Hi there,
I'm currently performing cytotoxicity tests using Chromium 51 (PerkinElmer) with a half-life of 27,7 days, let's say 28 days. If I have to add 21ul of the product to label 1x10*6 cells with 100uCi, how much product do I have to add to the same amount of cells to obtain the same effect 3 weeks later?
I think it is 21ul + (3/4)*21ul = 36,75ul
Is this calculation correct?
Thanks!
I don´t know your experiment. But you have to calculate the decay formula. Your rest activity will be after 21 days exp(-ln2x21/27.7). You take the reciprocal and get 1.691 times 21 ul or a total volume of 35.52 ul.
According to the description of your experiment your are adding a volumen of 21 ul which contains 100 uCi. Let`s say that today you have 100uCi/21ul. You need to know how much activity you will have after 3 weeks (21 days) in 21 ul. Applying the decay law: A = Ao exp (-ln(2)/half life*time) for your case you will have: A= 100 uci exp (-0.693/27.7*21) = 59 uCi in 21 ul. If you need to have the same radiation effect you need 100 uci. If yu have 59 uCi in 21 ul you have then 59/21 = 2.81 uCi/ul. You need 100 uCi, then you have to take a volume equal to 100 uCi/2.81 uCi/uL = 35.6 ul.
Hope you understand and I was helpful.
Your calculation is not rigorous, but you give him one brought nearer of the volume to add, owed that the elapsed time ( 21 days ) is approximately 3/4 of the half-time . More rigorous it is using the law of the radioactive decay
The answer provided by K. Isaac-Olive is the best, very similar to answer provided by H. Krieger.
You do not have to take only in consideration the hall-life but also the fact that radioactivity is more diluted. In real life I would add the initial 21 microliters in the smallest total volume and add 2 micrometers one day after calibration date ( written on the recipient) , 4 micrometers 2 days after calibration day, etc....
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