PROBLEM SET 2 EVOLUTIONARY BIOLOGY FALL 2016 KEY Mutation, Selection, Migration, Drift (20 pts total) 1) A small amount of dominance can have a major effect in reducing the equilibrium frequency of a harmful allele. To confirm this for yourself, imagine an allele that is nearly lethal when homozygous (s = 0.9), and that the frequency of the deleterious allele is maintained by mutation-selection balance with a mutation rate from the wild type normal allele to the deleterious allele of u = 6 X 10-6. Calculate the equilibrium frequency of the deleterious allele in the cases where (1) the allele is completely recessive and (2) the allele is partially dominant with h = 0.03. Compare these two equilibrium frequency estimates and discuss the implications for the load of deleterious mutations found in human populations. (3 pts) When the deleterious allele is completely recessive the equilibrium frequency is: 610 0.9 2.610 When the deleterious allele is partially recessive the equilibrium frequency is: 610 0.030.9 2.210 The frequency of the deleterious allele is much higher when it is completely recessive than when it is partially dominant. Even in this case where h is very small the equilibrium frequency is an order of magnitude smaller than for the completely recessive allele. The implication is that in human populations there are likely far more deleterious recessive alleles segregating (i.e., floating around) than there are partially dominant deleterious alleles. 2) In class we examined data from the Yorubas of Ibadan, Nigeria who have a high frequency of the sickle-cell anemia allele (S). (6pts total) a) Given the expected genotype frequencies from Bodmer & Cavalli-Sforza (1976), estimate the underlying allele frequencies for the sickle-cell (S) and normal (A) alleles? (1pt) The expected frequency of the genotypes is: SS = 187.4/12387 = 0.015 SA = 2672.2/12387 = 0.216 AA = 9527.2/12387 = 0.769 The allele frequencies of the S and A alleles are: p S = p 2 + ½(2pq) = 0.123 q A = 1 0.123 = 0.877
b) How do these frequencies compare to the equilibrium frequency you would expect in this population? (1 pt) The expected frequencies in Bodmer & Cavalli-Sforza are the same as the HW equilibrium expected genotype frequencies with allele frequencies of p S = 0.123 and q A = 0.877. Given heterozygote advantage selection at this locus there is a stable equilibrium frequency determined by the values of t = -0.86 and s = -0.12. To estimate these equilibrium frequencies: 0.86 0.877 0.12 0.86 0.12 0.122 0.12 0.86 Even though the SS genotype has very low fitness, with overdominant selection the S allele will persist at a fairly high frequency (12 %) in the population. c) What factors explain the maintenance of the highly deleterious S allele in these populations? (1pt) The normally deleterious sickle cell allele confers a fitness advantage to the heterozygous genotype in the presence of malaria. The homozygous normal allele is susceptible to malaria, and the homozygous sickle cell allele, while resistant to malaria, suffers a clear fitness deficit. Both homozygotes have lower fitness than the heterozygous genotype. This form of selection is Heterozygote Advantage (overdominance). Under this type of selection an intermediate stable allele frequency prevents the loss of either allele. d) In a second population of individuals from southeast Asia the frequencies of the S and A alleles are 0.25 and 0.75respectively. Assuming that these populations are in HW equilibriumwhat is the level of population subdivision (Fst) between these two populations?(2pts) p S q A 2pq Subpop 1 0.123 0.877 0.216 Subpop 2 0.25 0.75 0.375 Average frequency = 0.1865 0.8135 Average expected heterozgosity within populations H s = (0.216 + 0.375)/2 = 0.295 Total expected heterozygosity H t = 2(0.1865*0.8135) = 0.303 F st = (H t H s )/H t = (0.303-0.295)/0.303 = 0.027 (very little subdivision)
e) Assuming the necessary assumptions are met, what is the effective number of migrants(n m ) among these populations each generation? (1pt) F st = 1/(1+4N m ) =>N m = 9.2 3) In western Montana a Rancher wants to breed trophy Elk with large antlers so she maintains a herd of 200 Elk. The sex ratio in the census population (N a ) is 50:50 but elk are highly polygamous and it turns out that four (4) of the Bulls are monopolizing all of the females. The Rancher is concerned that the skewed breeding sex ratio may be lowering the genetic effective size of her herd. (3 pts total) a) What is the effective population size (N e ) of this herd? (1pt) 4 15.38 b) The Rancher s main concern is that genetic variation in this herd will be lost over time and this may compromise her breeding program. If the original heterozygosity in the herd is 0.80, estimate the amount of heterozygosity that will remain after 10 generations if the farmer maintains the herd at its current size and breeding sex ratio. (2 pts) The loss of heterozygosity after 10 generations is given by the following equation: 1 1 2 1 1 215.38 0.8 0.575 4) The Elk Rancher in the previous problem was motivated by research on antler size and hunting pressure conducted by students at the University of Montana. These students wanted to test whether selection from hunting pressure was changing the allele frequencies of the gene controlling antler size in Elk. They genotyped a large sample from a wild population before and after the hunting season. Use the genotyping data from their experiment (given below) to answer the following questions. Assume that Antler size is controlled by a single locus with two alleles (H & h). (8pts total) Large antlers = HH Medium antlers = Hh Small antlers = hh Genotype # Before Hunting Season # After Hunting Season HH 200 50 Hh 500 175 hh 300 275 Total (N) 1000 500
a) What is the survival rate for each genotype? (1pts) There are a couple of different ways to get estimates of fitness. One straightforward way is to first calculating the frequency of each genotype in the Before sample. Genotype Frequency in the Before Sample HH 200/1000 = 0.20 Hh 500/1000 = 0.50 hh 300/1000 = 0.30 And then generate expected numbers in the After sample if mortality is random using these frequencies. Genotype Expected in the After Sample (1pt) HH 500(0.20) = 100 Hh 500(0.50) = 250 hh 500(0.30) = 150 Then estimate the Survival Rate = Ratio of Observed to Expected (O/E). This is a measure of fitness. HH 50/100 = 0.5 Hh 175/250 = 0.7 hh 275/150 = 1.8 b) What form of selection is operating on this locus? (1pt) Directional c) What are the selection (s) and the dominance (h) coefficients? (3pts) To estimate the selection and dominance coefficients the survival rates need to be scaled so that the best genotype has a fitness of one and then using the general framework for directional selection the coefficients can be estimated. Relative Fitness = Survival Rate scaled to the best genotype HH 0.5/1.8 = 0.28 Hh 0.7/1.8 = 0.39 hh 1.8/1.8 = 1.00 hh Hh HH 1 1-hs 1-s s = 1-0.28 = 0.72 hs = 1 0.39 = 0.61 h = 0.61/0.72 = 0.85
d) Given continual hunting pressure, eventually the H allele will be driven to extremely low levels in these populations resulting in lower average antler size. To avoid this outcome game managers are considering introducing animals from Yellowstone Park where hunting is prohibited. In the national park population the frequency of the H allele is 0.6. Given the selection coefficient you estimated and a migration rate of m=0.3, what will the frequency of the H allele be in the hunted population at equilibrium? (3pts) If you assume that the national park can be represented by a continent in a continent-island model of migration-selection balance the predicted equilibrium frequency of the H allele in the hunted population is approximated by: 0.30.6 0.72 0.25 This result highlights that migration can offset even the strong selection that can be imposed by culling individuals from a managed population.