We transformed E coli cells with a plasmid modified to contain a 'virulence factor' which would allow growth on media containing the antibiotic kanamycin (Kan). The plasmid confers constitutive resistance to ampicillin (Amp). The bacterial experiment is about understanding whether such a 'virulence factor' confers physiological adaptation to Kan or whether the development of resistance can be explained by random mutations. For each independent transformation we re-suspended the cells from three colonies in Luria broth. For each suspension of cells we plated 100 microliters on a Kan plate. To estimate the number of cells seeded on each Kan plate we made serial dilutions that were plated on Amp plates and we counted the number of cells growing on them. From this we extrapolated how many cells had been seeded on the Kan plate. Then we normalised the Kan results for all the plates, assuming that every plate had been seeded with 10[5] cells. The table below shows the results. Using these observed numbers and assuming a Poisson distribution, what is the probability of observing plates with 2 Kanamycin resistant colonies? Select only one answer. Normalised number of colonies per Observed N of
We transformed E coli cells with a plasmid modified to contain a 'virulence factor' which would allow growth on media containing the antibiotic kanamycin (Kan). The plasmid confers constitutive resistance to ampicillin (Amp). The bacterial experiment is about understanding whether such a 'virulence factor' confers physiological adaptation to Kan or whether the development of resistance can be explained by random mutations. For each independent transformation we re-suspended the cells from three colonies in Luria broth. For each suspension of cells we plated 100 microliters on a Kan plate. To estimate the number of cells seeded on each Kan plate we made serial dilutions that were plated on Amp plates and we counted the number of cells growing on them. From this we extrapolated how many cells had been seeded on the Kan plate. Then we normalised the Kan results for all the plates, assuming that every plate had been seeded with 10[5] cells. The table below shows the results. Using these observed numbers and assuming a Poisson distribution, what is the probability of observing plates with 2 Kanamycin resistant colonies? Select only one answer. Normalised number of colonies per Observed N of
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O 5. 0.151
![We transformed E coli cells with a plasmid modified to contain a 'virulence factor' which would allow growth on media containing the antibiotic kanamycin (Kan). The plasmid confers constitutive
resistance to ampicillin (Amp). The bacterial experiment is about understanding whether such a 'virulence factor' confers physiological adaptation to Kan or whether the development of resistance can
be explained by random mutations.
For each independent transformation we re-suspended the cells from three colonies in Luria broth. For each suspension of cells we plated 100 microliters on a Kan plate. To estimate the number of
cells seeded on each Kan plate we made serial dilutions that were plated on Amp plates and we counted the number of cells growing on them. From this we extrapolated how many cells had been
seeded on the Kan plate. Then we normalised the Kan results for all the plates, assuming that every plate had been seeded with 10[5] cells. The table below shows the results.
Using these observed numbers and assuming a Poisson distribution, what is the probability of observing plates with 2 Kanamycin resistant colonies? Select only one answer.
Normalised
number of
colonies per
Kan plate
0
6
12345 12
DRTNINOMOT
7
8
9
10
Observed
N of
11
plates
151
20
13
7
2
1
1
2
0
0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fef1d2d9f-af7d-4b72-8500-32049913a884%2F0d196f74-ef65-49f0-8c31-9536d31e90ef%2Fqdcm93w_processed.jpeg&w=3840&q=75)
Transcribed Image Text:We transformed E coli cells with a plasmid modified to contain a 'virulence factor' which would allow growth on media containing the antibiotic kanamycin (Kan). The plasmid confers constitutive
resistance to ampicillin (Amp). The bacterial experiment is about understanding whether such a 'virulence factor' confers physiological adaptation to Kan or whether the development of resistance can
be explained by random mutations.
For each independent transformation we re-suspended the cells from three colonies in Luria broth. For each suspension of cells we plated 100 microliters on a Kan plate. To estimate the number of
cells seeded on each Kan plate we made serial dilutions that were plated on Amp plates and we counted the number of cells growing on them. From this we extrapolated how many cells had been
seeded on the Kan plate. Then we normalised the Kan results for all the plates, assuming that every plate had been seeded with 10[5] cells. The table below shows the results.
Using these observed numbers and assuming a Poisson distribution, what is the probability of observing plates with 2 Kanamycin resistant colonies? Select only one answer.
Normalised
number of
colonies per
Kan plate
0
6
12345 12
DRTNINOMOT
7
8
9
10
Observed
N of
11
plates
151
20
13
7
2
1
1
2
0
0
![For each independent transformation we re-suspended the cells from three colonies in Luria broth. For each suspension of cells we plated 100 microliters on a Kan plate. To estimate the number of
cells seeded on each Kan plate we made serial dilutions that were plated on Amp plates and we counted the number of cells growing on them. From this we extrapolated how many cells had been
seeded on the Kan plate. Then we normalised the Kan results for all the plates, assuming that every plate had been seeded with 10[5] cells. The table below shows the results.
Using these observed numbers and assuming a Poisson distribution, what is the probability of observing plates with 2 Kanamycin resistant colonies? Select only one answer.
Normalised
number of
colonies per
Kan plate
0
1
2
3
4
5
6
7
8
9
10
11
12
16
60
79
83
Observed
N of
plates
151
20
13
7
2
1
1
1
2
0
3
0
1
1
1
1
1
E](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fef1d2d9f-af7d-4b72-8500-32049913a884%2F0d196f74-ef65-49f0-8c31-9536d31e90ef%2F83k0uv_processed.jpeg&w=3840&q=75)
Transcribed Image Text:For each independent transformation we re-suspended the cells from three colonies in Luria broth. For each suspension of cells we plated 100 microliters on a Kan plate. To estimate the number of
cells seeded on each Kan plate we made serial dilutions that were plated on Amp plates and we counted the number of cells growing on them. From this we extrapolated how many cells had been
seeded on the Kan plate. Then we normalised the Kan results for all the plates, assuming that every plate had been seeded with 10[5] cells. The table below shows the results.
Using these observed numbers and assuming a Poisson distribution, what is the probability of observing plates with 2 Kanamycin resistant colonies? Select only one answer.
Normalised
number of
colonies per
Kan plate
0
1
2
3
4
5
6
7
8
9
10
11
12
16
60
79
83
Observed
N of
plates
151
20
13
7
2
1
1
1
2
0
3
0
1
1
1
1
1
E
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