Team:HZAU-China/Biological

In this part, we will list the important biological processes and explain how the related molecules work, and then we can describe them using some equations. Because the timescales of many processes are separated, we use a quasi-steady-state approximation (QSSA) to reduce the number of dimensions in the systems in most cases. However, if we focus on some processes more than the final steady state, we may use other approximations such like prefactor method (Bennett et al., 2007)。

2.1 Transcription and translation

According to the Central Dogma, some DNAs can be transcribed into RNAs and some RNAs can be translated into proteins. These procedures sometimes can be regulated by some molecules like transcription factors or non-coding RNAs. The interactions among them can be understood by chemical reactions. We use reaction (1) to depict the interaction between transcription factor $TF$ (sometimes it forms a polymer, we omit this step) and inducible promoter $Pro$. \begin{equation} K_D=\frac{k_{-1}}{k_1}=\frac{[TF]^n\cdot [Pro]}{[Pro']}. \end{equation}

The equilibrium dissociation constant $K_D$ for the reaction can be calculated. \begin{equation} K_D=\frac{k_{-1}}{k_1}=\frac{[TF]^n\cdot [Pro]}{[Pro']}. \end{equation}

We assume that the DNA copy number is a constant n, \begin{equation} [Pro]+[Pro']=n. \end{equation} Then the $Pro'$ proportion depends on the concentration of $TF$, and this proportion also be regarded as the probability of the binding events. \begin{equation} P(\text{binding})=\frac{[Pro']}{n}=\frac{[TF]^n}{K_D+[TF]^n} \end{equation}

If the transcription factor is a activator, the gene is transcribed at the maximal transcription rate $\beta_1$ when the promoter is bound by the transcription factor. If the transcription factor is a repressor, the gene is transcribed at the maximal transcription rate $\beta_1$ when the promoter is free from the transcription factor.

In deterministic model, we use Hill function to describe the rate of production, which is $\beta_1$ times its occurring probability.

For transcription activation, the maximal transcription rate will occur if the activator bind to the promoter, so the Hill function is \begin{equation} f(x)=\beta_1\frac{[TF]^n}{K^n+[TF]^n}. \end{equation}

For transcription repression, the maximal transcription rate will occur if the repressor doesn't bind to the promoter, so the Hill function is \begin{equation} f(x)=\beta_1\frac{K^n}{K^n+[TF]^n}. \end{equation} $K$ is activation coefficient or repression coefficient, which is the same as $\sqrt[n]{K_D}$.

In addition, many genes have a nonzero minimal expression level, namely basal expression level. It can be described by adding a term $\beta_0$. And some genes that have a constitutive promoter cannot be described by Hill function. We just set their production rate to be a constant.

The translation and degradation processes in our model are based on following reaction: For translation process, the protein production rate is proportional to the corresponding mRNA concentration $K_{tl}\cdot mRNA$. $K_{tl}$ is determined by the efficiency and concentration of ribosome, the sequence of RBS and concentration of amino acids in the cell, which can be considered identical under the experiment condition. For degradation process, the degradation rate is proportional to the substrate. The proportion for mRNA is $K_{R}$ and the proportion for protein is $K_{P}$. So we list following equations, \begin{equation} \begin{aligned} \frac{dmRNA_{Y}}{dt}&=\beta_0+\beta_1\frac{[TF]^n}{K^n+[TF]^n}-K_{R}\cdot mRNA_{Y} \text{ (activation)}\\ \frac{dmRNA_{Y}}{dt}&=\beta_0+\beta_1\frac{K^n}{K^n+[TF]^n}-K_{R}\cdot mRNA_{Y} \text{ (repression)}\\ \frac{dY}{dt}&=K_{tl}\cdot mRNA_{Y}-K_{P}\cdot Y. \end{aligned} \end{equation}

2.2 RNA interaction

Different riboregulators have different regulation mechanisms to control the gene expression. Here what we used is a classical engineered riboregulator designed by Isaacs and his colleagues in 2004. This riboregulator can control the gene expression at post-transcription level. More details can be found in (a link in project input module)

We describe the RNA interactions by this reaction: which means \begin{equation} \frac{d[RNA\ duplex]}{dt}=k_2\cdot [taRNA]\cdot [crRNA]-k_{-2}\cdot [RNA\ duplex] \end{equation} This reaction is much faster than gene expression, so the equilibrium state can be reached quickly, \begin{equation} \begin{split} \frac{d[RNA\ duplex]}{dt}&=0 \\ [RNA\ duplex]_{st}&=\frac{k_2}{k_{-2}}\cdot [taRNA]\cdot [crRNA] \end{split} \end{equation}

The total mRNA of the recombinase has two forms, $crRNA$ and $RNA\ duplex$, resulting in another equation: \begin{equation} [crRNA]+[RNA\ duplex]=[mRNA_{Cre}]. \end{equation} Then we have \begin{equation} [RNA\ duplex]_{st}=[mRNA_{Cre}]\cdot\frac{[taRNA]}{K_m+[taRNA]}, \end{equation} where $K_m=\frac{k_{-2}}{k_2}$. So the protein production rate depends on not only the concentration of its corresponding mRNA but also the concentration of $taRNA$. This translation process with post-transcriptional control can be described by \begin{equation} \frac{d[Cre]}{dt}=K_{tl}\cdot [mRNA_{Cre}]\cdot\frac{[taRNA]}{K_m+[taRNA]}-K_{P}\cdot [Cre] \end{equation}

2.3 Processes related to AHL

In this section, we describe the 3OC6HSL synthesis, degradation and regulation. 3OC6HSL, a kind of AHL, is enzymatically synthesised by LuxI proteins from some substrates. Here, we use AHL to refer to 3OC6HSL.

For simplicity, we assumed that the amount of substrates is sufficient so that the 3OC6HSL synthesis rate is proportional to the LuxI protein concentration. And the degradation of 3OC6HSL can be divided into two parts. With the help of the AHL-degrading enzyme, AHL-lactonase (AiiA), the AHL can be degraded at a high rate, since the complex of AiiA and AHL is more easy to be degraded. Also, the AHL itself can be degraded at a lower rate.

Besides, AHL can regulate gene expression by binding the protein LuxR. The complex can activate luxpR promoter and repress luxpL promoter.

Although the binding events about AHL can be described by some differential equations, we apply QSSA when we focus on the gene expression process, because the timescale of these binding events is much less than the time scale of gene expression. So change of AHL and AHL-LuxR complex over time can be described by differential equations: \begin{equation} \begin{split} &\frac{d[AHL]}{dt}=k_3\cdot [LuxI]-k_4k_5\cdot [AHL] \cdot [AiiA]-k_6\cdot [AHL]-k_7\cdot [LuxR]^2\cdot [AHL]^2\\ &\frac{d[AHL\text{-}LuxR\ complex]}{dt}=k_7\cdot [LuxR]^2\cdot [AHL]^2-K_{P}\cdot [AHL\text{-}LuxR\ complex] \end{split} \end{equation}

2.4 DNA recombination

Ringrose and his colleagues have developed mathematic models to describe the kinetics of Cre and Flp recombination. The site-specific recombination process can be mainly divided into four steps as the following figure depicts: DNA binding, synapsis, recombination and dissociation. The detailed mathematical representation can be found in previous research (Ringrose et al., 1998), so we didn't list them here. We want to explain Cre-mediated inversion using the mutated Cre/loxP system we used prefers the forward reaction. The difference between the mutated loxP site and wild type loxP site only influences the DNA binding and dissociation processes. So we focus on these two steps. A Cre monomer binds to one half of a loxP site and then an asymmetrical homodimer is formed when a second Cre molecule binds to the other half of loxP. \begin{equation} \begin{split} lox+Cre\autorightleftharpoons{$k_8$}{$k_{-8}$} Cre\cdot lox\\ Cre\cdot lox+Cre\autorightleftharpoons{$k_9$}{$k_{-9}$}2Cre\cdot lox \end{split} \end{equation}

Because the concentration of intermediate products $Cre\cdot lox$ is very low, and $k_9>>k_{-8}>k_{-9}$, we can apply steady-state treatment to it, \begin{equation} \frac{d[Cre\cdot lox]}{dt}=k_8\cdot [lox]\cdot [Cre]-k_{-8}[Cre\cdot lox]-k_9\cdot [Cre\cdot lox]\cdot [Cre]+k_{-9}\cdot[2Cre\cdot lox]=0, \end{equation}

Figure 1: recombination mechanism

The second dissociation process is the slowest reaction within these four reactions. Hence, we ignore the last term $k_{-9}\cdot[2Cre\cdot lox]$. Then we get the steady-state of intermediate products $Cre\cdot lox$ \begin{equation} [Cre\cdot lox]_{st}=\frac{k_8\cdot [lox]\cdot [Cre]}{k_{-8}+k_9\cdot [Cre]}. \end{equation} So the reaction rate for $2Cre\cdot lox$ is \begin{equation} r_c(2Cre\cdot lox)=\frac{k_9}{k_{-9}}\cdot[Cre]\cdot\frac{k_8\cdot [lox]\cdot [Cre]}{k_{-8}+k_9\cdot [Cre]}=\frac{k_8k_9\cdot [lox]\cdot [Cre]^2}{k_{-8}k_{-9}+k_9k_{-9}\cdot [Cre]}. \end{equation}

References

Bennett, M. R., Volfson, D., Tsimring, L., & Hasty, J. (2007). Transient dynamics of genetic regulatory networks. Biophysical journal, 92(10), 3501-3512.

Ringrose, L., Lounnas, V., Ehrlich, L., Buchholz, F., Wade, R., & Stewart, A. F. (1998). Comparative kinetic analysis of FLP and cre recombinases: mathematical models for DNA binding and recombination. Journal of molecular biology, 284(2), 363-384.