Question

"Let's say you have a T4 phage stock solution that is 1.4x10^8 PFU/mL. How would you dilute this solution and add it to 500g of dirt so that each gram of mixed dirt would have less than 200 phage/gram of soil?"

Answer 1

It is given that 200 phage/gram of soil are required, therefore for 500 grams of soil, less than 500 x 200 = 100000 = 1 x 10⁵ phage for 500 grams of soil must be there.

Also, it is given that the stock solution of T4 phage has a concentration of 1.4 x 10⁸ PFU/mL, which means that 1.4 x 10⁸ phage are present in 1mL of stock solution, the dilution factor can be calculated as follows:

Final number of cells = initial cells x Dilution factor

1 x 10⁵ = 1.4 x 10⁸ x dilution factor

dilution factor = 1 x 10⁵ / 1.4 x 10⁸ = 7.14 x 10^-4 = 7.14/10⁴

Therefore, the stock solution will have to be diluted by approximately 10⁴ to 10⁵ times and then mixed with 500 grams of soil in order to have a concentration of 200 phage/gram of soil for 500 grams total soil.