Human-Assisted Spread of a Maladaptive Behavior in a Critically Endangered Bird

Conservation management often focuses on counteracting the adverse effects of human activities on threatened populations. However, conservation measures may unintentionally relax selection by allowing the ‘survival of the not-so-fit’, increasing the risk of fixation of maladaptive traits. Here, we report such a case in the critically-endangered Chatham Island black robin (Petroica traversi) which, in 1980, was reduced to a single breeding pair. Following this bottleneck, some females were observed to lay eggs on the rims of their nests. Rim eggs left in place always failed to hatch. To expedite population recovery, rim eggs were repositioned inside nests, yielding viable hatchlings. Repositioning resulted in rapid growth of the black robin population, but by 1989 over 50% of all females were laying rim eggs. We used an exceptional, species-wide pedigree to consider both recessive and dominant models of inheritance over all plausible founder genotype combinations at a biallelic and possibly sex-linked locus. The pattern of rim laying is best fitted as an autosomal dominant Mendelian trait. Using a phenotype permutation test we could also reject the null hypothesis of non-heritability for this trait in favour of our best-fitting model of heritability. Data collected after intervention ceased shows that the frequency of rim laying has strongly declined, and that this trait is maladaptive. This episode yields an important lesson for conservation biology: fixation of maladaptive traits could render small threatened populations completely dependent on humans for reproduction, irreversibly compromising the long term viability of populations humanity seeks to conserve.


Phenotypes
Let L be the set of mature females who were observed to lay eggs and let |L| denote its size. Thus, L is a subset of F.
We can use (2) recursively on the pedigree starting from a given founder genotypes and propose genotypes independently for each offspring given its parental genotypes and the model of inheritance.
Finally the probability of φ(u), the phenotype for an individual u, given its genotype γ(u) and the model of inheritance M c h is the following: Since φ(u) is unobservable if u is not an egg-laying female, i.e., u / ∈ L, we set Pr(φ(u)) = 1 as in [1] so that it will not affect the likelihood of the observable phenotypes.

Likelihood of Observed Phenotypes on Pedigree
We want to compute the probability of the inheritance model M c h and the founder genotypes γ(A) given G, the observed pedigree, and φ(V), the observed phenotypes over the nodes in G: Note that Pr (G | M c h , γ(A)) = Pr (G) due to the assumed independence of the model of inheritance and the pedigree and therefore cancels out from the posterior probability of M c h and γ(A) given by (4). We assume a uniform prior probability Pr (M c h , γ(A)) = 1 64 over the 2 × 2 × 4 2 = 64 possible models in the family: of vertices in G obtained by a breadth-first expansion from V 0 that satisfies the condition that for each ). For each t, by the law of total probability, And for a given γ(V t ) ∈ Γ Vt we have the convenient recursive structure, where, Due to this recursive likelihood decomposition over the increasing pedigree sequence {G[V t ]} T t=0 , we can use the following sequential Monte Carlo algorithm to compute the likelihood of interest by a straightforward adaptation of [2] in the spirit of [3] but over a nested sequence of pedigrees.

Sequential Monte Carlo Algorithm
• At t = 0, initialize all particles at the founder nodes to have the given founder genotypes.

Phenotype Permutation Test
Our statistical test requires us to compute P 0 (φ(V)) := P 0 ({φ o (u) : u ∈ V}), the probability of phenotypes under a null hypothesis H 0 that is independent of the pedigree and thereby devoid of any notion of genetic inheritance. Under the null hypothesis H 0 we model the phenotypes φ(V) using independent and identical Bernoulli(θ) random variables with the probability of laying rim eggs given by θ ∈ (0, 1).
In other words, the probabilities under H 0 for the two phenotypes 1 and 0 are P 0 (φ(u) = 1) = θ and u ∈ L} is modeled to be realized under H 0 with probability: For the unobservable case, i.e., when u ∈ V \ L, we set P 0 (φ(u)) = 1 so that P 0 (φ(V)) = P 0 (φ(L)). Therefore is equally likely. The set of such phenotypes can be constructed by first taking each permutation of the vertex labels in L and then assigning phenotye 1 to the first r (and phenotype 0 to the last − r) vertex labels of this permutation. The basic idea behind our phenotype permutation test is identical to that behind classical permutation test of the null hypothesis: two samples are drawn from the same distribution, except that we need meaningful test statistics that can take specific notions of inheritance on the observed pedigree (that relates vertices or individuals in L under an alternative hypothesis) into account. In particular, we would like the test statistic T = T (φ) to take larger values if the alternative hypothesis H a that specifies a law of inheritance on the observed pedigree becomes more likely. Then the P-value of rejecting H 0 in favor of a given alternative hypothesis of inheritance H a is given by We take T to be the likelihood under our Model 4, the model of inheritance with the highest posterior probability. This likelihood will become large when the alternate hypothesis of Model 4 is true and we can obtain its null distribution from the permuted phenotypes that are equally likely when the null hypothesis is true. Note that we cannot do a likelihood ratio test between the null model of independent Bernoulli trials and the alternate Model 4 since they are not nested. Finally we use a Monte Carlo estimate of the P-value from the proportion of test statistics that are greater than or equal to the observed test statistic in a random sample of 10000 phenotype permutations that are equally likely under H 0 .