Question

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 prevailed 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.

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.

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

A)

K = 0.693/half lige
    = 0.693/20
    = 0.03465 min-1

No = 1 cell

To saturate 5 L , N = 1010 cell/mL * 5000 mL= 5050,000 cell
use:
N = No*e^(K*t)
5050000 = 1*e^(0.03465*t)
0.03465*t= 15.43
t = 445.5 minutes
   = 7.4 hour
Answer: time to saturate : 7.4 hour

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B)

1 % for saturation means, N = 1% of 5050,000 cell = 50500 cell
use:
N = No*e^(K*t)
50500 = 1*e^(0.03465*t)
0.03465*t= 10.83
t = 312.5 minutes
   = 5.2 hour
Answer: time to reach 1 % saturation : 5.2 hour

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C)

No = 5050000 cell
for 20 L to get saturated, N = = 1010 cell/mL * 20000 mL = 20200000 cell

use:
N = No*e^(K*t)
20200000 = 5050000 *e^(0.03465*t)
0.03465*t= 1.386
t = 40 minutes
Answer: time to saturate 20 L: 40 minutes