Problem 22.1  A species sometimes interbreeds with another species; on average, 10% of matings are with the other species. The F₁ offspring from such matings have normal viability but are completely sterile. Offspring spend their early life together, with each family growing within its own patch. There is intense competition within patches, so that the same number emerges from each.

Problem 22.2  Two species diverged 1 Mya, but they have since hybridized at a constant rate: In each species, a fraction m = 5 × 10^–7 of breeding individuals come from the other species. The effective population sizes of each species and of the ancestral species are all N_e = 10⁵. Individuals reproduce annually, and so a year equals one generation. See Figure P22.1.

Problem 22.3  Approximately 9.3 Mb of autosomal sequences from human, chimpanzee, and gorilla were aligned; sequences from orangutan and macaque were also aligned to provide an outgroup. Sites were classified according to whether variants were found only in human (H), only in chimpanzee (C), only in chimpanzee and human (HC), and so on. The total numbers of sites were as follows:

These numbers have been corrected for multiple hits (see Chapter 27); so you can assume that these represent mutations that occurred only once in the genealogy.

The simplest hypothesis to account for these data is that the population ancestral to humans and chimpanzees separated T₀ years ago and that the population ancestral to gorillas split off from the ancestors of humans and chimpanzees a time T₁ before that. Assume that both ancestral populations had effective size N_e. Figure P22.2A shows an example of where the genealogy at a site (thin lines) is the same as the species’ phylogeny; the star marks a mutation ancestral to H and C, which would appear in class HC. Figure P22.2B shows an example where C and G are most closely related, which disagrees with the species’ phylogeny; the asterisk marks a mutation that would appear in class CG. This phenomenon is known as lineage sorting (see p. 628).

Problem 22.4  A simple model for the evolution of reproductive isolation in allopatry is that each population accumulates substitutions at a steady rate. These substitutions might occur by random drift or because they are positively selected, but they do not lead to any incompatibilities being expressed during the course of evolution within the diverging populations. Substitutions can occur at very many sites, and so we can assume that the substitutions in each population are different. There is a small chance p that any pair of alleles will cause incompatibility; each incompatibility that is expressed in a hybrid reduces its fitness by a factor (1 – s), where s = 10%. For example, if each population accumulates 100 substitutions, then there are 100 × 100 = 10,000 pairs of alleles that might prove incompatible. If p = 10^–3, then we expect 100 × 100 × p = 10 incompatibilities to arise, and we expect the fitness of hybrids to be reduced by a factor ~(1 – s)¹⁰ = 0.35.

Alleles might be dominant or recessive. Thus, if there is an incompatibility between two dominant alleles, A¹ and B¹, then the F₁ genotype A¹B⁰/A⁰B¹ will suffer, as will the three other genotypes A¹B⁰/A¹B¹, A¹B¹/A⁰B¹, A¹B¹/A¹B¹, which could appear in F₂. On the other hand, if A¹ and B¹ are both recessive, then only the double homozygote A¹B¹/A¹B¹ will express the incompatibility. Assume that most alleles are recessive, with only a proportion α = 0.01 of alleles being dominant; assume that p = 10^–3, as before. NOTE 22C

Problem 22.5  The chance that a single copy of a chromosome rearrangement will fix in a population with effective size N_e, despite selection s against heterozygotes, is close to

where N is the actual number of individuals. NOTE 22D

Problem 22.6  Under the simple model of Problem 22.5, the rate of chromosomal evolution is faster for more weakly selected changes, but, on the other hand, such changes contribute little to isolation.

Problem 22.7  The Dobzhansky–Muller model allows incompatibilities to evolve in allopatry, without being opposed by selection and without any difference in environment. Can such incompatibilities be maintained after divergent populations come back into contact?

To keep the model as simple as possible, we consider two unlinked genes (A, B) in a haploid population. In one population, an allele A^P increases relative fitness by s = 0.01 and replaces allele A^Q. In another population, an allele B^P replaces B^Q, with the same selective advantage. However, A^PB^P genotypes die. You can assume that there is no association between the genes (i.e., they are in linkage equilibrium). NOTE 22F

A population fixed for A^PB^Q receives a fraction m of its individuals from a population fixed for the opposite genotype, A^QB^P. As a result, booth genes are polymorphic, with p_A < 1 and 0 < p_B.

Problem 22.8  In Problem 22.7 we asked whether incompatibilities could persist despite gene flow. Here, we ask whether they could arise in the first place despite gene flow. Can incompatibilities evolve under the Dobzhansky–Muller model when two populations are connected by gene flow? As in Problem 22.4, assume that substitutions can occur at very many sites and that there is a small chance p = 10^–3 that any pair of substitutions are incompatible. Imagine two populations of N = 10⁴ diploid individuals that exchange genes at a very low rate, m = 10^–6 per generation, and that share the same environment. Favorable alleles arise and fix at a rate Λ = 0.02 per generation within each of the two populations. A new allele that fixes in one population will eventually spread to fix in the other, unless there is another allele there that is incompatible with it. Thus, if a stable genetic difference is to be established, then either the new allele has to be incompatible with a divergent allele that is already established in the other population or an incompatible allele has to fix in the other population before the new allele gets transmitted across.