Question

You are given a phage lysate and a culture of bacterial cells. You are asked to determine the titer (# of phage/mL). You make three serial dilutions of 100-fold and a final 10-fold dilution of the sample. After infecting the host with 0.5 mL of the last dilution and plating with top agar, the lawn of bacteria generate 40 plaques.

What was the titer of the original phage sample?

Explain the extent to which each of the following would/would not impact your answer to the question above.

a. A student infects the host with .75 mL of the final dilution

b. You accidently perform a final 100-fold dilution

c. Your lab partner infects the host with .5mL of the second dilution.

Answer 1

Number of plaques on the agar plate = 40 plaques per 0.5 mL
= 80 plaques per 1 mL

The final sample is prepared by the following dilutions
1. 100-fold
2. 100-fold
3. 100-fold
4. 10 fold

Dilution factor = 10-7

Concentration of the stock = 80 X 107 plaques per mL
= 8.0 X 108 plaques per mL

a. If 0.75 mL of the final dilution is used,
0.5 mL ---> 40
0.75 mL ---> ?
= (40*0.75)/0.5
= 60 plaques per 0.75 mL
It does not affect the original stock concentration.