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90. The population of E. coli in a cell culture doubles every 20 minutes. The population of P. aeruginosa in a cell culture doubles every 120 minutes. Which of the following is false?

A. When 240 minutes have passed, the population of E. coli will have doubled 12 times.
B. The growth rate of E. coli in its cell culture is six times as fast as the growth rate of P. aeruginosa in its cell culture.
C. When the population of P. aeruginosa has doubled 8 times in its cell culture, the population of E. coli will have doubled 42 times in its cell culture.
D. When E. coli has doubled 30 times in its cell culture, then P. aeruginosa will have doubled 5 times in its cell culture.

Sagot :

GinnyAnsweridn
Let's analyze each statement to determine its correctness:

### Statement OA:
"When 240 minutes have passed, the population of E. coli will have doubled 12 times."

- E. coli doubles every 20 minutes.
- Total time given: 240 minutes.
- Number of times E. coli doubles in 240 minutes: [tex]\( \frac{240 \text{ minutes}}{20 \text{ minutes}} = 12 \)[/tex].

This statement is true.

### Statement OB:
"The growth rate of E. coli in its cell culture is six times as fast as the growth rate of P. aeruginosa in its cell culture."

- Doubling time of E. coli: 20 minutes.
- Doubling time of P. aeruginosa: 120 minutes.
- The growth rate comparison: [tex]\( \frac{20 \text{ minutes}}{120 \text{ minutes}} = \frac{1}{6} \)[/tex].

E. coli doubles 6 times faster than P. aeruginosa, making this statement true.

### Statement OC:
"When the population of P. aeruginosa has doubled 8 times in its cell culture, the population of E. coli will have doubled 42 times in its cell culture."

- P. aeruginosa doubles every 120 minutes.
- When P. aeruginosa has doubled 8 times: [tex]\( 8 \times 120 \text{ minutes} = 960 \text{ minutes} \)[/tex].
- E. coli doubles every 20 minutes.
- Number of times E. coli will have doubled in 960 minutes: [tex]\( \frac{960 \text{ minutes}}{20 \text{ minutes}} = 48 \)[/tex].

According to the values above, E. coli will have doubled 48 times, not 42 times. Therefore, this statement is false.

### Statement OD:
"When E. coli has doubled 30 times in its cell culture, then P. aeruginosa will have doubled 5 times in its cell culture."

- E. coli doubles every 20 minutes.
- When E. coli has doubled 30 times: [tex]\( 30 \times 20 \text{ minutes} = 600 \text{ minutes} \)[/tex].
- P. aeruginosa doubles every 120 minutes.
- Number of times P. aeruginosa will have doubled in 600 minutes: [tex]\( \frac{600 \text{ minutes}}{120 \text{ minutes}} = 5 \)[/tex].

This statement is true.

### Conclusion:
The false statement is OC.
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