Problem box answers

Problem box 4.1 answer

The allele frequencies for each of the alleles in the paternal haplotype are obtained in Table 4.3. For tree 4865 there is only one possible paternal allele at each locus. The chance of any genotype having one copy of each paternal allele at each locus is:

A: (0.1216)2 + 2(0.1216)(1 - 0.1216) = 0.2284 B: (0.3971)2 + 2(0.3971)(1 - 0.3971) = 0.6365 C: (0.0761)2 + 2(0.0761)(1 - 0.0761) = 0.1464 D: (0.1905)2 + 2(0.1905)(1 - 0.1905) = 0.3447 E: (0.1250)2 + 2(0.1250)(1 - 0.1250) = 0.2344

The paternal allele is expected to occur in between 14 and 64% of possible genotypes for any individual locus. The probability of a random match at all five loci is 0.2284 x 0.6365 x 0.1464 x 0.3447 x 0.2344 = 0.0017, or in 17 out of 10,000 random genotypes. The probability of exclusion is then 1 - 0.0017 = 0.9983 while the probability of exclusion for a sample of 30 candidate parents is (0.9983)30 = 0.9502. There is about a 95% chance that there would not be a random match in a sample of 30 candidate parents; therefore, we have high confidence that 4865 is the true father of seed 25-1 from tree 989. For this offspring-maternal parent combination, the B locus is the least useful in resolving paternity since the frequency of the 106 allele is almost 40%. The 167 allele at the C locus is the most useful with a frequency of just over 7%.

Problem box 4.2 answer

Ht = — ^Hj where H is the observed frequency of heterozygotes in each of the n subpopulations.

The average observed heterozygote frequency is 0.22 or 22%.

frequencies in subpopulation i.

Hs = (2(0.0)(1.0) + 2(0.93)(0.07) + 2(0.17)(0.83) + 2(0.51)(0.49))/4

Ht = 2pq where p and q are average allele frequencies for all the subpopulations. Let f be the frequency of the fast allele and s the frequency of the slow allele so that f + s = 1. Then estimate the average allele frequency for the fast allele in the total population (the slow allele could be averaged as well):

whereas the frequency of the other allele is found by subtraction: s = 1 - 0.4025 = 0.5975.

We can now calculate the F statistics using Ht, HS, and HT.

There is no evidence for self-fertilization since these four populations have observed heterozygosities very close to that expected under random mating. Comparing the observed and expected heterozygosity for each population shows that subpopulations 9 and 43 have a slight excess of heterozygotes while subpopulation 68 has about a 10% deficit. These three deviations along with the zero deviation in subpopulation 1 all average out to approximately zero.

1 CT

There is less heterozygosity within the subpopulations than we expect under Hardy-Weinberg based on allele frequencies for the total population. This value reflects the substantial differences in subpopulation allele frequencies.

F HizHL

This is the deficit of heterozygosity caused by both non-random mating within populations and allele-frequency divergence among subpopulations. In this case almost all of the deficit in heterozygosity is due to allele frequency divergence among the subpopulations. The three fixation measures are related by

Using the values for FIS and Fst and then solving for Fit gives the same value that was determined by direct computation:

(1 - Fit) = (1 - 0.035)(1 - 0.526) (1 - FT) = (0.965)(0.474) (1 - Fit) = 0.4574 Fit = 0.543

Based on the data from all 43 subpopulations, Levin (1978) estimated FIS = 0.70, Fst = 0.80, and Fit = 0.80 in P. cuspidata.

Problem box 4.3 answer

The Wahlund effect shows that population structure causes heterozygotes to become less frequent and homozygotes to become more frequent by a factor proportional to the amount of allele frequency divergence among populations. Using the allele frequencies in Table 2.3 we can calculate the expected genotype frequencies for each locus with adjustment for population using equation 4.37 for homozygous loci and equation 4.38 for heterozygous loci:

D3S1358 2(0.2118)(0.1626)(0.95) = 0.0655 D21S11 2(0.1811)(0.2321)(0.95) = 0.0799 D18S51 (0.0918)2 + (0.0918)(0.9082)(0.05) = 0.0126 vWA (0.2628)2 + (0.2628)(0.7372)(0.05) = 0.0788 FGA 2(0.1378)(0.0689)(0.95) = 0.0181 D8S1179 2(0.3393)(0.2015)(0.95) = 0.1299 D5S818 2(0.3538)(0.1462)(0.95) = 0.0942 D13S317 2(0.0765)(0.3087)(0.95) = 0.0448 D7S820 2(0.2020)(0.1404)(0.95) = 0.0539

Assuming that the Amelogenin locus should not be affected by population structure, the expected frequency of the 10-locus genotype after adjustment for population structure is 0.0655 x 0.0799 x 0.0126 x 0.0788 x 0.0181 x 0.1299 x 0.0942 x 0.0448 x 0.0539 x 0.5 = 1.514 x 10"12 with an odds ratio of one in 660,501,981,506. Compare that with the expected genotype frequency of 1.160 x 10-12 and odds ratio of one in 862,379,847,814 assuming panmixia. After accounting for population structure this genotype appears more likely to occur by chance, although its expected frequency is still extremely rare.

Problem box 4.4 answer

The joint effects of drift and migration in the fixation index are expressed by

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