The growth of the HPV was monitored as follows: 10 9 cells and 10 6 viruses in o
ID: 64760 • Letter: T
Question
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.
1. 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?
2. What would be the total number of viruses after the last infectious cycle?
3. 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
1. In 0.1mL 67 plaques were observed in 10-5 dilution.
Number of plaques in 10-8 dilution = 67 x 10-5 / 10-1 x 10-8 = 67 x 10-5 x 109 = 67 x 104 or 6.7 x 105 plaques.
2. After first infectious cycle 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.
After final infectious cycle in 10-8 dilution = 1 x 104 / 10-8 = 1 x 104 x 108 = 1 x 1012 beads/mL
3. A microscopic field of vision revealed 20 beads and 150 virions after first infectious cycle in 1000x dilution.
1 x 1012 beads/mL; for 20 beads = 1 x 1012 x 20 = 2 x 1013
The number of viruses expected in10-7 dilution of the supernatant after the last infectious cycle = 2 x 1013 / 10-7
= 2 x 1013 x 107 = 2 x 1020
150 virions in 10-3 dilution (1000x)
In 10-7 dilution = 150 x 10-3 / 10-7 = 150 x 10-3 x 107 = 150 x 104 or 1.5 x 106 virions.
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