You have identified the causative agent of the illnesses as Listeria monocytogen

Question

You have identified the causative agent of the illnesses as Listeria monocytogenes and have laboratory data to support your conclusion. Now you will need to determine the bacterial load of the pathogen within the samples collected. You'd like to know whether the illness is caused by a low level exposure (low bacterial numbers) or a high level of exposure (high bacterial numbers). You will use serial dilutions and plate counts to quantify the number of bacteria in your sample. By using a "serial 10" dilution method, each new tube of diluted bacteria will be 10 times more diluted than the previous tube. A 1 mL sample of each of these dilutions can then be plated, and the resulting colony forming units counted to yield an estimate of the number of bacteria in your sample. You count 50 colony-forming units on the fourth plate that corresponds to the 10-4 dilution tube. How many bacteria would you estimate to be in the original sample? 50,000 bacteria per mL 5 times 10^5 bacteria per mL 5 times 10^5 bacteria total 50 times 10^-4 bacteria per mL

Explanation / Answer

The correct option is 5 * 105 cells / ml.

Since the bacterial count was found to be 50 and was counted on the fourth plate corresponding to 10-4 dilution.

The number is calculated as :

No: of bacterial cells counted * Dilution Factor = 50 * 10 -4

Since 1 ml of the sample was plated on the first plate, the count is calculated / ml

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