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Team:Paris Bettencourt/Modeling/Assisted diffusion/Back of the envelope calculation
From 2011.igem.org
Is this phenomenon important? Back-of-the-envelope calculation
Let's start with a rough estimation of the flux transmitted through a nanotube. We assume that 5% of the surface area of one bacterium could be transmitted through the nanotube to a second bacterium. An order of magnitude of the transmitted surface is 10-14 m2. A nanotube of the radius 100 nm (that is the mean value of radius taken from the article) with this surface would be 0.5 µm long. The nanotubes observed in Dubey and Ben-Yehuda's experiments were about 100 nm to 1 µm. Let's take for example a tube 0.2 µm long. That means that a total surface participating in the process is equivalent to 2.5*surfaces of the tube (the tube itself and the membrane parts adjacent to the tube). We can imagine the process to be split in two stages: the first is the establishment of contact between two bacteria, for that the equivalent of 1 surface of the tube will be used. The remaining 1.5 surface equivalents of the tube will be used to transport the liquid inside the tube, as it will be pulled along with the membrane. As the diameter of the tube is of order of 100 nm, we can neglect the complexity of the fluid movement inside. In our model we will consider the fluid to be moving with the membrane. So the volume of transmitted liquid is about 10-20 m3, whereas the cell volume is 5*10-18 m3. We consider that some molecules are produced in the first cell and they are uniformly distributed. With 10'000 molecules in the first cell, we can expect about 200 molecules transmitted through one nanotube to another cell. That number of molecules seems to be sufficient for detection.Click here to come back to Assisted diffusion section.
