HINT 19A: Find the change in the estimated mean, divided by the standard deviation of this estimate. (The variance of the mean is one-tenth the estimated variance among the ten replicate cages.) A change of more than about two standard deviations is unlikely to occur by chance.

HINT 19B: Begin by finding how the ratio of allele frequencies changes.

HINT 19C: The number of *P* alleles in a sample of *n* genes, *j*_{P}, follows a binomial distribution. The variance of allele frequency, *j*_{P}/*n*, is *pq*/*n*, where *p* is the allele frequency.

HINT 19D: See Box 15.1.

HINT 19E: See Figure 17.19B. Make (rough) quantitative estimates of the increase in ratio of allele frequency, the equilibrium frequency, and selection coefficient (see Table 17.1).

HINT 19F: Assume Hardy–Weinberg proportions.

HINT 19G: See page 504.

HINT 19H: Use the approximation that (1 – λ)^{t} ~ exp(–λ*t*) (Chapter 28).

HINT 19I: Equate the result from ii) to the fraction of sites with no change.

HINT 19J: The number of substitutions follows a **Poisson distribution** (Chapter 28).

HINT 19K: Think about the constraints imposed by the genetic code.

HINT 19L: The observed number of differences is less than the actual number, because multiple substitutions appear as a single change. The actual number can be estimated from the fraction of sites that stays unchanged, *Q*, which is exp(–2λ*T*).

HINT 19M: The product θ(θ + 1) ··· (θ + 11) = 39916800 θ + 120543840 θ^{2} + 150917976 θ^{3} + 105258076 θ^{4} + 45995730 θ^{5} + 13339535 θ^{6} + 2637558 θ^{7} + 357423 θ^{8} + 32670 θ^{9} + 1925 θ^{10} + 66 θ^{11} + θ^{12}.

HINT 19N: The only configuration of four alleles with as much or more homozygosity is 9:1:1:1.

HINT 19O: The difference between observed and expected nonsynonymous divergence can be attributed to positive selection.

HINT 19P: Recall Equation 17.2.

HINT 19Q: Recall Figure 19.17A,B.

HINT 19R: If two lineages both trace back to the single genome that carried the favorable mutation, without recombination, then they are identical by descent.

HINT 19S: Fitting a curve by eye will do.

HINT 19T: See page 426.

HINT 19U: Work out the diversity for pairs of genes that are linked to the same allele and that are linked to different alleles.

HINT 19V: Ignore the possibility that lineages coalesce more recently than 10^{5} generations ago (see p. 542).

HINT 19W: Recall Figure 18.6.

HINT 19X: Once a pair has been disrupted, a second substitution will occur that restores pairing. This may take the pair back to the original state or it may produce a new pair.

HINT 19Y: The rate of back mutation to original state at one site is equal to the rate at which the other site in the pair mutates to produce a new pairing.

HINT 19Z: The number of heterozygous loci follows a **binomial distribution**; see Box 28.5.

HINT 19AA: Use the method of Box 28.2.

HINT 19BB: Use the same method as that in ii); see Table 17.1.

HINT 19CC: Assume that selection is weak, so that allele frequency changes continuously through time.

HINT 19DD: Ignore the effects of random drift. Then, the ratio of allele frequencies (*p/q*) is proportional to (1 + *s*)^{t} ~ exp(*st*).

HINT 19EE: Because *A*^{U} is not directly selected, its frequency *within* the genetic backgrounds defined by *B*^{P}, *B*^{Q} does not change as a result of selection.

HINT 19FF: We saw in iii) that selection does not change *u*_{P}, *u*_{Q}. So, we need only look at the change due to recombination at meiosis.

HINT 19GG: Combine the results from i)–iv).

HINT 19HH: An approximate answer will do here. Think about how strong the association *u*_{P} – *u*_{Q} will be at the time when *B*^{P} gets to detectable frequency and sweeps to fixation.

HINT 19II: Average over the random positions at which these mutations arise. You can assume that the neutral gene is in the middle of the genome, *R*/2 from either end.