(Updated question) Problem with Fold-change mean compared to absolute data (in qPCR)

Question

For the problem below I am aware of the statistics involved but just can't get my finger around the following:

In biology we use qPCR to measure gene expression or basically number of mRNA copies. It does this by counting how many cycles of amplification (2^) it needs to reach a threshold. So to simplify everything lets assume that 1024 copies would require 10 cycles (2^10). If there are more copies it would take less cycles to reach the threshold, therefore 2024 copies would take 9 cycles, 512 copies 11 cycles etc.. Now imagine the following scenario:

We have the following samples:

Sample 1. 1024 copies of gene A and 4096 copies of gene B
Sample 2. 1024 copies of gene A and 16384 copies of gene B

Now we want to compare sample 2 to 1, with gene B in relation to gene A:

In absolute numbers this would be:

Sample 1. 4096 / 1024 = 4x more B
Sample 2. 16384 / 1024 = 16x more B
The average amount more B = (16 + 4) / 2 = 10x more B

Now with qPCR the same samples from above would look like the following:

Sample 1. Gene A 10 cycles, Gene B 8 cycles.
Sample 2. Gene A 10 cycles, Gene B 6 cycles.

Now using the methods standard used for qPCR data we first take the difference between Gene B and A. Followed by the mean of the difference and this is taken 2^.

Sample 1. 10 - 8 = 2 cycles
Sample 2. 10 - 6 = 4 cycles
The average amount more B = 2^((2 + 4)/2) = 8x more B

All publications / software tools etc will compute an average of 8x more B. And I am aware this originated from a log2 scale, however why does this statistically correct method differ in outcome from absolute numbers?

Update I oversimplified my question therefore below some more details of my "problem".

Condition 1 biorep 1. 1024 copies of ref and 1024 copies of gene X
Condition 1 biorep 2. 1024 copies of ref and 1024 copies of gene X
Condition 2 biorep 1. 1024 copies of ref and 4096 copies of gene X
Condition 2 biorep 2. 1024 copies of ref and 16384 copies of gene X

So to compute in absolute numbers the average amount more of gene X in condition 2 versus 1:

We can forget the reference gene here as cDNA amounts are the same, therefore: 
Condition 1 biorep 1 & 2: average (1024 + 1024) / 2 = 1024 copies
Condition 2 biorep 1 & 2: average (4096 + 16384) / 2 = 10240 copies
Condition 2 vs 1: average 10240 / 1024 = 10x more gene X in condition 2 vs 1

Now with qPCR CT values based on the numbers above:

Condition 1 biorep 1. Ref_CT 10 and GeneX_CT 10
Condition 1 biorep 2. Ref_CT 10 and GeneX_CT 10
Condition 2 biorep 1. Ref_CT 10 and GeneX_CT 8
Condition 2 biorep 1. Ref_CT 10 and GeneX_CT 6

Now using the official qPCR calculations:

Condition 1, Ref_CT mean (10 + 10) / 2 = 10
Condition 2, Ref_CT mean (10 + 10) / 2 = 10
Condition 1, GeneX_CT mean (10 + 10) / 2 = 10
Condition 2, GeneX_CT mean (8 + 6) / 2 = 7

Ref_deltaCT (10 - 10) = 0
GeneX_deltaCT (10 - 7) = 3

delta_deltaCT (3 - 0) = 3
Difference Condition 2 vs 1 = (2^3) = 8x

This equates to 8x more of Gene X on average in condition 2 compare to condition 1. Here now is the difference of 8x vs 10x. Also any program you will use to fill in these CT values it will result in a ratio of 8 compared to 10.

Is this perhaps as this method equating to 8 should only be used for technical replicates, and bioreps should use a different equation resulting into 10?

Answer 1

Sample 1. 4096 / 1024 = 4x more B

Sample 2. 16384 / 1024 = 16x more B

The average amount more B = (16 + 4) / 2 = 10x more B

That is not the right way to go about. You are calculating average expression of B in both samples. It does not mean there is 10x more B. There is just 4x more B in Sample-2 relative to Sample-1.

Gene-A does not change its expression between samples and can be used as a reference gene.

If you calculate by threshold cycle the fold change $ = 2^{8-6} = 4x$

See this post too.

The average is calculated between replicates, not between different experimental conditions.

Sometimes the average of two experimental samples is calculated — this is typically done in a MA-plot analysis which is done to filter out genes which show very high fold changes just because of their overall low expression (1:4 vs 100:300).

Answer to your Edit

You should not take the average (arithmetic mean) of Ct. They don't scale linearly with expression (and so is the converse).

For any non-linear function $f(x)$:

$$f(x+y) \ne f(x)+f(y)$$

$$f\left(\frac{x+y}{2}\right) \ne \frac{f(x)+f(y)}{2}$$

Furthermore, for a convex function:

$$f(E[x]) \le E[f(x)] $$

$$ Jensen's\ inequality $$

where $E[x]$ is expected value of $x$

$a^x$ is a convex function with respect to $x$ ($a$ is any real number > 1). So you can apply Jensen's inequality to Ct values and expression.