Question

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

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

The MPN of coliform organisms per 100 mL is 
 

• A. 1100000
• B. 110000
• C. 1100
• D. 110

Hint:

In the MPN method, use the given table to find the MPN index based on the positive results from each dilution. This method helps in determining the concentration of coliform organisms in water samples.

Answer 1

The Correct Option is B

We are given the results of the multiple-tube fermentation test for the detection of coliform organisms. The sample size, number of positive and negative results, and the corresponding MPN Index are provided. The MPN Index per 100 mL is determined using the combination of positive results from the different dilutions (0.01 mL, 0.001 mL, 0.0001 mL). The combinations of positive results for the three dilutions are: - For 0.01 mL sample, there are 5 positive results (i.e., all tubes tested positive).
- For 0.001 mL sample, there are 3 positive results and 2 negative results.
- For 0.0001 mL sample, there is 1 positive result and 4 negative results.
Based on the MPN Index table for the corresponding number of positive results:
- The combination 5 - 3 - 1 corresponds to an MPN index of 110,000. Thus, the MPN of coliform organisms in 100 mL is \( 110,000 \), which corresponds to option (B). Final Answer: 110000