any special properties or rules. All the rules you have learned about
graphing and manipulation of these general equations and how those functions look on a graph apply
here. These same equations come up over and over in biology, physics, chemistry and any other
science you take. The more familiar they are and comfortable you get recognising them, the easier all
this maths stuff will become.
So, one more time, we're going to show you the
1) general equations (that you should recognise from your high school maths classes!),
2) the specific formulation for its application to diffusion, and
3) the graph of the associated function
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Diffusion Part 2
December 2015
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Fick's First Law as a
function of
GRADIENT
Fick's First Law as a
function of DISTANCE
Fick's Second Law, or
Time to Diffuse
Biological Applications
So how does this really apply to biology??? Does it apply to biology at all??
We offer two examples of how the equations and mathematical approaches can help us understand
biological concepts. First, we talk about why rhinos (and other vertebrates) have lungs, then we show
how we can use maths to determine how quickly we can fight off a cold.
Why do rhinos have lungs and amoebas don't?
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This sounds like a question for kindergarten – why don’t amoebas have lungs? – That’s because
they’re too small! However let’s think about it a little more. What is the function of a lung?bringing oxygen into the body and expelling carbon dioxide. When I contract my diaphragm, my
lungs expand and air (with oxygen) gets sucked in. When I relax my diaphragm, the lungs contract
and air gets pushed out.
Another way to say this is that gases move into and out of lungs through bulk flow (the details are
different for other animals, who may have lungs, gills, spiracles, whatever). Do amoebas need to
exchange gases? Of course, they respire just like the rest of us. So why don’t amoebas require bulk
flow to exchange gases?
OK, you’ve probably figured out the answer, having just read an entire module on diffusion. Amoebas
are small, oxygen and carbon dioxide are small molecules, and we’ve said that diffusion works well
for small molecules over short distances. How small and how short? Recall that
T=Δx2/2D
A humongous amoeba might measure as much as 0.02 cm in diameter. That means distance from the
outside to the centre of the cell is at most 0.01 cm.
And here are some representative values of D:
Molecule and Medium
Oxygen in air
Oxygen in water
Protein in water
Phospholipids in water
D (cm2/sec)
1.0 x 10-1
1.8 x 10-5
1.0 x 10-6
1.0 x 10-8
We’ll use 1.8 x 10-5 as the diffusion coefficient for calculations in this section.
How long (at most) should it take oxygen to diffuse throughout the amoeba?


What is the value of x, the distance from the outside of the cell to the center of the cell?
If you are having problems on your calculator, work through the numerator first, then the
denominator. Then solve for T.
Answer: T = ΔX 2 / 2D = (0.01cm)2 / (2*1.8*10-5cm2/sec ) = 2.8sec
Diffusion is efficient in small organisms, but not big ones
So even a huge amoeba can diffuse oxygen through its body in about 5 seconds. If I could do that, I
probably wouldn’t bother with lungs either. A similar calculation shows that carbon dioxide can also
be diffused out of the cell without assistance.
Lots of other, larger animals do gas exchange without lungs – but they tend to be large in only one or
two dimensions, so that gases can diffuse through the smaller third dimension. Thus, flatworms,
nematodes, and many other kinds of worm-like animals don’t need complicated respiratory systems.
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How long would it take for oxygen to diffuse from the outside to the middle of a
mouse if its body has a 1 cm radius (Please, just ignore for the sake of this simplistic
argument that the mouse is more irregularly shaped than a cylinder and contains
numerous barriers to diffusion!!)

(1 cm)2 is 1000 times more than (0.01cm)2 in the amoeba calculation
Answer: T = ΔX 2 / 2D = (1cm)2 / (2*1.8x10-5cm2/sec ) = 27 778 sec = 7.7 hours hours
How about a rhino whose body had a radius of 75 cm?

(75cm)2 is 6625 times as big as the (1cm)2 in the mouse calculation.
Answer: T = ΔX 2 / 2D = (75cm)2 / (2*1.8x10-5cm2/sec ) =1.6*108 sec = 43 402 hours =
5 years
Whoah! Even this very tiny mouse cannot survive on diffusion alone, and with a rhinoceros, the
question is clearly preposterous!
Why we have lungs
So does that mean that diffusion has no place in gas exchange of larger organisms? Not at all, once
air is in the lungs, gases get exchanged through...none other than diffusion. The lung is
subdivided, ultimately, into millions of tiny sacs called alveoli, each of which has a thickness of only
0.5 to 1 micrometre. Each alveolus is surrounded by capillaries, and it takes blood about 1 second to
pass through these capillaries.
Let’s do the maths:
Plenty of time!
However, when you get a nasty cold, or a case of pneumonia, the lung tissue can swell and thicken
enough to endanger your ability to breathe!
There’s a second place where diffusion is vital in gas exchange: capillaries are the very thin, small
blood vessels through which gases are exchanged with cells. The cells making up the walls of the
capillaries are around 1 micrometre thick, so gases can diffuse directly through them in the same
amount of time as above (0.00028 seconds).
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Incidentally, nutrients -- such as glucose, amino acids and lipids -- and metabolic wastes also diffuse
to and from the capillaries. Capillaries in different parts of your body even differ in their permeability
– that is, the diffusion coefficient D varies. So, entire proteins can diffuse in and out of liver cells,
while almost nothing gets to or from the brain cells except oxygen and carbon dioxide.
How many macrophages does it take to kill a virus?
Some organisms move around more or less at random, so we can use diffusion to "model" how
these cells move. This works fine, as long as you accept the assumption / simplification that these
organisms are mindless little robots that walk around changing directions randomly. Sometimes this is
a reasonable concession to make in exchange for the simplicity and power of the diffusion model,
other times it is not. It’s a choice that you, the person implementing the model, need to make.
Here is an example of using diffusion to model a cell moving (adapted from Essential
Mathematical Biology by N. F. Britton):
The macrophage is a specialised cell that defends that body against foreign materials. Within the
lungs, macrophages engulf inhaled viruses and digest them. Macrophages move in a way that is very
similar to diffusion – they ooze along at a stately 3 micrometres per minute, but change directions
more or less randomly about every 5 minutes.
Consider an alveolus (diameter = 300 micrometres) that contains a single macrophage
and a single virus. Assume the macrophage and virus start at opposite ends of the
cell and the macrophage moves in a straight line (not similar to diffusion) towards the
virus. How long will it take for the macrophage to arrive?


Remember from physics: speed = distance / time!
Moving that equation around....we get: time = distance / speed!
Answer: time = total distance/distance per minute = 300/3 = 100 minutes
…..BUT….(see next page)
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December 2015
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You need a lot of macrophages - or one smart one!
So. It would take about an hour and a half for the macrophages to meet with and devour the hapless
bacteria. Except that the macrophage does not travel in a straight line. Instead it goes in a variety of
random-looking directions, resulting in a diffusion coefficient of 11 micrometres 2 / minute.
Therefore, the real time to viral annihilation is more like:
Assuming that the macrophage’s movement is similar to diffusion, how long will it
take the macrophage to reach the virus (on average)?
D = 11 µm2/min. Distance = 300 µm

(300 µm)2 / (2*11 µm2/min) = 4500 min
Answer: About 3 days.
You might say that this calculation is not realistic – even a macrophage does not move completely at
random. You would be right, the macrophage can home in on the virus to some extent (called
chemotaxis). Nevertheless, this simple model points out something important:


either you need many macrophages
or the macrophages need to be able to direct themselves towards the viruses.
As it happens, the answer in this case is a little of both.
There are many other systems, particularly in ecology, that use diffusion as a simple model of
organismal movement. The organisms modeled include forests (even though the trees don’t move
the forest will “move” via seed dispersal), insects, and even larger animals like rabbits. Diffusion is
one way that biologists are looking at the likely effects of large-scale stressors like global warming –
can oak trees disperse north fast enough to escape the pressures of a warmer world? However, these
models are mathematically more complicated than the diffusion we have been talking about: the
organisms do not only move around, they also reproduce.
Summary

In diffusion, the time to diffuse across a distance is given by ,
so that time to diffuse increases with the square of distance (consequence of Fick’s Second
Law).


Diffusion is efficient only at very short distances. If organisms need to transport materials
over longer distances, they need mechanisms such as lungs and circulatory systems.
Diffusion (and all the maths that goes along with it) can be used as an analogy to understand
the movements of cells and more complex organisms.
MathBench- Australia
Diffusion Part 2
December 2015
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Learning Outcomes


You should now be able to:
Explain how the rate of movement of a substance is proportional to the distance squared (Fick’s
Second Law).
Measure the time it takes for a substance to move from one area to another as a function of distance
and the permeability of the medium.