Introduction

Using the previously derived expressions from the ODE team we use the Mathworks Simulink package to derive solutions to our system for a range of parameters.

ODE Solution

Solving:

Integrating Factor:

Multiplying both sides by our integrating factor:

R plots

// #reading data
table <- read_excel("C:/Users/Ali/Desktop/igem/Wiki Files/table2.xlsx")

#Vectorizing Data

time <- table$`RFU/OD600`[c(3:15)]

time <- as.numeric(time)


x1 <-table$X__1[c(3:15)]

x1 <- as.numeric(x1)


x2 <- table$X__2[c(3:15)]
x2 <- as.numeric(x2)


x3 <- table$X__3[c(3:15)]
x3 <- as.numeric(x3)

x <- c(x1,x2,x3)

time_ <- c(time,time,time)
#plotting data vs time
#plot(c(time,time,time), c(x1,x2,x3), xlab = 'Time', ylab = 'RFU/OD600')

#Transforming variable

log_x = log(x)



plot(c(time,time,time), log_x, xlab = 'Time', ylab = 'log(RFu/OD600)')

#regression model

fit <- lm(log(x) ~ c(time,time,time))

#regression information
summary(fit)


#graphing best fit line
abline(fit, col='red')


#orginal data points

plot(c(time,time,time), x, xlab='Time', ylab='RFu/OD600')

#transformed prediction line

time_val <- seq(min(time),max(time), by = 13/38)

#prediction
lm2 <- exp(predict(fit,list(time=time_val)))


#plotting prediction

lines(time_val, lm2[c(1:39)], col="red")

R Analysis

R Analysis:
Call:
lm(formula = log(x) ~ c(time, time, time))

Residuals:
     Min       1Q   Median       3Q      Max
-0.58853 -0.15536  0.01303  0.19867  0.44055

Coefficients:
                    Estimate Std. Error t value Pr(>|t|)
(Intercept)          2.87199    0.21773   13.19 1.47e-15 ***
c(time, time, time)  0.15267    0.01142   13.37 9.74e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.2935 on 37 degrees of freedom
Multiple R-squared:  0.8285,	Adjusted R-squared:  0.8238
F-statistic: 178.7 on 1 and 37 DF,  p-value: 9.741e-16

Intercept represents the equilibrium value of LacILov, our intercept: