qPCR: Why is fold change and standard deviation calculated after transformation?

Question

I am analyzing data from a quantitative polymerase chain reaction (qPCR) using R. After cleaning the raw data, it looks something like this:

> dput(x)
structure(list(Reporter = c("FAM", "FAM", "FAM", "FAM", "FAM", 
"FAM", "FAM", "FAM", "FAM", "FAM", "FAM", "FAM", "VIC", "VIC", 
"VIC", "VIC", "VIC", "VIC", "VIC", "VIC", "VIC", "VIC", "VIC", 
"VIC"), Number = c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 
12L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L), A = c(22.19, 
22.24, 22.5, 22.54, 22.6, 22.59, 23.07, 23.46, 22.43, 22.74, 
24.09, 23.91, 24.52, 25.03, 25.25, 25.82, 25.13, 24.71, 25.34, 
25.85, 25.25, 26.15, 25.81, 25.29), B = c(21.72, 21.78, 22.86, 
22.73, 19.88, 20.07, 21.06, 21.06, 20.96, 21.11, 19.46, 19.43, 
24.75, 24.69, 25.64, 25.19, 23.76, 23.69, 24.35, 25.05, 24.1, 
23.81, 22.81, 23.13), C = c(21.37, 21.56, 20.07, 20.01, 21.17, 
21.08, 20.54, 20.36, 33, NA, NA, NA, 23.91, 24.31, 23.61, 23.88, 
24.33, 24.31, 23.37, 23.53, 33, NA, NA, NA), E = c(26.26, 27.33, 
25.93, 26.56, 25.76, 23.03, 24.72, 25.27, 24.43, 24.31, 26.98, 
23.33, 24.04, 25.02, 25.1, 25.1, 24.68, 25.48, 25.87, 26.22, 
25.35, 25.36, 25.11, 25.98), F = c(25.81, 26.9, 25.58, 26.61, 
25.06, 21.85, 23.59, 24.04, 23.19, 23.19, 25.17, 20.8, 24.12, 
24.26, 25.32, 25.25, 24, 23.78, 24.7, 24.48, 23.52, 23.87, 23.05, 
23.05), G = c(26.12, 27.02, 24.08, 25.15, 25.99, 23.18, 24.2, 
24.05, 33, NA, NA, NA, 23.47, 23.45, 23.7, 23.74, 24.46, 24.19, 
23.56, 23.53, 33, NA, NA, NA)), .Names = c("Reporter", "Number", 
"A", "B", "C", "E", "F", "G"), class = c("tbl_df", "tbl", "data.frame"
), row.names = c(NA, 24L))

I think that each well measures two genes: target which is 12 different genes (color FAM), and the reference or housekeeping gene, GAPDH (color VIC). Also, control is triplicate A, B, C (3 x 12 wells), and treatment is E, F, G (3 x 12 wells).

I have created a function ddCt_ for analyzing the data with the ddCt algorithm (Livak & Schmittgen, 2001). It takes one argument x which is a data.frame of the form exemplified above.

ddCt_ <- function(x) {
  # Subset x by control/treatment and target/reference
  # Then, calculate Ct averages for each triplicate
  TC <- x %>% 
    select(Reporter, Number, A, B, C) %>% 
    filter(Reporter == "FAM") %>% 
    rowwise() %>%
    mutate(Ct_Avg = mean(c(A, B, C), na.rm = TRUE))

  RC <- x %>% 
    select(Reporter, Number, A, B, C) %>% 
    filter(Reporter == "VIC") %>% 
    rowwise() %>%
    mutate(Ct_Avg = mean(c(A, B, C), na.rm = TRUE))

  TT <- x %>% 
    select(Reporter, Number, E, F, G) %>% 
    filter(Reporter == "FAM") %>% 
    rowwise() %>%
    mutate(Ct_Avg = mean(c(E, F, G), na.rm = TRUE))

  RT <- x %>% 
    select(Reporter, Number, E, F, G) %>% 
    filter(Reporter == "VIC") %>% 
    rowwise() %>%
    mutate(Ct_Avg = mean(c(E, F, G), na.rm = TRUE))

  # Normalize Ct of the target gene to the Ct of the reference gene
  dCt_control   <- TC$Ct_Avg - RC$Ct_Avg
  dCt_treatment <- TT$Ct_Avg - RT$Ct_Avg

  # Normalize dCt of the treatment group to the dCt of the control group
  ddCt <- dCt_treatment - dCt_control

  # Calculate fold change
  fc <- 2^(-ddCt)

  # Calculate avg and sd
  dCt_control_avg   <- mean(dCt_control)
  dCt_control_sd    <- sd(dCt_control)
  dCt_treatment_avg <- mean(dCt_treatment)
  dCt_treatment_sd  <- sd(dCt_treatment)

  # Create output
  df <- data_frame(
    Sample = 1:12, "dCt Control" = dCt_control, "dCt Treatment" = dCt_treatment, 
    "ddCt" = ddCt, "Fold Change" = fc, "dCt Control Avg" = dCt_control_avg, 
    "dCt Control SD" = dCt_control_sd, "dCt Treatment Avg" = dCt_treatment_avg, 
    "dCt Treatment SD" = dCt_treatment_sd
    ) %>% round(2)
  write.csv(x = df, file = "result.csv")
  saveRDS(object = df, file = "result.rds")
  df
}

However, I am not a biochemist or molecular biologist, so I am unsure of the basic concepts involved and the overall approach. So, my questions are:

1. Is the function correct?
2. Why is it important to calculate fold change and standard deviation after normalization?

Answer 1

1. The functions look correct, but calculate a few by hand and ensure they match. One thing I should note is that the subtraction of the Ct values usually happens before an average is made, since generally both are in the same well. In other word, subtract the reference Ct from the gene of interest Ct from the same well and then average the dCts. This order is important since one usually does a multiplexed qPCR.
2. You have to normalize to a reference gene to control for how much cDNA was used, since that will alter the Ct values. If you calculated the fold-changes without normalization then they could be purely due to using more/less cDNA in the reaction (i.e., the output would be meaningless).