Team:Paris Bettencourt/Modeling/Assisted diffusion
From 2011.igem.org
Assisted diffusion
Introduction to the model
When the cell "start the communication", a flow of phospholipids of the membrane can pass from the cell that have the highest membrane tension to the other one. To pass from one cell to another, phospholipids run around the tube.
The newly arrived phospholipids change the membrane tension of the bacterium and, under our hypothesis of constant volume, which is justified by the sugar layer that "forbids" the bacterium to grow. As a consequence they change the internal Laplacian pressure of the bacterium.
where R is the radius of bacterium, τ is a membranian tension.
A simple analogy of clothesline can help to understand what is happening. You need to pull more your clothesline to put more clothes on it. When you pull the rope by two ends you create a tension. More this tension is more weight (pressure) you can put on the rope.
All this will lead to establish the pressure difference at the tube extremities and we will get a Poiseuil flow. Constituents diluted in water will move from one cell to another unidirectionally and faster than simply diffusing. This is a fast process that we have named the "assisted diffusion"
When membranes behave like a 2D Van der Waals fluid
Trapped between the osmotic pressure and the cell wall, phospholipids particles are moving in a double layered environment. Their motion is constricted into a two dimensional motion, following the shape of the membrane. At the scale of the phospholipid, the motion can be approximated to a motion in 2D. Using statistical physics approach, we can say that the phase space has 4N dimensions: two dimensions of impulsion and two dimensions of coordinate for each particle(N is the number of particles in the system).
Can we describe the interaction between two phospholipids in the membrane? The question is complicated but we will demonstrate that by describing properly the phospholipid, it is reasonable to fit a Lennard-Jones potential energy to this interaction. The energy of binding is smaller and the speed of sliding of one phospholipid against another is slower than the internal vibration of the chemical bounds and the internal conformers rotation of the CH2 tails. Though, in Intermolecular Forces by Israelachvili(to be completed) the author shows that a phospholipid trapped in a bilayer can be aproximated by a section of cone. The section of cones is an individual, and the others are section of cone joining the previous one. The shape of the sections are giving the shape of the global structure.
Speaking in term of energy, once we have these cones, it is reasonable to fit a Lennard-Jones potential to this interaction. If the cones are interpenetrating, a sterical repulsion keeps the molecules apart, and the hydropatic interaction that joins one lipid with the other acts as an attractive force keeping the coherence of the membrane. We will discuss later about the value of the two coefficients.
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 experiments were about 100 nm to 1 µm.
Let's take for example a tube 0.2 µm long. That means that 2.5*of the surface of the tube will be used in the process. We can imagine it splited in two stages: the first is the establishment of contact between two bacteria, for that 1 surface of the tube will be used. The remaining 1.5 surface of the tube will be used to transport the liquid inside the tube. 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 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 then that some molecules are produced in the first cell and they are uniformally 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 to be detected.
The membrane tension calculation
Considering one membrane a sphere of the radius R, on which the phospholipid double layer is uniformally distributed, we can find approximately a part of the whole sphere surface per one phospholipid :
Back to the classical physics
To deduce the force exercising on the tube from membranian pressure, please refer to an explanation figure below :
Here T is a surface tension, N is a resulting force of the area of the tube attachment to the bacterium, Δα or α is the half of the angle with a vertex in the center of the bacterium and between the tube extremities. Only this zone is giving a non-compensated contribution to the resulting force.
So the force exercising on the tube is given by the formula :
where r is the tube radius, R is the bacterium radius.
At first approximation we can consider the process a serie of quasi equilibrium states, that it is rather slow that acceleration of the membrane is zero. We find the speed of the tube membrane at every moment of time out of the second Newton's law :
where ΔF is the difference of the forces applied to the extremities of the tube.
The number of phospholipids in two bacteria will change according to these equations :
where N1, N2 are numbers of phospholipids on the membrane of the first and the second bacterium corrispondingly, ΔS is the surface per one phospholipid one the tube.
Cutting the whole period of time of our process into laps of time dt we should solve the Poiseuille equation each time reevoluating the difference of pressure :
The Poiseuille equation :
where L is the total length of the nanotube, r is the distance from the center line of the nanotube, R is the radius of the nanotube.
The flow of the liquid passed through the nanotube is :
By integrating the last formula in time we will get the mass of the liquid passed from one bacterium to another :
where T is the time of the process which we can estimate like...
Getting the good parameters
It will take some time, but once you have the equation, from the size of a vesible in equillibium, when we know the number of phospholipids. The viscosity has probably been estimated by FRAP techniques. For the rest, we will find out.
Analysing the results
Can this process be an efficient way for passing molecules?
Conclusion