CONSTRUCTS

Well saturation Modeling

The design we made for our allergy detection kit is based on the emission of a GFP molecule. The way it is explained on other parts of the wiki are pretty detailed enough and it is pretty straight forward, but given that we are not only producing the allergens but also coating the well with the antibody some other problems arise. The question is how much scFv anti-His protein has to be coated onto the well to be enough for the plate reader to register a significant value? To answer this question, we must first find about the sensibility of different models of plate readers to GFP emission.

Afterwards, it is important to know how much concentrated our sample will need to be to coat the well as saturated as it is needed to have a fair response. To find this answer we did a geometrical analysis which considers the surface area of the bottom of the well, the projected shadow area of the protein to be coated and other parameters that will be listed step by step.

Making it short, we have a volume of scFv His in solution that will be coated into the bottom of the well, this coating will saturate a fraction of the bottom of the well and finally we have to take in account that not all of the molecules will be coated in the right orientation.

So first we have to define what well saturation is and how to calculate it. Well saturation can be defined as what fraction of the surface area of the bottom of the well is being occupied by the scFv His.

Where nact is the actual number of scFv His molecules and nmax is the maximum capacity of scFv His molecules that the well has.

The nmax is a constant that varies only in two cases, first case being when you are using another type of well plate, and second case being when coating another type of molecule different from scFv His (although, if another scFv is used, de changes are minimal), It is calculated using the surface area of the bottom of the well and the shadow projection area of the molecule as follows (Pettersen EF, 2004).

The nact depends on how many scFv molecules were on the coating volume at the start of the process, so in the end it depends on the concentration of scFv His molecules in the coating solution and the volume capacity of the well (Gasteiger E. H. C., 2005).

Another parameter that should be taken in account when doing this model is the fraction of the total molecules that would actually be coated with the active site spatially available (not facing bottom).

This parameter can be calculated measuring the total surface area of the molecule and the active surface area; finally, like when calculating the probability in a dice to get a desired number facing up, we calculate the probability of our active area not facing down. This probability is the complement of the active fraction. In other words, we are calculating the probability of the inactive area facing bottom and that will be the fraction of our coated molecules that will actually work for binding the His-tagged allergens.

For example the active area in our molecule is shown in red and the rest in gray.

In our case, we do know all of these parameters, so a simulation was run so that we could actually use the model.

Thanks to this model you can actually know the concentration you need in your sample in order to coat the well enough to have a readable GFP emission.

You will only need to know how much GFP emission you will like to read as a minimum. A calibration curve that relates fluorescence to number of your GFP molecules will tell you how much saturated your well needs to be (Gsat) taking in account the fraction of molecules that will be inactive due to its active area facing bottom.

Knowing this, you will only need to see how much you have to concentrate your sample of scFv His and concentrate it by a centrifugal filter. To help you in that procedure, we developed another mathematical model.

• f_sat=fraction of the area from the bottom of the well that is filled with scFv His molecules
• n_act=actual number of scFv His molecules in the well
• n_max=maximum number of scFv His molecules in the well
• 〖SA〗_well=Surface area of the bottom of the well
• 〖SPA〗_His=Area from the shadow projection of our scFv His molecule
• d_His=diameter of scFv His molecule
• 〖MW〗_His=Molecular weight of scFv His
• [scFv His]=concentration of scFv His in a solution
• N=Avogadro^' s number
• 〖Vol〗_well=volume capacity of a well
• f_act=active surface fraction of the coated molecules
• A_act=area of the active part of the molecule
• A_Total=Total surface area of the molecule
• G_sat=desired well saturation
• GFP need=amount of GFP molecules needed for a plate reader to detect

Concentration Modeling

Centrifugal filters work using centrifugal acceleration to create a pressure difference between the upper and the lower side of the tube. Centrifugal filters have different parameters that affect their functionality; these parameters include: membrane surface area, initial volume capacity, material the membrane is made of, and molecular weight cut off (MWCO). Also, the solution to be filtered has some effect on the filtration. The composition of the solution, its density, and viscosity, along with the operation conditions all has some effect on the process. Nevertheless, we did not find any model that described its function and decided to model it ourselves.

To model the functionality of a filter, and almost any other process, the important thing is a correct geometrical analysis. The geometrical arrangement of the filter tube 1) and what happens at the filtering area 2) is described at the following image.

Some of the parameters are variable and depend on time or filtrated volume; others could be considered as constant or even ignored. For example, the pressure that has effect on the flow across the filtration membrane it’s not constant; it depends on the amount of volume left to be filtered. Taking in account hydrostatic pressure and its differences along time is determinant for the precision of the model.

Then, taking in account that the volume inside the tube will be continuously filtered the pressure acting on the “not filtered yet” side of the filter, thanks to Pascal’s principle, can be described as following.

The height “h” of the liquid in the upper chamber of the tube depends on how many volume has not been filtrated yet “V1”. But, as it can be seen on the diagram, the geometry is irregular and the tubes do not come with the specification on the width, length, or depth on the filtrate part, so the function between the “h” and “V1”is a step function.

This function presents its difference at “V0” because this is the approximate amount of volume that we are calling “dead volume”, which represents the volume in the non-cylindrical challenge. At this dead volume, the relation between h and V1 is modeled just by a constant because of the lack of information. After the dead volume, at the V_0

This volume itself depends on the filtrated volume “V2” because, logically, as V2 decreases, V1 increases with a direct negative proportionality.

The interesting part comes when modeling V2. This part has to be analyzed carefully as many parameters are included. From the previous image where the filtering area was detailed, we can deduce that the force that lets the movement across the membrane occur is mainly the pressure change, and the forces that oppose to the movement are mainly the resistance from the membrane itself and the resistance created by the accumulation of matter in the surface of the membrane, called cake. So, step by step we will add our parameters to the model.

• The velocity at which V2 flows across the membrane is proportional to the area “A” and to the pressure difference P2-P1.



• Also, this flow experiments some resistance from the cake and the membrane itself, this makes it to be inversely proportional to the resistance.



• The resistance is composed by two different terms. First we have to consider the viscosity and the filter membrane, which opposes a constant resistance that depends on A to the flow. This resistance Rf includes the porosity of the membrane, the size of the pore (MWCO) and the material it is made of, being different for every kind of filter and independent from the solution to be filtered.



• Finally, the other term that opposes resistance is the formation of a cake. The cake is the accumulation of matter that happens through the filtration, this depends on the volume that has already been filtered V2, the viscosity, and a proportionality constant that includes the particles in the solution to be filtered, their density, their weight, and porosity.



• This leaves us with a final differential equation as following.



• To facilitate its usage, it is appropriate to include variables of concentration and magnification. The initial concentration of the desired solute “conci”, will be magnified by the filtration by a factor “xconc” depending on how much time the filtration occurred. Finally, the user can stop the filtration whenever the concentration at the end “conf” (which is the one of interest in our case) reaches its desired value, and this is determined as follows.

With the model already done, it is important to remember what does every variable mean and what weaknesses our model has.

Nomenclature

• P_1=Pressure acting over the filter membrane
• ρ=Solution^' s density
• a_c=Centrifugal acceleration
• h=Height of the solution at V_1
• V_1=Volume of the solution above the filter
• P_o=Atmospheric pressure
• V_0=Dead volume,or volume below the cylindrical part of the filter tube
• c=Proportionality constant that includes all geometrical relations in the irregular (non-cylindrical part of the filter tube)that relate V_0 to h
• r=Radius of the cylindrical part of the filter tube
• h_0=Dead height,or height below the cylindrical part of the filter tube
• V_2=Filtered volume
• A=Area of the filter membrane
• P_2=Pressure after the filter membrane
• R_f=Resistance to flow from the filter membrane. includes the porosity of the membrane, the size of the pore (MWCO)and the material it is made of
• μ=Viscosity of the solution
• k=Proportionality constant that relates the filtered volume V_2 to the resistance of the cake.It includes the particles in the solution to be filtered,their density,their weight,and porosity
• conci=initial concentration
• xconc=magnifying factor
• concf=final concentration

For the correct usage of this model, it is important to know its limitations and the physical assumptions that were made. First of all, some parameters were taken as constant. Density of the fluid, the area of the filtering membrane and viscosity were used as they didn’t change. We know density should change as some water is lost and the composition of the solution changes, as the solution becomes more concentrated. This same phenomenon has some effect on the viscosity. The area of the membrane is not constant; it changes when the volume V1 is below the dead volume V0, as the filter membrane covers all of the dead height h0. Also, with this comes the supposition that implies that all of the membrane’s area is receiving an equal amount of pressure, which is not true and ideally should be modeled with an integrated function, Another assumption is that the height h, ideally, we should be used in the form of a step function, but for this model, for simplicity and because our model simulator cannot incorporate step functions, we are not considering the dead volume or height. Finally, this model supposes that no protein of interest passes through the filter, it is important to make this assumption because it is not possible to predict exactly how much protein of interest will be lost even if the MWCO is known, as the protein structures can bypass this molecular weight limit.

Let’s say, for example, that we have our scFv His solution with an initial concentration of 0.0002mg/ml and we need to saturate our well to 0.4518. This means you will need to concentrate your sample until it gets to 0.001, this is 5 times as much as the original solution. Knowing this, you only have to run the model and look for when the parameter xconc reaches 5, this will give you the time needed (163 seconds in our example) and you will be ready to run your process!

The example described above is shown below using some real data. Fm, fkg are just some factors for unit conversion (Shacham, 2004).

• d(V2)/d(t) = A * A * (-1) * (P2 - P1) / (u * k * V2 + u * A * Rf)
• d(V1)/d(t) = -(A * A * (-1) * (P2 - P1) / (u * k * V2 + u * A * Rf))
• h = V1 / (3.14159265 * r * r)
• V1(0) = 5
• V2(0) = 0
• t(0) = 0
• A = 6
• r = 1.5
• P2 = 101325 / fm * fkg
• fm = 100
• fkg = 1000
• P1 = p * ac * h / fm + 101325 / fm * fkg
• p = 1.2
• ac = 3000 * 9.8 * fm
• u = 3 / 100
• k = 1000000
• Rf = 100000000
• t(f) = 180
• xconc = (V1 + V2) / V1
• conci = 0.0002
• concf = xconc * conci