A researcher sets up the following dilution to determine phage titer of a sample.

For Tube A, 0.5 mL stock is added to 4.5 mL diluent. For Tube B, 1 mL from Tube A is mixed with 9 mL diluent. For Tube C, 0.5 mL from Tube B is added to 9.5 mL diluent.

From Tube C, 0.025 mL of this is used to set up a plaque assay with a susceptible host. After incubation, 88 plaques are counted. What was the phage titer in the stock?

Just type in the big number below without units, Canvas doesn't allow exponents. However, please use scientific notation on your assignments and exams.

Question

A researcher sets up the following dilution to determine phage titer of a sample.

For Tube A, 0.5 mL stock is added to 4.5 mL diluent. For Tube B, 1 mL from Tube A is mixed with 9 mL diluent. For Tube C, 0.5 mL from Tube B is added to 9.5 mL diluent.

From Tube C, 0.025 mL of this is used to set up a plaque assay with a susceptible host. After incubation, 88 plaques are counted. What was the phage titer in the stock?

Just type in the big number below without units, Canvas doesn't allow exponents. However, please use scientific notation on your assignments and exams.

BlackTom AI · Accepted Answer

Dilution factors multiply and plaque count = (stock concentration in PFU/mL) × (volume plated in mL) × (overall dilution).

1) Relevant concepts
- Each dilution step reduces concentration by the fraction transferred/total volume; overall dilution = product of stepwise dilutions.
- Plaques counted reflect the number of infectious phage particles plated: plaques = concentration × plated volume × dilution.
- Solve for stock concentration: concentration = plaques / (plated volume × dilution).

2) Step-by-step reasoning
- Compute each tube’s dilution relative to the stock and multiply to get the overall dilution to Tube C.
  Tube A: 0.5 mL stock + 4.5 mL diluent → total 5.0 mL. Dilution = 0.5/5 = 1/10.
  Tube B: 1 mL from A + 9 mL diluent → total 10 mL. Dilution relative to A = 1/10. So relative to stock = (1/10)×(1/10) = 1/100.
  Tube C: 0.5 mL from B + 9.5 mL diluent → total 10 mL. Dilution relative to B = 0.5/10 = 1/20. So overall dilution relative to stock = (1/100)×(1/20) = 1/2000.

3) Calculations (all intermediate steps)
- Plated volume = 0.025 mL from Tube C.
- Number of plaques (88) = C_stock × (plated volume) × (overall dilution)
  88 = C × 0.025 × (1/2000)
- Solve for C:
  C = 88 × 2000 / 0.025
  2000 / 0.025 = 2000 × (1/0.025) = 2000 × 40 = 80,000
  C = 88 × 80,000 = 7,040,000

4) Why this is correct
- The calculation accounts for each dilution step and the exact plated volume; the algebra isolates the stock concentration consistent with the definition of PFU/mL.

5) Common wrong approaches (why other answers would be incorrect)
- Omitting the 1/20 dilution from B→C or treating Tube A or B incorrectly (e.g., thinking A is 1:9 rather than 1:10) changes the overall dilution and leads to large errors.
- Forgetting to convert plated volume or using 25 µL without converting units would also produce wrong numeric scaling.

6) Conclusion
The computed phage titer in the stock is 7040000, which matches the provided correct answer.

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