How many 1100 dilutions are necessary to yield a countable p

How many 1:100 dilutions are necessary to yield a countable plate from a stock culture of Clostridium perfringens with 3.2x10^7 cfu/ml?

Solution

Answer : Two of 1:100 dilutions

Countable number of colonies are 30-300

Let’s assume we need 300 colonies

Given:

Dilution factor ?

Number of colonies= 300

Volume of sample = 1 mL to the plate

CFU/ml = 3.2x10^7

cfu/ml = (no. of colonies)/ dilution factor x volume of culture plate

dilution factor= (no. of colonies)/ [CFU/mL x volume of culture plate]

=300/3.2x10^7 x 0.1

=300/3.2 x 10^6

=93.2 x10^-6

Or ~ 100 x 10^-6

It can also be written as 10^-4

Therefore to get 300 colonies we need two times 1:100 dilutions before plating.