Question

Part A

Escherichia coli (E. coli) can under optimal conditions double every 20.0 minutes. How long would it take a single cell to saturate a 5.0 L culture assuming that optimal growth conditions prevaled and that no cells died? The maximum cell denisty in a saturated culture is 10¹⁰ cells mL^-1. Enter your answer to the nearest tenth of an hour.

Part B

Consider the information in Part A and decide how long it would take for the 5.0 L culture to attain 1.0% saturation? Again, enter your answer to the nearest tenth of an hour.

Part C

Imagine that just after the culture described above became saturated a researcher poured the culture into 15.0 L of fresh growth media. How long would it take for this 20.0 L (total volume) cuture to become saturated, again, assuming that optimal conditions prevailed and that no cells died. Enter your answer to the nearest minute.

Answer 1

First you can construct a table like this:

time (min)	No cells
0	1
20	2
40	4
60	8
80	16
100	32
120	64

The growth is exponential, as you can see in the graph (time Vs Cells), so you can make an exponential regression, by calculating the logarithm of the population:

the final equation is:

For part A:

First you need to convert 5 L of saturated culture into cells using the density:

Now using the formula above you can calculate the time for saturating the media:

For Part B:

You can consider either a 1% of saturation of the 5L culture or 1% of th density of saturation, I am going to use the 1% of the density:

Again converting that density, into the 5L culture:

And using the exponetial equation you can calculate the time:

For the Part C you need to do again the regression because when you poured the culture you already have 5050000cells(from the saturated 5L) so you need to calculate using this in your time 0.

time (min)	cells
0	5.05x106
20	10.1x106
40	20.2x106
60	40.4x106
80	80.8x106

you got the following two graphics:

So the equation becomes:

the equation fits:

Now you need to calculate the total number of cells in a 20 L saturated culture:

Now using the fitted equation: