Team:HZAU-China/Comparison

In this part, we want to demonstrate some advantages of our design by making quantitative comparisons. These advantages include safety, energy efficiency and stability.

3.1 The post-transcriptional control ensures lower leakage

Firstly, we don't expect that our designed processing modules will alter its function by some noises of environment. So we design a coherent feedforward loop to filter noise. Here we use $x$ to represent a general input signal. As we mentioned before, the dynamic of the input module with post-transcriptional control can be described by \begin{equation} \begin{aligned} \frac{d[mRNA_{Cre}]}{dt}&=\beta_0+\beta_1\frac{[x]^n}{K^n+[x]^n}-K_{R}\cdot [mRNA_{Cre}]\\ \frac{d[taRNA]}{dt}&=\beta_0+\beta_1\frac{[x]^n}{K^n+[x]^n}-K_{R}\cdot [taRNA]\\ \frac{d[Cre]}{dt}&=K_{tl}\cdot [mRNA_{Cre}]\cdot\frac{[taRNA]}{K_m+[taRNA]}-K_{P}\cdot [Cre]. \end{aligned} \end{equation}

If there is no post-transcriptional control, this process can be described by \begin{equation} \begin{aligned} \frac{d[mRNA_{Cre}]}{dt}&=\beta_0+\beta_1\frac{[x]^n}{K^n+[x]^n}-K_{R}\cdot [mRNA_{Cre}]\\ \frac{d[Cre]}{dt}&=K_{tl}\cdot [mRNA_{Cre}]-K_{P}\cdot [Cre]. \end{aligned} \end{equation}

We compare the expression dynamic of Cre in either case at different level of input signal $x$. This comparison reveals that the riboregulator ensures lower leakage but doesn't affect the expression at a high input level. We also design some experiment to validate the model, the results are consistent with this model.

Figure 2: dynamic of the input module

(some data will be showed here)

3.2 The mutated Cre/loxP system determines the inversion direction

Secondly, once the engineered cells receive a sure signal, they will process information as we designed but not the other way around. So we must ensure the direction of the DNA invertion. In other words, the site-specific recombination can be regarded as a unidirectional one. To this end, we choose Cre recombinase and a pair of mutant lox sites, lox66 and lox71 to rearrange DNA sequence. Here, we explain why the forward reaction rate is higher than the reverse one. The mutant site will have a lower affinity for Cre. The binding event mainly depends on the rate of free diffusion, but the dissociation rate will be high if the binding strength is weak. According to the equation \begin{equation} r_c(2Cre\cdot lox)=\frac{k_8k_9\cdot [lox]\cdot [Cre]^2}{k_{-8}k_{-9}+k_9k_{-9}\cdot [Cre]}, \end{equation} we know that the second dissociation event is more significant than the first one. So mutant lox site like lox66 and lox71 can choose a wise way to benefit the formation of dimer. Here we assume that lox66 and lox71 will have higher $k_{-8}$. However, double mutant loxP site like lox72 will have high dissociation rate at both steps. For simplicity, we assume that the affected dissociation rate is $\epsilon$ times the original one. $\epsilon>1$.

After binding event, the two loxP-bound dimers associate to form a tetramer, and recombination proceeds via a Holiday Junction intermediate. Therefore, we can compare $r_c(2Cre\cdot loxP)\cdot r_c(2Cre\cdot lox72)$ with $r_c(2Cre\cdot lox66)\cdot r_c(2Cre\cdot lox71)$ to see what kind of synapsis is easy to form. \begin{equation} \begin{split} r_c(2Cre\cdot loxP)\cdot r_c(2Cre\cdot lox72)&=\frac{k_8^2k_9^2\cdot[lox]^2\cdot [Cre]^4}{({\epsilon}^2k_{-8}k_{-9}+\epsilon k_9k_{-9}[Cre])(k_{-8}k_{-9}+k_9k_{-9}[Cre])}\\ r_c(2Cre\cdot lox66)\cdot r_c(2Cre\cdot lox71)&=\frac{k_8^2k_9^2\cdot[lox]^2\cdot [Cre]^4}{({\epsilon}k_{-8}k_{-9}+k_9k_{-9}[Cre])^2} \end{split} \end{equation}

For $\epsilon>1$, \begin{equation} \begin{aligned} \frac{r_c(2Cre\cdot loxP)\cdot r_c(2Cre\cdot lox72)}{r_c(2Cre\cdot lox66)\cdot r_c(2Cre\cdot lox71)}&=\frac{({\epsilon}k_{-8}k_{-9}+k_9k_{-9}[Cre])^2}{({\epsilon}^2k_{-8}k_{-9}+\epsilon k_9k_{-9}[Cre])(k_{-8}k_{-9}+k_9k_{-9}[Cre])}\\ &=1-\frac{(\epsilon-1)(k_9k_{-9}[Cre]^2)+\epsilon(\epsilon-1)(k_{-8}k_9k_{-9}^2[Cre])}{({\epsilon}^2k_{-8}k_{-9}+\epsilon k_9k_{-9}[Cre])(k_{-8}k_{-9}+k_9k_{-9}[Cre])}\\ &<1 \end{aligned} \end{equation}

Under the condition that the $Cre\cdot lox$ is much more easy to get a Cre monomer rather than loss a Cre monomer, this proportion can be approximately equal to $\frac{1}{\epsilon}$, because \begin{equation} \begin{aligned} r_c(2Cre\cdot lox)&=\frac{k_8k_9\cdot [lox]\cdot [Cre]^2}{k_{-8}k_{-9}+k_9k_{-9}\cdot [Cre]} &\approx\frac{k_8\cdot [lox]\cdot [Cre]}{k_{-9}} \end{aligned} \end{equation}