Polynomial-Time Statistical Estimation of Species Trees Under Gene Duplication and Loss

Abstract

Phylogenomics—the estimation of species trees from multi-locus datasets—is a common step in many biological studies. However, this estimation is challenged by the fact that genes can evolve under processes, including incomplete lineage sorting (ILS) and gene duplication and loss (GDL), that make their trees different from the species tree. In this paper, we address the challenge of estimating the species tree under GDL. We show that species trees are identifiable under a standard stochastic model for GDL, and that the polynomial-time algorithm ASTRAL-multi, a recent development in the ASTRAL suite of methods, is statistically consistent under this GDL model. We also provide a simulation study evaluating ASTRAL-multi for species tree estimation under GDL. All scripts and datasets used in this study are available on the Illinois Data Bank: https://doi.org/10.13012/B2IDB-2626814_V1.

A Additional Proofs

1.1 A.1 Proof of Theorem 1: case (2)

In case (2), assume that $T^{\mathcal {Q}} = (((A,B),C),D)$, let R be the most recent common ancestor of A, B, C (but not D) in $T^{\mathcal {Q}}$, and let I be the number of gene copies exiting R. As in case 1), it suffices to prove (2) almost surely. Let $i_{x} \in \{1,...,I\}$ be the ancestral lineage of $x \in \{a,b,c\}$ in R. Then

$$\begin{aligned} \mathbf {P}'_I[q = ab|cd]&= \mathbf {P}'_I[i_a = i_b] + \mathbf {P}'_I[q = ab|cd \text { and }i_a, i_b, i_c \text { all distinct}]. \end{aligned}$$

(9)

On the other hand,

$$\begin{aligned} \mathbf {P}'_I[q = ac|bd]&= \mathbf {P}'_I[i_b \ne i_a = i_c] + \mathbf {P}'_I[q = ac|bd \text { and } i_a, i_b, i_c \text { all distinct}], \end{aligned}$$

(10)

with a similar result for $\mathbf {P}'_{I}[q=ad|bc]$. By symmetry again, the last term on the RHS of (9) and (10) are the same. This implies

$$\begin{aligned}&\mathbf {P}'_I[q = ab|cd] - \mathbf {P}'_I[q = ac|bd] = \mathbf {P}'_I[i_a = i_b] - \mathbf {P}'_I[i_b \ne i_a = i_c]\\&= x - (1-x)\mathbf {P}'_{I}[i_{a}=i_{c}|i_{a}\ne i_{b}] = x - (1-x)\frac{1}{I} \equiv g(x), \end{aligned}$$

where $x = \mathbf {P}'_{I}[i_a = i_b]$. This function g attains its minimum value at the smallest possible of x, which by Lemma 1 is $x = 1/I$. Evaluating at $x = 1/I$ gives

$$g(1/I) = \frac{1}{I} - \frac{1}{I} + \frac{1}{I^2} = \frac{1}{I^2} > 0,$$

which establishes (2) in case (2).

1.2 A.2 Proof of Theorem 2

First, we prove consistency for the exact version of ASTRAL. The input to the ASTRAL/ONE pipeline is the collection of gene trees $\mathcal {T} = \{t_i\}_{i=1}^k$, where $t_i$ is labeled by individuals (i.e., gene copies) $R_i \subseteq R$. For each species and each gene tree $t_i$, we pick a uniform random gene copy, producing a new gene tree $\tilde{t}_i$. Recall that the quartet score of $\widetilde{T}$ with respect to $\widetilde{\mathcal {T}} = \{\tilde{t}_i\}_{i=1}^k$ is then

$$ Q_k(\widetilde{T}) = \sum _{i=1}^k \sum _{\mathcal {J} = \{a,b,c,d\} \subseteq R_i} \mathbf {1}(\widetilde{T}_{ext}^{\mathcal {J}},\tilde{t}_i^{\mathcal {J}}). $$

We note that the score only depends on the unrooted topology of $\widetilde{T}$. Under the GDL model, by independence of the gene trees (and non-negativity), $Q_k(\widetilde{T})/k$ converges almost surely to its expectation simultaneously for all unrooted species tree topologies over S.

For a species $A \in S$ and gene tree $\tilde{t}_i$, let $A_i$ be the gene copy in A on $\tilde{t}_i$ if it exists and let $\mathcal {E}_i^A$ be the event that it exists. For a 4-tuple of species $\mathcal {Q} = \{A,B,C,D\}$, let $\mathcal {Q}_i = \{A_i,B_i,C_i,D_i\}$ and $\mathcal {E}_i^{\mathcal {Q}} = \mathcal {E}_i^A\cap \mathcal {E}_i^B\cap \mathcal {E}_i^C\cap \mathcal {E}_i^D$. The expectation can then be written as

(11)

as, on the event $(\mathcal {E}_1^\mathcal {Q})^c$, there is no contribution from $\mathcal {Q}$ in the sum over the first sample.

Based on the proof of Theorem 1, a different way to write $\mathbf {E}[ \mathbf {1}(\widetilde{T}_{ext}^{\mathcal {Q}_1},\tilde{t}_1^{\mathcal {Q}_1}) \,|\,\mathcal {E}_1^\mathcal {Q}]$ is in terms of the original gene tree $t_1$. Let a, b, c, d be random gene copies on $t_1$ in A, B, C, D respectively. Then if q is the topology of $t_1$ restricted to a, b, c, d,

From (5), we know that this expression is maximized (strictly) at the true species tree $\mathbf {P}'[q = T^{\mathcal {Q}}]$. Hence, together with (11) and the law of large numbers, almost surely the quartet score is eventually maximized by the true species tree as $k \rightarrow +\infty$. This completes the proof for the exact version.

The default version is statistically consistent for the same reason as in the proof of Theorem 3. As the number of MUL-trees sampled tends to infinity, the true species tree will appear as one of the input gene trees almost surely. So ASTRAL returns the true species tree topology almost surely as the number of sampled MUL-trees increases.

1.3 A.3 Proof of Theorem 3: case (2)

In case (2), assume that $T^{\mathcal {Q}} = (((A,B),C),D)$ and let R be the most recent common ancestor of A, B, C (but not D) in T. We want to establish (8) in this case. For $i=1,2,3$, let $\mathcal {N}_{AB|CD}^{\mathcal {Q},\{i\}}$ (respectively $\mathcal {N}_{AC|BD}^{\mathcal {Q},\{i\}}$) be the number of choices consisting of one gene copy from each species in $\mathcal {Q}$ whose corresponding restriction on $t^{\mathcal {Q}}$ agrees with AB|CD (respectively AC|BD) and where, in addition, copies of A, B, C descend from i distinct lineages in R. We make five observations:

• Contributions to $\mathcal {N}_{AB|CD}^{\mathcal {Q},\{2\}}$ necessarily come from copies in A and B descending from the same lineage in R, together with a copy in C descending from a distinct lineage and any copy in D. Similarly for $\mathcal {N}_{AC|BD}^{\mathcal {Q},\{2\}}$

• Moreover $\mathcal {N}_{AC|BD}^{\mathcal {Q},\{1\}} = 0$ almost surely, as in that case the corresponding copies from A and B coalesce (backwards in time) below R.

• Arguing as in the proof of Theorem 1, by symmetry we have the equality $\mathbf {E}_I[\mathcal {N}_{AB|CD}^{\mathcal {Q},\{3\}}] = \mathbf {E}_I[\mathcal {N}_{AC|BD}^{\mathcal {Q},\{3\}}].$

• For $j \in \{1,\ldots ,I\}$, let $\mathcal {A}_j$ be the number of gene copies in A descending from j in R, and similarly define $\mathcal {B}_j$, $\mathcal {C}_j$. Let $\mathcal {D}$ be the number of gene copies in D. Then, under the conditional probability $\mathbf {P}_I$, $\mathcal {D}$ is independent of $(\mathcal {A}_j, \mathcal {B}_j, \mathcal {C}_j)_{j=1}^I$. Moreover, under $\mathbf {P}_I$, $(\mathcal {C}_j)_{j=1}^I$ is independent of $(\mathcal {A}_j, \mathcal {B}_j)_{j=1}^I$.

• Similarly to case 1), by symmetry we have $X^{=} \equiv \mathbf {E}_I[\mathcal {A}_{j_1} \mathcal {B}_{j_1}] = \mathbf {E}_I[\mathcal {A}_{1} \mathcal {B}_ {1}]$, $X^{\ne } \equiv \mathbf {E}_I[\mathcal {A}_{j_1} \mathcal {B}_{k_1}] = \mathbf {E}_I[\mathcal {A}_1]\mathbf {E}_I[\mathcal {B}_1]$ for all $j_1, k_1$ with $j_1\ne k_1$. Define also $X = I X^{=} + I(I-1) X^{\ne }$, $Y \equiv \mathbf {E}_I[\mathcal {C}_{1}]$ and $Z \equiv \mathbf {E}_I[\mathcal {D}]$.

Putting these observations together, we obtain

$$\begin{aligned}&\mathbf {E}_I[\mathcal {N}^{\mathcal {Q}}_{AB|CD}] - \mathbf {E}_I[\mathcal {N}^{\mathcal {Q}}_{AC|BD}]\\&\qquad = \mathbf {E}_I[\mathcal {N}^{\mathcal {Q},\{1\}}_{AB|CD}] + \mathbf {E}_I[\mathcal {N}^{\mathcal {Q},\{2\}}_{AB|CD}] - \mathbf {E}_I[\mathcal {N}^{\mathcal {Q},\{2\}}_{AC|BD}]\\&\qquad = I X^{=} Y Z + I(I-1) X^{=} Y Z - I(I-1) X^{\ne } Y Z\\&\qquad > 0, \end{aligned}$$

where we used Lemma 2 on the last line.