QUESTION 16 1. Through PCR, we have determined the PER3 genotypes for a class of students as follows: H4/H4 = 125 individuals; H4/H5 = 85 individuals and H5/H5=24 individuals. Considering the Hardy Weinberg equilibrium and calculating the Chi-square statistic, what is the overall conclusion based on the results of the Chi-square test? 1. The observed number of individuals per genotype is compatible with the assumption that the class may represent a population in Hardy Weinberg equilibrium 2. The observed number of individuals per genotype is compatible with the assumption that the class represents a population in Hardy Weinberg equilibrium 3. The observed number of individuals per genotype is not compatible with the assumption that the class represents a population in Hardy Weinberg equilibrium 4. The observed number of individuals per genotype is compatible with the assumption that the class may not represent a population in Hardy Weinberg equilibrium QUESTION 17 1. 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). What is the function of the heat shock in the transformation protocol? 1. To stimulate cells to uptake the DNA 2. To increase cell survival 3. To sterilise the solution to avoid contamination 4. To solubilise the DNA to make transformation possible QUESTION 18 1. 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). Assume you were given competent cells of known transformation efficiency (TE). Assume TE= 1x10^8 (note 10^8 means 10 to the power of 8). You want to have about 100 colonies on the P-20 plate. How many nanograms of plasmid should you use in the transformation reaction? 1. 0.05 2. 200.00 3. 5.00 4. 20.00 5. 0.50 6. 50.00 QUESTION 19 1. 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 four serial dilutions that were plated on Amp plates (1 - 4) 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. Consider one Kan plate with 8 colonies. The Amp dilutions plate it is associated with is Amp(3) with 200 colonies. What is the expected number of colonies on the Kan plate once it is normalised to 10^5 seeding cells? 1. 8 2. 100,000 3. 2 4. 200 5. 4 6. 20 QUESTION 20 1. 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. Normalised number of colonies per Kan plate Observed N of plates 0 156 1 110 2 85 3 71 4 15 5 2 6 1 7 0 8 2 9 0 10 3 11 0 12 1 16 1 60 1 79 1 83 1 Using these observed numbers and assuming a Poisson distribution, what is the probability of observing plates with 3 Kanamycin resistant colonies? 1. 0.500 2. 0.271 3. 0.174 4. 0.083 5. 0.062