From a bacterial suspension you spread 0 1 mL onto a petri d

From a bacterial suspension, you spread 0 1 mL onto a petri dish and gel 40 colonies. From the same suspension, you spread 0.2 mL and gel 120 calories. The suspension contains _____ viable bacteria per ml. From a bacterial suspension showing 100 bacteria in 10 boxes of a Petroff-Hauser Coining Chamber, you remove 0.1 ml and place it into 9 9 ml sterile diluents, mix, and repeat this by transferring from the second lube 0.1 mL into 9 9 mL sterile diluents in another tube. From the last tube, you plate 0.1 mL and you produce 100 colonies on a spread plate. The percent viability of the suspension is _____%. From a bacterial suspension, you spread 0.1 mL onto a petri dish and get 40 colonies. From the same suspension, you spread 0 2 mL and get 120 colonies. The suspension contain _____ viable bacteria per mL

Solution

Answer:

1. The problem can be approached in 2 ways:

a. A strong argument has been made to take the dilution providing the larger number of CFU in the countable range. This approach minimizes two concerns: errors in the estimates increase with increasing serial dilutions, and errors in the estimates increase with decreasing plate counts. Use of the smaller dilution (e.g., 1:10 vs. 1:100) could be justified from this perspective.

So, working with the 0.2 ml sample:

Dilution = 5 times = 1/5

Number of viable bacteria = 5 * 120 = 600

b. Calculating the number of viable bacteria/mL for each dilution separately and averaging the two estimates.

Number of viable bacteria for the 0.2mL sample = 600

Number of viable bacteria for the 0.1mL sample = 10 * 40 = 400

So, average number of viable bacteria/mL of the sample = (600+400)/2 = 1000/2 = 500

2. Number of bacteria/mL of the sample = (Total cells counted * 2.0 * 10⁷ * dilution factor)/ (Number of small squares counted)

[The value 2.0 * 10⁷ comes from the depth of the measuring chamber being used. It can be different with change in depth size]

= 100 * 2 * 10⁷ / 10 = 20 * 10⁷ = 2 * 10⁸

1st dilution: 1/100

2nd dilution: 1/100. So, the combined dilution now = 10^-4

From the 2nd dilution, 0.1 mL is plated. So, total dilution now = 10^-5

Therefore, number of viable bacteria from this sample = 100/10^-5= 10⁷

So, percent viability = (10⁷/ 2*10⁸)*100 = 0.5 *100/10 = 5%