Calculate Ct/Cq from amplification curves

[ewallace]

We want an open-source and tested algorithm to calculate Ct/Cq from raw amplification curves.

A threshhold method is probably best, see T Nolan, J Huggett, E Sanchez, Good practice guide for the application of quantitative PCR (qPCR), LGC (2013).

The underlying principle of the threshold method is that information related to the target quantity
is available in the log-linear phase of the amplification curve. By simply detecting the cycle where
the log-linear phase of the amplification curve crosses an arbitrary threshold level, information
about relative target quantities in different samples is obtained.

It is recommended to set the threshold as follows:

• Sufficiently above the background fluorescence baseline to be confident of avoiding having
  any amplification curve cross the threshold prematurely due to background fluorescence, and
• As low as possible to ensure that the threshold crosses at the log-linear phase of the
  amplification curve where it is unaffected by the plateau phase.

See also chipPCR paper, who have an implementation in R based on S4 classes. As they say, "Pre-processing in most commercial cyclers is a black box, which restrains reproducible research".

[ewallace]

Also see System-specific periodicity in quantitative real-time polymerase chain reaction data questions threshold-based quantitation, which reviews existing methods and recommends "workers should either use scale-insensitive Cqs or normalize their growth curves to constant amplitude before applying the constant threshold method."

[ewallace]

Discussed 4 September 2020: we are prioritising features downstream of Cq calculation for now, looking towards rOpensci submission. We might make Cq calculation a higher priority later, for example if requested by reviewers.

[DimmestP]

Elliott said that previous members of Teuta Pilizota's lab were looking into raw amplification curves and may have python code to calculate Dt/Cq (Dario Miroli an old PhD student was working on it apparently)