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The Question

Two yeast cells were placed into a special container to which food was continually added, to keep it at a constant concentration. All other factors were set for optimal yeast growth (for example, temperature, oxygen, and pH). The population was sampled every hour for 21 hours and the results of the estimated population size were recorded in the table below.

Time (hour), Number of yeast cells

(0, 2)

(1, 10)

(2, 15)

(3, 20)

(4, 40)

(5, 60)

(6, 100)

(7, 190)

(8, 260)

(9, 350)

(10, 450)

(11, 530)

(12, 580)

(13, 600)

(14, 600)

(15, 600)

(16, 600)

(17, 600)

(18, 600)

(19, 600)

(20, 600)

What population growth model does the population appear to follow? Write out the mathematical equation it follows.

The Attempt

Looking at the values, I recognize the carrying capacity is reached at hour $13$ with $600$ cells, signifying a logistic growth model. The formula given to me for logistic growth is:

$\frac{dN}{dt} = rN[\frac{(K - N)}{K}]$

From what I have researched:

$dN$ = population size change

$dt$ = time interval

$r$ = (max) growth rate

$N$ = starting population size

$K$ = carrying capacity

So:

$dN = 600 - 2 = 598$?

$dt = 21$ hours

$r = $?

$N = 2$

$K = 600$

Plugging the values in gives me:

$598/21$ = $r2[(600 - 2)/600]$

Update 25/01/19

With some help, I have discovered how to determine the r/rmax value and have the following formula:

$r = (90+100)/2$

$r = 95/hour$

$dN/dt = 95N[(K - N)/K]$

With $N= 2$

$dN/dt=95(2)[(K-2)/K]$

and if $K$ value is carrying capacity, $K = 600$

$dN/dt=95(2)[(600-2)/600]$

$dN/dt=190(598/600)$

$dN/dt=190(0.997)$

$dN/dt=189?$

However, the question asks me to "Write out the mathematical equation it follows." So am I supposed to leave a value unknown?

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    $\begingroup$ The coefficients of the solution to this differential equation are usually derived by regression. $\endgroup$ – Roland Jan 25 '19 at 12:50
  • $\begingroup$ @Roland I was able to make some progress with help. Am I on the right track? $\endgroup$ – Maria Jan 25 '19 at 16:45
  • $\begingroup$ Can you tell how you got the $r$ value, or rather the $r/rmax$ you refer to? $\endgroup$ – stochastic13 Jan 26 '19 at 8:12
  • $\begingroup$ @SatwikPasani From what I understood, the r value refers to the maximum growth rate value, which happens around the 9th hour. So I was directed to average out the growth rate between hours 8 and 9 (90) and 9 and 10 (100) so (90+100)/2 = 95 $\endgroup$ – Maria Jan 27 '19 at 0:11
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I think your problem arises from misunderstanding the $r$ coefficient. I've tried to explain it below, let me know if you need more clarification.

Instantaneous vs Average Growth

Differential equations (like the logistic equation you write) talk about the instantaneous drowth, $dt$ is infinitesimally small, as opposed to your working where you are subtracting the population at two distinct time points and dividing to find $r$, which will only be an average $r$ for that time period. If we were to plot population vs time, we can understand $\frac{dN}{dt}$ as the slope of the equation, and it is apparent that it is changing with time. So, you cannot use distinct time points to find out R, since that will only give you an average $r$.

enter image description here

Logistic Curve Fitting

You can solve the differential equation to get $N$ as a function of time. The ideal way to do this is to try and fit a logistic curve to the data. This way, we already know $K$, $N_0$, and out job is to fit a curve to our data by trying different $r$ values to see which $r$ fits our data the best.

$$N(t) = \frac{KN_0e^{rt}}{K + N_0(e^{rt}-1)}$$

There is a formal way to do this cure fitting. Have a look here, here and here. There will definitely be better resources than these, but I could find these on a cursory google search.

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  • $\begingroup$ Actually, for intervals of time where the slope remains approximately constant (i.e. the central part or exponential part of the growth), you can use the finite approximation of the derivative, taking two pints and dividing by the time interval between them. That will be an approx. of the exponential growth rate in that interval. $\endgroup$ – Alejandro Vignoni Mar 26 at 22:55
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The growth of the yeast can be studied with a Logistic model (i.e. differential equation) as follows: $$ \frac{dX}{dt}=\mu X \left(1 - \frac{X}{X_{max}} \right) $$ This is an ordinary differential equation that tells you how the population of yeast is changing with time (in fact is telling you how the concentration of Yeast $X$ changes with time). The two parameters in the equation are the specific growth rate $\mu$ and $X_{max}$ the carrying capacity following the Verlhust model. We could also write the equation following your notation: $$ \frac{dN}{dt}=r N \left(1 - \frac{N}{K} \right) $$ where $r (\mu)$ is the specific growth rate, $K (X_{max})$ is the carrying capacity, and $N$ is the number of elements in the population.

Note that this is a dynamic model that you need to solve (i.e. integrate the differential equation) to be able to compare with your experimental data. This model tells you how any population of this time behaves not only your Yeast in the mentioned experiment.

The solution of this model is the following Logistic equation: $$ N(t) = \frac{K}{1+\frac{K-N_0}{N_0}e^{-rt}} $$ Where $N_0$ is the initial number of Yeast cells. If you fit this equation (i.e. with least-squares) you get the following values for your experiment (with their 95% confidence intervals):

K = 603.6 (594.3, 612.9)

N0 = 2.108 (1.179, 3.036)

r = 0.6729 (0.6196, 0.7262)

And the following curve. Logistic model curve fitting (using your data) $r$ is the growth rate (in $hr^{-1}$) that is related with the doubling time by the following formula: $$ t_d = \frac{ln 2}{r} = 1.0301 hr $$

Check also this answer about growth models

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