Answer:
Explanation:
A. What is the doubling period of E. coli under optimum conditions?
b. If you start a culture at 5 PM with one E. coli cell, how many will you have at 8 AM the following morning, assuming that the bacteria experience optimum conditions all night? Show your work!
c. In fact, bacteria use up all the nutrients in standard medium by the time they reach a density of about [tex]10^{10}[/tex] cell/ml. In order to feed all the bacteria in part (b), how many liters of medium would you need? To put this in perspective, remember 1 liter is about 1 quart!
ANSWER
a. The doubling period of E. coli under optimum condition is 20 minutes.
b. The time from 5PM to 8AM is 15 hours.
Assuming the bacteria experiences optimum growth, that means its doubling time is 20 minutes.
In one hour, it would have doubled 3 times ([tex]\frac{60 minutes}{20 minutes}[/tex])
In 15 hours, it would have doubled [15*3] times = 45 times
That implies 45 generations
The total number of generations after n generation is calculated using the formula:
[tex]2^{number of generations}[/tex] * initial number of bacteria
since number of generation is 45, and initial number of bacteria is 1
=[tex]2^{45}[/tex] *1
=35184372088832
=3.5*[tex]10^{13}[/tex]
Therefore, there will be 3.5*[tex]10^{13}[/tex] E. coli cells
c. If [tex]10^{10}[/tex] cells requires 1ml of nutrient
Then 3.5*[tex]10^{13}[/tex] cells require:
[tex]\frac{3.5*10^{13} }{10^{10} }[/tex] * 1ml
=3500ml
To convert to liter
1000ml = 1 liter
3500ml = [tex]\frac{3500}{1000}[/tex]
=3.5 liters
You will need 3.5 liters of medium
