Team:TU Delft-Leiden/Modeling/Curli

Curli Module

The goal of our project for the conductive curli module is to produce a biosensor that consists of E. coli that are able to build a conductive biofilm, induced by any promoter, in our case a promoter that gets activated in the presence of DNT/TNT. The biofilm consists of curli containing His-tags that can connect to gold nanoparticles. When the curli density is sufficiently high, a dense network of connected curli fibrils is present around the cells. Further increasing the amount of curli results in a conductive pathway connecting the cells, thereby forming conductive clusters. Increasing the amount of curli even further, sufficiently curli fibrils are present to have a cluster that connects the two electrodes and thus have a conducting system.
The goal of the modeling of the curli module is to prove that our biosensor system works as expected and to capture the dynamics of our system. So, we want to answer the question: "Does a conductive path between the two electrodes arise at a certain point in time?" However, we not only want to answer the question if our system works as expected qualitatively, but we also want to make quantitative predictions about the resistivity between the two electrodes of our system in time.

The conductive curli module has different dynamics on different length scales:

• The behavior of the system on the gene level, that is the dynamics of the activation of the promoter and the dynamics of the production of proteins needed for curli growth.
• The behavior of the system on the cell level, that is the curli production of each cell in time.
• The behavior of the system on the colony level, that is the change of the resistivity between the two electrodes of our system in time.

To capture the dynamics of our system, we have implemented a three-layered model, consisting of the gene level layer, the cell level layer and the colony level layer.
The gene level layer is used to determine characteristic parameters that will be used in the cell level layer. Subsequently, the cell level layer is used to determine characteristic parameters that will be used in the colony level layer. Lastly, the colony level layer is used to determine if our system works as expected, ie. determine if a conductive path between the two electrodes arises at a certain point in time, and to determine the change of the resistivity between the two electrodes of our system in time. A figure of our three-layered model is displayed below.

summary of the conclusions

Gene Level Modeling

We will start with the modeling of the expression of curli on the gene level. Proteins that are dedicated to the curli formation are CsgA/B/D/E/F/G [1]. CsgA is the main building block of the curli. When produced, this protein is secreted out of the cell by the CsgEFG complex. In the absence of CsgB, there is no curli formation, since the CsgA proteins remain unpolymerized. CsgB is the starting block of the curli fibrils and connect the cell membrane to the first CsgA protein in the curli fibril. Once CsgB is located on the outside of the cell surface, the CsgA can polymerize onto the starting curli fibril.
In the constructs we made in the wet lab, CsgA is continuously being produced. However, in our constructs the CsgB gene is placed under the control of a landmine promoter, activated by either TNT or DNT reference to landmine. So, when the cells get induced by TNT or DNT, CsgB protein production will get started and CsgA will already be present in the system, as CsgA is continuously being produced.

Extensive Gene Level Modeling

to be written, low priority

Simplified Gene Level Modeling

Though the model described above, providing that all rates are known, has a more accurate (though still simplified) representation of the curli assembly system, we have chosen to decrease the complexity further to the bare essentials, as most of the production rates cannot be found in literature. Measuring the accurate rates in the wet lab is, within the scope of this project, infeasible and therefore, we constructed a model that only includes the rate limiting step of the system as this will mostly determine the dynamics of the system.
First of all, we investigated if the diffusion of the CsgA and CsgB proteins to their final destination is the rate limiting step in curli formation. From the literature and the wet lab, we know that the system response to the induction by TNT or DNT is in the order of hours [reference]. If diffusion is the rate limiting step, it would mean that CsgA and CsgB proteins would pile up inside and outside the cell, because it takes a long time for them to travel to their final destination, the end of a growing curli fibril and the outer membrane, respectively. A quick calculation shows that after one second, the displacement of a spherical particle with radius \(r = \ 10 \ nm\) is 6.6 μm due to Brownian motion in liquid water at room temperature using equation 1; many times the bacterial radius! Hence, we conclude that diffusion is not rate limiting [4].

$$ \bar{x}^2 = \ \frac{k_b T t}{3 \pi \eta r} \tag{1}$$

What we do expect to be the rate limiting step for curli formation is the large amount of CsgA and CsgB proteins that have to be produced. Hence, we expect the production rate of one of these proteins to be the rate limiting step. Instead of including the intermediate steps, we have implemented the production of the CsgA and CsgB proteins with one reaction and associated production rate each. These rates have to be measured in the lab. We will use the following system of equations:

$$ \emptyset \xrightarrow{p_{A}} \ CsgA_{free} \tag{2} $$ $$ \emptyset \xrightarrow{p_{B}} \ CsgB \tag{3} $$ $$ CsgA_{free} + \ CsgB \xrightarrow{k} \ CsgA_{curli} + \ CsgB \tag{4} $$

Reactions 2 and 3 represent the production of CsgA and CsgB proteins, respectively. Equation 4 represents the growing of a curli fibril, where a curli fibril reacts with a free CsgA protein to become part of the curli. In reality, this reaction only happens at the end of the curli fibrils. In our model, we assume a homogeneous concentration of all the substances and we cannot discriminate between curli subunits. It is theoretically possible to model the system as an infinite amount of possible reactions that can take place to increase a curli fibril with length i to length i+1 at rate k [7]. However, we are merely interested in the growth rates of the curli, since the distribution of the curli length will follow from the model at the cell level. Therefore, we decided to model the growing of curli at the gene level as reaction 4. We assume that each CsgB protein is the start of a curli fibrils, thus the concentration of CsgB equals the concentration of curli. We can do this, because we showed that the diffusion of CsgA and CsgB proteins to their final destination is not the rate limiting step. Therefore, nearly all the CsgB proteins will be the beginning of a curli fibril in reality and our assumption is valid.
So, in reaction 4 we let a free CsgA protein react with a curli fibril to a CsgA protein that is part of that curli and the curli itself again, as it is immediatily again availible for the next reaction with a free CsgA protein to grow even more. Therefore, curli growth is dependent on the rate k and the concentration of \(CsgA_{free}\) and CsgB.

Writing reactions 2-4 into differential equations results in:

$$ \frac{d}{dt} [CsgA_{free}] = \ p_{A} - \ k [CsgA_{free}][CsgB] \tag{5.1} $$ $$ \frac{d}{dt} [CsgB] = \ p_{B} \tag{5.2} $$ $$ \frac{d}{dt} [CsgA_{curli}] = \ k [CsgA_{free}][CsgB] \tag{5.3} $$

Fortunately, this system can be solved analytically. To do this, we need the initial conditions. Say the CsgB promoter is activated at \(t= \ 0\). At this time there are no curli present, so \([CsgB]|_{t=0} = \ [CsgA_{curli}]|_{t=0}= \ 0\). However, the CsgA promoter is continuously active, so we expect to have an initial concentration \(A_0\) of free CsgA proteins at time \(t= \ 0\).

The solution to equation 5.2 is trivial:

$$ [CsgB] = \ p_B t \tag{6}$$

Substituting this into equation 5.1 results in:

$$ \frac{d}{dt} [CsgA_{free}] = \ p_{A} - \ K p_B [CsgA]t \tag{7} $$

It can easily be proven that a first order differential equation of the form

$$ y(t)' + \ f(t)y(t) = \ g(t) $$

has a solution of the form

$$ y(t) = \ e^{-F(t)} \int{g(t) e^{F(t)} dt} + \ y_0 e^{-F(t)} $$

where \(F(t)= \int{f(t) dt}\). In our case, \(f(t) = \ k p_B t\) and \(g(t) = \ p_A\). This yields equation 8.

$$ [CsgA_{free}] = \ p_A e^{\frac{-k \ p_B t^2}{2}} \int{e^{\frac{k \ p_B t^2}{2}} dt} + \ C_{1} e^{\frac{-k \ p_B t^2}{2}} = \ p_A e^{\frac{-k \ p_B t^2}{2}} \int_{0}^{t}{e^{\frac{k \ p_B \tau^2}{2}} d\tau} + \ C_{2} e^{\frac{-k \ p_B t^2}{2}} \tag{8} $$

One with a keen eye may recognize the Dawson function (equation 9):

$$ D_+ (x) = \ e^{-x^2 } \int_{0}^x{e^{y^2} dy} \tag{9} $$

As in our case, \(x^2 = \ k p_B t^2 \) and \(y^2 = k p_B \tau^2 \) and equation 10 obtained.

$$ [CsgA_{free}] = \ \frac{p_A D_+ (t\sqrt{\frac{k \ p_B}{2}})}{\sqrt{\frac{k \ p_B}{2}}} + \ C_{2} e^{\frac{-k \ p_B t^2}{2}} \tag{10}$$

Using the boundary condition \([CsgA_{free}]|_{t=0}= \ A_0\), the expression for the concentration of free CsgA proteins becomes:

$$ [CsgA_{free}] = \ \frac{p_A D_+ (t\sqrt{\frac{k \ p_B}{2}})}{\sqrt{\frac{k \ p_B}{2}}} + \ A_0 e^{\frac{-k \ p_B t^2}{2}} \tag{11}$$

Now, we can fill in equations 11 and 6 into equation 5.3, which gives us equation 12.

$$ \frac{d}{dt} [CsgA_{curli}] = \ k p_B t \frac{p_A D_+ (t\sqrt{\frac{k \ p_B}{2}})}{\sqrt{\frac{k \ p_B}{2}}} + \ A_0 e^{\frac{-k \ p_B t^2}{2}} \tag{12} $$

For the parameters \(p_{A}\), \(p_{B}\), \(k\) and \(A_0\), we have estimated the following values explain:

Table 1: Parameters used to obtain quantitative results from the analytical solution for the curli production.

Parameters	Value	Unit
\(\boldsymbol{p_{A}}\)	\(1.0 \cdot 10^{-10}\)	\(\frac{1}{Ms}\)
\(\boldsymbol{p_{B}}\)	\(3.4 \cdot 10^{-12}\)	\(\frac{M}{s}\)
\(\boldsymbol{k}\)	\(4.0 \cdot 10^{4}\)	\(\frac{1}{Ms}\)
\(\boldsymbol{A_0}\)	\(6.0 \cdot 10^{-6}\)	\(M\)

Plotting equation 12 with the parameter values in table 1 yields the graph shown in figure 1. insert caption

Figure 1 shows a steady production of CsgB. \(CsgA_{curli}\) concentration at \(t= \ 0\) is zero as expected, since there is no CsgB at that point. In the next few hours, \(CsgA_{curli}\) concentration peaks. We think that this is due to the high concentration of \(CsgA_{free}\) that is present at \(t= \ 0\). In figure 2, curli growth as function of time is plotted for different initial concentrations of \(CsgA_{free}\). insert caption

We conclude the following from figure 2:
Firstly, as expected, curli growth stabilizes to a rate equal to \(p_{A}\) after approximately 2 hours, independent of the initial concentration of \(CsgA_{free}\), \(A_0\).
Secondly, increasing the initial concentration of \(CsgA_{free}\), \(A_0\), increases the height of the peak. Even with zero initial \(CsgA_{free}\) concentration, a small peak can be found at one hour. This is a consequence of \(CsgA_{free}\) build-up when the CsgB concentration is still very small.
Thirdly, during the first two hours, few CsgB proteins are present in the system. We therefore expect that the length of the curli fibrils that started in the first few hours are much longer than the fibrils that started at later times.

write some conclusion about rate limiting step and which information we gained from this level and what will be used in the subsequent levels

Cell Level Modeling

Now that the growth rate of curli and production of CsgB protein as function of time is obtained, the conductivity as a function of time can be computed. The relevant length scale is the cell length, or the micrometre scale. The approach we used for this is relatively simple:

• We discretize the amount of curli subunits (\(CsgA_{curli}\) in the gene level model) and CsgB proteins that have to be added for each time step.
• At each time step, we add more curli subunits to growing curli fibrils. Also, we add more new curli fibrils to the model.
• From the density of the curli fibrils around the cell as a function of the radius, we calculate the conductive radius of the cell.

Discretization of Gene Level Model

We have discretized equations 5.2 and 12 in N time steps. These give the expected number of new CsgB proteins and curli subunits for each time step, as we plotted the solution of these two equations in figures 1 and 2. From these figures we determine the expected number of new CsgB proteins and curli subunits for each time step. However, a fundamental assumption in deterministic modeling is that the concentration is continuous. In reality, the amount of added curli subunits is discrete, since we cannot add half a curli subunit.
Furthermore, in the gene level model we did not take into account the statistical variation of gene transcription and adding of curli subunits; sometimes less and some times more curli subunits are added with respect to the expected value. To include this in the cell level model, we drew the amount of new curli subunits from a Poisson distribution where λ equals the expected amount of added subunits.
So, for each time step we now have \(B_n\) new CsgB proteins and \(C_n\) new curli subunits, where \(C_n\) varies for each time step, as it is drawn from a Poisson distribution. A assumption of this distribution is that the time at which a new curli subunit is added, is uncorrelated to the time at which the previous curli subunit was added, we think this is a fair assumption. Note that the cell level model we made, accounts for the stochasticity of adding curli subunits, but not for the stochasticity of gene expression, so for the production of CsgB protein. The value \(B_n\) and the Poisson distribution are determined from figures 1 and 2. say something about the time steps, how much time represents each step and determine Bn and Cn from figures

Building the Curli Fibrils

Firstly, \(B_n\) CsgB proteins are added to our model that mark the starting points for new curli fibrils. These new curli fibrils are located at random points on a sphere with radius r, which represents the cell. The radius r is chosen such that the volume of the cell is\(\ \sim 1.1 \ \mu m^3\) [5]. A CsgB protein is modeled by a line of length 4 nm that points radially outward, perpendicular to the cell surface [source]. In reality, the distribution of CsgB on the cell surface is not uniformly distributed [6]. However, we assumed uniformly distributed CsgB to keep our model prehensile. This is a point that may be used to further improve the model.

Next, \(C_n\), which is drawn from the Poisson distribution, where λ equals the expected amount of added curli subunits, new curli subunits are added to curli fibrils by repeating the following process \(C_n\) times:

• Firstly, a random curli fibril is selected, e.g. curli number k. A curli fibril is represented by a 3 (the x, y and z coordinates) by l+1 matrix, where l is the amount of curli subunits of the curli fibril and the origin is chosen to be the center of the sphere. Thus, by storing the ending coordinates of each curli subunit, we know the starting and end coordinates of each curli subunit. The curli subunits are modeled by a line of length 4 nm [source].
• Secondly, the polar angle in spherical coordinates of the last curli subunit is computed, \(\theta_{1}\).
• Thirdly, the new curli subunit has a small angular deviation with respect to the previous one. This polar angle \(\theta_{2}\) is chosen from a Gaussian distribution with parameters N(0,σ). σ is chosen such that the persistence length, the distance over which a fibril has bend by \(90^{\circ}\) and has ‘lost’ its directional information, is 4 µm. The azimuthal angle ϕ is completely random between 0 and 2π radians, and chosen from an uniform distribution.
• Fourthly, for the new curli subunit for which we determined \(\theta_{2}\) and ϕ, the polar angle is determined to be \(\theta_{1} + \theta_{2}\). We now know the length of the new curli subunit (4 nm), its polar angle and its azimuthal angle. Subsequently, we add it to the previous curli subunit of the fibril and calculate the ending coordinate of the added curli subunit from its length, polar angle and azimuthal angle and the ending coordinate of the previous curli subunit. This calculated ending coordinate of the added curli subunit is stored in the matrix that represents the curli fibril.

The angular deviation σ is a critical parameter in our model. Increasing this value increases the flexibility of our curli, where decreasing this value increases the stiffness of the curli. This is shown in figure 3. If the length of one subunit is 4 nm and the total persistence length is 4 µm, then \(\sigma = \ 3.47^{\circ}\). Furthermore, we think that it is justified to add the curli subunits one at a time to a random curli. We expect no discrimination of the CsgA proteins for binding to a large or small curli or one that has recently gotten a new curli subunit. figure caption

An illustrative view of what our cell looks like during the adding of curli subunits is shown in figure 4. This figure is created when just a few curli were added (\( \sim 1/2 \ hour\)). A similar figure after \(t = \ 10 \ hr\) would look like a fuzzy ball of curli. figure caption

Now that we have a model of a cell with growing curli, we want to extract relevant data for the colony level modeling. Ideally, the resistance as function of radius and time would be calculated by looking at connections between the curli fibrils. However, this requires insight of the behavior of the curli on the nanoscopic scale. For instance, what is the conductivity of a single curli fibril with gold nanoparticles and what is the critical distance between the fibrils that make them connect? After an extensive literature study, we have decided to simplify this model. Furthermore, when interactions between the curli fibrils have to be taken into account, the model gets too computationally expensive.

[write something about the part where we tried the percolation on this level], low priority

To have a reasonable computational time, we decided to extract our parameters for the colony level modeling from the curli density around the cell. Figure 5 contains a histogram with the amount of curli subunits as a function of the cell radius after 10 hours. [add figure] Note how no curli are found below the actual cell radius. It can be seen from the figure that there is a large peak, followed by a plateau. When this histogram is observed in time, you would notice that at first large curli are being created. Figure 6 shows the length of all curli after 10 hours. figure caption

Looking back at figures 1 and 2, the fact that, as can be seen in figure 6, the first curli are much longer that the later ones, can be explained by the fact that there is relatively large curli growth in the beginning, because few CsgB have been produced and therefore, only a few curli fibrils are available for CsgA proteins. After a couple of hours there are more CsgB proteins, thus more curli fibrils, but CsgA protein production does not increase. Therefore, the ratio [CsgA]/[curli fibrils] is much smaller than in the beginning and each curli will grow much slower. A consequence of this is that the ‘newer’ curli fibrils are much shorter. The first histogram of figure 7 shows the amount of curli as function of radius after a growth time of 10 hour. The second histogram shows the curli density as function of radius after a growth time of 10 hour. [add figure]

Conductive Radius of the Cell

We think that a reasonable approximation of the conductivity is the density of the curli around the cell as a function of the radius (see figure 8 in the middle, the blue line). When the density is higher, there are more gold particles, thus higher conductivity. In our simplest model we say that there is a critical density \(\rho_{crit}\) of curli that is needed to have conductivity. The density \(\rho_{curli}\) decreases as function of the radius. The largest radius where \(\rho_{curli} > \rho_{crit}\), we call the conductive radius \(r_{cond}\). With only this simple approximation we can calculate some interesting properties of our system: the time at which we expect percolation to happen and the resistivity of our system. Though this approximation seems to be rather arbitrary, we do have some reasoning for this: [add caption]

• First of all, the goal of this parameter is to get information about our system that will be calculated in colony level modeling. We use this parameter in colony level modeling to find connections between cells. To have a continuous path from one electrode to the other electrode, we must have a lot of cells that are connected to each other. In order to know when cells are connected to each other, we have to assume that everything at a certain radius from the cell is conductive; for this radius we use the critical density \(\rho_{crit}\). However, for this to be true the fibrils on one side of the cell must be connected to the fibrils on the other side. The Percolation Theory prescribes that this is a sharp transition as a function of the density, so we can choose \(\rho_{crit}\) in such a way that we are very sure that everything at \(\rho_{crit}\) from the cell is conductive.
• While the precise value of \(\rho_{crit}\) may be unknown and should be measured, we think that we can still get plenty of information about the qualitative behaviour of our system in advance. Figure 8 at the bottom shows the conductive radius \(r_{cond}\) as function of time using \(\rho_{crit}\) as shown in figure 8 in the middle as the red line. Increasing or decreasing \(\rho_{crit}\) would result in a similar \(r_{cond}\) as function of time. Hence, the qualitative behaviour is preserved.
• Due to the simplifications that we made in order to be able to model our system, we cannot include interactions or cluster forming between the curli themselves. Using \(\rho_{crit}\), we have an elegant way to filter out modeling errors.

Note that figure 8 at the bottom shows a discrete increase of the conductive radius. This is because \(\rho_{curli}\) is dependent on the radius, and because we run a computer simulation, the radius is therefore discretized. Moreover, if we rerun the script to calculate the conductive radius, we expect some variation in the curve, as \(\rho_{curli}\) is subject to stochasticity. Therefore, we have repeated our simulations on the cell level many times in order to get statistically valid results for the mean and standard deviation of \(r_{cond}\).
When we ran our simulation 100 times, we got the results displayed in figure 9. Figure 9 displays the curli density at \(\ t= \ 2 \ hours\) for all cells in the left figure. The orange line represents the average of the simulations. It can be concluded that the intercellular variation is relatively small. This makes sense, since the relative deviation of stochastic processes decreases with the sample size. In the right figure, the mean and standard deviation of the curli density as a function of the radius is shown. insert caption

Another interesting figure to look at, is the density for different times, shown in figure 10. This figure shows that, corresponding with what we have seen previously, \(\rho_{curli}\) decreases as a function of the radius. Also, it decreases faster as a function of the radius in the first two hours. After two hours, we can see that the curli density increases only for small r, as mainly short curli are added to the system. This agrees with our previous results. insert caption

We use the data from figure 10 to make a first approximation for \(\rho_{crit}\). However, the exact value of \(\rho_{crit}\) has to be measured. We set \(\rho_{crit}\) equal to 1% of the maximum \(\rho_{curli}\) observed in figure 9. So, we set \(\rho_{crit} = \ 10^3 \ \frac{number \ of \ curli}{\mu m^{2}}\). The following figure is obtained. insert caption and check if rho_crit value is correct

Again, the orange line represents the mean conductive radius. A sharp increase in the conductive radius can be observed for \(t < 1 \ hour\), and after \(t = \ 1 \ hour\) the conductive radius increases slowly. The cellular variation in the second regime is relatively large, as all the lines except the orange line in figure 11 represent single simulations of our system. Different values of \(\rho_{crit}\) result in different characteristic curves for \(r_{cond}\) as shown in figure 12. insert caption

From figure 12, we conclude that low values of \(\rho_{crit}\) result in a sharp increase of \(r_{cond}\) followed by a steady, slow increase of \(r_{cond}\) in time. During the steady, slow increase of \(r_{cond}\) in time, the cellular variation is relatively large. For high values of \(\rho_{crit}\), there is a delayed sharp increase of \(r_{cond}\) and less cellular variation.

make a fit for \(\rho_{curli}\)
from this fit determine \(\rho_{crit}\) (transition value between the two different fits)
plot \(r_{cond}\) for this found value of \(\rho_{crit}\)

Colony Level Modeling

The goal of the modeling of the curli module is to prove that our system works as expected and to capture the dynamics of our system. The product we aim for is a chip where two parallel electrodes are a distance w apart. Between the electrodes, cells will grow and start building curli in the presence of DNT/TNT. Then, we will measure the conductivity of the resulting biofilm, which is related to the amount of DNT/TNT. Since even with bound gold nanoparticles the conductivity of the curli is very low, the chip is designed such that the electrodes are as long as possible.
The first question we are interested in is: can we prove that our system works as expected? So, does a conductive path between the two electrodes arise at a certain point in time? We do this by modeling the curli growth on the colony level; each cell is now visualized and has curli growth. First we have to make some approximations. Since the cells are grown on a chip, we assume that the cells and curli grow on a surface. This reduces our problem from 3D to 2D. This saves much computational time and memory. For this model, we take a chip of 500 by 500 µm. The next approximation is that the cells are already present when they are induced by DNT/TNT, we neglect cell growth. In our model, E. coli are present with a density of \(\rho_{cell}\). Furthermore, we assume there is no spatial correlation between the cells; hence we place them at random on our chip.

Now that we have designed our chip and placed cells on it, we let them increase their conductive radius in time, corresponding with our findings on the cellular level. A connection is created from one electrode to the other electrode when there is a conductive path between them. Conductive paths consists of cells that have a connection between each other, cells connect when there is an overlap between their conductive radius. This problem is very similar to problems in percolation theory. From this, we can make conclusions about how our system works in an experimental setting. However, we not only want to answer the question if our system works as expected qualitatively, but we also want to make quantitative predictions about the resistivity between the two electrodes of our system in time. Therefore, we used graph theory to translate the cells on the chip to a graph and used an algorithm from graph theory to calculate the resistivity between the two electrodes.

Percolation

So, we now have designed our chip as a 500 by 500 µm square with an electrode on the left and right side. The electrodes are placed parallel to the y axis on x = 20 µm and x = 480 µm. On this chip, we place cells randomly with a density of \(\rho_{cell}\). Subsequently, we increase the conductive radius of each cell in time, corresponding with our findings on the cellular level, see figure 11. A connection is created from one electrode to the other electrode when there is a conductive path between them, so when there is percolation. Conductive paths consists of cells that have a connection with each other, cells have a connection with each other when there is an overlap between their conductive radii. A simulation of our resulting model is shown in figure 13.

We have stochasticity in our model, as we place the cells randomly with a density of \(\rho_{cell}\) on the chip. Therefore, we simulated our model 100 times and for each point in time we checked if there was percolation. Since we will only get a yes (1) or no (0) response, we have a Bernoulli process [reference]. This enables us to find the chance of percolation at each time point, shown in figure 14 as the yellow line. The yellow line shows a sharp transition between 1.5 and 2 hours. We also plotted the variance of this Bernoulli process as the green line. Since this is a Bernoulli process, the variance is exactly equal to p(1-p). The variance must be as low as possible to get trustworthy measurement results, as in that case the transition from no percolation to percolation is as sharp as possible.
We assumed in our model that \(r_{cond}\) is the same for each cell at each point in time. However, figure 11 shows that there clearly is some cellular variation in \(r_{cond}\). Therefore, we changed our model; the conductive radius of each cell can now deviate from the mean \(r_{cond}\) with the standard deviation as found in figure 12. We simulated our resulting model again 100 times and for each point in time we checked the chance of percolation, see figure 14 as the blue line. Fortunately, the resulting curve is very similar to the curve without cellular variation in \(r_{cond}\) (yellow line). This means that cellular variation has little influence on the chance of percolation at each point in time. Therefore, the results of our model are robust to cellular variation and it is likely that many factors that could increase the cellular variation, e.g. different CsgA or CsgB protein production rates, are relatively unimportant.

Resistivity

Now, we uses graph theory to translate the cells on the chip to a graph and use an algorithm from graph theory to calculate the resistivity between the two electrodes.
First, we have to get a quantitative measure for the conductivity between two cells. To do this, we will quantify the overlap of two conducting spheres, where we assumed that the conducting spheres represent cells surrounded by curli filaments. We subdivide the overlapping region in infinitesimal volumes \(dV\). The infinitesimal conductivity of such an infinitesimal volume is given by:

$$ d \sigma (y) = \ \frac{\rho_1}{r_1} dV \frac{\rho_2}{r_2} dV \tag{}$$

The factor \( 1/r \) is introduced to account for the conductivity of the wires itself, which is inversely proportional to the length of the conducting wire. [source: Narinder Kumar (2003). Comprehensive Physics XII. Laxmi Publications. pp. 282–. ISBN 978-81-7008-592-8.] Further away from the cell, the wires need a longer distance to go to the cell. Since we want to know the strength of the connection between the cells, we have to include this factor. For a straight line this is inversely proportional to the distance. For a single curli fibril, this relation does not hold. However, we assume that the curli density is high, thus there are many connections between the curli. Then there is a pathway from the origin to \( r \) roughly proportional to the distance from the cell. To find the total conductivity, we integrate on both sides. To account for the fact that both volume elements \(dV\) are the same, we make use of the Dirac-delta function \(\delta_3\) [source]. This gives us the following:

$$ \sigma (y) = \int{ \frac{\rho_1(\vec{r_1})\rho_2(\vec{r_2})}{r_1 r_2}\delta_3(\vec{r_2}-f(\vec{r_1}))d^3\vec{r_1}d^3\vec{r_2}} \tag{} $$

The Dirac delta allows us to remove the \(\vec{r_2}\) dependence by expressing these in \(\vec{r_1}\). The still undetermined relation between \(\vec{r_1}\) and \(\vec{r_2}\) is given by \(\vec{r_2} = f(\vec{r_1})\). Applying this removes one of the two volume integrations. Using spherical coordinates, the resulting single volume integration can be written as:

$$ \sigma (y) = \int_{r_0}^{r_{max}} \int_0^{\theta_{max}(r)} \int_0^{2\pi} \rho(r_1)\rho_2(f(r_1))\frac{r_1}{f(r_1)} \sin(\theta_1) d\phi_1 d\theta_1 dr_1 \tag{} $$

Here we have made use of the fact that the density \(\rho\) is only dependent on \(r\) and not on \(\phi\) and \(\theta \). The integral over \(\phi_1\) is trivial and gives us a multiplication factor of \(2 \pi\):

$$ \sigma (y) = \ 2 \pi \int_{r_0}^{r_{max}} \int_0^{\theta_{max}(r)} \rho(r_1)\rho_2(f(r_1))\frac{r_1}{f(r_1)} \sin(\theta_1) d\theta_1 dr_1 \tag{} $$

Now that we have reduced our integration to two dimensions, we will work out \(f(\vec{r_1})\). To do this, we introduce the vector from the origin of cell 1 to the origin of cell 2, \(\vec{y}\). This allows us to express \(\vec{r_2}\) in terms of \(\vec{y}\) and \(\vec{r_1}\):

$$ \vec{r_2} = \ \vec{y} - \ \vec{r_1} = \begin{bmatrix}y \\0\\ \end{bmatrix} - \begin{bmatrix} r_1 \cos(\theta_1) \\r_1 \sin(\theta_1)\\ \end{bmatrix} \tag{} $$

Now it is straightforward to express \(r_2\) in terms of \(y\), \(r_1\) and \(\theta_1\):

$$ r_2 = \ |\vec{r_2}| = \ \sqrt{(y - r_1 \cos(\theta_1))^2 + \ r_1^2 \sin^2(\theta_1)} \tag{} $$

Plugging this in yields the following integral:

$$ \sigma (y) = \ 2 \pi \int_{r_0}^{r_{max}} \int_0^{\theta_{max}(r)} \frac{\rho(r_1)\rho_2 \left( \sqrt{(y - r_1 \cos(\theta_1))^2 + r_1^2 \sin^2(\theta_1)}\right) r_1 \sin(\theta_1)}{ \sqrt{(y - r_1 \cos(\theta_1))^2 + r_1^2 \sin^2(\theta_1)}} d\theta_1 dr_1 \tag{} $$

We will now have a closer look at the boundary values for \(r_1\) and \(\theta_1\). We want to integrate over the entire space. Therefore, \( \theta(max) = \pi \) and \( r_{max}=\infty \). By introducing no cut-off radius, we are able to take into account the possibility of having by chance a very large conductive radius. Here we have approximated our cells as points in space. Hence \( r_0 =0 \).

We will now use the previously [link] found fact that the curli density can be described as:

$$ \rho(r) = \ C_{1}e^{-\frac{r}{C_{2}}} \tag{} $$

Plugging in the boundary values and our expression for \(\rho(r)\), we find the following expression for the conductivity between two cells:

$$ \sigma (y) = \ 2 \pi C_{1}^2 \int_{0}^{\infty} \int_0^{\pi} \frac{e^{-\frac{r_1}{C_{2}}} e^{-\frac{ \sqrt{(y - \ r_1 \cos(\theta_1))^2 + \ r_1^2 \sin^2(\theta_1)}}{C_{2}}} r_1 \sin(\theta_1)}{\sqrt{(y - r_1 \cos(\theta_1))^2 + r_1^2 \sin^2(\theta_1)}} d\theta_1 dr_1 \tag{} $$

This integral looks very complicated, but don't panic! It can algebraically be simplified with some substitutions. We can rewrite this integral by moving all terms independent of \( \theta \) out of the integral over \(\theta_1\). Furthermore, using that \( \sin^2 (\theta_1) + \cos^2(\theta_1) = 1 \) we get.

$$ \sigma (y) = \ 2 \pi C_{1}^2 \int_{0}^{\infty} r_1 e^{-\frac{r_1}{C_{2}}} \int_0^{\pi} \frac{e^{-\frac{ \sqrt{y^2+r_1^2-2yr_1 cos( \theta_1 ) }}{C_{2}}} \sin(\theta_1)}{ \sqrt{y^2+r_1^2-2yr_1 \cos( \theta_1 ) }} d\theta_1 dr_1 \tag{} $$

Now we must recognize that we can substitute \( x= cos(\theta_1) \) such that \( dx = -\sin(\theta_1) d\theta_1 \). This results in:

$$ \sigma (y) = - \ 2 \pi C_{1}^2 \int_{0}^{\infty} r_1 e^{-\frac{r_1}{C_{2}}} \int_1^{-1} \frac{e^{-\frac{ \sqrt{y^2+r_1^2-2yr_1 x }}{C_{2}}}}{\sqrt{y^2+r_1^2-2yr_1 x }} dx dr_1 \tag{} $$

In the second integral we recognize something of the form \( \int \frac{e^{-\sqrt{a+bx}}}{C_2\sqrt{a+bx}} dx \) with \( a= \frac{y^2+r_1^2}{C^2_2} \) and \(b=-\frac{2yr_1}{C^2_2} \). Substituting \( h= \sqrt{a+bx} \) with \( dx= \frac{2h}{b} dh \) yields:

$$ \int_1^{-1} \frac{e^{-\sqrt{a+bx}}}{C_2\sqrt{a+bx}} dx = \frac{2}{bC_2} \int_{\sqrt{a+b}}^{\sqrt{a-b}} e^{-h} dh= \frac{-2}{bC_2} (e^{-\sqrt{a-b}}- \ e^{-\sqrt{a+b}})$$

Now \(a\) and \(b\) can be substituted:

$$ \int_1^{-1} \frac{e^{-\sqrt{a+bx}}}{C_2\sqrt{a+bx}} dx = \frac{C_2}{yr_1} \left( e^{-\frac{\sqrt{y^2+r_1^2+2yr_1}}{C_2}} - e^{-\frac{\sqrt{y^2+r_1^2-2yr_1}}{C_2}} \right)$$

Hence, the entire integral now becomes

$$ \sigma (y) = \frac{ 2 \pi C_{1}^2 C_2 }{y} \int_{0}^{\infty} e^{-\frac{|y-r_1|+r_1}{C_2} } - e^{-\frac{y+2r_1}{C_2} } dr_1 \tag{} $$

Solving the second integral is fairly easy:

$$ \sigma (y) = \frac{ 2 \pi C_{1}^2 C_2 }{y} \int_{0}^{\infty} e^{-\frac{|y-r_1|+r_1}{C_2} }-e^{-\frac{y+2r_1}{C_2} } dr_1 = \frac{ 2 \pi C_{1}^2 C_2 }{y} \left( \int_{0}^{y} e^{-\frac{y}{C_2}} dr_1 +\int_{y}^{\infty} e^{-\frac{2r_1-y}{C_2}} dr_1 -e^{\frac{-y}{C_2}}\int_0^{\infty} e^{-\frac{2r_1}{C_2} } dr_1 \right) \tag{} $$

Which brings us to the final result: $$ \sigma (y) = \ 2 \pi C_{1}^2 C_2 e^{-\frac{y}{C_2}} \tag{} $$

solving this integral numerically, storing values in table
giving the conductivity as weights to the adjacency matrix
from these weights (conductivity) determine if cells are connected or not
from the resulting adjacency matrix, calculate the resistivity of the system using graph theory

References

still has to be made