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.
