I spiked some clam tissue with stock phage and recovered 2.6E6 cp/uL using digital PCR.
here are all my #s:
digital PCR - 2.6E6 cp/uL
amount used in dPCR reaction- 5uL
Amount eluted for extraction- 50uL
Amount used for extraction- 400uL (phage stock)
is this correct?
2.6E6 cp/uL x 50uL = 1.3E8 cp / 400uL = 3.25E5 cp/uL
thanks!
Can you clarify if the Digital PCR value of 2.6E6 cp/µL is the concentration readout from the dPCR instrument? Because I'm pretty sure that is higher than the detectable range for dPCR.
The standard way to get to concentration of phage in the stock before extraction would be as follows:
- Concentration readout from dPCR (cp/µL) * volume of dPCR (µL) = amount of template (cp)
- Amount of template (cp) / volume of extract added to dPCR (µL) = concentration of template in extract (cp/µL)
- Concentration of template in extract (cp/µL) * volume of eluent (µL) = amount of template extracted (cp)
- Amount of template extracted (cp) * recovery (%) = amount of template spiked (cp)
- Amount of template spiked (cp) / volume of spike (µL) = concentration of spike (cp/µL)
You don't mention knowing the recovery, so assuming that it is 100% (spoilers: it won't be) and your dPCR reaction volume is 20 µL (a guess assuming you are using a Bio-Rad QX200) equations look like this:
- 2.6E6 cp/µL * 20 µL = 5.2E7 cp total template in PCR
- 5.2E7 / 5 µL = 1.0E7 cp/µL in the extract
- 1.0E7 * 5.2E8 cp extracted
- 5.2E8 * 1.0 = 5.2E8 cp spiked
- 5.2E8 cp / 400 µL = 1.3E6 cp/µL in the original spiked sample