Question

A water sample is analyzed for coliform organisms by the multiple - the fermentation method. The results of the confirmed test are as follows:

Sample size (mL)	Number of positive results out of 5 tubes	Number of negative results out of 5 tubes
0.01	5	0
0.001	3	2
0.0001	1	4

The most probable number (MPN) of coliform organisms for the above results is to be obtained using the following MPN Index.

MPN Index for Various Combinations of Positive Results when Five Tubes used per Dilution of 10.0 mL, 1.0 mL, and 0.1 mL
Combination of positive tubes	MPN Index per 100 mL
0 – 2 – 4	11
1 – 3 – 5	19
4 – 2 – 0	22
5 – 3 – 1	110

The MPN of coliform organisms per 100 mL is

1. 110000
2. 1100000
3. 110
4. 1100

Answer 1

Correct Answer - Option 1 : 110000

Concept:

Procedure to find MPN:

• Select a series where three tubes each have positive results. (not necessary but recommended).
• Use sample size corresponding to which the MPN value is larger.
• The MPN index and confidence limit are given corresponding to the sample size of 10 mL, 1 mL, and 0.1 mL.

So if we are selecting sample size as 1 mL, 0.1 mL, and 0.01 mL, then the series of the sample used is one-tenth of the 10, 1, and 0.1 sample size, therefore multiply the MPN index value and confidence limit by 10 and similarly so on.

Calculations:

Since only three sample dilution has been taken.
We will search for a combination of +ve tubes as 5-3-1 in 0.01, 0.001 & 0.0001 ml from the table.
Since the MPN chart given MPN value of 110 per 100 ml for 10 ml, 1 ml & 0.1 ml dilutions than the result so obtained is multiplied by \(\frac {10\ ml}{0.01\ ml}\) = 1000, to obtain MPN per 100 ml for 0.01, 0.001, 0.0001 ml dilutions.

Hence, MPN per 100 ml = 110 × 1000 = 110000.

Most probable number (MPN):

MPN stands the foremost probable number obtained in multiple tube fermentation tests. In this test, the water sample of different dilution ratios is mixed with lactose broth (nutrient for coliforms) and is incubated for 48 hours. During incubation, coliforms consume lactose broth and reduce it to acids and gasses. The presence of acid or gases in a sample tube indicates a positive test result otherwise negative.

Hence, No of positive tubes = No of tubes in which acid or gasses is detected