2
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I totally agree with your answer, I think I over simplified my question, above I edited my original post to clarify a bit more my question. $\endgroup$ – Pim Nov 6 '14 at 8:21
  • $\begingroup$ The answer you give in your edit is also the answer I expected. However all (commercial) tools assume that the Ct values are on a linear scale and therefore you take the arithmetic mean of the Ct, resulting into 8x. And as you take 2^ of Ct the absolute values are not. This means you would have to take the geometric mean of 4096 and 16384, which is 8192 (which is 8x 1024). Apparently all published methods consider the absolute values not to be on a linear scale, and actually the Ct values are. Which biologically speaking is difficult to grasp. $\endgroup$ – Pim Nov 6 '14 at 13:47
  • $\begingroup$ If someone is interested in average Ct then they may very well do it. But we are not interested in that are we. Independently the values are not in any scale. They scale with other value as per some function. So expression scales exponentially with Ct. So $ Expression = f(Ct)$. I should clarify about it. Edited $\endgroup$ – WYSIWYG Nov 7 '14 at 16:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.