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

Answer:

The answer is "$\bold{4.0 \times 10^7 \ \frac{CFU}{mL}}$"

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.

$\to 72+ \frac{65}{2} = 68.5$

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

All of the dilutions you also made:

$\to \frac{7}{17} \times \frac{1}{60} \times \frac{1}{1000} \\\\= \frac{7}{1020000}\\\\= 6.8 \times 10^{-6}$

$\to \frac{14 \ ml \ soup}{14 \ ml} + 20\ mL$

$\to \frac{CFU}{mL} = \frac{Colonial \ number}{Dilution \ Factor \times Volume \ plated}$

            $= \frac{68.5}{6.8 \times 10^{-6} \times 0.25 \ mL} \\\\ = 4.0 \times 10^7$