Team:HUST-China/Modelling/DDE Model

We had to validate the feasibility of our supposed genetic pathway in the first place. We wanted to know if our engineered cells could generate periodical signal, if it's stable against environment changes, and how its period can be adjusted. To solve these issues, we build a DDE model.

Methods

1. Establish DDEs based on mass action law;
2. Investigate reasonable parameters set from previous researches;
3. Discuss DDEs' stability through examining their characteristic equations;
4. Determine the most effective factors in changing period;

Establishing DDEs

Fig 1.The pathway of genetic oscillator

The Arabinose Operon and the lac Operon is the core to the functioning of the oscillator. With the presence of Arabinose, dimeric AraC \cite{AraC_dimer}can induce the expression of downstream gene; On the other hand, with minor presence of IPTG, tetrameric LacI \cite{lac_tetramer}may suppress the expression of downstream gene. However, AraC and LacI cannot combined with promoters simultaneously. Therefore, we have: $$D+2a\overset{k_1}{\underset{k_{-1}}\rightleftharpoons} D_1$$ $$D+4r\overset{k_2}{\underset{k_{-2}}\rightleftharpoons} D_2$$ where $D$ stands for ratio of promoters which don't combine with any protein among all promoters, a for AraC protein(activator), r for LacI protein(repressor), $D_1$ for ratio of operons combined with AraC dimer, $D_2$ for ratio of operons combined with tetrameric LacI, $k_1$,$_{-1}$,$k_2$ and $k_{-2}$ are reaction rate constants. We assumed that comparing with numbers of operons, that of protein is significantly large enough to ignore whose changes brought by combination of protein and operons. According to law of mass action, we had: $$\frac{dD}{dt}=-k_1Da^2+k_{-1}D_1-k_2Dr^4+k_{-2}D_2$$ $$\frac{dD_1}{dt}=k_1Da^2-k_{-1}D_1$$ $$\frac{dD_2}{dt}=k_2Dr^4-k_{-2}D_2$$ With $a,a_0,r,r_0 > 0$, we found that eigenvalues $\lambda < 0$. Thus, when $t \rightarrow \infty$ we had$\frac{dD}{dt}=\frac{dD_1}{dt}=\frac{dD_2}{dt} = 0$. We also have $D+D_1+D_2 = 1$, so we arrive at: $$D=\dfrac{k_{-2}}{k_2r^4+k_{-2}+\frac{k_1k_{-2}a^2}{k_{-1}}}$$ $$D_1 = \frac{k_1a^2}{k_{-1}}D$$ $$D_2 = \frac{k_2r^4}{k_{-2}}D$$ Denote $a_0 = \frac{k_{-1}}{k_1}$ and $r_0 = \frac{k_{-2}}{k_2}$, we had: $$D = \dfrac{1}{1+\frac{a^2}{a_0}+\frac{r^4}{r_0}}$$ $$D_1 = \dfrac{a^2}{a_0(1+\frac{a^2}{a_0}+\frac{r^4}{r_0})}$$ $$D_2 = \dfrac{r^4}{r_0(1+\frac{a^2}{a_0}+\frac{r^4}{r_0})}$$ What's worth mentioning is that both $a_0$ and $r_0$ are constant that are related to IPTG(mM) and Arabinose(%). $$a_0 = \dfrac{(6.25+ara^2)(1+\frac{iptg^2}{3.24})}{100ara^2}$$ $$r_0 = \dfrac{1}{2000000(\dfrac{0.19}{1+{\frac{iptg}{0.035}}^2+0.01})}$$ During the transcription process, we had: $$D_1\xrightarrow{k_3}R_{a/r}$$ $$D_2\xrightarrow{k_4}R_{a/r}$$ $$D\xrightarrow{k_5}R_{a/r}$$ where $R_{a/r}$denotes mRNAs of either AraC or LacI. $k_3$, $k_4$ and $k_5$ are transcriptional reaction rates constants. For convenience, we assumed that transcription rate for either AraC or LacI are exactly the same. During the translation and folding processes, we have: $$R_a\xrightarrow{t_a}a_{uf}$$ $$R_r\xrightarrow{t_r}r_{uf}$$ $$a_{uf}\xrightarrow{k_{fa}}a$$ $$r_{uf}\xrightarrow{k_{fr}}r$$ where $a_{uf}$ and $r_{uf}$ are unfolded proteins, $t_a$ and $t_r$ are translational reaction rate constants while $k_{fa}$,$k_{fr}$ are folding rates constants. In degradation process, we had: $$R_{a/r}\xrightarrow{d_{a/r}}\varnothing$$ $$a_{uf}\xrightarrow{\lambda f(x)}\varnothing$$ $$r_{uf}\xrightarrow{f(x)}\varnothing$$ $$a\xrightarrow{\lambda f(x)}\varnothing$$ $$r\xrightarrow{f(x)} \varnothing $$ where $d_{a/r}$, $f(x)$ and $\lambda f(x)$ are degradation rate constants.
According to law of mass reaction, we had: $$\frac{dR_a}{dt} = copy_a(k_3D_1+k_4D_2+k_5D)-d_{a/r}R_a$$ $$\frac{dR_r}{dt} = copy_r(k_3D_1+k_4D_2+k_5D)-d_{a/r}R_r$$ $$\frac{da_{uf}}{dt} = t_aR_a-k_{fa}a_{uf} - \lambda f(x)a_{uf}$$ $$\frac{dr_{uf}}{dt} = t_rR_r-k_{fr}r_{uf} - f(x)r_{uf}$$ $$\frac{da}{dt} = k_{fa}a_{uf} - \lambda f(x)a$$ $$\frac{dr}{dt} = k_{fr}r_{uf} - f(x)r$$ where $copy_a$ and $copy_r$ are plasmid copies that are transfected into E.coli.
Transcriptional and translational processes of genes take time and consequently, protein that combined to promoters can be seen as those started transcription process before a specific time interval $\tau$. Thus we converted three ODEs into DDEs: $$D = \dfrac{1}{1+\frac{a(t-\tau)^2}{a_0}+\frac{r(t-\tau)^4}{r_0}}$$ $$D_1 = \dfrac{a(t-\tau)^2}{a_0(1+\frac{a(t-\tau)^2}{a_0}+\frac{r(t-\tau)^4}{r_0})}$$ $$D_2 = \dfrac{r(t-\tau)^4}{r_0(1+\frac{a(t-\tau)^2}{a_0}+\frac{r(t-\tau)^4}{r_0})}$$

Table 1-1 Parameters and variables used in DDEs

Parameters and Variables	Meaning	Value
$a_0$	Dissociation rate constant of AraC binding and unbinding with promoters	$\dfrac{(6.25+ara^2)(1+\frac{IPTG^2}{3.24})}{101ara^2}$
$r_0$	Dissociation rate constant of LacI binding and unbinding with promoters	$\dfrac{1}{2000000\frac{0.19}{1.0+(\frac{IPTG}{0.035})^2+0.01}}$
$D$	Ratio of promoters which don't combine with any protein among all promoters	$\dfrac{1}{1+\frac{a^2}{a_0}+\frac{r^4}{r_0}}$
$D_1$	Ratio of operons combined with AraC dimer	$\dfrac{a^2}{a_0(1+\frac{a^2}{a_0}+\frac{r^4}{r_0})}$
$D_2$	Ratio of operons combined with tetrameric LacI	$\dfrac{r^4}{r_0(1+\frac{a^2}{a_0}+\frac{r^4}{r_0})}$
$a$	AraC protein(activator)
$r$	LacI protein(repressor)
$R_a$	mRNA of AraC
$R_r$	mRNA of LacI
$a_{uf}$	Unfolded AraC
$r_{uf}$	Unfolded LacI
$copy_a$	plasmid containing araC copies that are transfected into E.coli	50
$copy_r$	plasmid containing lacI copies that are transfected into E.coli	25
$k_3$	Transcriptional reaction rates constants of $D_1$	196/min
$k_4$	Transcriptional reaction rates constants of $D_2$	0/min
$k_5$	Transcriptional reaction rates constants of D	5.6/min
$d_{a/r}$	Degradation rate constants	10.54/min
$t_a$	Translational reaction rate constants of araC	90/min
$t_r$	Translational reaction rate constants of lacI	90/min
$k_{fa}$	Folding rate constants of AraC	0.9/min
$k_{fr}$	Folding rate constants of LacI	0.9/min
$f(x)$	Degradation rate constants	$\frac{1080}{0.1+X}/min$
$\lambda f(x)$	Degradation rate constants	$\frac{2887.92}{0.1+X}/min$
$\tau$	Time delay because of the transcription and translation time	2min

Results

We solved these DDEs with R language. To make it more realistic, we also simulated the situation in which lag obeys a specific Gaussian distribution, and the lag $\tau$ changes in every certain interval. The results are below.

(a) A numeric solve of AraC

(b) 5 random tests numeric solve of AraC

Fig 2.(a)A numeric solve of AraC when lag $\tau = 2$min, Arabinose concentration is 5%, IPTG concentration is 1mM, time interval is 0.1min. (b)numeric solve of AraC concentration versus time of 5 random tests, when Arabinose concentration is 0.7%, IPTG concentration is 10mM, and $\tau \sim (2.0,0.3^2)$.

The period of this particular solve is 49.0minutes. The numeric solve of DDEs shows that the supposed oscillator is feasible. On the other hand, interval between every adjacent peak is different in a single random test, thus period is calculated by average intervals. Even so, the average period of each random test is different from each other: $T_1$= 43.95min, $T2$= 47.65min, $T_3$= 40.625min, $T_4$ = 39.375min, $T_5$ = 45.975min. Also, the amplitude of each curve is different. The random solve suggests that extern works should be concerned on forcing the period to be the same. However, due to limited time, we did not have such plan. To analyze whether it is stable against environment changes.

Stability of DDEs

Basing on the equations, since applying their Taylor series makes the equations formed in a linear one keeping the topology of the solution to the original equations, the characteristic equations can be presented: $$E(\mu) = \left(\mu+\lambda\frac{\gamma}{(Ce+a)^2}\right)(\mu+d_{a/r})\left(\mu+k_{fa}+\lambda\frac{\gamma}{(Ce+a_{uf})^2}\right)\\ -k_{fa}T_acopy_a(k_3E_1+k_4E_2+k_5E_3) = 0$$ $$\begin{cases} E_1 = \dfrac{\frac{2a}{a_0}e^{-2\mu \tau}\left(1+\frac{r^4}{r_0}e^{-4\mu \tau}\right)}{\left(1+\frac{r^4}{r_0}e^{-4 \mu \tau}+\frac{a^2}{a_0}e^{-2 \mu\tau}\right)^2}\\ E_2 = \dfrac{2ae^{-2 \mu\tau}\frac{r^2}{r_0}e^{-2 \mu\tau}}{\left(1+\frac{r^4}{r_0}e^{-4 \mu\tau}+\frac{a^2}{a_0}e^{-2\mu\tau}\right)^2}\\ E_3 = \dfrac{\frac{2a}{a_0}e^{-2\mu \tau}}{\left(1+\frac{r^4}{r_0}e^{-4 \mu\tau}+\frac{a^2}{a_0}e^{-2\mu\tau}\right)^2} \end{cases}$$ If there were at least one periodical solution, the equation should have at least one imaginary root.
Denote$ \begin{cases} c_1 = \lambda\frac{\gamma}{(Ce+a)^2}\\ c_2 = d_{a/r}\\ c_3 = k_{fa}+\lambda\frac{\gamma}{(Ce+a)^2}\\ c_4 = k_{fa}T_a(k_3E_1+k_4E_2+k_5E_3)copy_a \end{cases} $
Namely, $E(\mu) = (\mu+c_1)(\mu+c_2)(\mu+c_3)-c_4 = 0$ has imaginary root(s).
Set $\mu = iy$, $y\in R$ Thus $iy(c_1c_2+c_2c_3+c_1c_3-y^2)-(c_1+c_2+c_3)y^2+c_1c_2c_3 = c_4$ is required for imaginary root(s).
Namely, $ \begin{cases} y(c_1c_2+c_2c_3+c_1c_3-y^2) = Im(c_4) \\ -(c_1+c_2+c_3)y^2+c_1c_2c_3 = Re(c_4) \end{cases} $
Namely $\|y(c_1c_2+c_2c_3+c_1c_3-y^2\|^2+\|-(c_1+c_2+c_3)y^2+c_1c_2c_3\|^2=\|c_4\|^2$
Set $x=y^2$
Thus the existence of periodical solution is equal to: $$x^3+(c_1^2+c_2^2+c_3^2)x^2+(c_1^2c_2^2+c_2^2c_3^2+c_1^2c_3^2)x+c_1^2c_2^2c_3^2 \\ =(k_{fa}T_acopy_a)^2\left(\dfrac{\left(\frac{2a}{a_0}\right)^2\left(k_3^2+k_5^2\right)}{\left(1+\frac{r^4}{r_0}+\frac{a^2}{a_0}\right)^4}+\dfrac{\left(\frac{2ar^2}{r_0}\right)^2\left(r^4k_3^2+k_4^2\right)}{\left(1+\frac{r^4}{r_0}+\frac{a^2}{a_0}\right)^4}\right)$$ has at least one positive real root. Denote p, q, $\delta$ as coefficients related to parameters in DDEs. Due to the limited space, these coefficients can be found in supplementary data.
According to the Cardano's Formula, If $q<0$, $\delta>0$,then there exists a center of oscillation. In this case, derivative of period with respect to both Arabinose and IPTG are not 0.
If $q = 0$, $p<0$, then also exists a center of oscillation. In this case, derivative of period with respect to both Arabinose and IPTG are 0.
In fact, in our case - when Arabinose $\in [0,10] $ and IPTG $\in [0,10]$ - belongs to the first aspect, which means both Arabinose and IPTG contribute to the period.

Parameter Range and Sensitivity Analysis

To see how different concentration of both IPTG and Arabinose can affect the period of the oscillator , we solve the DDEs in various values of Arabinose and IPTG using R language. Both Arabinose and IPTG $\in [0,10] $, which is proved to be stable in the previous section.

(a)AraC period map

(b)AraC period Contour

Fig 1.(a)AraC period map with IPTG from 0mM to 10mM, step is 1mM, Arabinose from 0% to 10%, step is 0.1%. (b)Contour of period concerning Arabinose and IPTG. In both figures, color shows the values of period.

The period map can be divided into two areas according to IPTG concentration: 'mountain'(0mM~5mM) and 'plain' (5mM~10mM). These areas can be more clearly seen in contour. The difference between largest and smallest period is approximately 6 minutes, which is insignificant comparing with the scale of period. In 'mountain' area, when IPTG concentration is fixed, the period increases alongside with Arabinose concentration; on the other hand, when Arabinose concentration is fixed, the period increases at first when IPTG concentration rises, then it decreases when IPTG concentration keep on rising. However, in 'plain' area, the period remain steady against either IPTG or Arabinose change.
Due to the computational expenses and our limited computing power, we set a rather large step. Though coarse, we can still grab the big picture of how IPTG and Arabinose can affect the period of the oscillator. We also examined a specific area in a small step size.

Fig 2.AraC period of area whose IPTG is 0~2mM, step is 0.1mM and Arabinose is 4~5%, step is 0.1%. The edge is less sharp than the one above.

By examining specific area in a smaller step size, we found that surface of period is actually rather smooth, which suggests that a large step size does not limit its representativeness, since spline interpolation is used in plotting the discrete data and high accuracy is guaranteed by smoothness of the function.
Then we examined the derivative of AraC's period with respect to both Arabinose and IPTG.

(a)Derivative of AraC's period with respect to Arabinose

(b)derivative of AraC's period with respect to IPTG

Fig 3. derivative of AraC's period with respect to both Arabinose and IPTG. Derivatives are calculated using the data from the bigger step period map.

These figures further suggest the asymmetry behavior of Arabinose and IPTG in terms of affecting AraC's period. Lastly, we examined the role of lag $\tau$ in the period of AraC.

Fig 4.AraC's period against lag $\tau$

In these figures, we could clearly see that AraC's period increases linearly as lag increases. To sum up, the range of period is rather limited while $\tau$ is fixed. In other word, it's rather stable against Arabinose and IPTG changes. On the other hand, we can see that $\tau$ has the biggest influence on AraC's period, which should be concerned firstly while adjusting the period of the oscillator.

References

Team:NTU-Singapore/Modelling/Parameter
Jesse Stricker et al., 2008, Supplementary Information From A fast, robust and tunable synthetic gene oscillator, Nature 456, 516-519