Question

Dilution Practice Problem #5 You are given the following bacterial dilution scheme: ml 1 ml 1 ml 1 ml 1 ml CODILE 9ml 9ml 9ml 0.1 ml 0.1 ml colonies) (colonies colonies) Individual tube dilutions: BCD Dilution Factor: Total tube dilutions) IDE AB Plated dilution factor 1. Which plate is the countable plate? (30-300 countable range) 2. Calculate the titer of the original culture in cfu/ml.

Answer 1

Answer :)

The individual tube dilution is the dilution used in the particular tube.

In the first tube, 1 mL of bacterial culture is added to 4 mL. Therefore, the dilution is 1:5. In other tubes, the dilution is 1:10 because each tube gets 1 mL from the previous tube into 9 mL of a fresh tube. Therefore, in each tube except tube A will have 1:10 dilution.

In the dilution factor, dilutions in each tube will increase by a dilution factor from the previous tube.

In the first tube (A) where dilution is 1:5, the dilution factor is 0.2.

In the next tube (B), the dilution is 1:10, which is a 10^-1 dilution factor. But, this tube gets a 0.2 diluted sample from the previous tube. Therefore, the total dilution in the second tube will be 0.2 x 10^-1.

That is how in the third tube (C) the dilution factor would be 0.2 x 10^-2.

The tube D will have a 0.2 x 10^-3 dilution factor.

The tube E = 0.2 x 10^-4.

The tube F = 0.2 x 10^-5.

Plated dilution factor will consider only in G, H, and I. in the tube D the dilution factor is 0.2 x 10^-3 in which 0.1 mL is taken to spread in the plate. Therefore, the plate gets 1:10 dilution (10^-1 factor).

Net dilution factor in the plate G = 0.2 x 10^-3 x 10^-1

Net dilution factor in the plate G = 0.2 x 10^-4

That is how, Net dilution factor in the plate H = 0.2 x 10^-5

Net dilution factor in the plate H = 0.2 x 10^-6

Net dilution factor in the plate I = 0.2 x 10^-7

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