4. A researcher wants to determine how much virus she has isolated. She dilutes

Question

4. A researcher wants to determine how much virus she has isolated. She dilutes her virus stock 100 times to make a diluted virus stock. This diluted virus stock is again diluted 100 times to make a working stock of the virus. Then she performs a dilution series of the working stock. She prepares a serial dilution of 3, 1:10 dilutions of her already working stock. She takes 100 HL of each tube from her dilution series and performs a plaque assay. The 10 plate has 80 plaques, the 102 has 4 and the 10 has none. Use these data to determine pfu/ml expressed in correct notation. (2 pts)

Explanation / Answer

Ans. Let the viral load in stock solution = X PFU/mL

# First dilution: Given- first dilution factor, DF = 1/100 = 0.01

PFU count after 1st dilution = Original PFU count x Immediate DF = X PFU/mL x 0.01

                                                = 0.01X PFU/mL

# Second dilution: Given- second dilution factor, DF = 1/100 = 0.01

PFU count after 2nd dilution = PFU count after 1st dilution x Immediate DF

= 0.01X PFU/mL x 0.01 = 0.0001X PFU/mL

# Third dilution: Given- third dilution factor, DF = 1/10 = 0.1

PFU count after 3rd dilution = PFU count after 2nd dilution x Immediate DF

= 0.0001X PFU/mL x 0.1 = 0.00001X PFU/mL

# Plating:

#1st plating: 100.0 uL = 0.100 mL of 1st dilution gives 80 plaques.

So,

            PFU count per mL inoculum = 80 plaques / 0.100 mL = 800 PFU/mL

We have, PFU count after 1st dilution = 0.01X PFU/mL

So,

            0.01X PFU/mL = 800 PFU/mL

            Or, X = 800 / 0.01 = 80000

Therefore, PFU count of the stock = 80000 PFU/mL

#2nd plating: 100.0 uL = 0.100 mL of 2nd dilution gives 4 plaques.

So,

            PFU count per mL inoculum = 4 plaques / 0.100 mL = 40 PFU/mL

We have, PFU count after 2nd dilution = 0.01X PFU/mL

So,

            0.0001X PFU/mL = 40 PFU/mL

            Or, X = 40 / 0.0001 = 400000

Therefore, PFU count of the stock = 400000 PFU/mL

Conclusion: Note that there are several experimental errors like non-homogenous solutions, clumping of viral particles, etc. that often yield a lower PFU count even if proper serial dilution is performed. However, there is no such errors that would yield a higher PFU count. Therefore, the maximum value of calculated PFU count shall be treated as highest PFU count.

# Therefore, viral load in stock solution = 400000 PFU/mL = 4.0 x 105 PFU/mL

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