Question

Dilutions. In the following example: a. What is the titer of the virus? b. Is the...

Dilutions.

In the following example:
a. What is the titer of the virus?
b. Is the virus titer high enough of the newly arrived virus shipment to be used as a control in the experiment?
c. Show your work

Note: The virus titer needed as a control in the experiment was at least 1 x 10^7 viruses per ml.

The control virus stock arrived from the ATCC distributor arrived frozen and was labeled as 3x10^8 viruses/ml (equal to 300,000,000 viruses/ml or “three times ten to the power of eight viruses per ml”).

The stock was thawed on ice and aliquoted into ten fresh cryovials and frozen at -80 degrees. One aliquot was checked for titer.

The dilution was made as you have done in class in three bottles, A, B and C.

One ml of the thawed aliquot of the virus control was put into 99 ml in bottle 1 and swirled to mix.

With a fresh pipette, one ml of bottle one was placed into bottle 2 and swirled.

The procedure was repeated as one ml of bottle B was placed in bottle C and swirled.

Bottle C was plated onto two plates. Plate #1 had 1.0 ml of Bottle C added to the top agar (containing the cell host of the virus) and overlaid on the plate. Plate #2 had 0.1 ml of bottle C added to the top agar and plated.

After incubation, plate #1 had 551 plaques and plate #2 had 52 plaques.

Describe or provide a picture of your calculation:

What was the titer of the virus aliquot?

Can the virus shipment be used as a control?

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Answer #1

The titer of the virus stock when it came from ATCC distributor was 3x 10 8/ ml.

So, when 1ml was taken, from the aliquot of this stock and the volume was adjusted to 100ml, that means it was diluted 100 times, then the resulting titer would be 3x 10 8 / 100= 3x 10 6/ml.

If bottle 2 and bottle 3 also contained 99ml of diluant, where the virus was added, then that means the virus solution was diluted 100 times twice more, then the resulting virus solution in bottle C, will be 3x 102/ml= 300/ml

When, it was again diluted for plate 2, the resultant titer was 30/ml of virus particle.

The titer of the virus particle in the provided shipment was 3x 10 8/ ml, so it could be used as a control, as it is mentioned that  1x 107 / ml would be enough, to use as control.

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