Question

A 50 mL sample of soup was found to be contaminated with E. coli. A dilution of the soup was prepared by adding 14 mL of the soup to 20 mL of water. This was then diluted 1/60, and finally 10-3. Then .25 was added in to 2 plates. The first plate had a colony count of 72 and second had 65.

To do the dilution series does the 50 need to be added?

Answer 1

The answer is "
`\bold{4.0 * 10^7 \ (CFU)/(mL)}`"

Step-by-step explanation:

The very first thing we must do is figure out where the plate was really a countable sheet. By default, at least 30 and 300 colonists must have been on a computable plate. It is deemed incorrect beyond and below this range. We also will pick the studded 0.25 size, leading towards 72 and 65 colonies. Now calculate the double average.

`\to 72+ (65)/(2) = 68.5`

We need to calculate its dilution factor for both the second item. Increase for all of this

All of the dilutions you also made:

`\to (7)/(17) * (1)/(60) * (1)/(1000) \\\\= (7)/(1020000)\\\\= 6.8 * 10^(-6)`

`\to (14 \ ml \ soup)/(14 \ ml) + 20\ mL\\\\`

`\to (CFU)/(mL) = (Colonial \ number)/(Dilution \ Factor * Volume \ plated)\\\\`

`= (68.5)/(6.8 * 10^(-6) * 0.25 \ mL) \\\\ = 4.0 * 10^7`