Team:NWU-CHINA-A/Model

Overview

We aim to develop mathematical models that act as a bridge between the theory of design and our biological work, to further optimize our experiments, understand our results and bring our ideas closer to reality.

Our E.coli is used to product PhaP-AMPs, a kind of fusion protein has antibacterial activity. For that, we simulated the process of PhaP-AMPs expression under IPTG induction and analyse their antibacterial effect quantitatively.

Our Goals:

1. Explore the optimal expression method in PhaP-AMPs product

2. Show the antibacterial effect of PhaP-AMPs

Production of PhaP-AMPs

Our fermentation productions PhaP-AMPs has antibacterial activity. They may cause product inhibition to engineering E.coli. So by assuming cells growth follows Aiba equation, specific growth rate can be represented as:

$\mu =\mu _m\frac{S}{K_S+S}e^{-K_{Ip}P}=\frac{1}{X}\frac{dX}{dt}$(1)

In equation (1), $X$refers to cell concentration, $S$refers to substrate concentration and $P$is product concentration. $\mu $is specific growth rate, as well as $\mu _m$is maximum specific growth rate and $K_S$ is half-velocity constant, $K_{Ip}$ is product inhibition constant.

Product formation kinetics is expressed by Luedeking-Piret equation as:

$\frac{dP}{dt}=\alpha \frac{dX}{dt}+\beta X$ (2)

Because product is independent of cell growth, can be regarded as$ \alpha =0$, so equation (1) can be reformed into equation (3) as follows.

$\frac{dP}{dt}=\beta X$ (3)

Our product, PhaP-AMPs are intracellular protein. Perforating at the plasma membrane of bacteria is their bactericidal mechanism. They can make the content flow out and the bacteria die.

Assumptions:

1.The cell growth and product is uniform.

2.Before induction, there is no leakage expression.

3.When product is in cell, it has no inhibition effect to bacteria.

4.After product concentration reach MIC (minimum inhibitory concentration), PhaP-AMP overflow and take inhibition effect to bacteria.

A set of ordinary differential equations(ODE) is used to describe the changes of substrate, biomass and product at each stage in medium.

$$ \begin{cases} \mu =\mu _m\frac{S}{K_S+S}=\frac{1}{X}\frac{dX}{dt},t\left( P \right) \leqslant t\left( MIC \right)\\ \mu =\mu _m\frac{S}{K_S+S}e^{-K_{IP}P}=\frac{1}{X}\frac{dX}{dt},t\left( P \right) \geqslant t\left( MIC \right)\\ \frac{dP}{dt}=0,t\left( X \right) \leqslant t\left( Xa \right)\\ \frac{dP}{dt}=\beta X,t\left( X \right) \geqslant t\left( Xa \right)\\ -\frac{dS}{dt}=\frac{1}{Y_{X/S}^{*}}\frac{dX}{dt}+\frac{1}{Y_{P/S}}\frac{dP}{dt}+mX\\ \end{cases} $$

$m$ is maintenance coefficient, $Y^*_{\frac{X}{S}}$ is growth yield coefficient and $Y_{\frac{P}{S}}$ is product yield coefficient.

In equation (3), coefficient $\beta =\frac{dP}{Xdt}$ , namely the average protein production per cell in unit time, so we built up a expression model in single cell.

$DNA\xrightarrow{\text{ }a_p\text{ }}mRNA$

$mRNA\xrightarrow{\text{ }b_p\text{ }}protein$

$mRNA\xrightarrow{\text{ }r_m\text{ }}\varphi $

$protein\xrightarrow{\text{ }r_p\text{ }}\varphi $

$\frac{dmr}{dt}=a_p-r_mmr$

$\frac{dPr}{dt}=mrb_p-r_pPr$

$mr$ expresses the number of mRNA per cell and $Pr$ is number of PhaP-AMP per cell.$a_p$ is promoter activity, $r_m$ is degradation rate of mRNA, $b_p$ is average translation rate and $r_p$ is degradation rate of protein.

At the initial time, the number of mRNA and protein is approximated as zero. $mRNA|_{t=0}=0$

$protein|_{t=0}=0$

Figure 1-1 mRNA per cell changes over time

The curve of mRNA-t shows the number of mRNA per cell rising and entering a stable stage around 8 rapidly in 30 seconds. So the number of mRNA per cell can be considered as constant in the whole process of PhaP-AMP expression.

Figure 1-2 PhaP-AMPs per cell changes over time

The curve of PhaP-AMPs-t shows the amount of PhaP-AMPs is stable in about four hours, and gene expression enters a steady state around 2.85$\times $ 106.

After analyzing for individual bacteria, we need to perform further modeling analysis to the whole reaction system.

$\beta =\frac{dP}{Xdt}=c\frac{dPr}{dt}$

$c$ is a constant, namely the mass of a PhaP-AMP molecular per the average mass of a single cell.

PhaP-AMP c(mg)
PhaP-DCD-1L 4.1910e-5
PhaP-HD5 3.9867e-5
PhaP-HD5d5 3.9651e-5
PhaP-HBD3 4.1096e-5
PhaP-P5 3.7824e-5
PhaP-P6.2 3.8090e-5

Use a web site to calculate molecular weight of PhaP-AMPs

(http://www.bio-soft.net/sms/prot_mw.html)

Phase 1: Escherichia coli culture (no IPTG induction)

Figure 2-1
Figure 2-2
Figure 2-3

Although the inflection points in the figure2-1 and figure2-2 appear sharp, when magnified them, the inflection points are still graded naturally.

Figure 2-4

Go through the simulation of cell growth and substrate consumption in 10 hours, we find that the quality of substrate is down to zero in 5 hours, then mass of cell keeps constant without considering the natural cell death. And the specific growth rate is close to maximum in about 3 hours. After that, the mass of cell quickly grows and substrate consumes rapidly. The simulation result is instructive to our experiment. We start the induction with IPTG at third hour.

Phase 2: Induction expression with IPTG

Figure 3-1 the mass of cell changes over time in 6 hours after induction
Figure 3-2 the quality of substrate changes over time in 6 hours after induction
Figure 3-3 the concentration of product changes over time in 6 hours

We simulate the concentration of cell, substrate and product changes over time in 6 hours after IPTG induction and find that substrate has been completely consumed before the concentration of product reaches MIC in one hour’s induction.(The molecular weight of PhaP-HD5 is 24kDa. And the MIC of PhaP-HD5 to E.coli is 2.5μM, namely 60 mg/L in reference.) So there is no obvious inhibitory effect on engineer bacteria. Then E.coli do not grow because of inadequate substrate.

In order to achieve higher yield and more sustainable production, batch culture will be replaced by continuous culture in future. So we simulate the process of continuous culture. Feeding concentrate is 10g/L (Luria-Bertani medium) and the speed of feeding is the same amount as product flowing. The reactor volume is 1 L.

$yield=product_{inside}+product_{outside}$
$yield=product_{inside}+product_{outside}$

When the feeding rate is slow, it can not satisfy the growth and production of engineer bacteria. But when the feeding rate is too fast, the cells are diluted rapidly and level of product become lower. It also cause waste of substrate. When feeding speed is around 0.1 mL/s, yield is higher and it can keep fermenting longer in the reactor.

Antibacterial effect

Assumptions:

1.The bacteria distributes evenly on solid medium.

2.The concentration of PhaP-AMPs distributes evenly in zone of inhibition

3.At the border of inhibition zone, the concentration of PhaP-AMPs is MIC.

We use $l$ as the diameter of inhibition zone. So the area of inhibition zone can be expressed as:

$S_0=\pi \text{(}\frac{l}{2}\text{)}^2=\frac{\pi}{4}l^2$

The amount of bacterium in inhibition zone can be expressed by multiplying bacteria concentrate of inhibition zone method $c_0$ by area of inhibition zone.

$n_{bac}=c_0S_0=\frac{\pi c_0}{4}l^2$

$c_0=10^7cm^{-2}$

Total amount of PhaP-AMPs is equal to sum of the left part in inhibition zone and the consumption part for bacteria in inhibition zone.

$v_0c_x=c_{MIC}S_0h+n_{bac}\alpha $

Volume of a drop solution: $v_0=0.05ml$ Height of solution: $h=0.01cm$

Titer of PhaP-AMPs (consumption quality of PhaP-AMPs per bacteria): $\alpha =\frac{\varDelta n_{Pr}}{n_{bac}}$

Integrate these equation above, we can find the relationship between diameter of inhibition zone and PhaP-AMPs concentration. And we can get titer of PhaP-AMPs.

$l^2=\frac{4v_0}{\pi \text{(}c_{MIC}h+c_0\alpha \text{)}}c_x$

Figure 4-1 fitting of Antibacterial effect
Parameter Description Value Unite Reference
$a_p$ Promoter activity 1.29 $molecules/s$ U. Alon.,2007
$r_m$ Degradation rate of mRNA 0.16 $molecules/s$ U. Alon.,2007
$b_p$ Average translation rate 6 $molecules/s$ U. Alon.,2007
$r_p$ Degradation rate of Protein 3.334e-4 $molecules/s$ U. Alon.,2007
$\mu _m$ Maximum specific growth rate 1.4705e-4 $s^{-1}$ [2]
$K_S$ Half-velocity constant 0.0149 $g/L$ [2]
$m$ Substrate maintenance constant 2.3333e-6 $s^{-1}$ [2]
$Y^*_{\frac{X}{S}}$ Growth yield coefficient 0.4297 $g*g^{-1}$ [2]
$Y_{\frac{P}{S}}$ Product yield coefficient 192.1599 $g*g^{-1}$ [2]

Reference

[1] Zelić B, Vasić-Racki D, Wandrey C, Takors R. Modeling of the pyruvate production with Escherichia coli in a fed-batch bioreactor. Bioprocess Biosyst Eng. 2004 Jul;26(4):249-58. doi: 10.1007/s00449-004-0358-0. Epub 2004 Apr 15. PMID: 15085423.

[2] 马晓轩. 基因重组大肠杆菌高密度发酵生产类人胶原蛋白及其分离纯化[D].西北大学,2006.