Question

21 A sample has a density of 1.37x 10 CFUmL a. What sample volume should yield a countable plate? b. Which two dilution tubes could be used to produce this sample volume? How? A sample has a density of 7.9 × 10) CFU/mL. a. What sample volume should yield a countable plate? b. Which two dilution tubes could be used to produce this sample volume? How?

Answer 1

Solution:-

I think this is wrong. The sample volume used would be 10^-3. You went the wrong way. If you used 10^-7, you would have .0137 CFUs on the countable plate.

Use this formula: dilution factor= counted number of CFUs/volume plated (mL) x original CFU/mL

137 counted number of CFUs/ 1 mL x 1.37x10^5 = 0.001= 10^-3

Now we need to know which two dilution tubes can produce a 137 CFU countable plate. We already know the dilution factor is 0.001 or 10^-3. 1/10 x 1/100= 1/1000 or 10^-1 x 10^-2= 10^-3

How are we going to do this? Say we have 1 mL of the original volume and therefore 1.37x10^5 original CFUs. We take that 1 mL of original CFUs and dilute to a tube of 9 mL of distilled water. This gives us our first dilution tube 1/10 or 10^-1. Now we can take .1 mL from the first dilution tube and put it into 9.9 mL of distilled water. This gives us our second dilution tube 1/100 or 10^-2.

Or we can flip it around. Take .1 mL from the original CFU broth and put it in 9.9 mL of distilled water, then take 1 mL of the first dilution tube and put it in 9 mL of distilled water. The latter might be better because we are conserving the original broth, but both should give us a countable plate of 137 CFUs.

Hopefully this is right and is helpful!

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