please answer question 10- 12 The growth of the HPV was monitored as follows: 10
ID: 64727 • Letter: P
Question
please answer question 10- 12
The growth of the HPV was monitored as follows: 109 cells and 106 viruses in one milliliter were incubated under appropriate conditions. Following a first infectious cycle, a direct count and a plaque assay were performed. In the case of the direct count, the mixture was diluted by a factor of 1 000X after which beads were added to obtain a final concentration of 1 X 104 beads/mL. A microscopic field of vision revealed 20 beads and 150 virions. In the case of the plaque assay, 67 plaques were observed on a plate on which 0.1 mL of a 10-5 dilution was assayed.
What was the initial multiplicity of infection?
According to the direct count, what was the volume observed in a field of vision?
According to the direct count, what was the number of virus/mL after the first infectious cycle?
According to the direct count, what was the burst size after one infectious cycle?
According to the plaque assay, what was the number of virus/mL after the first infectious
cycle?
According to the plaque assay, what was the burst size after one infectious cycle?
According to these results, what percentage of the viruses obtained after the first infectious cycle are non-infectious?
What was the new multiplicity of infection (of infectious viruses) after the first growth cycle?
How many infectious cycles, including the first one, could be completed under the above described conditions?
After the last infectious cycle, a 10-8 dilution was made on the supernatant. If 0.1 mL of this dilution is applied to a monolayer of cells, how many plaques would you expect?
What would be the total number of viruses after the last infectious cycle?
How many viruses would you expect to observe in the field of vision of the above described microscope for a 10-7 dilution of the supernatant after the last infectious cycle?
Explanation / Answer
Multiplicity of infection is the average number of viruses that are infecting each cell.
Here 106 viruses are infecting 109 cells, so 106/109 = 0.97 is MOI
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