You added 0.1ml stock bacteria to 9.9ml media (tube 1). After mixing you transferred 0.1ml diluted bacteria and mixed with 9.9ml media (tube 2). You took 0.1ml diluted bacteria from tube 2 and...

You added 0.1ml stock bacteria to 9.9ml media (tube 1). After mixing you transferred 0.1ml diluted bacteria and mixed with 9.9ml media (tube 2). You took 0.1ml diluted bacteria from tube 2 and plated. Next day, you counted 100 colonuies. in the plate. What is the concentration of bacteria in the original (stock) container?

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mvcdc | Student, Graduate | (Level 2) Associate Educator

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The best way to solve this problem is to identify the dilutions first.

The first dilution involves adding 0.1 ml of stock to 9.9ml media. This dilution is `1/(0.1 + 9.9) = 1/10` .

The second dilution involves getting 0.1ml of the diluted bacteria, and transferring it to 9.9ml of a second media. This dilution is again `1/(0.1 + 9.9) = 1/10` .

Total dilution then is:

`1/10 * 1/10 = 1/100 = 10^(-2)`

The last step is to identify the number of colonies, and the amount that gave rise to these colonies. The amount plated is 0.1ml of the final dilution, which gave rise to 100 colonies. This means that we have:

`(100 c o l o n i e s)/(0.1ml) = 1000 (c o l o n i e s)/(ml-p l a t e d)`

This means that for every mililitre of the diluted solution plated, we get 1000 colonies. To get the original number of colonies in the stock solution, we simply divide this value by the total dilution:

`(1000 ( c o l o n i e s)/(m l - p l a t e d))/(10^(-2)) = 1000 times 10^(2) (c o l o n i e s)/(ml)`

This means that in the original stock solution, there are 1000*100 = `1 times 10^5`colony-forming units per ml.

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