Previous Results:

• The plaque assays from the 3rd purification (9/26) were examined. The control plate had a few spots for bacterial growth, indicating the sample was contaminated.
• All three plaque assays (10^0, 10^-1, and 10^-2) were all positive with plaques.

Objective:

• Correctly calculation the titer of the plate and prepare for a webbed plate

Procedure:

• The 3rd purification plate containing the 10^0 dilution was examined and the number of plaques totaled 15 on the plate
• To calculate the titer of the dilution the following equation was used:
  • (# pfu / volume used in microL) x (10^3 mciroL / mL) x dilution factor
  • (15 / 10 microL) x (10^3 microL / mL) x 10^0 = 1.5 x 10^3 pfu/mL
  • Low titer
• The diameter of 5 plaques were measured under the dissecting scope, then they were averaged
  • (1 mm + 0.8 mm + 1.5 mm + 0.8 mm + 1 mm) / 5 = 1.02 mm
• The diameter of the plate was measured to be 85 mm
• The areas of both the plaques and the plate were found
  • Plaque Area = 0.82 mm^2
  • Plate Area = 5674.5 mm^2
• It was found that the number of plaques needed to web the plate would be 6920.12 plaques
  • 5674.5 mm^2 / 0.82 mm^2 = 6920.12 plaques
• Then the volume of the lysate needed to web the plate was found using the following equation: (# of plaques) / (titer) = volume needed
  • 6920.12 plaques / (1.5 x 10^3 mL) = 4.61 mL

Results:

• After calculations it was determined that 4.61 mL of lysate will be needed to web the plate

Next Steps:

• During next lab, a webbed plate will be made using the remaining sample of lysate from the 10^0 dilution made in lab (9/24). If results are positive, then the plate will be flooded to collect the phage to move on to the next procedure.