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Dominance, Epistasis and the Genetics of Postzygotic Isolation
Michael Turellia and H. Allen Orrba Section of Evolution and Ecology and Center for Population Biology, University of California, Davis, California 95616
b Department of Biology, University of Rochester, Rochester, New York 14627
Corresponding author: Michael Turelli, Center for Population Biology, University of California, 1 Shields Ave., Davis, CA 95616., mturelli{at}ucdavis.edu (E-mail)
Communicating editor: P. D. KEIGHTLEY
| ABSTRACT |
|---|
The sterility and inviability of species hybrids can be explained by between-locus "Dobzhansky-Muller" incompatibilities: alleles that are fit on their "normal" genetic backgrounds sometimes lower fitness when brought together in hybrids. We present a model of two-locus incompatibilities that distinguishes among three types of hybrid interactions: those between heterozygous loci (H0), those between a heterozygous and a homozygous (or hemizygous) locus (H1), and those between homozygous loci (H2). We predict the relative fitnesses of hybrid genotypes by calculating the expected numbers of each type of incompatibility. We use this model to study Haldane's rule and the large effect of X chromosomes on postzygotic isolation. We show that the severity of H0 vs. H1 incompatibilities is key to understanding Haldane's rule, while the severity of H1 vs. H2 incompatibilities must also be considered to explain large X effects. Large X effects are not inevitable in backcross analyses but ratherlike Haldane's rulemay often reflect the recessivity of alleles causing postzygotic isolation. We also consider incompatibilities involving the Y (or W) chromosome and maternal effects. Such incompatibilities are common in Drosophila species crosses, and their consequences in male- vs. female-heterogametic taxa may explain the pattern of exceptions to Haldane's rule.
IN a landmark article, ![]()
This model of the evolution of postzygotic isolation was later elaborated by ![]()
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This article has two purposes. First, we hope to fill a vacuum in the theory of speciation. Although Dobzhansky's experimental approach has been followed in many studies of postzygotic isolation, there has been no attempt to predict the pattern of fitness differences seen across the hybrid genotypes produced. Are certain genotypes predictably less fit than others? Here, starting from the Dobzhansky-Muller model, we present a theoretical analysis of the expected fitness of different hybrid backcross and F2 genotypes. This analysis differs from our previous ones (![]()
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This fuller analysis forces us to deal with several subtleties of the Dobzhansky-Muller model. Analysis of Dobzhansky-Muller interactions is more difficult than it first appears, as these interactions involve both dominance and epistasis: the severity of the interactions between two loci depends on the dominance of each incompatible allele. In addition, the dominance at one locus might well depend on the genotype at the other locus. In our previous work, we simplified our picture of Dobzhansky-Muller incompatibilities to render the analysis tractable and intuitive (e.g., when studying Haldane's rule, we assigned all dominance effects to the X-linked locus involved in an X-autosomal incompatibility). Here we present a complete model of two-locus Dobzhansky-Muller incompatibilities.
We use this model to address the only two known patterns in the genetics of speciation: Haldane's rule and the "large X effect" (![]()
Our second aim is to determine if the patterns of postzygotic isolation seen across hybrid genotypes in genetic studies of speciation are compatible with our model and to suggest new experimental tests.
| GENETIC MODELS OF POSTZYGOTIC ISOLATION |
|---|
We follow ![]()
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Suppose that a hybrid genotype suffers n distinct incompatibilities and that the ith incompatibility contributes ei > 0 to the breakdown score. (Note that specific loci may contribute to more than one incompatibility.) The fitness of this genotype is
![]() |
(1) |
where W = 0 if the breakdown score S =
ni=1 ei
C. Our analyses apply to taxa in which either males or females are the heterogametic (XY or ZW) sex. Because most relevant data come from Drosophila, however, we assume male heterogamety and note those occasions where female heterogamety alters our conclusions. Among individuals of identical genotype (e.g., F1 hybrid males), some are often fertile and others sterile (e.g., ![]()
Formally, the assumption that many incompatibilities contribute to hybrid sterility/inviability is central to our analysis. Without loss of generality, we can assume that the critical threshold is C = 1. When a large number, n, of incompatibilities is required to reach this threshold, the effect of each must be roughly proportional to 1/n. Hence, the variance of the breakdown scores must also be at most proportional to 1/n as the breakdown score is simply the sum of n random variables, each having mean on the order of 1/n and variance on the order of 1/n2. If we assume that n is large, we can approximate the fitness of a genotype by evaluating the function V in (1) at the expected breakdown score. This simplification is used throughout.
A model of two-locus incompatibilities:
The central idea of our analysis is that hybrids can experience three different types of two-locus Dobzhansky-Muller incompatibilities, depending on whether the incompatible alleles at the interacting loci are homozygous or heterozygous. Imagine that one species' genotype is AAbb and the other aaBB, where a and b are incompatible in hybrids and A and B denote ancestral (and hence compatible) alleles. Further imagine that there are many such pairs of loci. We label the incompatibilities according to the number of loci that are homozygous for incompatible alleles. In an F2 hybrid genotype, certain incompatibilities might be homozygous-homozygous (e.g., aabb), which we denote H2; each such incompatibility contributes, on average, h2 to the hybrid breakdown score. Other incompatibilities might be homozygous-heterozygous (aaBb or Aabb), denoted H1; each contributes, on average, h1 to the breakdown score. All remaining incompatibilities are heterozygous-heterozygous (AaBb), denoted H0; each contributes, on average, h0 to the breakdown score. Note that any locus homozygous for a compatible allele (AA or BB) cannot contribute to hybrid breakdown.
This last assumption merits comment. If the substitutions causing hybrid problems are driven by natural selection within species, one might expect that genotypes combining homozygous ancestral alleles at one locus with selectively favored alleles at another would enjoy enhanced fitness relative to the ancestral genotype, AABB. We assume, however, that any such effects are negligible compared with the deleterious Dobzhansky-Muller effects. Indeed these positive effects must, on average, be small compared to the Dobzhansky-Muller effects, or hybrids would not suffer large net fitness losses (relative to the parental means).
Initially, we consider autosomal-autosomal, X-autosomal, and X-X interactions, but ignore maternal effects and the Y. We also assume dosage compensation such that hemizygosity at X-linked loci is equivalent in fitness effect to homozygosity. (This assumption is not needed for our analysis of Haldane's rule, which centers on incompatibilities that either do or do not involve a hemizygous X chromosome; these correspond to H1 vs. H0 incompatibilities, irrespective of dosage compensation.) To compare the breakdown scores of different genotypes, we assume that there are n two-locus incompatibilities in a reference genotype that carries one set of autosomes and one X from each parental species, as in F1 females. These incompatibilities are assumed to be randomly scattered throughout the genome. For normalization purposes, we use this reference genotype even when considering hybrid males.
To find the expected breakdown score for any particular hybrid genotype, we merely need to know the proportion of the genome that is homozygous (or hemizygous) from species 1 (p1), the proportion that is homozygous (or hemizygous) from species 2 (p2), and the proportion that is heterozygous for material from the two species (pH, where pH = 1 - p1 - p2). A simple combinatoric argument shows that, on average, a fraction p1p2 + (p1 + p2)pH + p2H of the n incompatibilities in the reference genotype appear in our hybrid genotype. In particular, our hybrid suffers an average of np1p2 H2 incompatibilities, n(p1 + p2)pH H1 incompatibilities, and np2H H0 incompatibilities. (These calculations take into account the fact that incompatibilities are generally "asymmetric"; i.e., if introgression of allele A1 from taxon 1 into taxon 2 lowers fitness, the reciprocal introgression of A2 into taxon 1 does not, e.g., ![]()
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(2) |
This simple result forms the foundation of our analyses. We now use it to consider the two best known phenomena in the genetics of speciation: Haldane's rule and the large X effect.
Haldane's rule for inviability:
The case of hybrid inviability is simpler than that of sterility as the same genes likely affect the viability of both hybrid males and females (![]()
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(3) |
Let gX be the fraction of all loci causing incompatibilities that are X-linked (cf. ![]()
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(4) |
In the limiting case in which none of the genome is X-linked (gX = 0), (3) and (4) show that hybrid males and females have the same expected breakdown scores and Haldane's rule for inviability cannot arise (by this mechanism). As noted by ![]()
More generally, (3) and (4) show that if gX > 0, Haldane's rule is expected, i.e., E(Sm) > E(Sf), whenever
![]() |
(5) |
Condition (5) allows for X-X incompatibilities that afflict only females and is, not surprisingly, somewhat more stringent than our previous condition, h0/h1 < 1/2 (or equivalently d < 1/2), which ignored X-X incompatibilities (![]()
![]()
0.2, (5) requires h0/h1 < 4/9. As expected, (5) reduces to h0/h1 < 1/2 when X-X incompatibilities are ignored. Condition (5) also implies that whenever F1 males show reduced viability, "unbalanced" F1 females, who inherit both X's from one species, should be less viable than normal F1 females, a fact confirmed by experiment (![]()
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When there are unequal rates of evolution between the X and autosomes, gX no longer equals the proportion of the genome that is X-linked. If, for instance, the X evolves faster than the autosomes, as suggested by ![]()
Haldane's rule for sterility:
There are good reasons for thinking that different loci will affect hybrid male vs. female fertility. First, mutagenesis experiments within D. melanogaster show that the overwhelming majority of steriles afflict one sex only, while most lethals affect both sexes (reviewed in ![]()
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This opens the door to a possibility that does not arise with hybrid inviability: when different loci affect the two sexes, we have no guarantee that substitutions ultimately afflicting males will occur at the same rate as those afflicting females. Indeed there is evidence that alleles causing sterility in hybrid males accumulate faster than those causing sterility in females, at least in Drosophila and mosquitoes (![]()
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Incompatibilities affecting hybrid female vs. hybrid male sterility may differ in both number and average effects. Let nf (nm) denote the number of incompatibilities in our reference genotype that affect hybrid females (males). We assume that the average effects of H0 and H1 incompatibilities are h0 and h1 for females and
0 and
1 for males. Thus, the expected breakdown scores of F1 females and males (from Equation 2) are
![]() |
(6a) |
and
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(6b) |
where gX denotes the fraction of all loci involved in male-sterilizing incompatibilities that are X-linked.
Under this model, Haldane's rule for sterility is expected [E(Sm) > E(Sf)] whenever
![]() |
(7) |
To simplify the notation, we let
= nm
1/(nfh1) and assume that the ratio of effects for H0 vs. H1 incompatibilities is the same in females and males, i.e.,
=
d0. Note that
quantifies the relative cumulative effects of incompatibilities contributing to hybrid male vs. hybrid female sterility, taking into account both their numbers and average effects. Criterion (7) shows that both faster-male evolution, as measured by
> 1, and dominance, as measured by d0, can contribute to Haldane's rule. We can express (7) either as a bound on
or a bound on d0, namely
![]() |
(8a) |
or
![]() |
(8b) |
As before, the constraint (8b) on the dominance parameter d0 is somewhat more restrictive when X-X incompatibilities are considered than the analogous expression (B2) of ![]()
= 1, condition (8b) reduces to (5), i.e., the condition for inviability. When none of the genome is X-linked (gX = 0) (8a) becomes
> 1. In this case, dominance cannot play a role in Haldane's rule and faster-male evolution is required.
In general, both faster-male evolution and dominance might contribute to Haldane's rule for sterility in male-heterogametic species. To assess their relative contributions, one can consider the ratio of male to female breakdown scores. This ratio, which must exceed 1 for Haldane's rule in male-heterogametic species, is
![]() |
(9) |
Ratio (9) reveals that Haldane's rule is promoted by faster-male evolution (large
) and by greater recessivity of the factors causing hybrid sterility (small d0). Both effects can be seen in Fig 1. If
is sufficiently large, Haldane's rule will occur for any level of dominance. Conversely, in female-heterogametic species, R must be <1 to produce Haldane's rule; this is clearly made more difficult by faster-male evolution (
> 1). The resulting conditions correspond to
< 1 in a male-heterogametic species. As illustrated in Fig 1, extreme recessivity is required to produce Haldane's rule in this case.
|
Comparative analyses support the idea that faster-male evolution acts against Haldane's rule for sterility in female-heterogametic species: Haldane's rule is more common for sterility than for inviability in male-heterogametic species, while the reverse appears true in female-heterogametic species, as expected if faster-male evolution acts in both groups (![]()
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Our analyses of Haldane's rule did not require us to consider homozygous-homozygous incompatibilities, which cannot appear in F1 hybrids. Thus, our previous simple "dominance" analyses (![]()
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The large X effect:
As noted by ![]()
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Consider a backcross analysis of male sterility in which fertile F1 females are crossed to males from taxon 1. When assessing the effect of "foreign" loci, note that the introgressed segments of X2 are necessarily hemizygous, while the introgressed autosomal segments are heterozygous. To quantify the X effect, we require an explicit comparison between X-linked and autosomal introgressions. Although discussions of the large X effect have been mostly qualitative, at least two quantitative criteria can be used. A large X effect might be declared if a hemizygous X introgression has (1) a greater effect than a heterozygous autosomal introgression that is twice as large; or (2) more than twice the effect of an equal-sized heterozygous autosomal introgression. The first criterion requires fewer assumptions, but the second (used by ![]()
We first use criterion 1, comparing X-linked and autosomal introgressions from species 2 into a background homozygous for species 1. For the X introgression, let p2 = q, where q is the fraction of the haploid genome that is introgressed, and p1 = 1 - q. For the autosomal introgression, let pH = 2q (i.e., it is twice as large as the X introgression) and p1 = 1 - 2q. The expected breakdown scores are
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(10a) |
and
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(10b) |
The X introgression has a greater effect when E(SX) > E(Sauto), which requires q[(1 - q)(h2 - 2h1) + 2q(h1 - 2h0)] > 0 or
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(11) |
Our dominance-theory explanation of the large X effect (![]()
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(12) |
which constrains only the relative effects of H0 and H1 incompatibilities. But (11) shows that H2 interactions are also important. The natural extension of our "dominance" constraint (12) is
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(13) |
i.e., homozygous-homozygous incompatibilities have more than twice the deleterious effect of heterozygous-homozygous ones. The important point is that when dominance conditions (12) and (13) are met, so is (11).
The analysis above is idealized. Real backcross analyses involve comparing many genotypes, some of which carry the "foreign" X (or portions of it) and some of which do not. We thus extend our analysis to more realistic situations in which the "background" for an introgression includes heterozygous material (i.e., pH > 0). In this case, we compare E(SX) = E(S|p2 = q, pH = r, p1 = 1 - r - q) with E(Sauto) = E(S|pH = r + 2q, p1 = 1 - r - 2q). Again, (12) and (13) suffice to give E(SX) > E(Sauto). Thus, no matter what the rest of the hybrid genome looks like, introgressing part of the X will lower hybrid fitness more than introgressing twice as much autosomal material when (12) and (13) hold.
Criterion 2 is trickier to apply: while it is easy to compare the average breakdown scores of X vs. autosomal introgressions, we must actually compare the fitness effects of X vs. autosomal introgressions. But these are known only if we know the function V that maps breakdown score onto fitness. For simplicity, then, we assume that the fitness function is nearly linear over the relevant range of values, so that comparing breakdown scores themselves suffices. If fitness declines faster than linearly (e.g., ![]()
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We again first compare X-linked vs. autosomal introgressions from taxon 2 into a background homozygous for taxon 1. The breakdown score is initially zero in both cases, and we want to compare E(SX) = E(S|p2 = q, p1 = 1 - q) with 2E(Sauto) = 2E(S|pH = q, p1 = 1 - q). With (2), we see that E(SX) > 2E(Sauto) if
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(14) |
If we impose our standard "dominance" condition (12) on h0/h1, we see that (14) is satisfied when
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(15) |
For small introgressions (q
0), this reduces to (13), h1/h2 < 1/2. For larger introgressions, it is more restrictive. But even if an X-linked introgression involves, say, 20% of the genome, (15) requires h1/h2 < 4/9 to produce a large X effect, which is only slightly more stringent than (13).
Next we consider introgressions into backgrounds in which some of the autosomal genome is already heterozygous. We want to compare E(
SX)
E(S|p2 = q, pH = r, p1 = 1 - r - q) - E(S|pH = r, p1 = 1 - r) with E(
Sauto)
E(S|pH = r + q, p1 = 1 - r - q) - E(S|pH = r, p1 = 1 - r ). Using (2), we see that E(
SX) > 2E(
Sauto) if
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(16) |
For small introgressions (q
0), conditions (12) and (13) are again sufficient. We assume that both (12) and (13) hold and ask what additional constraints on h1/h2 and h0/h1 are imposed by (16). In general, these depend both on the size of the introgressions, q, and on the extent to which the genetic background is heterozygous, r. Condition (13) implies that the left-hand side of (16) decreases as q increases. Thus, assuming q
0.2, the most restrictive conditions on h1/h2 and h0/h1 occur with q = 0.2. By also considering the left-hand side of (16) as a function of r, one can show that (16) is always satisfied if
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(17) |
Our key biological conclusion is that large X effects, by either criterion 1 or 2, are not automatic consequences of comparing hemizygous X with heterozygous autosomal substitutions. Instead, large X effects imply that the alleles causing hybrid sterility and inviability are fairly recessive. Indeed, large X effects are essentially guaranteed if condition (17) is met (in Drosophila). The more extreme the recessivity of the incompatible alleles, the more pronounced the large X effect. The "faster X" mechanism of ![]()
Comparisons among other backcross genotypes:
There are two different classes of backcross to consider, depending on whether one crosses F1 hybrid males or females to a parental species. When F1 males are backcrossed, the resulting progeny all inherit an X and a complete set of autosomes maternally. In this case, H2 incompatibilities are impossible, and (2) reduces to
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(18) |
where pH is the fraction of the genome that is heterozygous and d0 = h0/h1. When these incompatibilities, rather than those involving the Y or maternal effects, dominate hybrid performance, the lowest fitness results when
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(19) |
For d0 < 1/2, the right-hand side of (19) is between 0.5 and 1.0. Thus we predict that introgressions of "intermediate" size will have the most devastating effects on hybrid fitness.
When F1 females are backcrossed, the resulting male progeny can inherit an X that is incompatible with the haploid set of autosomes inherited paternally. This can produce H2 incompatibilities so that all three terms in (2) may be nonzero, precluding simple predictions analogous to (19).
Other incompatibilities affecting hybrids:
Data from several well-known hybridizations have shown important effects of Y-X (e.g., ![]()
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|
Given their frequent roles, it is important to incorporate Y and maternal effects into our analysis. This can be accomplished via Equation 1, by assuming that Y and maternal incompatibilities simply add to the total breakdown score, which we now denote ST. Keeping with our earlier notation, we denote the cumulative breakdown score attributable to X-X, X-autosome, and autosome-autosome incompatibilities by S. Additional contributions from Y-associated and maternal-zygotic incompatibilities are denoted SY and SMZ. We discuss each in turn.
Y-linked incompatibilities:
We consider both male-heterogametic (XY) and female-heterogametic (ZW) taxa, but, for ease of discussion, refer to the sex-limited sex chromosome as the Y and its partner as the X. For simplicity, we focus on E(SY). Our qualitative predictions do not require that the number of Y-associated incompatibilities is large. Indeed the "large n" assumption is clearly implausible for Y-associated incompatibilities, as (1) the Y carries few complementation groups, at least in D. melanogaster and D. hydei (![]()
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Because the Y is hemizygous, we treat the Y-linked partner in any incompatibility as effectively homozygous. We do not distinguish between Y-X and Y-autosome incompatibilities and assume that each occurs in proportion to the fraction of the genome that is X-linked vs. autosomal. Let nY denote the number of incompatibilities between Y-linked loci from taxon 1 and loci in a complete haploid set, including an X, from taxon 2. In F1 and backcross genotypes, these incompatibilities can occur in two forms, depending on whether the non-Y partner is homozygous or heterozygous. Incompatibilities involving homozygous partners have average effect y2, whereas those involving heterozygous partners have average effect y1.
Because the Y is largely heterochromatic, we have no a priori basis for estimating the fraction of all incompatibilities that involve this chromosome and therefore no basis for drawing quantitative conclusions about SY vs. S. But, by considering E(SY) alone, we can still draw interesting, albeit qualitative, conclusions. If the Y is inherited from taxon 1 and the source of the rest of the nuclear genome is described by the fractions p1, p2, and pH, we get
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(20) |
For F1 males (or F1 females in female-heterogametic species), we expect that p2 = gX and pH = 1 - gX. Thus,
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(21) |
Obviously, SY = 0 for the homogametic sex. Thus, Y-associated incompatibilities always afflict only the heterogametic sex and promote Haldane's rule (![]()
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Y-associated incompatibilities also relax the constraints on the dominance, h0/h1, required for Haldane's rule for sterility, particularly in female-heterogametic species. For two reasons, this effect is likely to be especially important in taxa having small sex chromosomes. First, in the absence of Y effects, the upper bound on h0/h1 needed for Haldane's rule when faster-male evolution acts is proportional to gX when gX is small (see 8b). Thus, in female-heterogametic species with relatively small X's (on the order of 10% of the genome or less), such as birds (![]()
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Given this expected disproportionate effect of the Y in small-X taxa, it is interesting to note that Haldane's rule for sterility shows only a single exception in Lepidoptera and birds (![]()
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Next, we consider the role of the Y in the large X effect. Consider a study of F1 male sterility in which F1 females are backcrossed to taxon 1. As above, we can compare the values of SY produced by an X-linked introgression of size q vs. an autosomal introgression of size 2q. These are
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(22a) |
and
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(22b) |
respectively. Hence, an X introgression yields a larger Y-associated contribution to ST if
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(23) |
Note that if incompatibilities involving the Y are highly recessive (i.e., y1/y2 << 1/2), the relative effects of Y-autosomal interactions will be negligible unless the "foreign" autosomal segments are homozygous. Thus, our analysis suggests a bias toward finding Y-X incompatibilities vs. Y-autosomal. Indeed, in the best-known case of a Y-autosomal incompatibilitythat between the Y of D. arizonae and a region of the fourth chromosome of D. mojavensisflies that are homozygous for the incompatible autosomal segment are sterile whereas flies that are heterozygous show normal fertility (![]()
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Maternal-zygotic incompatibilities:
Again we focus on E(SMZ) but recognize that the data suggest that maternal incompatibilities may involve few factors. These incompatibilities arise from interactions between loci in two different diploid genomesthat of the mother and her offspringand they may generally affect viability only as maternal control of development is surrendered fairly early in development (![]()
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Maternal-zygotic incompatibilities may occur in three forms depending on whether the incompatible alleles are homozygous or heterozygous. To distinguish these incompatibilities from those acting within the offspring genome, we denote them by M0, M1, and M2. There may be a qualitative distinction between the two types of M1 incompatibilities (depending on whether the mother or offspring is homozygous for an incompatible allele), but we assume they have equal average effects. We assume that the average effect of Mi incompatibilities is mi for i = 0, 1, 2 and that there are nMZ maternal-zygotic incompatibilities between a cytoplasm produced by a taxon 1 mother and her F1 hybrid daughters. Obviously, these would all be M1 incompatibilities involving homozygous maternal loci and heterozygous loci in the offspring. We assume that a fraction gX of these incompatibilities involve loci on the zygotic X. With backcross analyses involving hybrid mothers, a wide range of incompatibilities can appear. We focus on only those that occur in hybrids with taxon 1 mothers.
If the offspring genome is characterized by the fractions p1, p2, and pH, as before, the expected contribution to their breakdown score from maternal-zygotic interactions is
![]() |
(24) |
This reveals a qualitative difference between the consequences of these incompatibilities in male-heterogametic vs. female-heterogametic taxa.
In male-heterogametic species, the only relevant difference between F1 females and males is that females suffer from M1 incompatibilities between the paternal X and the maternal cytoplasm, whereas males do not (![]()
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![]() |
(25a) |
and
![]() |
(25b) |
since pH = 1 in females, but pH = 1 - gX and p1 = gX in males. Thus females in male-heterogametic taxa suffer more from maternal incompatibilities than males; and incompatibilities involving the X can give rise to exceptions to Haldane's rule in these taxa. Indeed, ![]()
In contrast, in female-heterogametic species, F1 males have pH = 1, but F1 females have pH = 1 - gX and p2 = gX. Thus,
![]() |
(26a) |
and
![]() |
(26b) |
This shows that maternal-zygotic interactions contribute to Haldane's rule for inviability in female-heterogametic speciesas the heterogametic sex gets its cytoplasm from one species and its X from anotherwhenever M2 incompatibilities are more severe than M1. It would be surprising if this very weak "dominance" constraint is not satisfied.
Thus, in female-heterogametic species, dominance and maternal effects act in concert to promote Haldane's rule, whereas they act in opposition in male-heterogametic species. Maternal effects may, therefore, explain both the prevalence of exceptions to Haldane's rule for viability in Drosophila (and many of these exceptions appear evolutionarily independent; ![]()
![]()
Complex genetic interactions:
We have focused on two-locus incompatibilities as they capture the essence of the Dobzhansky-Muller mechanism and are easily modeled. But hybrid inviability and sterility may be produced by more complex interactions involving three or more loci (e.g., ![]()
| HYBRIDIZATION DATA |
|---|
While the above theory makes several predictions, existing data do not yet allow critical tests. Our purpose here therefore is merely to show (1) what predictions can be made; and (2) how these predictions can be tested with hybrid backcross data.
Data from introgressions:
A large body of introgression data supports our assumption that H2 incompatibilities are more severe than H1. ![]()
![]()
D. buzzatii-D. koepferae:
These species obey Haldane's rule for sterility. They have N = 6 chromosomes, with four autosomes and an X, all of roughly equal size, and a tiny sixth. ![]()
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0.3/5 = 0.06) essentially never cause male sterility, whereas introgressions of >40% of a large autosome (pH
0.4/5 = 0.08) almost always do. These introgressions correspond to p1 = 1 - pH in Equation 2 and the expected breakdown scores are given by (18). These autosomal data imply that
![]() |
(27a) |
and
![]() |
(27b) |
In contrast, ![]()
0.04/5 = 0.008 and p2 = 1 - p1
0.992). Indeed, they argue that introgressions as small as 1% of the X cause sterility (i.e., p1
0.002). These hemizygous introgressions cause all H2 incompatibilities, implying that
![]() |
(28) |
Combining (27a) and (28), we get (0.056)h1 < (0.002)h2 or
![]() |
(29) |
This suggests a level of recessivity for hybrid steriles comparable to that of within-species lethals in D. melanogaster (![]()
H0 vs. H1 incompatibilities:
We have less direct evidence about the magnitudes of h0 and h1. However, ![]()
Hybrid backcross analyses:
We now turn to traditional backcross/F2 analyses. Our approach is statistical: we pool all hybrid backcross or F2 genotypes that have the same p1, p2, and pH. Consider, for instance, a backcross between two Drosophila species possessing five roughly equal-sized chromosome arms, one of which is the X. We pool all data obtained when a single autosomal arm is introgressed, or two autosomal arms are introgressed, etc. We appreciate that experiments repeatedly show that particular chromosomes have large effects on backcross fitness while others do not, but our primary goal is not to explain the detailed outcomes of particular species crosses but to search for statistical regularities.
In a species cross, any of the incompatibilities discussed above might act. While we can make some predictions about the relative roles of X-autosomal vs. X-X incompatibilities, we have no theory allowing us to predict how often, for instance, Y-linked or maternally acting genes might contribute toor even dominatepostzygotic isolation. Thus we concentrate on cases in which Y and maternal effects are absent or small.
We make one further simplification. The above theory requires that we know p1, p2, and pH. Unfortunately, most backcross studies in Drosophila involve taxa obeying Haldane's rule, so that backcrosses must proceed through F1 females. Because females recombine, markers remain associated with only some (inexactly known) chromosome region. We do not, therefore, know p1, p2, and pH. This difficulty does not arise when backcrosses are performed through F1 males, as there is no recombination in Drosophila males. Thus we consider only backcrosses that proceed through F1 males. We know of two relevant studies: D. mojavensis-D. arizonae (formerly arizonensis) and D. hydei-D. neohydei. We focus on the first for purposes of illustration.
D. mojavensis-D. arizonae:
Backcross hybrid males are sterile if they have a Y from D. arizonae and are homozygous for the fourth chromosome of D. mojavensis. These species have five chromosomes of roughly equal size (including the X), and a dot sixth chromosome that we will ignore. To avoid the complications of Y-linked incompatibilities, we focus on the backcross of F1 males (from D. mojavensis mothers) to D. arizonae females. Table 2 of ![]()
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The data are given in Table 2 along with the predicted breakdown scores from our two-locus analysis, Equation 18, and the three-locus analysis from the Appendix Because of small samples, there is only one statistically significant jump between adjacent fractions in the tablethat from pH = 0.2 (6/24) to pH = 0.4 (27/29) (P < 10-5).
It is worth noting that our large n assumption likely does not hold here. This is suggested by several lines of evidence. First, XmYa AmAa F1 males (where A denotes a haploid set of autosomes) are fertile, whereas XaYm AaAm males are all sterile. As both F1 males have the same expected breakdown scores, this difference reveals heterogeneity in the numbers of incompatibilities between "replicate" X chromosomes. Second, Zouros and collaborators have shown that hybrid male sterility is caused by X-autosome and autosome-autosome incompatibilities, not Y effects. But the data in Table 2 of ![]()
Despite this, it seems worth asking if the pooled data in the first column of Table 2 agree with our predictions (given in the last two columns), where we assume that higher expected breakdown scores will be associated with greater sperm immotility.
First consider the two-locus predictions. From (18), as pH increases, the expected breakdown score, E(S), rises to a maximum of nh1/[4(1 - d0)] at pH = 1/[2(1 - d0)] (see Equation 19), then falls to nh1(0.16 + 0.64d0) at pH = 0.8. Because we ignore Y effects, the expected breakdown score for males with pH = 0.8 in Table 2 is the same as for the sterile XaYm AaAm F1 males. Indeed, both are sterile, as they suffer the same incompatibilities. Note that if d0 < 2/7, E(S) for pH = 0.6 is larger than E(S) for pH = 0.8. As expected, the pH = 0.6 males are also all sterile. The lack of statistical power precludes more detailed tests, but the data suggest the kind of inferences possible. For instance, if the difference between the fractions of males with motile sperm for pH = 0.4 and pH = 0.8 were statistically significant, we could conclude that the corresponding breakdown scores must satisfy 0.24 + 0.16d0 < 0.16 + 0.64d0, so that d0 > 1/6.
Now consider the three-locus predictions. The most interesting difference between the two- and three-locus predictions is that the latter implies that the largest breakdown score occurs for smaller pH than in the two-locus case. This can be seen in the breakdown scores for pH = 0.2 vs. pH = 0.8. In the two-locus model, E(S) is always larger for pH = 0.8. In contrast, assuming that d0 = d1 = d, the three-locus model implies that E(S|pH = 0.8) < E(S|pH = 0.2) if d < 0.197. Thus the observation that males from this cross with pH = 0.8 are less fit than those with pH = 0.2 suggests that most of the D. mojavensis/arizonae incompatibilities act more like two- than three-locus ones (or that these incompatibilities are less recessive than suggested by our inferences from other data).
| DISCUSSION |
|---|
Our understanding of speciation has been characterized by several steps in which large but nebulous problems have been reduced to smaller but sharper ones. During the modern synthesis, for instance, "the origin of species" was largely reduced to "the origin of reproductive isolation" (![]()
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Fortunately, the Dobzhansky-Muller mechanism is simple enough that it can be captured in mathematical models. Here we have presented a complete model of two-locus Dobzhansky-Muller interactions. To simplify our analysis, we assumed that the number of incompatibilities is large and that individual incompatibilities contribute linearly to a "breakdown score," such that higher scores lead to lower fitness (see Equation 2). Either assumption may be incorrect and more general models could be constructed. Increased generality would, however, lead to more ambiguous predictions, dependent on a proliferation of parameters whose values are unknown. We have also assumed that dosage compensation renders the effects of hemizygotes equivalent to those of homozygotes. This assumption is irrelevant to our analysis of Haldane's rule, as F1 individuals do not experience H2 incompatibilities and all of their H1 incompatibilities involve the hemizygous X (or Z) chromosome. Our assumption is, however, critical to our analyses of backcrosses. Fortunately, essentially all of the relevant data come from Drosophila in which dosage compensation occurs. For taxa like birds and lepidoptera, in which dosage compensation appears absent (![]()
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We have used our model to address several questions in the genetics of speciation, including Haldane's rule and the large X effect.
Haldane's rule:
Table 3 summarizes the forces hypothesized to contribute to Haldane's rule. We indicate whether each might act for hybrid sterility and/or inviability and in male and/or female heterogametic taxa. We first consider dominance alone.
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Roughly speaking, our analysis shows that Haldane's rule arises if the factors causing postzygotic isolation act as partial recessives, i.e., if H1 incompatibilities are somewhat more than twice as severe as H0 ones. This condition emerges if either the same genes affect males and females or if male and female incompatibilities evolve at the same rate.
Fortunately, we possess data allowing rough inferences about the magnitude of h0/h1. First, in two Drosophila hybridizations, "normal" F1 females (who suffer H0 incompatibilities only) are viable, while "unbalanced" females carrying an attached-X stock (and who thus suffer some H1 incompatibilities) are inviable (![]()
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With hybrid sterility, our analysis is more complex, as different loci appear to affect males vs. females, allowing for the possibility that male- and female-expressed genes evolve at different rates (![]()
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There is now considerable evidence that faster-male evolution occurs for genes causing postzygotic isolation (reviewed in ![]()
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Second, the relative rates of accumulation of male-sterilizing vs. female-sterilizing substitutions can be easily overestimated from the observed excess of hybrid male over female steriles. ![]()
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The reason is simple. If there have been Kf substitutions at female-expressed genes and Km at male genes and all incompatibilities involve pairs of loci, the expected numbers of hybrid female vs. male incompatibilities are
![]() |
(30) |
where p is the probability that two diverged genes are incompatible (![]()
(Km/Kf)2 and the ratio of male-to-female substitution rates is
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(31) |
Thus, if genetic analysis suggests nm/nf
6, the ratio of substitution rates is only RK
2.4. Moreover, this figure must be an overestimate as we have assumed that all incompatibilities result from two-locus interactions. If, instead, incompatibilities are due exclusively to three-locus interactions, the rate of male-to-female evolution would be roughly the cube root of nm/nf, which is yet smaller (
1.8). The point is thatif male incompatibilities are more common than female and complex incompatibilities are commonthe ratio of male-to-female substitution rates must be much smaller than the observed ratio of male-to-female incompatibilities. This reflects the fact that Dobzhansky-Muller incompatibilities "snowball," accumulating at least as fast as the square of the substitution rate (![]()
Third, the fact that autosomal regions cause male sterility more often than female sterility when introgressed into a foreign species does not necessarily reflect faster evolution of autosomal male-expressed than female-expressed genes. The reason is that introgressions confront different genetic backgrounds in males and females. Autosomal introgressions confront a "foreign" Y chromosome in males but not females. To the extent that the Y plays an important role in hybrid male sterilityand our review of the literature strongly suggests it doesone would expect more male than female sterility in introgression experiments even if male and female autosomal genes evolve at the same rate.
Last, while proponents of faster-male evolution often cite the fact that male reproductive tract proteins evolve faster than nonreproductive ones, recent studies indicate that both male and female reproductive tissues evolve at high rates. Indeed, ![]()
We are not suggesting that faster-male evolution for hybrid sterility does not occur. It almost certainly does [it is hard to see how else one could explain the fact that taxa lacking a hemizygous X obey Haldane's rule for sterility (![]()
Our extensions to our basic modelincorporating Y and maternal effectsalso have important bearings on understanding Haldane's rule. The consequences of Y-linked incompatibilities are simple: Y effects always promote Haldane's rule in both male- and female-heterogametic species. Moreover, Y effects may have disproportionately greater effects in taxa, like birds and Lepidoptera, that have relatively small X chromosomes. Our review of genetic analyses of postzygotic isolation in Drosophila leaves no doubt that Y effects are very common, at least for male sterility. The consequences of maternal effects are more subtle: maternal-zygotic incompatibilities contribute to Haldane's rule in female-heterogametic species (as the XY sex gets its cytoplasm from one species and its X from another) but work against Haldane's rule in male-heterogametic species (as the XX sex gets its cytoplasm from one species and one X from another). This asymmetry suggests that maternal effects might explain both the prevalence of exceptions (for viability) in Drosophila (![]()
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The "faster X" hypothesis of ![]()
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Large X effect:
Our analysis shows that large X effects are not an inevitable consequence of backcross analysis. Substitution of a hemizygous X does not invariably lower hybrid fitness more than twice as much as substitution of a similarly-sized heterozygous autosome. Instead, large X effects arise if the genes causing postzygotic isolation are fairly recessive. Here "recessive" refers to two comparisons. Because backcross hybrid males suffer from H0, H1, and H2 incompatibilities, the ratios h0/h1 and h1/h2 are both relevant. This highlights the shortcomings of previous attempts, including ours, to understand the implications of dominance in Dobzhansky-Muller incompatibilities. Because such interactions involve both dominance and epistasis, no single dominance parameter fully captures the behavior of hybrid lethals/steriles. While this did not affect our earlier analyses of Haldane's ruleas F1 hybrids cannot suffer H2 incompatibilitiesit plays a key role in backcross hybrids and thus in discussions of the large X effect.
As noted previously, we have qualitative information about h1/h2 from the introgression experiments of ![]()
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Evidence bearing on the dominance theory may also emerge from quantitative trait loci (QTL) studies performed for quite different reasons. Interspecific QTL analyses often uncover distorted segregation ratios at marker loci (e.g., ![]()
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The weight of the evidence suggests, then, that both h0/h1 and h1/h2 are fairly small. If so, the dominance theory provides a powerful explanation of Haldane's rule and the large X effect for both inviability and sterility in all species having heteromorphic sex chromosomes.
In sum, we believe our analysis sheds considerable light on the genetics of speciation by postzygotic isolation. Our model's one obvious merit is that it is firmly grounded on the known genetic mechanism underlying postzygotic isolationDobzhansky-Muller incompatibilities. Such a model would seem, therefore, to provide a more solid basis for future work than the abstract theories and verbal speculations that all too often characterized discussions of speciation.
| FOOTNOTES |
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This article is dedicated to our pal King Coyne on the occasion of his fiftieth birthday. ![]()
| ACKNOWLEDGMENTS |
|---|
We thank N. H. Barton, A. Betancourt, J. A. Coyne, C. Jones, R. Haygood, P. D. Keightley, X. R. Maside, D. Presgraves, E. Zouros, and two anonymous reviewers for helpful comments and discussion. M.T. thanks the University of Edinburgh for providing an excellent sabbatical research environment. This research was supported in part by National Science Foundation grant DEB 9527808 (M.T.) and by National Institutes of Health grant GM51932 and the David and Lucile Packard Foundation (H.A.O.).
Manuscript received September 1, 1999; Accepted for publication December 21, 1999.
| APPENDIX |
|---|
THREE-LOCUS INCOMPATIBILITIES
The text focuses on the simple case of two-locus epistatic incompatibilities; but hybrid dysfunction may be caused by more complex interactions involving three or more loci. To explore the robustness of our conclusions, we consider three-locus interactions here. We denote the interacting loci A, B, and C and use subscripts to indicate species identity.
Three interacting loci produce five distinct types of incompatible genotypes, all of which are assumed to consist of two alleles from one taxon that are incompatible with a third allele from the second taxon. We again assume incompatibility effects are asymmetric, in that if allele A1 contributes to an incompatibility when introgressed into taxon 2, the reciprocal introgression of allele A2 into taxon 1 has no deleterious effect. The five types of incompatibilities are labeled according to the number of loci that are homozygous for alleles involved in the incompatibility, with the exception that there are two distinct types of incompatibilities involving two homozygous loci; these are distinguished by using H2 vs. H1,1. The types of incompatibilities are
- H0, all three loci heterozygous: A1A2B1B2C1C2;
- H1, one locus homozygous: e.g., A1A2B1B2C1C1;
- H2, two loci homozygous for alleles from one taxon: e.g., A1A2B1B1C1C1;
- H1,1, two loci homozygous for alleles from different taxa: e.g., A1A1B1B2C2C2; and
- H3, all three loci homozygous: e.g., A2A2B1B1C1C1.
Let hi denote the average effect of Hi incompatibilities for i = 0, 1, 2, 3; and let h1,1 denote the average effect of H1,1 incompatibilities.
Again let n denote the number of three-locus incompatibilities that occur in a reference genotype containing one X chromosome and one complete set of autosomes from each of the hybridizing taxa. When calculating the average number of these incompatibilities that will afflict any specific hybrid genotype, it is important to note that there are two types of three-locus incompatibilities: those in which two alleles from taxon 1 interact negatively with an allele from taxon 2 and those in which an allele from taxon 1 interacts negatively with two alleles from taxon 2. More complex interactions (involving reciprocal effects of both alleles at a locus) are expected to occur infrequently and will be ignored. Among the n incompatibilities, on average half should fall into each of the two types.
To find the expected breakdown score for any particular genotype of F1, backcross, or F2 hybrid, we need to know the proportion of the genome that is homozygous (or hemizygous) from species 1 (p1), the proportion homozygous from species 2 (p2), and the proportion heterozygous for material from the two species (pH, where pH = 1 - p1 - p2). The expected breakdown score is
![]() |
(A1) |
The logic of the derivation will be illustrated by considering the term proportional to p21p2. Let I1 denote the portion of the genome characterized by p1 and let I2 denote the portion characterized by p2. Of the n three-locus incompatibilities in the reference genotype, a fraction 3p21p2 will, on average, involve two loci in I1 and one locus in I2. Of these, on average, half will involve two alleles from taxon 1 and half will involve two alleles from taxon 2. By definition, two of the alleles in the incompatibilities under consideration must reside in I1. Since I1 is inherited from taxon 1, none of the second class of incompatibilitiesinvolving two alleles from taxon 2can contribute to the p21p2 term. Of the first class of incompatibilitiesinvolving two alleles from taxon 1only one-third will have both taxon 1 alleles in I1. Thus, on average, only one-sixth of the 3np21p2 three-locus incompatibilities involving two loci from I1 and one from I2 will occur in these hybrids, and all of them will be H3 incompatibilities. Each term in (A1) can be derived similarly. We now use (A1) to determine conditions for Haldane's rule.
Haldane's rule for inviability:
For F1 females, all loci are heterozygous (pH = 1), and thus
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(A2) |
In F1 males, p1 + p2 = gX, the fraction of the genome that is X-linked, p1p2 = 0, and pH = 1 - gX; so
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(A3) |
Thus Haldane's rule occurs on average [i.e., E(Sm) > E(Sf)] whenever
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(A4) |
When gX is very small, this reduces to h0/h1 < 1/2. In general, the bound on h0/h1 depends on h2/h1. When h1/h2
1/2, it suffices to have
![]() |
(A5) |
For Drosophila species with ~20% of the genome X-linked, (A5) yields h0/h1 < 0.46, whereas our two-locus criterion (5) yields h0/h1 < 0.44.
We can obtain conditions for Haldane's rule for sterility and the large X effect by straightforward extensions of the arguments presented in the text.
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