lentiviral transduction and determine viral titer

I was reading the nature protocol article published by Joung et al. in 2017 and it seems that during lentiviral transduction by spininfection, the authors suggest we should seed 3x106 cells/well in a 12-well plate format (step 53). I understand that we need to begin with LOTS of cells during transduction as lots of them may die due to virus, but 3 million cells still seem very high to me. I usually do 3X105 cells/well. Can anyone please explain to me why we need this many cells?

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Tan Krakenover 4 years ago

3x106 293FT cells.

They use 5x105 cells in 6-well plate to test virus titer

Fuchsia Kelpieover 4 years ago

Yes 3 million cells is a lot for each well of a 12-well plate, but the cells are only that dense for 1 day during the transduction, and then we expand them. We want to keep the cells dense during the spinfection because this drastically reduces the number of 12-well plates we need for a genome-scale screen. If you are working with cell types that cannot be kept in dense conditions, e.g. stem cells, then you can reduce the number of cells for both the titer and screen spinfections.

we also use 3 million cells for viral titer if we are spinfecting. The titer and screen transductions need to be the same so the titer during the screen transduction is accurate. The 500k cells in a 6-well plate for viral titer is used for stem cells, and during the screen we will also use 500k cells in a 6-well plate.

Aquamarine Wyrmover 4 years ago

This answered my question perfectly.

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