Lab 6 - Faecal contamination

Question 1

If pathogenic microbes are present in very low numbers in water, how is contamination assessed?

Answer 1

Detect microbes known to occur normally in GI tract as an indicator

Question 2

Which bacteria are included in faecal coliform bacteria, and which one is included in the Australian standard?

Answer 2

• Citrobacter, Enterobacter, Escherichia, Klebsiella
• Australian standard = E. coli

Question 3

Why is E. coli the most sensitive indicator of faecal pollution?

Answer 3

1. Present in gut of humans and mammals
2. Doesn’t generally inhabit other environments
3. Numerically dominant amongst coliforms
4. Tests are specific, easy, cheap, well established and widely used

Question 4

Why are dilutions included in the process?

Answer 4

In case of high numbers of bacteria in the sample which would form confluent growth on agar instead of colonies

Question 5

Explain the shortened method of water testing used in this experiment

Answer 5

1. Water sample dilutons passed through filter
2. Filter paper transferred to m-FC plates selective for coliform bacteria
3. Incubate at 44C selective for thermotolerant coliform E. coli
4. Stains react to acid formation by faecal coliforms = detect and number
5. Inocualate colonies into peptone water and incubate
6. Kovac’s reagent to test for indole produced by E. coli’s tryptophanase

Question 6

What stains are in the m-FC plates and what do they do?

Answer 6

React to acid formation during lactose fermentation by faecal coliforms

Question 7

How is the step involving Kovac’s reagent able to select for E. coli only?

Answer 7

E. coli has tryptophanase while most coliforms don’t. It cleaves tryptophan and produces indole detected by Kovac’s reagent by turning red

Question 8

How do you calculate the faecal coliform count (CFU/100ml)?

Eg. 280 colonies on 10^-5 plate and 100uL spread on plat

Answer 8

Divide number of colonies by sample volume filtered, then multiply by 100, then multiply by DF

Eg. 280 x 0.1mL or 10 bc 10 x 100uL in 1 mL = 2800
2800 x DF of 10^5 = 280 000 000
2.8 x 10^8 cells/mL

Question 9

What is Australia’s standard for water testing?

Answer 9

No E. coli in any 100mL sample of water

Question 10

What is a limitations of this method of water testing?

Answer 10

• Not practical to test for pathogens in every water sample collected
• Coliforms could be from environment

Question 11

Explain how Kovac’s reagent is used to determine positive/negative results

Answer 11

When added to peptone water tubes, they should turn red when positive