You spread 0.1 mL volume of a 10^(-5) dilution onto a nutrient agar plate. After 24 hours of incubation at 37°C, there were 223 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^(-7) dilution from this same stock of bacteria. (2 points)

The correct answers are 2.23 * 10^8 CFU/ml and 2 colonies.

Explanation:

Based on the given information, 0.1 ml is the amount of bacterial culture plated, 10^-5 is the dilution factor and the number of bacterial colonies produced is 223.  

A) 223 is the number of colonies produced when 0.1 ml of the culture is plated. Therefore, the number of colonies produced when 1 milliliter of bacterial culture plated us (223/0.1)*1 = 2230

The calculation of the CFU/ml is done by using the formula,  

CFU/ml = Number of colonies per ml plated / dilution factor

Thus, 2230/10^-5

= 2230 * 10^5 or 2.23 * 10^8 CFU/ml

B) The number of colonies, which would grow on a plate, which is inoculated with 0.1 ml volume of 10^-7 dilution from the similar bacterial stock will be calculated as,  

CFU/ml = Number of colonies per ml plated/ dilution * volume plated.  

2.23 * 10^8 CFU/ml = Number of colonies per ml plated / 10^-7 * 0.1

Number of colonies per ml plated = 2.23 * 10^8 * 0.1 / 10^7 = 2.23 or 2 colonies.  

option number one would be youre best option

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