Diffusion in Biology

Module 3: Modelling with Randomness

The relation $\langle x^2 \rangle = 2Dt$ has a direct consequence for living cells: diffusion is fast over short distances and very slow over long ones. The numbers on this page come from (berg1993random?).

Time to diffuse a distance

Invert $\langle x^2 \rangle = 2Dt$ from Mean Squared Displacement to get the typical time to cover a distance $x$ along one axis:

\[ t \approx \frac{x^2}{2D}. \]

(berg1993random?) uses $D = 10^{-5}\ \text{cm}^2/\text{s}$ for a small molecule in water at room temperature. The code below reproduces his two examples: the width of a bacterium ($10^{-4}$ cm) and the width of a test tube (1 cm).

D = 1e-5  # cm^2/s, small molecule in water (Berg 1993, Chapter 1)

for label, x_cm in [("bacterium, 1 micrometre", 1e-4), ("test tube, 1 cm", 1.0)]:
    t = x_cm**2 / (2 * D)
    print(f"{label:25s} t = {t:9.2e} s = {t / 3600:8.2e} h")

bacterium, 1 micrometre   t =  5.00e-04 s = 1.39e-07 h
test tube, 1 cm           t =  5.00e+04 s = 1.39e+01 h

The first time is about half a millisecond and the second about 14 hours, as in (berg1993random?). The distance grew by $10^4$ and the time by $10^8$.

Consequences

There is no “diffusion velocity”. The ratio $\sqrt{\langle x^2 \rangle}/t = \sqrt{2D/t}$ depends on the observation time (berg1993random?).
Over micrometres diffusion moves small molecules within milliseconds. Over centimetres it takes hours, so transport over large distances needs flow or active mechanisms. Berg’s example is perfume: with $D \approx 10^{-1}\ \text{cm}^2/\text{s}$ in air, diffusion alone would take about a month to cross a crowded room, so convection does the work (berg1993random?).
A random walker explores a small region thoroughly and returns to the same points many times before it wanders off (berg1993random?). Chapter 3 of the same book computes how fast molecules diffuse to a receptor or a cell (“diffusion to capture”).
Swimming bacteria are self-propelled random walkers. Their motion is treated in (berg1993random?).

(britton2003essential?) covers biological motion from the modelling side, including taxis and travelling waves.

Exercise

A protein such as lysozyme has $D \approx 10^{-6}\ \text{cm}^2/\text{s}$ in water (berg1993random?). Estimate the time it needs to cross a cell of diameter $10\ \mu\text{m}$ and to travel along a nerve axon of length 1 m. What does the second number tell you about how neurons must transport proteins?

Solution

D = 1e-6  # cm^2/s
for label, x_cm in [("cell, 10 micrometres", 1e-3), ("axon, 1 m", 100.0)]:
    t = x_cm**2 / (2 * D)
    print(f"{label:22s} t = {t:9.2e} s = {t / (3600 * 24 * 365):8.2e} years")

cell, 10 micrometres   t =  5.00e-01 s = 1.59e-08 years
axon, 1 m              t =  5.00e+09 s = 1.59e+02 years

Crossing the cell takes about half a second. Diffusing along a 1 m axon would take over a century, so axons must use active transport. This is an order-of-magnitude argument; it ignores crowding and binding, which slow diffusion further.

--- title: "Diffusion in Biology" subtitle: "Module 3: Modelling with Randomness" format: html --- The relation $\langle x^2 \rangle = 2Dt$ has a direct consequence for living cells: diffusion is fast over short distances and very slow over long ones. The numbers on this page come from @berg1993random [Chapter 1]. ## Time to diffuse a distance Invert $\langle x^2 \rangle = 2Dt$ from [Mean Squared Displacement](mean-squared-displacement.qmd) to get the typical time to cover a distance $x$ along one axis: $$ t \approx \frac{x^2}{2D}. $$ @berg1993random [Chapter 1] uses $D = 10^{-5}\ \text{cm}^2/\text{s}$ for a small molecule in water at room temperature. The code below reproduces his two examples: the width of a bacterium ($10^{-4}$ cm) and the width of a test tube (1 cm). ```{python} D = 1e-5 # cm^2/s, small molecule in water (Berg 1993, Chapter 1) for label, x_cm in [("bacterium, 1 micrometre", 1e-4), ("test tube, 1 cm", 1.0)]: t = x_cm**2 / (2 * D) print(f"{label:25s} t = {t:9.2e} s = {t / 3600:8.2e} h") ``` The first time is about half a millisecond and the second about 14 hours, as in @berg1993random [Chapter 1]. The distance grew by $10^4$ and the time by $10^8$. ## Consequences - There is no "diffusion velocity". The ratio $\sqrt{\langle x^2 \rangle}/t = \sqrt{2D/t}$ depends on the observation time [@berg1993random, Chapter 1]. - Over micrometres diffusion moves small molecules within milliseconds. Over centimetres it takes hours, so transport over large distances needs flow or active mechanisms. Berg's example is perfume: with $D \approx 10^{-1}\ \text{cm}^2/\text{s}$ in air, diffusion alone would take about a month to cross a crowded room, so convection does the work [@berg1993random, Chapter 1]. - A random walker explores a small region thoroughly and returns to the same points many times before it wanders off [@berg1993random, Chapter 1]. Chapter 3 of the same book computes how fast molecules diffuse to a receptor or a cell ("diffusion to capture"). - Swimming bacteria are self-propelled random walkers. Their motion is treated in @berg1993random [Chapter 6]. @britton2003essential [Chapter 5] covers biological motion from the modelling side, including taxis and travelling waves. ::: {.callout-note} ## Exercise A protein such as lysozyme has $D \approx 10^{-6}\ \text{cm}^2/\text{s}$ in water [@berg1993random, Chapter 1]. Estimate the time it needs to cross a cell of diameter $10\ \mu\text{m}$ and to travel along a nerve axon of length 1 m. What does the second number tell you about how neurons must transport proteins? ::: ::: {.callout-tip collapse="true"} ## Solution ```{python} D = 1e-6 # cm^2/s for label, x_cm in [("cell, 10 micrometres", 1e-3), ("axon, 1 m", 100.0)]: t = x_cm**2 / (2 * D) print(f"{label:22s} t = {t:9.2e} s = {t / (3600 * 24 * 365):8.2e} years") ``` Crossing the cell takes about half a second. Diffusing along a 1 m axon would take over a century, so axons must use active transport. This is an order-of-magnitude argument; it ignores crowding and binding, which slow diffusion further. :::