What is the concentration of a suspension that grew 84 colon

What is the concentration of a suspension that grew 84 colonies on a plate made with 0.1 ml from a culture with a dilution of 1 : 10,000?

Solution

Ans. Given,

            Concertation of stock suspension = X

            Volume of inoculum = 0.100 mL                   

            Plate (colony) count = 84 colonies (cfu)

Step 1: Calculate plate count per mL inoculum

            Since 0.100 mL inoculum gives 84 colonies (cfu)

            Or,       1.00 mL           -           (84 cfu / 0.100 mL)

= 840 cfu/mL

So, concentration of the inoculum = 840 cfu/mL

Step 2: Calculating back to [Original suspension]

Concentration of original suspension = concentration of inoculum / dilution factor

                                                = (840 cfu/mL) / (1: 10000)

                                                = 8400000 cfu/mL

                                                = 8.4 x 10⁶ cfu/mL

Hence concentration of original suspension = 8.4 x 10⁶ cfu/mL

Approach 2: A 1:10000 dilution means that 1.0 ml of the original suspension is diluted to 10000 mL. So,

I. Total number of cells in 10000 mL diluted solution (inoculum) is equal to the number of cells in 1.0 mL original suspension.

II. 1.0 mL of diluted sample (inoculum gives) 840 colonies.

So, total cfu in 10000.0 mL inoculum = concentration of inoculum x total volume of made upto

                                                = (840 cfu/ mL) x 10000.0 mL

                                                = 8.4 x 10⁶ cells/ mL

So, total number of cfu in 10000.0 mL inoculum = 8.4 x 10⁶ cells/ mL

Also, according to point 1, total number of cells in 1.0 mL original suspension = total number of cells in 10000.0 ml diluted sample = 8.4 x 10⁶ cells

Thus,

Concertation of original suspension = No. of cells / volume

                                                = 8.4 x 10⁶ cells / 1.0 mL

                                                = 8.4 x 10⁶ cells/ mL