1) (10 points) Annotate the README.md file in your logistic growth repo with more detailed information about the analysis. Add a section on the results and include the estimates for N0, r and K (mention which .csv file you used).*

Answer

in this analysis we investigated the dynamics of population growth for a culture of Escherichia coli, with the aim of estimating inital population size, growth rate and carrying capacity. For this analysis I used the first dataset : growth_data <- read.csv("/cloud/project/experiment1.csv")

Firstly, I constructed two plots, of population size (N) against time (t), with no model fitted to the data. The code for the first plot was as follows

ggplot(aes(t,N), data = growth_data) + geom_point() + xlab("t") + ylab("y") + theme_bw()

On the 2nd plot, I log transformed the y axis (N), the code is as follows

ggplot(aes(t,N), data = growth_data) + geom_point() + xlab("t") + ylab("y") + scale_y_continuous(trans='log10')

These plots suggest that the popuation shows logistic growth, where it increases exponentially with time until it reaches carrying capacity and plateaus. During the period of logistic growth, population size is relatively small and resources relatively abundant.

From this point, I use knowledge of these plots, along with the differential equation that relates N at time t with the starting N (N0), the growth rate (r) and the carrying capacity (K), in order to estimate these parameters.

r was estimated by isolating the exponential part of the growth curve. This was arbritraily determined to be when t < 1500. I subset the data by this value and then constructed a linear model.

data_subset1 <- growth_data %>% filter(t<1500) %>% mutate(N_log = log(N))

model1 <- lm(N_log ~ t, data_subset1) summary(model1)

The output gave the estimate for coefficent t to be 0.0100086, this represents the slope, and therefore r The intercept value was given as 6.8941709, this is N0 because it is the population size at the start.

Then, I estimated K by isolating the part of the curve that represents the population at carrying capacity. Once again, I subset the data, and constructed a linear model.

data_subset2 <- growth_data %>% filter(t>2500)

model2 <- lm(N ~ 1, data_subset2) summary(model2)

The output gave an intercept value of 6.00e+10, which is k

To summarise, these were the parameter estimates:

N0 <- exp(6.8941709) #intercept of the 1st regression e^N0

r <- 0.0100086 #gradient of the 1st regression

K <- 6.00e+10 #intercept of 2nd regression

We can add these parameters to the following model:

logistic_fun <- function(t) { N <- (N0Kexp(rt))/(K-N0+N0exp(r*t)) return(N) }

We can see how the predictions (red line) fit the model graphically:

ggplot(aes(t,N), data = growth_data) + geom_function(fun=logistic_fun, colour="red") + geom_point() + scale_y_continuous(trans='log10')

2) (10 points) Use your estimates of N0 and r to calculate the population size at t = 4980 min, assuming that the population grows exponentially. How does it compare to the population size predicted under logistic growth?

Answer

Under the assumption of expontential growth: N(t) = N0e^rt Population size ( at 4980 minutes) = exp6.8941709e ^ (0.0100086 x 4980) = 4.370846e+24 N(t) = 4.370846e+24

Under the assumption of logistic growth: N(t) = K + 0 · t Population size (at 4980 minutes) = 6.00e+10 N(t) = 6.00e+10

This is significantly lower than the population size predicted under the exponential model as it assumes that the population is eventually limited by the resources in the growth medium of the culture and therefore reaches a fixed carrying capcity.

3.(20 points) Add an R script to your repository that makes a graph comparing the exponential and logistic growth curves (using the same parameter estimates you found). Upload this graph to your repo and include it in the README.md file so it can be viewed in the repo homepage.

The code can be found in my logistic growth repository, under Question 3 code.R