What is the most accurate way to calculate a child's genetic predisposition to a phenotype, given both parents' genotypic predispositions?

Assume trait X is 50% heritable and controlled by an arbitrarily large number of SNPs. The trait is normally distributed in the population. Parent 1 is in the 80th percentile of genetic predisposition to trait X. Parent 2 is in the 60th percentile of genetic contribution to trait X. Will the child's predisposition be on average the 70th percentile?

Due to regression to the mean and given 50% heritability, will the best estimate of the child's phenotype be in the 50 + ((70-50)*0.5) = 60th percentile?


1 Answer 1



OP specifically drew my attention to the use of the genetic prediction, rather than phenotype.

I think that the answer remains more or less the same, except that instead of the actual phenotype distribution, you have the distribution of genetic predictions.

I don't think that quantiles are appropriate for prediction tasks.

Original answer

I believe that the answer is "no" because the percentile is a nonparametric description of the distribution. I believe that the form of your calculation is right, but the values you've plugged in are wrong.

For example, 1.755m is ~50th percentile male human height (from here), 1.78m is ~60th, 1.795m is ~70th, 1.815 is ~80th, and 1.85 is ~90th. Note that the intervals between those quantiles is not regular, so your estimate will be accordingly biased if you act like the quantile matches the value of the heritable trait.

(Sorry about the dumb rounding, I was working from the calculator, which is somewhat imprecise.)

So your midparent estimate for height would be something like 1.755 + (.0425 * .5) = 1.77625m

In other words: use the actual phenotypic measurements (or whatever transformation the heritability estimate comes from), not quantiles.

  • $\begingroup$ are you sure? I'm referring to percentile of genetic risk, not the observed phenotype, when I say percentile, until the last step $\endgroup$
    – BigMistake
    Oct 26, 2023 at 20:28
  • $\begingroup$ @BigMistake- ah, I see. I think that my answer remains the same, except that the phenotype in question would have to be whatever the genetic risk value is. It's still a quantile of the underlying distribution, it's just that rather than the phenotype you have a genetic risk measurement. Depending on the structure of the prediction task, there may well be additional biases that require a different form for the prediction, but I can't comment on that. $\endgroup$ Oct 26, 2023 at 20:53
  • $\begingroup$ Does this change if we assume the population genetic predisposition for the phenotype normally distributed? I think in this case, the percentile might yield the same result $\endgroup$
    – BigMistake
    Oct 27, 2023 at 2:07
  • $\begingroup$ @BigMistake not sure that I understand. When you perform a quantile transformation of data (percentile is a 100-quantile transformation), you fundamentally change the shape of the data to a uniform distribution. Normal -> uniform data transformation will not preserve underlying normality, so you will not preserve the moments of the underlying distribution, not even the mean. It will only work if the underlying data is also uniformly distributed. I'd suggest playing with some data to investigate this if it doesn't make sense. $\endgroup$ Oct 27, 2023 at 16:36
  • $\begingroup$ Exactly. Under the assumption that the data is already normally distributed, transforming it into percentiles (now uniformly distributed) doesn't affect the calculation's validity $\endgroup$
    – BigMistake
    Oct 27, 2023 at 21:10

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