Since I'm using different cell types I'm interested in using 3 housekeeping genes (HKG) in order to have a robust normalization. Normally in the results you should have a standard deviation from the replicates you have per condition/cell type. However, I also do replicates of the HKG. How can I include the standard deviation between HKG replicates in the final relative expression of the gene of interest?
The only thing that matters is: how much do the measures of biological replicates vary?
The "measures" we are talking about are normalized expression values, the normalization is done to some kind of loading control. And as "loading control" you may decide to use the mean Ct value from k reference genes. The result is one single value per target gene and individual (biological replicate): a normalized expression value. It does not matter how you eventually quantify the loading. Taking the average of k reference genes will reduce the variance of the estimate of the "loading value", what is a good thing.
As long as you are not interested in the performance of the method but in the biological effect (of some treatment or condition or whatever), the only thing that counts is the variability between such measures between different individuals. It is irrelevant what proportion of variability between such measures might be attributable to technical variance. You know that the technical variance is lower when you average the Ct values of several reference genes. That's the reason why using several reference genes is better than using only one.
Maybe from another perspective:
If you used only a single reference gene, you would not even think about the fact that the value measured for this gene is also variable. You have no way to estimate this variability, but it still exists. But you don't need to care, because whatever it is, it will be part of the variability of your result anyway. So if measuring n individuals per group (giving you n varying results per group) is sufficient to see a systematic difference that is clearly larger than the "noise" then you solved your problem.
Simply use the average Ct of your 3 HKGs per sample.
PS: not every housekeeping gene is a valid reference gene, and a valid reference gene doesn't need to be a housekeeping gene. For qPCR it is important to have a valid reference gene.
So the standard deviation of the average of the HKGs shouldn't be taken into account?
The only thing that matters is: how much do the measures of biological replicates vary?
The "measures" we are talking about are normalized expression values, the normalization is done to some kind of loading control. And as "loading control" you may decide to use the mean Ct value from k reference genes. The result is one single value per target gene and individual (biological replicate): a normalized expression value. It does not matter how you eventually quantify the loading. Taking the average of k reference genes will reduce the variance of the estimate of the "loading value", what is a good thing.
As long as you are not interested in the performance of the method but in the biological effect (of some treatment or condition or whatever), the only thing that counts is the variability between such measures between different individuals. It is irrelevant what proportion of variability between such measures might be attributable to technical variance. You know that the technical variance is lower when you average the Ct values of several reference genes. That's the reason why using several reference genes is better than using only one.
Maybe from another perspective:
If you used only a single reference gene, you would not even think about the fact that the value measured for this gene is also variable. You have no way to estimate this variability, but it still exists. But you don't need to care, because whatever it is, it will be part of the variability of your result anyway. So if measuring n individuals per group (giving you n varying results per group) is sufficient to see a systematic difference that is clearly larger than the "noise" then you solved your problem.
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