Model of PET degradation by Chlamydomonas reinhardtii

A C. reinhardtii which expresses and secretes the enzymes PETase and MHETase could pose as a solution for the problem of micro-plastic polluted water. Nevertheless, the viability of PET degradation by C. reinhardtii at a larger scale is yet unknown. Models of biological systems allow us to design experiments in silico that are difficult to reproduce in vivo and give us special insights into the role that parameters might play in the given biological system. Therefore, to assess the efficiency of PET degradation by C. reinhardtii, a model of PET degradation in continuous culture of C. reinhardtii was designed.

The overall goal of the model is to determine the time needed to degrade 1 mg of PET. As we are building a bioreactor for C. reinhardtii, it is imperative to know the best parameters that have to be fulfilled by our bioreactor and our algae to achieve the successful degradation of PET. The expression rate, secretion rate and kinetics of the enzymes, such as also the cultivation density, influence the degradation rate of PET in the bioreactor. Based on this assumption, the model was designed to take these factors into account. The model was programmed in Tellurium (Choi et al., 2018) and encompasses six reactions. The reactions are as listed on Fig. 1 and Tab. 1.

Assumptions and Hypothesis

The reaction rate of the PETase enzyme is known to be one of the main limiting factors in PET degradation. It is a slow enzyme and because of this reason there have been efforts to optimize it (Ma et al., 2019). This was one of our main concerns while designing and programming our model. Our hypothesis was, that the process of PET degradation would be a slow process that is yet unviable for industrial application and that its speed could eventually be regulated to a certain degree by biological parameters and cultivation parameters. We decided to focus our simulations on two main parameters, one of them external to the biology of the cell, and the other of biological nature. The external parameter that we assumed could influence the degradation of PET was the cultivation density of C. reinhardtii. A higher cultivation density would lead to a higher concentration of the secreted enzymes PETase and MHETase and thus to a faster PET degradation. The biological parameter that we chose to variate was the enzyme kinetics of the PETase. By increasing the kinetics of the enzyme PETase by a factor of 1000, a significantly faster degradation of PET is expected. The arbitrary value of 1000 was chosen as an extremely optimistic optimization of the PETase to examine the effect of such a substantial change to the kinetic parameters.

Results: Variation of the Cultivation Density

As a first approach, the cultivation density of the algae was varied to examine its effect on the degradation rate of PET. For the simulation presented in Fig. 2, the cell volume to culture volume ratio of 1:10 was examined. This means that of 10 ml culture, you would have 1 ml of cells. The time needed to degrade 1000 µg (or 1 mg) of PET was extracted from this simulation, leading to a total degradation time of 8200 s, or 2,27 hours. This is a very optimistic simulation for the cultivation of the alga C. reinhardtii because the cultivation density of 1:10 is very difficult to achieve and sustain. Even with this optimistic calculation, a PET plastic bottle weighing 40 g would need approximately 10 years to be completely degraded. According to Klein et al. (2105), the water of the Rhein river in Germany is polluted with micro-plastic up to 1 g per kg. This would mean that to clean one liter of water, assuming that all the micro plastic is PET, 94,5 days would be needed. To analyse a more realistic cultivation density, a simulation for a cultivation ratio of 1:100 was made, as can be seen on Fig. 3. This simulation led to the degradation of 1 mg of PET in 82000 s, or 22,7 h. According to this simulation, it would take approximately 100 years for a 40 g PET bottle to be completely degraded. These simulations show that even after altering the cultivation density parameter, an efficient degradation of PET on industrial scale is not in sight.

Results: Improving the PETase Kinetics by Factor 1000

The results for the first simulations, shown in Fig. 2 and Fig. 3, confirmed that the PETase kinetics are slow, especially in comparison to the kinetics of the MHETase. On the aforementioned graphs, the curve describing the concentration of MHET stays constant at a value converging to zero. This means that the MHETase immediately converts all MHET to terephthalate (TPA) and ethylene glycol (EG), demonstrating a big difference between the kinetics of the PETase and MHETase. Because the enzyme PETase is much slower than MHETase, MHET can not be accumulated but is rather immediately degraded to TPA and EG. Advanced methods to improve enzymes, for example by directed evolution, have shown that it is possible to drastically improve the kinetics of an enzyme. To assess the effect of an optimistically enhanced and improved PETase on the degradation rate of PET, two simulations were made, where the kinetics of the PETase were improved by a factor of 1000. We are aware that such an optimization of the PETase would be very difficult, if not impossible to achieve. Nevertheless, the drastic improvement of the enzyme for these simulations is expected to serve as an exaggerated case that shows under what conditions, if any, PET degradation is applicable at an industrial scale.

One simulation was made for the cultivation density ratio of 1:10 and the other one for a ratio of 1:100. The results can be seen on Fig. 4 and Fig. 5 respectively. For the simulation shown in Fig. 5, the time needed to degrade 1mg of PET was 260 s, or 4,3 min, which is a substantial improvement to the simulations discussed in the previous section. According to this simulation, the time needed for the degradation of a 40 mg plastic bottle would be approximately 119 days. Still, the problem of achieving and sustaining a cultivation ratio of 1:10 prevails. Therefore, the simulation was repeated for a cultivation ratio of 1:100 with the optimized PETase by a factor of 1000 (Fig. 5). The results of this simulation show that for this cultivation setup a time of 2600 s, or 43 min, is needed to degrade 1 mg of PET. The time needed to degrade a 40 mg PET bottle would be approximately 3,3 years.

Formulation of the model

For describing the interplay between vessel depth and culture density, we start from the two equations (Huisman et al., 2002) on the right. For other relevant mathematical derivations, see (Barbosa et al., 2004). Equation (1) provides us with the light intensity gradient inside the culture vessel due to algae cell concentration \(c_X\) and is well known as Beer-Lambert law. Important is the mins sign on the right hand side of the equation. \(\mathrm{d}I\) is antiproportional to \(\mathrm{d}z\), meaning that the intensity is declining along the light path \( z \). This will allow us to draw conclusions on the design of our setup pretty fast. Because we want to treat algae cells and other molecules, the Lambert-Beer law is written as a sum over all species contributing to the absorption along the z-axis.

Equation (2) can be understood by looking at the single terms being added to each other on the right hand side. It equates the temporal change of \(c_X\) to the sum of positive (growth with production rate per surface area \(P\) ) and negative (dilution \(D\) of the culture and maintenance \(\mu_e\) contributions. Dilution rate is the rate at which we pump out culture from the vessel. The maintenance rate is necessary because cells require energy to maintain processes such as osmoregulation, reducing our maximum achievable value (see [9] for details). For (2) it is also important to note that productivity per surface area \(p\) is a function of specific productivity \( \tilde p \), which expresses the amount of carbon produced per cell. By expressing \(\tilde p \) as a function of light intensity and calculating \(p\) as an integral over the light path, we can derive further conclusions already. At first, we will take a look at the Beer-Lambert law and derive some parameters important to our model.

For the sake of clarity, precise mathematical derivations for this model are given in a PDF supplement. We are still trying our best to make the model as clear as possible while trying to develop the mathematical approaches from a physical point of view.

\begin{equation} \frac{\partial}{\partial z} I(\lambda, z, t) = -I(\lambda, z, t) \cdot \sum_{i=1}^{n} \epsilon_i(\lambda)_ \cdot c_i \tag{1} \end{equation} \[\newcommand\T{\Rule{0pt}{1em}{.3em}} \begin{array}{r l} I(\lambda, z,t): & \text{Light Intensity in} \: \frac{\mu mol}{m^2 \cdot s} \\ \epsilon(\lambda): & \text{specific light attenuation coefficient in} \: \frac{L}{mol \cdot cm} \\ c_i: & \text{concentration of \(i\)-th species} \: \frac{mol}{L} \\ z: & \text{light path in}\:cm \\ \lambda: & \text{wavelength in}\:nm \end{array} \]

\begin{equation} \frac{\mathrm{d}}{\mathrm{dt} } c_X= \frac{Y}{z} \cdot p(\tilde p, c_X) - D \cdot c_X - \mu_e \cdot c_X \tag{2} \end{equation} \begin{array}{r l} c_X: & \text{cell concentration in}\: \frac{mol}{L} \\ p(\tilde p, c_X): & \text{production rate per surface area}\: \frac{mol \: carbon}{m^2 \cdot h} \\ Y: & \text{cell yield per carbon fixed in}\: \frac{mol \: cells}{mol \: carbon } \\ \mu_e : & \text{maintenance rate in} \: \frac{1}{h} \\ \tilde p(I) & \text{specific productivity as a function of light intensity}\: \frac{mol \: carbon}{mol \: cells \cdot mm^3 } \end{array}

\begin{equation} p(\tilde p, c_X) = \int_0^z \tilde p \cdot c_X \, \mathrm{d} z \end{equation} \begin{array}{r l} c_X: & \text{cell concentration in}\: \frac{mol}{L} \\ p(\tilde p, c_X): & \text{production rate per surface area}\: \frac{mol \: carbon}{m^2 \cdot h} \\ Y: & \text{cell yield per carbon fixed in}\: \frac{mol \: cells}{mol \: carbon } \\ \mu_e : & \text{maintenance rate in} \: \frac{1}{h} \\ \tilde p(I) & \text{specific productivity as a function of light intensity}\: \frac{mol \: carbon}{mol \: cells \cdot mm^3 } \end{array}

\begin{equation} I(\lambda, z, t) = -I_0(\lambda, t) \cdot \exp[-(\epsilon_X \cdot c_X+bg)z] \tag{1a} \end{equation} \[\newcommand\T{\Rule{0pt}{1em}{.3em}} \begin{array}{r l} I_0(\lambda, t): & \text{light intensity upon entrance into vessel}\\ \epsilon_X: & \text{light attenuation coefficent for algae } X \\ bg: & \text{background turbidity} \: \frac{1}{cm} \end{array} \] where \begin{equation} bg = \sum_{i=1}^{n} \epsilon_i(\lambda)_ \cdot c_i \end{equation}

\begin{equation} c_X^{\ast}= \frac{1}{\epsilon_x} \cdot (\frac{ln(\frac{I_{in}}{I_{out}})}{z} - bg) \end{equation}

Arriving at first conclusions

The solution of equation (1) requires evaluation of a well known integral (cf [10]). As a solution we get the exponential decay term on the left, describing the light intensity along the distance traveled \(z\) at a wavelength \( \lambda \). Because there are not only algae in our culture but also other compounds, we split the sum in a term dependent on algae concentration and one containing the rest, the background turbidity \( bg \). The equation shows that algae in a cultivation vessel are subject to a light gradient.
Before we continue, we can already extract first information: Because the main feature of turbidostat cultivation is constant turbidity, we can incorporate this into our equations. Keeping optical density constant translates to constant cell number, meaning \( \frac{\mathrm{d}}{\mathrm{d}t} c_X = 0 \). Concentration does not change with time. so we can rearrange equation (1a) to get an expression for the steady state cell concentration \(c_X^{\ast} \).
From this equation we see that the achievable steady state cell concentration is inversely proportional to the depth of our cultivation vessel \(z\). This is one of the reasons why we decided to build a flat panel device for cultivation. A dense culture can be more suitable for experiments that aim to produce a protein or other compound of interest.

To quantifiy the growth of cells, we need to calculate \( \tilde p \). We already gave the equation above: The productivity per surface area is given as the integral of product of specific productivity per cell times concentration over the light path of our vessel.
For this approach to sufficiently describe our setup, we need to make sure that there is no intensity gradient along the other coordinates. We realized this in our setup by using an array of LEDs spaced such that light can be regarded as travelling mainly in z-direction.
To solve our question, we used a table in [9] that provides a review of possible functions for \( \tilde p \). What we already know is that for the light limited chemostat, \( \tilde p \) should only be a function of light intensity. To realize this, one also should use cultivation media tailored to this need ([9]).
For our first approach, we are choosing the Monod saturation kinetics with a characteristic asymptotic shape and a value for intensity at half maximum \(H\).

\begin{equation} p(\tilde p, c_X) = \int_0^z \tilde p(I) \cdot c_X \, \mathrm{d} \tilde z \end{equation} \begin{array}{r l} & \text{productivity per unit area as an integral over light path} \end{array} \]

\begin{equation} \tilde p(I) = p_{max} \cdot \frac{I(\lambda, z, t)}{I(\lambda, z, t)+H} \end{equation} \begin{array}{r l} & \text{Monod-type growth curve} \\ p_{max}: & \text{maximum specific productivity in} \: \frac{mol \, carbon}{mol \, cells \cdot h} \\ H: & \text{intensity for which} \: \tilde p(H) = \frac{p_{max}}{2} \end{array}