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Modeling

Magnetosome can be used in different perspective and it shows a good variety in different application. Three applications has been modelled in different perspective. Let me show the great variety of using magnetosome:

1) Protein extraction:

Magnetosome can be used in microscopic point of view. We try to model his efficiency to bind with different proteins and use GFP-nanobody for immunoprecipitation. The main purpose of this modelling is to stimulate the binding dynamics of a fixed concentration of magnetosome and GFP-nanobody in different initial concentration of antigens.

Various conditions and parameters:

Fixed quantity	quantity	Volume of the mixture	1000ul

Parameters	quantity
Molecular Weight of magnetosome	6.89 x108 g
Number of GFP-nanobody per magnetosome	362
association rate constant kon(Marta H. Kubala†, 2010)	8.84x104M-1s-1
dissociation rate constant koff(Marta H. Kubala†, 2010)	1.24x10-4s-1

Condition	quantity
Amount of magnetosome	1.5mg
Weight of GFP-nanobody	negligible
Initial molarity of antigen (GFP)	Varying from 0 to 1.6E-6M
Initial amount of GFP:GFP-nanobody complex	0

First, the molarity of the magnetosome is needed to calculate since the amount of magnetosome and its molecular weight are known,

Molarity of Magnetosome=(1.5mg/6.89 x108 g)/(1ml)=2.18 x10-9M

There are 362 GFP-nanobody per each magnetosome, the molarity of GFP-nanobody is:

Molarity of GFP-nanobody = 2.18 x10-9M x 362 = 7.78 x10-7M

After that, a software called Simbiology is used in MATLAB and it help us to model and stimulate the dynamics of the association and dissociation between molecules. By constructing a model about the mathematics relationship between molecules. enmolecules,reaction process can be stimulated.

For Forward reaction (association) rate:

kon *[GFP-nanobody]*[antigens]

For backward reaction (dissociation) rate:

koff *[GFP-nanobody-antigens-complex]

Net reaction rate:

kon *[GFP-nanobody]*[antigens] - koff *[GFP-nanobody-antigens-complex]

Note: kon, koff are the reaction rate constant described in the parameters table.

By using the function of SimBiology, we stimulate the dynamic of the system with the initial concentration of antigen from 0M to 1.6E-6M with an interval of 2E-7M.

From figure 2, we can see that when the molarity of antigen below that of GFP-nanobody (7.78 x10-7M), it becomes the limiting reagent, and the final molarity of the nanobody-antigen complex equals the initial molarity of antigen, vice versa.

Another observation is that, as the molarity of antigen increase, the reaction (i.e. the formation of nanobody-antigen complex) goes equilibrium quicker. This can be explained by the increased forward reaction rate, which depends on the molarity of GFP-nanobody and antigen as well.

2) Microbial Fuel cell

In a Microbial Fuel cell, the chemical enrgy is transformed into the electrical energy through a cascade of electrochemical reaction. The mutated nitrogenase in Azobacter will produce hydrogen gas by the side reaction and break down into hydrogen ion due to the existence of hydrogenase. Alternatively the electrons can be transferred transferred to the oxidized mediator molecules that transfer them to the electrode. By using magnetosome, the distance between the bacteria and electrode will be decreased and it reduce the diffusion distance of the oxidized mediator. This approach increase the current density and increase efficiency of of generating electricity in Mircrobial fuel cell.

In this model, current density distribution in hydrogen-oxygen fuel cell is studied. It includes the ful coupling between the mass balances at the anode and cathode, the momentum balances in the gas channel, the gas flow in the porous electrodes, the balance of the ionic current carried by the mediator and an electronic current balance.

The fuel cell in the cathode and anode is counterflow and it shows that the hydrogen-rich anode gas is entering from the left. The electrochemical reaction in the cell are give below:

Anode: H2+ 2e--> 2H+

Cathode: 1/2O2+ 2e- ->O2-

This model includes different process that shows below:

• Electronic charge balance (Ohm’s law)

• Ionic charge balance (Ohm’s law)

• Butler-Volmer charge transfer kinetics

• Flow distribution in gas channels (Navier-Stokes)

• Flow in the porous GDEs (Brinkman equations)

• Mass balances in gas phase in both gas channels and porous electrodes (Maxwell-Stefan Diffusion and Convection)

Assume the Butler-Volmer charge transfer kinetics describe the charge transfer current density and the first electron transfer is used to be rate determining step, at the anode, hydrogen is oxidized to form hydrogen ion.

i0,a =the anode exchange current density (A/m2)

ch2 is the molar concentration of hydrogen

ch+ is the molar concentration of water

ct the total concentration of species (mol/m3)

ch2,ref and ch2,ref is the reference concentrations (mol/m3)

F is Faraday’s constant (C/mol)

R the gas constant (J/ (mol•K))

T the temperature (K)

η the overvoltage (V)

For the cathode:

At the anode’s inlet boundary, the potential is fixed at a reference potential of zero. At the cathode’s inlet boundary, set the potential to the cell voltage, Vcell. The latter is given by

where Vpol is the polarization. In this model, φeq,a Δ = 0 V and φeq,c Δ = 1 V , and you simulate the fuel cell over the range 0,2 V Vcell ≤ ≤ 0,95 V by using Vpol in the range 0.05 V through 0.8 V as the parameter for the parametric solver.

Results: The following figure shows the hydrogen mole fraction in the anode at a cell polarization of 0.5.V

The following figure shows the oxygen mole fraction in the cathode:

For the following figure, it shows power output as a function of cell voltage. The maximum power-output for this unit cell is about 940 W/m2

3) Water Treatment

Theoretically Power of Turbine:

Pth = ρ q g h

The theoretically power available from a flow of 1m^3/s water falling 100m:

Pth = (1000kg/m^3) (1m^3/s)(9.81ms^-2)(100m)

= 981 kw

Since the work done is equal to the power flow through 100m in 1s.

P=W/t

Therefore, W=981 kJ

After that, we would like to calculate the force created by the water flow. Since the work done is equal to the multiple of force and displacement that the water flow through.

W=Fs

Therefore, the force act on each magnetosome is

F= 981000/100=9810 N A straight wire electromagnet is needed to use to hold the magnetosome tightly with a strong magnetic force. The magnetic force produced by the electromagnet Fm is equal or bigger to the force created by the water flow Fwf since we need to hold the magnetosome tightly. Fm=Fwf Assume that the electromagnet with 1000 turns of wires and operate with 10A of current 1.5 meters from a piece of metal, Force = ((N x I)^2 x k x A) / (2 x g^2) The cross-sectional area of the electromagnet = 351 meters-square.