Question

The original cell density in a sample is 4.6 x 109 CFU/mL. Which sample volume would yield a countable plate? (Express your answer as 10-x mL)

Answer 1

Use the formula:

dilution factor= counted number of CFUs/volumeplated (mL) x original CFU/mL

= 4.6×10^9 counted number of CFUs/ 1 mL x 4.6E12= 0.001= 10^-3

Now we need to know which two dilution tubes can produce a 4.6×10^9 CFU countable plate. We already know that the dilution factor is 0.001or 10^-3. 1/10 x 1/100= 1/1000 or 10^-1 x 10^-2= 10^-3

Say we've got 1 mL of the original volume and therefore 4.6× 10^9 original CFUs. We take 1 mL of original CFUs and dilute them into a tube of 9 mL of distilled water. Thisgives us our first dilution tube 1/10 or 10^-1. Now we can take it.1mL from the first dilution tube and put it into 9.9 mL of distilledwater. This gives us our second dilution tube 1/100 or 10^-2.

Or we can flip around it. Take .1 mL from the original CFU brothand put it in 9.9 mL of distilled water, then take 1 mL of thefirst dilution tube and put it in 9 mL of distilled water. Thelatter may be better because we keep the originalbroth, but both should give us a countable plate of 4.6×10^9 CFUs.