A nutrient agar plate labeled 105 mL produced 154 colonies a

A nutrient agar plate labeled 10–5 mL produced 154 colonies after incubation.

a. What was the cell density in the original sample?

b. What combination(s) of volumes and dilution factors could have been used to inoculate this plate? (note two possibilities)

Solution

A. As dilution is 10^-5. And it produced 154 colonies. Then THE the cell density in original sample = 154 / 10^-5

= 154 x 10^5

= 15400000 cells/ ml

B. serial dilution procedure used to get this plate. Initial culture 1 ml taken. To this 9 ml saline added. It gives 10^-1 dilution. From this dilution, take 1 ml. To this add 9 ml saline. Gives 10^-2 dilution. In this way for 10^-4 also produced. From 10^-4 we can get 10^-5 dilution in 2 ways. Either 1:9 or 10:90 ratios of 10^-4 dilution and saline. These 10^-5 dilution gives the above said plate.