MathBench- Australia Diffusion Part 2 December 2015 page 1 Cell Processes: Diffusion Part 2: Fick s Second Law URL: http://mathbench.org.au/cellular-processes/time-to-diffuse/ 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:
MathBench- Australia Diffusion Part 2 December 2015 page 2 If D = 0.00001 cm 2 /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 = Δx 2 / 2D Δx = 0.01 cm! Answer: (0.01cm) 2 / (2 * 0.00001cm 2 / sec)) = 5 sec How about 0.1 cm? Answer: (0.1cm) 2 / (2 * 0.00001cm 2 / sec)) = 500 sec What about 1 cm? Answer: (1cm) 2 / (2 * 0.00001cm 2 / 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 over inflation 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,
MathBench- Australia Diffusion Part 2 December 2015 page 3 1. Use the equation T = Δx 2 /2D 2. Put distance (delta x) on the x-axis 3. Put time (T) on the y-axis 4. Assume D = 0.00001 cm 2 /sec So, what does the graph look like? Remember that T = Δx 2 / 2D is a quadratic equation, analogous to y = ax 2 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
MathBench- Australia Diffusion Part 2 December 2015 page 4 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?
MathBench- Australia Diffusion Part 2 December 2015 page 5 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=Δx 2 /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 D (cm 2 /sec) Oxygen in air 1.0 x 10-1 Oxygen in water 1.8 x 10-5 Protein in water 1.0 x 10-6 Phospholipids in water 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*10-5 cm 2 /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.
MathBench- Australia Diffusion Part 2 December 2015 page 6 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*1.8x10-5 cm 2 /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*1.8x10-5 cm 2 /sec ) =1.6*10 8 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).
MathBench- Australia Diffusion Part 2 December 2015 page 7 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)
MathBench- Australia Diffusion Part 2 December 2015 page 8 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 µm 2 /min. Distance = 300 µm (300 µm) 2 / (2*11 µm 2 /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 page 9 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.