Hello everyone,
I am currently investigating potential lethal alleles in canine genome, specifically with SNPs.
My goal is to identify SNPs where one homozygous genotype is completely absent from the sampled population, and determine whether this absence can be explained simply by low allele frequency or whether it may indicate negative selection (e.g., embryonic lethality or reduced viability of a homozygous genotype).
For each SNP, I have genotype counts in the form:
Example 1:
AA = 0 AB = 348 BB = 310
Example 2:
AA = 12 AB = 646 BB = 0
I am currently using the HardyWeinberg package in R and applying the exact test as follows:
library(HardyWeinberg)
geno <- c(AA, AB, BB)
HWExact(geno, alternative = "greater")
My understanding is that:
alternative = "greater"tests for excess heterozygotes.- A deficit of one homozygous class (AA or BB) should manifest as an excess of heterozygotes.
- Therefore, a one-sided test may be more powerful for detecting the specific pattern expected under recessive lethal alleles.
However, I have received conflicting advice:
Some suggest that
alternative = "greater"is the most appropriate choice because my biological hypothesis specifically predicts a deficit of homozygotes and an excess of heterozygotes.Others suggest using
alternative = "two.sided"as the primary Hardy-Weinberg test and then separately evaluating the deficiency of homozygotes by comparing observed and expected genotype counts.
My questions are:
For the specific purpose of detecting candidate lethal alleles characterized by missing homozygotes, is
HWExact(geno, alternative = "greater")the most appropriate statistical test?Would it be preferable to use the two-sided exact test and then quantify homozygote deficiency separately?
Is there a more standard approach in population genetics or livestock genetics literature for identifying putative lethal alleles from genotype count data?
If one homozygous genotype is completely absent (AA = 0 or BB = 0), would you recommend reporting an additional effect-size metric such as:
(Expected Homozygotes - Observed Homozygotes) /
Expected Homozygotes
alongside the Hardy-Weinberg p-value?
Any advice, references, or examples from studies that used Hardy-Weinberg tests to identify candidate recessive lethal alleles would be greatly appreciated.
Thank you!
1 answer
alternative = "greater" is the right direction, but the p-value isn't the output you want. Run your own examples through: example 1 gives p(A) = 348/1316 = 0.264, so expected AA is about 46 in 658 animals. Example 2 gives p(B) = 0.491 and roughly 159 expected BB. Zero observed against those is astronomically significant, so every genuine hit will have an unusable p-value and nothing will rank. Your (E-O)/E metric has the same problem, since it is exactly 1.0 whenever O is 0. The quantity that actually separates low allele frequency from selection is the expected homozygote count itself: 0 out of 0.8 expected is nothing, 0 out of 159 is a candidate. Rank on that and use the p-value only as a filter.
The thing I'd worry about more than the test choice: example 2 is 646 heterozygotes out of 658 with one homozygous class absent, and that pattern is more often a genotyping artefact than biology. Collapsed segmental duplications and paralogous sequence variants make every individual look heterozygous, and both dog arrays and dog WGS produce plenty of them. Your screen selects precisely for that signature, so the top of your list will be dominated by it unless you filter first. Check allele balance within the hets (real hets sit near 0.5, collapsed paralogs often sit off near 0.33 or 0.67), check for elevated coverage over the site, and check whether neighbouring SNPs show the same pattern - artefacts cluster, real lethals don't.
For the literature framing, the dairy cattle work is what people cite: VanRaden et al. 2011 (J Dairy Sci) found recessive lethal haplotypes on exactly this logic, expected homozygotes versus zero observed.
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