POPULATION GENETICS AND MICROEVOLUTIONARY THEORY FINAL EXAMINATION (Write your name on every page. One point will be deducted for every page without your name!) 1. Briefly define (5 points each): a) Average Excess of gamete type i The average excess is the average genotypic deviation of all individuals who received a copy of gamete type i. b) Industrial melanism Industrial melanism is the evolution of dark coloration in various species (mostly insects) in response to a darkening caused by air pollution of the background color upon which the species rests or lives. c) R1 insert R1 inserts are a type of retrotransposon found in Drosophila that specifically insert into the 28S ribosomal DNA and disrupt transcript processing, leading to the functional inactivation of an 18S/28S rdna unit bearing such an insert.
d) Malthusian parameter NAME The Malthusian parameter measures fitness in a population with overlapping generations as a rate of increase (or decrease) in absolute time. It is a function of both the number of offspring produced and how fast they are produced. e) Meiotic drive Meiotic drive is the excess recovery of one of a pair of alleles in the gametes of a heterozygous individual, thereby violating Mendel s first law of 50:50 segregation in heterozygotes. f) Hemoglobin C Hb C is an allele due to a missense mutation in the 6 th codon at the locus coding for the β-chain of hemoglobin (full credit for just this). The resulting protein is associated with high resistance to malaria when homozygous, but no when heterozygous with the A allele. Individuals who are S/C heterozygotes suffer from hemolytic anemia, although it is not generally as severe as that suffered by S/S homozygotes.
2. (25 points) Describe how natural selection, the ecological factors that limit or do not limit the absolute output from a niche or habitat, and the amount of gene flow (consider only a panmictic population and an island model) interact to influence the maintenance of genetic polymorphism in a coarse-grained, spatially heterogeneous environment. Under panmixia when the ecological factors that limit the absolute output from a niche are fixed in a manner unrelated to the locus under study (soft selection), coarse grained spatial heterogeneity broadens the conditions for maintenance of genetic polymorphism (harmonic overdominance, as shown by the equations on the equation sheet). (5 points). However, when the output from a niche is proportional to the average fitness of the genotypes at the locus of interest (hard selection), then the condition for polymorphism is arithmetic overdominance, the same as in a constant environment (see equations on equation sheet), so coarse grained spatial heterogeneity plays no special role. (5 points) Under an island model with soft selection, the conditions for polymorphism are greatly broadened. Just one niche favoring a rare allele such that the common homozygote has a fitness less than the heterozygote bearing the rare allele minus the gene flow parameter is sufficient to maintain that allele in the population. In the absence of such a niche, a broader type of harmonic mean overdominance is then sufficient (5 points). These features are shown by the equation: There exists at least one niche such that w i <1! m OR " i 1 [ ] c i / 1! (1! w i ) /m <1 Under an island model with hard selection, the conditions for polymorphism are greatly broadened. Just one niche favoring a rare allele such that the common homozygote has a fitness in the local niche weighted by (1-m) + the arithmetic fitness weighted by m that is less than the heterozygote bearing the rare allele minus the gene flow parameter is sufficient to maintain that allele in the population. In the absence of such a niche, a broader type of harmonic mean overdominance is then sufficient (5 points). These features are shown by the equation: There exists at least one niche such that! i "1# m where! i = (1# m)w i + m$ z i w i OR $ i 1 [ ] z i / 1# (1#! i ) /m <1 i As gene flow becomes more restricted, the conditions for protection converge for both the hard and soft selection cases. Hence, coarse grained spatial heterogeneity is a powerful force for maintaining polymorphisms when gene flow is very restricted regardless of the ecological conditions limiting niche output. (5 points)
3. (15 points) A certain species of insect overwinters as an egg. The eggs hatch in the Spring, and for the next three months the larvae feed and grow until pupation. The adults emerging from the pupal stage have no mouth parts, and spend their entire adult stage finding mates and laying eggs before starving to death. The eggs then remain quiescent until next Spring. Genetic variation at an autosomal locus with two alleles has no impact on the egg, larval, or pupal stages, but does influence the life history parameters for the adult stage as shown below: Genotype: AA Aa aa Adult Life History given it survived to adulthood l x b x m x l x b x m x l x b x m x Day 1: 0.9 0 1 0 1 0 Day 2: 0.5 40 0.9 10 0.8 15 Day 3: 0 0 0.5 22 0.4 30 Day 4: - - 0 0 0 0 Assuming that the frequency of A is initially 0.5, that selection is the only evolutionary force in operation, and that the population is randomly mating at this locus, what is the frequency of the A allele in the next generation (round all answers to 3 decimal points)? Because the generation length is fixed to be one year for all individuals, the appropriate fitness measure is the net reproductive rate, R 0 = max age!! x m x b x (3 points) x= 0 The net reproductive rates for the adults are: 20 for AA, 20 for Aa, and 24 for aa (3 points). To calculate the frequency of A in the next generation, use the equation!p = p w a A (3 points). The average fitness is 21 under random mating (1 point), and the average excess can be calculated from the equations g ij = G ij! µ and a i =! p j g ij (2 points). The average excess of A is -1 (1 point). Hence, delta p is -0.024 (1 point) and p is 0.476 (1 point). j
4. The total phenotypic variance of fitness in a population is 6, and its heritability is 0.2. The average fitness is 10. a) (4 points) Assuming discrete generations and an inbreeding system of mating with f=0.1, what is the average fitness in the next generation. The value of f is not used in any calculations because phenotypic variance, average fitness and heritability are always calculated using the genotype frequencies (not necessarily HW) and allele frequencies found in the population (1 point). One then uses the equations!w = " 2 a w and h2 = σ 2 a /σ 2 p (2 points) to obtain delta wbar = (0.2)6/10 = 0.12 and that wbar in the next generation is 10.12 (1 point). b) (2 points) Allow selection to operate on the above population until a selective equilibrium is reached. What is the heritability of fitness at this equilibrium? From Fisher s Fundamental theorem (equation!w = " 2 a ), the equilibrium occurs when the additive w variance is 0, which means from equation h 2 = σ 2 a /σ 2 p that h 2 = 0 at equilibrium (2 points) c) (4 points) A trait that contributes to fitness in this population has an optimal value of 5; that is, individuals with a value of 5 have the highest fitness over all possible values of this trait. Under what conditions will natural selection produce an equilibrium population whose average trait value is 5? From the equation w eq = w(x eq ) + 1 2 w! (x eq )" 2 eq (x) the equilibrium population will only have an average trait value of 5 when either! 2 eq (x) = 0 ; that is, the trait has no phenotypic variance at equilibrium (2 points), or if w!!(x eq ) = 0 ; that is, the trait is linearly related to fitness (2 points).
5. Two widely separated populations of a species are subject to a genome scan of SNP variation. One chromosome shows the following patterns for the metrics of expected heterozygosity within a population and the pairwise f st between the two populations (dashed lines indicate the thresholds for statistical significance of outlier regions from the genome-wide average): a) (10 points). Is there evidence of natural selection acting on this chromosome within one or both of these populations? If so, describe the likely type(s) of selection, the approximate chromosomal position(s) at which selection is acting, and the population(s) within which selection is acting. A significant reduction in expected heterozygosity in a genomic region indicates a recent selective sweep (positive, directional selection) (4 points). Such a region is found in population 1 near chromosome region 3 (3 points) and in population 2 near chromosome region 7 (3 points) Continue on next page if needed.
b) (10 points). Is there evidence for local adaptation in these data? If so, how strong is this evidence? Two chromosome regions (3 and 7) show an excess of differentiation through f st and these same two regions show evidence of differential selection (the two top graphs and part a). This is compatible with the hypothesis of local adaptation (5 points). However, the background f st is significantly different from 0, implying some sort of restricted gene flow between these two populations (1 point). Hence, different mutations arising in different locales but being adaptive to the same environment could establish local areas of differentiation even if the two mutants were compatible with one another, as illustrated in the lectures (2 points). Alternatively, the two selective sweeps could be adaptive to the same environment, but in an antagonistic, mutually exclusive manner (such as HbS and HbC as adaptations to malaria) (1 point). Hence, the evidence for local adaptation is NOT definitive (1 point).
6. (15 points) An experimenter puts Drosophila populations through two experimental evolution treatments for 50 generations. In a monogamous treatment, the females and males are singly mated; in the polygynous treatment females are mated with a new male every 2 days. Eggs are collected from the first 10 days, reared, and then adults are chosen randomly to start the next generation. The graph above shows survivorship of female flies from the end of experiment in three tests with males from either the same or the other treatment: 1. Monogamous females tested with a monogamous male 2. Monogamous females tested with males from the polygynous treatment 3. Polygynous females tested with polygynous males Label each test in the blanks above with its corresponding survivorship line on the graph (a, b, or c). Explain your reasoning, first in terms of sexual selection, then tie your reasoning to senescence theory. c, a, b top-to-bottom (1 pt each). Males may evolve to increase mortality of females if it increases the male s own reproduction, for example by making her produce more early young (2 pts). However, this cannot happen in the monogamous treatment where any decline in female reproduction is also a decline in her mate s reproduction, so maximum survival (c) is expected in test 1. Polygynous males can evolve to harm females but polygnous females will also evolve to resist this harm, so females in test 3 will have higher survivorship than females in test 2 who were not selected for resistance (6 pts). Senescence evolves because early reproduction is more strongly selected than late reproduction. Here, genes in the polygynous males especially benefit from inducing greater female early reproduction, even if this causes greater female mortality, because fewer of the females later eggs will he his (4 pts).
7. A neutral mutation occurs in an autosomal, tandemly-duplicated gene family consisting of 200 paralogous copies per chromosome in an ideal population of size 50,000 with discrete generations. The probability of one paralogue converting another to its state is 0.000002 per generation. a). (5 points) Given that this mutation goes to fixation in all orthologous and paralogous copies, how many generations is this expected to take? # 4N when! > 1 2N % Use Expected Time To Fixation = $ % 2! when! " 1 2N & (1 pt) with N=50,000 (1 pt) and α=0.000002 (1 pt) Then, α=0.000002 < 1/(2N) = 0.00001 (1 pt), so expected time to fixation is 2/α= 1,000,000 generations (1 pt) b). (5 points) Assume now that the molecular conversion rate of paralogues is greatly increased to 0.00002. How many generations is the fixation of the neutral mutation in all homologues and paralogues expected to take? # 4N when! > 1 2N % Use Expected Time To Fixation = $ % 2! when! " 1 2N & (1 pt) with N=50,000 (1 pt) and α=0.00002 (1 pt) Then, α=0.00002 > 1/(2N) = 0.00001 (1 pt), so expected time to fixation is 4N= 200,000 generations (1 pts) c). (5 points) In which case, a) or b), does the molecular rate of paralogue conversion play the more important evolutionary role in fixation time. Justify your answer. It plays the more important role in a) (2 pts) because when the molecular conversion rate is low, it is the rate limiting step to fixation in the tandem family and not genetic drift (3 pts).