hi. i want to know, how to calculate the final OD of the bacteria, when:
OD 600nm~ 0.5 nm,
MOI 1: 100
number of cells : 2 X 10^5 per well.
What is the correct formula to get the final OD of bacteria, so that I can make a dilution and used that to infect the cell?
for MOI 1:100 (assuming one bacterium per 100 host cells). you need to add 2 X 10^3 bacterial cells to 2 X 10^5 host cells.
usually 1 O.D of bacteria has 8 X 10^8.
1. adjust your bacterial suspension to 1 O.D .
2.After which you have to dilute to 4 X 10^4 times ( serial dilution of 1/10 vol four times, followed by a single dilution of 1/4 vol ) then your dilution has 2 X 10^4 bacterial cells/ml.
3. add 100ul of this bacterial suspension into each well. if you think 100ul is higher, dilute the 1 O.D suspension 2 X 10^4 times and add 50ul to each well.
note: use sterile PBS for dilution.
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