Baculovirus titration

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This protocol is not optimized yet. For example, adding SF+ cells before or after the virus seems to make a big difference… more troubleshooting is needed.

Material

Serial dilute the virus stock as follows :

Virus input SF900II+ 10% FCS Dilution 20µl of virus 180µl -1 20µl of -1 180µl -2 20µl of -2 180µl -3 20µl of -3 180µl -4 20µl of -4 180µl -5 150µl of -5 1350µl -6 150µl of -6 1350µl -7 150µl of -7 1350µl -8 150µl of -8 1350µl -9 150µl of -9 1350µl -10 150µl of -10 1350µl -11 150µl of -11 1350µl -12 150µl of -12 1350µl -13

Transfer 100µl from each dilution (-6 to -13) into a 96 well plate, flat bottom, repeat 10 times to get the following :

1 2 3 4 5 6 7 8 9 10 A -6 -6 -6 -6 -6 -6 -6 -6 -6 -6 B -7 -7 -7 -7 -7 -7 -7 -7 -7 -7 C -8 -8 -8 -8 -8 -8 -8 -8 -8 -8 D -9 -9 -9 -9 -9 -9 -9 -9 -9 -9 E -10 -10 -10 -10 -10 -10 -10 -10 -10 -10 F -11 -11 -11 -11 -11 -11 -11 -11 -11 -11 G -12 -12 -12 -12 -12 -12 -12 -12 -12 -12 H -13 -13 -13 -13 -13 -13 -13 -13 -13 -13

Procedure

Add 100µl of SF+ cells at 0.1 million/ml per each well. Incubate shaking at 200rpm for 7 days. Collect 100µl and infect again 100µl of SF+ cells at 0.1 million/ml. Check for signs of infection after 3-4 days. Put parafilm around the 96 well plate to prevent evaporation

Titer example : Dilution (10−X) 7 8 9 Positive rate/10 wells 10/10 4/10 0/10 Total positive wells 10 4 0 Total positive rate 10/10 4/10 0/10 Positive well % 100 40 0 PD 0.83333 TCID50 1.46780E − 08 TCID50/ml 6.81292E + 08 PFU/ml 4.70092E + 08

The dilution that would have given 50% infection lies between 10−7 and 10−8 in this example. This information is needed to calculate the proportionate distance (PD), which allows calculation of the titer: PD = (A−50)/(A−B), where the A is the % response above 50% and B is the response below 50%. In this example: PD=(100−50)/(100−40)=0.83333. The dose that would have given a 50% infection, the TCID50, is then calculated using the formula: Log TCID50 = log of the dilution giving a response greater than 50% – the PD of that respond. Thus, log TCID50 = −7−0.8333 = −7.8333. Therefore: TCID50 = 10−7.8333 =1.46780 × 10−8

The titer of the virus (pfu/ml) can then be calculated: 1/ TCID50 = 1/1.46780 × 10−8 =6.81292 × 107 TCID50/ml = 6.81292 × 107/0.1 (0.1 refers to 100 µl of virus used to infect the wells). =6.81292 × 108 This can be converted to pfu/ml by using the relationship pfu = TCID50 × 0.69. So the titer of the example virus is: 6.81292 × 108 × 0.69 = 4.70092 × 108 pfu/ml.


MOI = ratio of virions / cells In theory, when MOI = 2, more than 80% of the cells get infected right away.