You spread 0.1 mL volume of a 10^(-6) dilution onto a nutrient agar plate. After 24...
You spread 0.1 mL volume of a 10^(-6) dilution onto a nutrient agar plate. After 24 hours of incubation at 37°C, there were 280 colonies of bacteria on the plate.
A.) What is the original concentration (OCD) of bacteria in the stock sample this dilution came from? (2 points)
B.) Using the OCD value from part A, determine the number of colonies that would be expected to grow on a plate that is inoculated with 0.1 mL volume of 10^(-8) dilution from this same stock of bacteria. (2 points)
Show your work for both.
Homework Answers
A - Original concentration (OCD) = Number of colonies/ dilution * volume plated in ml.
Number of colonies = 280
dilution = 10-6
volume plated in ml. = 0.1
OCD = 280 / 10-6 * 0.1
= 280 * 106 * 10 = 2.80 * 100*106 * 10 = 2.8 * 109 CFU/ml
B -
dilution = 10-8
volume plated in ml. = 0.1
2.8 * 109 = number of colonies /10-8 * 0.1
Number of colonies = 2.8*109 *10-8 * 0.1 = 2.8 or around 3.
Number of expected colonies will be between 2 or 3.
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