Question

A culture of Staphylococcus is diluted as follows:
(1) 20mL are added to 80mL of water.
(2) 10uL from (1) are then added to 9.99mL of water.
(3) A 10^-2 dilution is made from tube # (2).
(4) 100uL from (3) are plated for a pour plate and incubated.
There were 34 colonies counted on one quarter of the plate following incubation.
a) What was the overall dilution?
b) How many cfu/mL were present in the original culture?
c) How many milliliters of water is needed to make a 10^-3 dilution using 1000 uL from the original culture?

Answer 1

Given that no of colonies on one quarter of plate= 34. Considering homogeneous colonies in each quarter, total no of colony present on plate= 34*4=136 colonies.

Calculation of dilution factor:

In step 1) 20mL was added to 80mL of water. Hence dilution is (20ml+80ml)/20ml= 5 fold

(2) 10uL from (1) was then added to 9.99mL of water. Hence dilution is

(0.01ml+ 9.99ml)/ 0.01ml= 1000 fold

(3) A 10-2 dilution is made from tube (2). Hence dilution is 100 fold

a.) Therefore overall dilution= 5x 1000x 100= 5x10⁵ fold

b.) cfu/mL present in the original culture is calculate by dividing the total number of colonies obtained on plate by volume actually plated multiplied with the degree of the dilution.

CFU/ml= n / (s) x (1/dilution fold)

where, n= total number of colonies obtained on plate

              s= volume of sample plated in ml

Here, sample plated = 100ul= 0.1ml

So, CFU/ml= 136/ (0.1)x(1/5x10⁵)

CFU/ml = (136 x 10⁵)/0.02= 6800x 10⁵= 6.8 x 10⁸

CFU/ml=6.8 x 10⁸

c) To make a 10^-3 dilution, let us assume that X ml of water is needed.

10^-3 dilution =1000 fold dilution.

1000 uL =1ml from the original culture

(1ml+ X ml of water)/ 1ml= 1000 fold

1ml+ X ml of water= 1000ml

X ml of water = 1000-1= 999ml

Hence 999ml of water is needed.