You carried out a 10-fold dilution series. You started with tube 0 which contained an overnight culture of E. coli. You estimated the original culture contained 6 x 108 cell/ml using McFarland nephlemeter standards. You plated 0.1 ml volumes of a number of the dilutions. Which set of plates (provide the plate dilution) did you expect to contain the significant counts?

To determine the set of plates that would likely contain significant counts, let's analyze the dilution series and calculate the expected count on each plate.

Given information:
- Tube 0: Initial concentration of 6 x 10^8 cells/ml
- Total volume plated on each plate: 0.1 ml

The 10-fold dilution series implies that each subsequent tube is diluted by a factor of 10 compared to the previous one. Therefore, we can calculate the dilution factor for each tube as follows:

Tube 0 (original culture concentration): Dilution factor = 1 (no dilution)
Tube 1: Dilution factor = 1/10
Tube 2: Dilution factor = 1/10^2
...
Tube 10: Dilution factor = 1/10^10

To estimate the count on each plate, we multiply the concentration of cells in the original culture (tube 0) by the dilution factor for that specific tube. Since we are plating 0.1 ml of each dilution, we need to multiply the result by 0.1 ml to convert to plate counts.

Plate dilution and expected count:
- Plate from tube 0: No dilution (concentration remains the same as tube 0)
- Expected count: 6 x 10^8 cells/ml x 0.1 ml = 6 x 10^7 cells

- Plate from tube 1: 1/10 dilution
- Expected count: (6 x 10^8 cells/ml) x (1/10) x 0.1 ml = 6 x 10^7 cells

- Plate from tube 2: 1/10^2 dilution
- Expected count: (6 x 10^8 cells/ml) x (1/10^2) x 0.1 ml = 6 x 10^6 cells

- Plate from tube 3: 1/10^3 dilution
- Expected count: (6 x 10^8 cells/ml) x (1/10^3) x 0.1 ml = 6 x 10^5 cells

- Plate from tube 4: 1/10^4 dilution
- Expected count: (6 x 10^8 cells/ml) x (1/10^4) x 0.1 ml = 6 x 10^4 cells

- Plate from tube 5: 1/10^5 dilution
- Expected count: (6 x 10^8 cells/ml) x (1/10^5) x 0.1 ml = 6 x 10^3 cells

- Plate from tube 6: 1/10^6 dilution
- Expected count: (6 x 10^8 cells/ml) x (1/10^6) x 0.1 ml = 6 x 10^2 cells

- Plate from tube 7: 1/10^7 dilution
- Expected count: (6 x 10^8 cells/ml) x (1/10^7) x 0.1 ml = 6 x 10^1 cells

- Plate from tube 8: 1/10^8 dilution
- Expected count: (6 x 10^8 cells/ml) x (1/10^8) x 0.1 ml = 6 cells

Based on the calculations, the plates with notable counts would be:
- Plate from tube 0 (no dilution): Expected count of 6 x 10^7 cells
- Plate from tube 1 (1/10 dilution): Expected count of 6 x 10^7 cells
- Plate from tube 2 (1/10^2 dilution): Expected count of 6 x 10^6 cells

These plates would likely contain significant counts compared to the other plates in the series.