You have a culture that contains 46 x 109 cfuml When you pla

You have a culture that contains 4.6 x 109 cfu/ml. When you plate 300 l of a certain dilution you get 95 colonies. What dilution was plated? Be sure to show your work. Use diagram where they would be helpful.

Solution

Ans. Given,

            Concertation of stock culture = 4.6 x 10⁹ cfu/ mL

            Volume of inoculum = 300 µL = 0.300 mL                 ; [1 mL = 1000 µL]

            Plate count = 95 colonies

Step 1: Calculate plate count per mL inoculum

            Since 0.300 mL inoculum gives 95 fcu

            Or,       1.00 mL           -           - (95 cfu / 0.300 mL)

= 316.66 cfu/mL

= 317 cfu/mL [to nearest whole number]

So, concentration of the inoculum = 317 cfu/mL

Step 2: Final dilution required =

concentration of stock culture / concertation of final solution (here, inoculum)

= (4.6 x 10⁹ cfu/ mL) / (317 cfu/mL)

= 1.45 x 10⁷    

Thus, the final dilution is 1.45 x 10⁷. That is, 1.0 mL of the stock solution was diluted to 1.45 x 107 mL to produce inoculum.