Question

This viral growth curve was generated as follows: 1.0 ml of a 10-5 dilution of a viral preparation was added to 5 X 10^8 cells at time 0 (final volume: 1.0 ml). A plaque assay done with 0.1 ml of the 10-5 dilution of the original viral preparation gave 100 PFU.

11.What was the M.O.I at point A?

12.If the burst size was 100, what was the viral concentration at point B?

13.How many additional infectious cycle can be completed after the first cycle?

14.At the end of the last cycle, a 10-1 dilution was done. 0.1ml of this dilution was mixed with counting beads to obtain a final concentration of 10^9 beads/ml. Approximately how many viral particles do you expect in a field of vision containing 10 beads?

Answer 1

First question-

Numbers of pfu = 100 over 10^-5 dilution in 0.1 ml

Therefore total numbers of pfu in 0.1 ml = 100 x10⁵

For 1 ml volume total numbers of pfu =100 x10⁵ x 10

For 1 ml volume total numbers of pfu =1 x10⁸

Multiplicity of infection (MOI) = Numbers. of pfu ÷ Numbers. of cells

Numbers of cells = 5 x 10⁸

MOI = 1 x10⁸ ÷ 5 x 10⁸

MOI = 0.2