The growth of the HPV was monitored as follows: 10 9 cells and 10 6 viruses in o
ID: 64758 • 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. According to the plaque assay, what was the burst size after one infectious cycle?
2. According to these results, what percentage of the viruses obtained after the first infectious cycle are non-infectious?
3. What was the new multiplicity of infection (of infectious viruses) after the first growth cycle?
4. How many infectious cycles, including the first one, could be completed under the above described conditions?
Explanation / Answer
1. Burst size = no. of phage progeny produced per cell.
No. of phage per ml is 67 *10-5/0.1= 67 *10-4
post dilution pfu density * burst size = pfu/ml
so, burst size = 67*10-4 /104 = 6.7 *10-7
*** hope it helps....
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.