Production and Conversion Rate Simulink Model

To understand how we get psicose in a deep view, replication, transcription and translation are involved to describe the synthesis of psicose and thus come to our production on a large scale. As we all know, the enzyme will degrade due to the mutation of coding sequence during DNA replication and the transcription error.

Although the rate mutation and transcription error are one in 100 million and one in 10 thousand, the yield of psicose may not decrease because even errors happen in the process of replication and transcription, the type and order of amino acid may not reverse as well as the function of our enzyme which lead to the same catalytic rate of the original one.

However, the butterfly effect tells us that we cannot ignore the small change in our system, even it is little enough for us to ignore. Considering the mutation rate and transcription error, the production of psicose is closely related to the cycling times, error rate and original production. The production of psicose is a function with variables of cycling times, error rate and original production. $$Prd(t)=Prd(t,Times,errRate,oriPrd)$$ Where Times is the cycling times of psicose, errRate is the rate of replication and transcription error, oriPrd is the original production of our system. The errRate is influenced by the replication error rate and the transcription error rate. Meantime, the more replication and transcription errors we get, the lower or higher production we will get. The error rate function is $$errRate(t)=repRat(t)\cdot trsRat(t)$$ Where the replication error rate depends on the length of D-psicose 3-epimerase coding sequence, the multiplication of single error rate define the replication and transcription rate as follows $$repRat(t)=srepRat^{len}(t)$$ The length of mRNA is one third of the cDNA length, here we get $$trsRat(t)=\sqrt[3]{strsRat^{len}(t)}$$ Although errors can result in the decrease of production, the increase of production is also promoting on account of the uncertainty of variation. For a natural evolution system, we consider that the probability of production decrease and increase is equal, decided by the error rate per cycling period. Therefore, the average production growth rate is distributed normally with the mean of 0 and the variance of error rate. The production growth rate is shown as follows $$grsRate \sim N(0,errRate)$$ If we set X as production growth rate,$\sigma^2$ as error rate, then we will get $X \sim N(0,\sigma^2)$ And the probability density function is $$f(x\mid 0,\sigma^2)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{x^2}{2\sigma^2}}$$ The probability of production growth rate is $$P(x_1 < grsRat < x_2)=F(x_2)-F(x_1)=\int_{x_1}^{x_2}f(x)dx$$ The original production is also vital when we simulate the production of our system by reason that the larger our system is, the more psicose we will get in our manufacturing and it makes no sense to compare a certain system with another in a totally different scale. So the standardization and the estimation of original production for psicose is necessary.

There are many variables which have impacts on the original production, such as the quantity of E.coli, the growth rate of E.coli, the conversion rate of psicose and the total amount of substrate which determines the scale of the production system. The original production rate is as follows
$$\frac{\text{d}oriPrd}{\text{d}t}=N\alpha C_{sub}$$ Where $oriPrd$ is the original production of the psicose production system, $N$ is the number of E.coli in the system, $\alpha$ is the conversion rate for D-psicose 3-epimerase, $C_{sub}$ is the concentration of substrate, which represents the concentration of fructose in our production system for psicose.

As for the growth of E.coli, we use Logistic equation to describe the process of E.coli proliferating. $$\frac{\text(d)N}{\text{d}t}=r_1N(1-\frac{N}{r_2})$$

Meanwhile, the concentration of substrate is changing during the process of fermentation in our psicose-production system. We assume the concentration of substrate changes little and can be processed as a constant in order to simplify our process and to find out the discipline of production change, as well as simulating our system in directed evolution method. The concentration of substrate is as follows: $$C_{sub}\approx C_{sub}(t_0)=constant$$ In this way, we can get the original production of psicose from the very beginning. The original production of psicose can be calculated as a constant as below $$oriPrd=\lim_{t_0\rightarrow 0}\int_0^{t_0}\alpha N(t)C_{sub}(t)dt$$ Finally, based on the probability distribution function, the variation of production is defined and we can compare the psicose production in natural system and directed evolution system.

Microfludics Model

In our Lab tour, we find that it is extremely complicated to use pipette to prepare and transfer solutions, especially gradient concentration solutions. High throughput methods to get the gradient concentration of solutions are well needed. In this case, we made a hardware by using the principle of microfluidics and then we simulate it whether it can give us different concentration by using the microfluidics device before the microfluidics chip is finally made.

To get the downstream concentration of high concentration and low concentration, particle collision is used to demonstrate the downstream concentration.

When the fluid flows into the vertical channel of the microfluidics chip, the number of the solute is equal and both are half of the original quantity. In this case, the concentration of the both sides or both directions are the same with the half original concentration. $$m_3=m_4=\frac{1}{2}m_1\qquad m_5=m_6=\frac{1}{2}m_2$$ $$n_3=n_4=\frac{1}{2}n_1\qquad n_5=n_6=\frac{1}{2}n_2$$ $$c_3=c_4=c_1\qquad c_5=c_6=c_2$$ When the liquid flows to the corner of the channels, the momentum is considered as a constant due to the orthogonal relation between the vertical channel and the horizontal channel. According to the momentum conservation theorem, we can get $$m_4\overrightarrow{v_1}+ m_5\overrightarrow{v_2}=(m_4^{'}+m_5^{'})\overrightarrow{v_3}+\int{f(n,l)dt}$$ Regard the process of two solutions converging together as particles colliding, most particles changes their direction of motion, and flow vertically down. While some particles still remain inside the tube. Take these as the loss at time $t$. Assume the probability function of collision loss is $p(n)$, for the loss is in proportion with the amount of particles, the convergence concentration $c_7$ $$c_7=(1-p(n_2))+c_2(1-p(n_1))$$ Hence the local concentration can be determined by the former concentrations. Assume the particle amounts as $n_7$
Take $ m=knc$ into equation 1, we can get $$m_4\overrightarrow{v_1}+ m_5\overrightarrow{v_2}=kn_7c_7\overrightarrow{v_3}+\int{f(n{4,5},ll)dt}$$

Therefore, $$n_7=\frac{m_4 \overrightarrow{v_1} +m_5 \overrightarrow{v_2} - \int{f(n,l)} dt}{kc_7 \overrightarrow{v_3}}$$ $$m_7 = kn_7 c_7$$ And $(m,n,c)$ at the convergence point can be fully determined. To simplify the calculation, consider the resistance $\int f(n,l)dt$ as a constant, and the speed inside the channel keeps constant either.

For the channel at the first column at each row, $$ c(i, 1)=c(1,1)$$ For channels the last column at each row, $$ c(i, i+1) = c(1,2) $$ For the concentration at a certain exit $$ c(i, j)= c(i-1,j-1)\times (1-p(n(i-1,j)) + c(i-1,j) \times (1-p(n(i-1,j-1)) $$ Then by the iterative approach, we can calculate the value of $n$, which is the constraint for the corners and crosses of the channels $$n(i,j)=\frac{m(i-1,j-1)\times \vec{v}+m(i-1,j)\times \vec{v}-\int f(n,l)dt}{kc(i,j)\vec{v}}$$ To solve our model ,the Navier-Stokes Equations are adopted with the assumption of momentum conservation and mass conservation. The results are based on the assumptions as follows:

Assumptions:
1. Newton Fluids: incompressible, its density and viscosity unaffected by concentration change
2. The channel is smooth and the phase change tension can be ignored
3. For the liquid near the channel walls, the velocity in y direction and z direction are 0, and x direction are not zero
4. Due to the microfluidics phase is mainly by the features of the solution, the grativity is ignored and the internal force is ignored.

Then we can get a set of equations:
$$\rho\frac{\partial v}{\partial t}+\rho v \triangledown v=-\triangledown P+\mu \triangledown ^2 v$$ Where, $\rho$ is the density of the solution, $v$ is the velocity, $P$ is the pressure intensitys of the system,$\mu$ is the viscosity of the liquid,$\triangledown$ is Hamiltonian operator: $$\triangledown =\frac{\partial}{\partial x} i+\frac{\partial}{\partial y} j \frac{\partial}{\partial z} k $$ We can get the velocity vector based on knowing the pressure intensity of a certain point in the channel. We apply the velocity vector into the C-D equation, then we can get the concentration $c$ of a certain point: $$\frac{\partial v}{\partial t}+Dv\cdot \triangledown c=D\triangledown^2 c$$ Where $D$ is the diffusion coefficient of the liquid and $c$ is the concentration of the solution. Taking $m,n,c$ we have mentioned, we can get the concentration distribution as follows:

Hairpin Coupling Model

Hairpin coupling model is set up with the help of OUC, we have a good time discussing our hairpin coupling model. Thanks to our collaboration, we can set up a hairpin coupling model as a fresh man. At the very beginning, the most important thing to set up the hairpin coupling model is to determine how can the hairpin structure couple the upstream and downstream expression.

The hairpin structure is the same with the stem-loop structure and forms when there is a similar strand of DNA or RNA due to the base complementary pairing principle. When the hairpin is in the gene and between two coding sequences, the expression levels of these two coding sequence will change because of the hairpin structure. What’s more important is that how can we determine the coupling efficiency based on the hairpin structure.

The hairpin structure is between the two coding sequences and we can use ordinary differential equation to describe the dynamics of genetic circuit. The expression level of protein can be defined as follows: $$\frac{\text{d}[Protein_1]}{ \text{d}t}=k_1-\delta_1 [Protein_1]$$ $$\frac{\text{d}[Protein_2]}{ \text{d}t}=k_2-\delta_2 [Protein_2]$$ When the protein expression level is stable, the expression level of two protein can be defined as: $$Protein_1^{Sta}=\frac{k_1}{\delta_1}$$ $$Protein_2^{Sta}=\frac{k_2}{\delta_2}$$ Then the coupling efficiency can be defined as follows: $$cpEff=\frac{ Protein_2^{Sta}}{ Protein_1^{Sta}}=\frac{\frac{k_2}{\delta_2}}{\frac{k_1}{\delta_1}}=\frac{k_2 \cdot \delta_1}{k_1 \cdot \delta_2}$$ Where $k_1$ means the transcription rate of $protein_1$ and $k_2$ represents the transcription rate of $protein_2$. Due to the research of OUC, we can calculate $k_1$, $k_2$ of coding sequence using a thermodynamics model.

The translation rate of the coding sequence can be defined as the following formula in statistical thermodynamics because with the decrease of Gibbs free energy, the system will become more stable and the ribosome will be more likely binding DNA. Then the $k_1$ can be calculated as $$k_1 \varpropto e^{-\beta \cdot \Delta G_{total1}}$$ We make the formula into an equation: $$k_1 = k_1^’ e^{-\beta \cdot \Delta G_{total1}}$$ Where $\beta$ is the Boltzmann constant, which is $0.45 \pm 0.05 mol/kcal $, $k_1^’$ is the coefficient of the Gibbs free energy function.

The $\Delta G_{total1}$ is the total binding free energy between the ribosome and $5’$ UTR, and can be calculated as follows: $$\Delta G_{total1} =\sum_i \Delta G_i$$ Where $\sum_i \Delta G_i$ means the Gibbs free energy from the different sites interacting with each other, including the free energy of mRNA-rRNA complex, the non-coding sequence between the coding sequence, the free energy of the ribosome-mRNA complex and the free energy of tRNA-ribosome complex.

The free energy above can be calculated by Nucleic Acid Package which is a software for nucleic acid structure and free energy design. Thus, the five parts of free energy can be calculated by the software.

For the calculation of transcription rate $k_2$, we add related factors into $k_2$ as follows to describe the influence of hairpin structure: $$k_2 \varpropto e^{-\beta \Delta G_{total2}}+r_{reinit}$$ Similarly, we make the formula above as an equation: $$ k_2 =k_2^’[ e^{-\beta \Delta G_{total2}}+r_{reinit}]$$ Where $k_2^’$ is the coefficient of the Gribbs free energy function and the hairpin influence function. And it is easy to calculate $\Delta G_{total2}$ due to the formula above: $$\Delta G_{total2} =\sum_i \Delta G_i$$

As for $r_{reinit}$, the downstream gene is likely to be expressed in a high level because the distance of two genes are short while the downstream gene is also more likely to be terminated because of the more complex hairpin structure it has. So $r_{reinit}$ can be calculated as follows: $$r_{reinit}=k_{dis} \cdot k_{com} e^{-\beta \Delta G_{total2}}$$ Where $\Delta G_{total2} =\sum_i \Delta G_i -\Delta G_{coupling}F_{coupling}$, $k_{dis} $ means the distance between two coding sequences with the hairpin structure and $k_{com}$ means the complexity of the hairpin structure. From the research of OUC, we can know $k_{dis}$ is the same with the intergenic distance dependence as follows: $$k_{dis}=\left\{\begin{matrix} {0.0072 \pm 0.0048 \qquad 0 \le d \le 25}\\ {0.0220 \qquad d=-4}\\ {0.0072+(0.0004)\cdot (d+10) \qquad -10\le d \le 25} \end{matrix}\right.$$ Where $d=n_{nt}-3$ and $n_{nt}$ is the number of single nucleotides between the starting site and the terminating site. For the different hairpin structure, the value of $k_{com}$ is different and is defined as the length of the hairpin structure: $$k_{com}= k_{com}^’ e^{-\beta \Delta Len G_{total2}}$$ Where $k_{com}^’$ is the coefficient of the hairpin complex function and $Len$ is the length of hairpin structure determined by the complementary base pairing. Considering $F_{coupling}=\frac{1}{1+C\cdot e^{- \beta \cdot \Delta G_{total2}}}$ as is shown in the hairpin coupling model by OUC, we can get the coupling efficiency as follows: $$cpEff==\frac{k_2 \cdot \delta_1}{k_1 \cdot \delta_2}=\frac{ \delta_2 \cdot k_2^{'} [ e^{-\beta \Delta G_{total2}}+r_{reinit}]}{\delta_1 \cdot k_1^{'} e^{-\beta \cdot \Delta G_{total1}}}$$

Market Prediction Model

In our market model, we’d like to analyze the relationship between the choice of psicose and the market scale. Then, the market od psicose is predicted.

Results and Discussion