Question

You are measuring the titer (i.e. plaque-forming units (PFUs) of the Nutty Fan Virus within the bloodstream of Michigan fans, and you measure the following numbers from blood drawn from one Wolverine (Michigan) fan. In each case below, you have plated 0.01 mL of the blood or diluted blood on a plate containing a confluent monolayer of tissue culture cells that are susceptible to the virus. You allowed the plates to incubate overnight, and upon arriving at the lab early the next morning you count the plaques on each plate. Your results are the following:

- the plate with the undiluted blood sample resulted in too many plaques to count
- the plate with the 10-fold dilution of the blood sample resulted in an estimated 300 plaques
- the plate with the 100-fold dilution of the blood sample resulted in approximately 32 plaques
- the plate with the 1000-fold dilution of the blood sample resulted in exactly 2 plaques counted
- the plate with the 10,000-fold dilution of the blood sample resulted in no detectable plaques

Based on these data alone, the approximate titer of the Nutty Fan Virus within the bloodstream of the single Wolverine (Michigan) fan you tested is:

______________________________________ plaques/mL.

Answer 1

Viable plate count = 30 to 300 plaques

So, we will consider only 10 fold and 100 fold dilutions.

PFU per mL = #plaque / (volume plated × dilution)

Volume plated = 0.01 mL

For 10 fold dilution,

Dilution = 1/10 = 10^-1

PFU per mL = 300 / (10^-1 × 0.01) = 300000 = 3 × 10^5 PFU per mL

For 100 fold dilution,

Dilution = 1/100 = 10^-2

PFU per mL = 32 / (10^-2 × 0.01) = 32 × 10^4 = 3.2 × 10^5 PFU per mL

Average count = (3 × 10^5) + (3.2 × 10^5) / 2 = 3.1 × 10^5 PFU per mL

Answer is 3.1 × 10^5 plaques per mL

Please rate high.