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:
1. the plate with the undiluted blood sample resulted in too many plaques to count.
2. the plate with the 10-fold dilution of the blood sample resulted in too many plaques to count.
3. the plate with the 100-fold dilution of the blood sample resulted in approximately 3000 plaques.
4. the plate with the 1000-fold dilution of the blood sample resulted in 298 plaques counted.
5. the plate with the 10,000-fold dilution of the blood sample resulted in 32 plaques counted.
Which of the following best describes the titer of the Rabid Wolverine Virus within the bloodstream of the single Buckeye (Ohio State) fan you tested?
a. 3 x 107 plaques miL.
b. 3 x 101 plaques / mL.
c. 3 x 109 plaques mlL.
d. 3 x 103 plaques miL.
e. 3 x 105 plaques / mL.

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cjmejiab cjmejiab · Answer 1 · 2020-05-14T02:51:12+02:00

Answer:

Viable plate count = 30 to 300 plates

see take as a limit between dilutions of 10 times and 100 times.

PFU per ml = #plate / (silver volume × dilution)

Silver volume = 0.01 mL

we solve for dilution 10 times

Dilution = 1/10 = 10 ^ -1

We replace

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

we solve for dilution 100 times

Dilution = 1/100 = 10 ^ -2

we replace

UFP per ml = 32 / (10 ^ -2 × 0.01) = 32 × 10 ^ 4 = 3.2 × 10 ^ 5 UFP per ml

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

The result is 3.1 × 10 ^ 5 plates per mL