Team:TU Delft-Leiden/Modeling/Curli/Colony

Curli Module: The Colony Level

summary of the conclusions

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 and at which time does this happen? 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 electrodes are placed parallel to the y axis on x = 20 µm and x = 480 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. The cell density we use is \( 2 \cdot 10^4 \) cells \( mm^2 \). We'd like to model higher cell densities and larger chips. However, the memory cost of the solution increases with the amount of cells squared and, even when the code is neatly vectorized, the computational time increases drastically more.

We have come up with two different approaches. The first is that we let the cells increase their conductive radius in time, according with our findings on the cellular level (figure 12). 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 with only a yes or no answer, but we also want to make 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. The conductivity between the cells is computed from an integral that we have set up starting with formula 13.

Percolation

So, we now have designed our chip as a 500 by 500 µm square with an electrode on the left and right side. 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 12. 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. In practice we have programmed this by comparing the distances of all cells with all other cells with the conductive radius. A simulation of our resulting model is shown in figure 14. Percolation is computed by applying an algorithm that can find clusters of connected cells. When one of the clusters connects both electrodes, we have percolation.

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. We will only get a yes (1) or no (0) response. This enables us to find the chance of percolation at each time point, shown in figure 15 as the yellow line. The yellow line shows a sharp transition between 1.5 and 2 hours. Since this is a Bernoulli process, [reference], 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.
At first we assumed in our model that \(r_{cond}\) is the same for each cell at each point in time (figure 12 red line). However, figure 12 shows that there clearly is some cellular variation in \(r_{cond}\). Therefore, we also added a feature to 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 15 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.

Influence of the chip geometry on the point of percolation

To further investigate the point of percolation we have varied the shape of our chip. We have decreased the relative distance between the electrodes by making our chip 250 µm x 500 µm, where the electrodes are 250 µm apart. From the result, shown in figure 15.5, it can be seen that our system behaves in correspondence with a percolation problem. The system is smaller, therefore the transition toward percolation is less sharp. This suggests that we want to increase the area size of our system. A larger area results in a shaper transition, thus a lower uncertainty. Other simulations with a chip of 1000 µm x 500 µm show a shaper transition.

Resistivity

To calculate the conductivity as function of time we repeat the following steps:

• Place our cells on our chip.
• Compute the conductivity between the cells.
• Compute the conductivity between the electrodes.

Compute the conductivity between the cells

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:

$$ \sigma (y) = \ 2 \pi C_{1}^2 C_2 e^{-\frac{y}{C_2}} \tag{} $$

For future research, we could extend our models such that the cellular variation is included. If \( \rho_1(r) = \ C_{1}e^{-\frac{r}{C_{2}}} \) and \( \rho_2(r) = \ C_{3}e^{-\frac{r}{C_{4}}} \) then the conductivity between the two electrodes, using the approach as described above is.

$$ \sigma (y) = \ \frac{4 \pi C_{1}C_3 C_2^2 C_4^2}{y \left( C_2^2 - C_4^2 \right)} \left( e^{-\frac{y}{C_2}} -e^{-\frac{y}{C_4}} \right) \tag{} $$

Compute the conductivity between the electrodes

Now, we use 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.

Results

We have calculated the conductivity as function of time for various different dimensions of our plate. The distances of the electrodes is varied from 250-500-1000x500 \( \mu m \). The result is shown in figure 16. The process is, for the smaller chips repeated 10 times to say something about the variance. For the 1000x100 \( \mu m \) plate it is only repeated thrice, for a single curve already takes over twelve hours to compute.

From figure 16 we can draw a couple of conclusions.

It seems that the conductivity increases exponentially over time. We expect that even after a long time, there is low conductivity [paper]. This means that before that time, it is hard to measure changes in conductivity.

The response-curve of the system is independent of the shape of our plate. The blue lines that have the distance between the electrodes doubled compared to the green line also has half the conductivity (and a quarter of the red lines). Thus the conductance is inversely proportional to the distance between the electrodes. This is precisely what you would expect if you see the system as a single resistor.

Increasing the chip size decreases the relative uncertainty of the response. The red lines are much further apart (also relatively) than the blue lines. This makes sense from a physical point of view, since we're dealing with larger samples. It is then more insensitive to the randomness due to the placement of the cells.

Our findings are in accordance to the behaviour expected from a random resistor network with percolation theory, where the conductivity increases exponentially after percolation.

It is impossible to observe a point of percolation in this. This is because we have made a continuous model.

Other simulations shows us that indeed, the conductivity scales linearly with the length of the electrodes. If we want to make a design of our system and have as high conductivity as possible, we want to decrease the distance between the electrodes as much as possible. At the same time, the total area of the chip should be reasonable large to reduce the effects of the stochastic behaviour of the system.

References

still has to be made

Click in the figure to move to the corresponding page.