where the sum is over all loops by which the same gene might be passed down by different routes to the individual in question; fA is the probability of identity between the two genes of the ancestor in each loop.
HINT 15A: After one generation, there is a 1/3 chance that there are no copies, and 2/3 that there are two; proceed in the same way to the next generation, keeping track of the various possibilities. Basic probability is explained in Chapter 28.
HINT 15B: Consider the chance of loss, Q, as the sum over two possibilities: no offspring or two offspring.
HINT 15C: Start by considering the expected number of copies after 1 minute.
HINT 15D: The form of the distribution is shown in Figure 15.3A, for a starting frequency of p0 = 0.5.
HINT 15E: You need to find the expected proportion of heterozygotes formed by random mating within a population with allele frequency p, E[2pq]. This is derived in Box 15.1 as 2(p0q0 – var(p)) = 2p0q0(1– 1/2N)t). If two individuals randomly chosen from different subpopulations were crossed, what would be the chance that their offspring would be heterozygous?
HINT 15F: Think about the decline in genetic variance.
HINT 15G: At equilibrium, the reduction in genetic variance due to drift must balance the increase in genetic variance due to mutation.
HINT 15H: The relation 1 + x + x2 + ... + xt – 1 = (1 – xt)/(1 – x) may be useful.
HINT 15I: See Box 15.2.
HINT 15J: You need to find the variance of this distribution, which is defined as the average squared deviation from the mean. So, first find the mean and then the average of the squared deviations from this.
HINT 15K: The coefficient of kinship is the chance that two randomly chosen genes, one from each individual, are IBD. The gene from the daughter is equally likely to have come from the father as from the queen.
HINT 15L: The formula in Box 15.3 gives the probability of identity by descent as
where the sum is over all loops by which the same gene might be passed down by different routes to the individual in question; fA is the probability of identity between the two genes of the ancestor in each loop.
HINT 15M: The expectation of an exponential, exp(lt), is the inverse of its rate, l; see Chapter 28.
HINT 15N: The rate until the first coalescence between three lineages is three times the rate between two, or 3/(2Ne), because there are three pairs that can coalesce. Similarly, the rate of the first coalescence between four genes is 6/(2Ne), because there are six pairs that can coalesce.
HINT 15O: Note that with n lineages, there are n(n – 1)/2 different pairs of lineages.
HINT 15P: The chance that a lineage does not coalesce with one other lineage in time T is exp(–T/2Ne), and the chance that it does not coalesce with k others is exp(–kT/2Ne).
HINT 15Q: Use the expected time to coalescence.
HINT 15R: The variance of a Poisson distribution equals its mean, whereas the variance of an exponential distribution equals the square of the mean; see Chapter 28.
HINT 15S: Assume that since the bottleneck, the population has been so large that coalescence is negligible but that there has been time for mutations to accumulate.
HINT 15T: The nucleotide diversity can be found by finding the average number of differences between every pair of individuals. However, because there are 190 pairs, this is a tedious method. It is quicker to work out the total heterozygosity by summing 2pq across all the sites and then divide by the length of sequence.
HINT 15U: The numbers of copies of the four possible two-locus genotypes for the first two rows are
HINT 15V: The problem can be simplified by deleting all rows that are the same and all columns that are the same—these give redundant information.
HINT 15W: Look for pairs of sites (i.e., pairs of rows) that show all four possible combinations 00, 01, 10, 11. Can these be found on a single genealogy under the infinite sites model?
HINT 15X: Work out the total length of the genealogy first.
HINT 15Y: Think of the balance between the coalescence and recombination rates.