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NAME_______________________________ EXAM#_______ 1. (15 points) Next to each unnumbered item in the left column place the number from the right column/bottom that best corresponds: 9 Lemba 1) additive genotypic deviation 2) broad-sense heritability 5 collared lizards 3) genotypes collectively shared by the individuals of a deme 4) together describe a normal distribution. 6 dn = do(1-2m)n 5) apprehended using dental floss 6) approaches zero over many generations 19 coronary artery disease (CAD) 7) effectively treated by angioplasty regardless of patients' genotypes 8) average value of 2pq measured within subpopulations of the total 14 neutrality theory population 9) Y-chromosomal haplotypes and mitochondrial-DNA haplotypes 15 environmental variance coalesce to different geographic populations. 10) describes the genetic divergence between populations caused by 30 HT genetic drift 11) This method permits measurement of epistatic variance involving two 11 candidate-locus approach or more loci. 12) the proportion of an individual's phenotype attributable to 28 isolation by distance environmental factors 13) explains why long-term rates of protein evolution are proportional to 4 mean and variance population sizes 14) explains why the pseudogene for beta hemoglobin evolves more 26 norm of reaction rapidly than the active gene 15) the proportion of populational phenotypic variation not explained by 20 Fisher's fundamental theorem of natural selection 16 phenylketonuria modeled genetic variation for a quantitative trait 16) illustrates inheritance of a response to dietary environment 17) population of gene copies collectively shared by the individuals of a deme 17 gene pool 18) maintain geographically widespread populations connected by gene flow throughout forests of the Missouri Ozarks 1 breeding value 19) No single genetic or environmental factor is both necessary and sufficient for producing a particular quantitative phenotype. 20) Rate of evolution by selection equals the additive genetic variance in fitness of a population. 21) a range of phenotypes collectively expressed by all genotypes at a locus in a population 22) a founder population used to find the chromosomal locus for Huntington's disease. 23) Rate of evolution by natural selection is influenced by a population's system of mating. Occurs 24) illustrates heritability of mental retardation in human populations 25) A high heritability indicates close correspondence between parents and offspring for both parameters. 26) a range of phenotypes associated with a particular genotype in interaction with diverse environmental and genetic backgrounds 27) a method using genomic scans to identify chromosomal segments affecting a quantitative trait 28) Geographic distance among populations is the best predictor of their amount of genetic divergence. 29) Geographically distant populations do not exchange genes and form separate evolutionary lineages. 30) probability that a pair of gametes sampled at random from a total population contains two different alleles at a particular locus. 1 NAME_______________________________ EXAM#_______ 2. (16 points). Two human demes of equal size are surveyed for MN blood types with the following results: M MN N Total Deme 1 38 324 638 1000 Deme 2 245 510 245 1000 a. (2 points) Calculate the frequencies of the M and N alleles in demes 1 and 2. Deme 1: freq(M) = (38 + 162)/1000 = 0.2; freq (N) = 1 - 0.2 = 0.8 Deme 2: freq (M) = (245 + 255)/1000 = 0.5; freq (N) = 1 - 0.5 = 0.5 b. (6 points) Test the hypotheses that each deme has Hardy-Weinberg genotypic frequencies. Be sure to give your chi-square values and degrees of freedom, and state whether the hypotheses of Hardy-Weinberg proportions are statistically rejected. (Critical Chi-square values at alpha = 0.05 for df 1 through 5: 1 df = 3.84, 2 df = 5.99, 3 df = 7.81, 4 df = 9.49, 5 df = 11.07) Deme 1: expected M = (0.2)2(1000) = 40; expected MN = 2(0.2)(0.8)(1000) = 320; expected (N) = (0.8)2(1000) = 640. Chi-square = (38-40)2/40 + (324-320)2/320 + (638-640)2/640 = 0.1 + 0.05 + 0.006 = 0.156 HW not rejected. Deme 2: expected M = (0.5)2(1000) = 250; expected MN = 2(0.5)(0.5)(1000) = 500; expected (N) = (0.5)2(1000) = 250. Chi-square = (245-250)2/250 + (510-500)2/500 + (245-250)2/250 = 0.1 + 0.2 + 0.1 = 0.4 HW not rejected. c. (4 points) Calculate the FST value for these 2 populations. Hs = [2(0.2)(0.8) + 2(0.5)(0.5)]/2 = 0.41; HT = 2(0.35)(0.65) = 0.455 FST = (0.455 - 0.41)/0.455 = 0.099 d. (2 points) Based on your answer to part c, are the populations compared more or less similar genetically than expected for two randomly-chosen human demes? Average value for comparisons of human populations = 0.15, so these populations are slightly more similar than expected on average. e. (2 points) Suppose that these demes begin to exchange migrants symmetrically so that 10% of each population moves to the other one each generation. What will be the difference in frequency of the M allele between the populations after one generation of gene flow? dn = do(1-2m)n = 0.3[1-2(.1)] = 0.24 2 NAME_______________________________ 3. EXAM#_______ 3 (5 points) After getting the sequencing results from the T. saxatilus experiment, you calculated FST values for burned and unburned glade areas. What are these two resulting FST values? What do they indicate about the balance between genetic drift and gene flow acting on these populations and the effect of the experimental treatment on these evolutionary forces? FST is 0.039 for the glades in the burned area and 0.075 for glades in the unburned areas. Both values indicate that gene flow greatly dominates the genetic structures of these populations regardless of burning. The burning appears to increase the prevalence of gene flow over genetic drift. 4. (6 points) Name and describe (including using formulas if appropriate) the two methods we used for calculating heritability in our artificial selection experiment with Brassica rapa. Be sure to define any symbols you use. response to selection: mean trichome # in G2 - mean trichome # of G1 mean trichome # of parents- mean trichome # of G1 parent offspring regression: need to find the slope of the regression line when regressing mother trichome # (x axis) versus offspring trichome # (y axis) heritability = 2 x slope x standard dev. of parents standard dev. of offspring 5. (6 points) Based on our Evolve simulations, describe one difference and one similarity in the results when selection favors a recessive allele versus a dominant allele. Assume in both cases that the favored allele starts at a very low frequency in the population, and that only two alleles are possible at the hypothetical locus. similarity: In both cases (selection for recessive or dominant allele), the allele with the selective advantage becomes fixed in the population. differences: (1) With selection for a dominant allele, the time to reach an allele frequency of 0.5 (or fixation) was faster (fewer generations). or (2) With selection for a dominant, the time to reach allele frequency 0.5 was the same in every trial (selection was so strong we saw no effect of genetic drift). NAME_______________________________ EXAM#_______ 6. (14 points) For each of the following discoveries regarding human population structure and history, indicate how the conclusions were inferred from analyses of molecular genetic variation using (1) FST values and/or (2) haplotype trees, as appropriate. a. The balance between gene flow and genetic drift among human populations on a worldwide basis is comparable on average to that measured among Kenyan impala populations. FST values both ~ 0.15. b. Isolation by distance is a prevalent feature of human evolution both currently and in the past. FST How- Isolation by distance (IBD) shows a pattern of FST values between population pairs show positive correlation with geographic distance between the pairs (i.e) plot FST vs distance Haplotype trees How- IBD shows a pattern of older haplotypes having a larger geographic distributions than younger haplotypes. This is observed in many nuclear and mtDNA markers covering different scales of evolutionary time. c. Three major expansions of populations out of Africa into Europe and Asia occurred over the past 1.7 million years. Haplotype trees How- Observe younger widespread haplotypes in Europe and Asia coalesce to ancestral haplotypes found in Africa at three different points in human evolutionary history (FST values are not relevant to this conclusion.) d. A recent range expansion from northeastern Asia established native American populations, which initially remained isolated from other human populations. Haplotypes trees How- Observe young widespread haplotypes in native Americans coalesce to older less widespread haplotypes in northeastern Asia indicating a recent range expansion from Asia into the Americas with no subsequent gene flow prior to recent times. (FST values cannot distinguish recent range expansion followed by isolation from ongoing gene flow) e. A traditional hypothesis that humans form three distinct biological races is refuted by comparative genetic data. FST How- has overall low value; Shows IBD pattern from part b; No discrete breaks in genetic variation among populations that correspond to “races” (i.e. relative dissimilarity of Melanesians and Africans from FST distance tree) Haplotype trees How- They show IBD pattern from part b; None of the lineages of haplotypes (evolutionary lineages) in the tree from the root correspond to “race” categories. 4 NAME_______________________________ EXAM#_______ 7. (10 points) For each of the following statements regarding linkage disequilibrium, indicate whether the statement is true or false. For false statements, explain why the statement is false. a. In large, randomly-mating populations, linkage disequilibria are unlikely to occur between loci located more than 1 cM apart on a linkage map. True b. Genome scans for chromosomal locations of quantitative-trait loci using SNP markers rely upon occurrence of linkage disequilibria between quantitative trait loci and genetically linked SNP loci. True c. Populations that have undergone a recent bottleneck in size, including founder events, often have higher average levels of linkage disequilibrium than their larger source populations because of the consequences of genetic drift for haplotype frequencies. True d. Haplotype trees are a crucial means of analysis for explaining linkage disequilibria occurring among sites within a small chromosomal region in which crossing-over is rare or nonexistent. True e. Variable genetic markers that are closely linked (within 1 cM of each other) almost always show high linkage disequilibrium. False - A figure shown in lecture depicting levels of linkage disequilibria between pairs of SNP markers from a Costa Rican founder population showed that only a fraction of the closely linked markers showed high linkage disequilibrium. Also, the variable sites within the apoE gene region are tightly linked but not all pairs of sites showed high linkage disequilibrium. 5 NAME_______________________________ EXAM#_______ 8. (18 points) The following data are used for an analysis of variance in LDL cholesterol levels (measured in mg/dl of blood serum) associated with genotypes at the ApoE locus. Allele frequencies at the ApoE locus are: allele 2 (0.1), allele 3(0.8) and allele 4(0.1). Genotype: 2/2 2/3 2/4 3/3 3/4 4/4 H-W freq. .01 .16 .02 .64 .16 .01 LDL-chol . 76 84 90 100 100 100 (mean = 97) a. (4 points) Calculate the genotypic deviations for each genotype. What are the units of measurement ? Genotype: 2/2 2/3 -21 -13 2/4 -7 3/3 3/4 4/4 3 3 3 all in mg/dl b. (2 points) Indicate the genetic variance for LDL cholesterol in this population, including units of measurement (answer may be left in polynomial form). .01(-21)2 + . 16(-13)2 + .02(-7)2 + .64(3)2+ .16(3)2+ .01(3)2 = 4.41 + 27.04 + 0.98 + 5.76 + 1.44 + 0.09 = 39.72 mg2/dl2 c. (4 points) Calculate the average excesses for alleles 2, 3 and 4. Include units of measurement. Allele 2 = .1(-21) + .8(-13) + .1(-7) = -2.1 -10.4 -.7 = -13.2 mg/dl Allele 3 = .1(-13) + .8(3) + .1(3) = -1.3 + 2.4 + .3 = 1.4 mg/dl Allele 4 = .1(-7) + .8(3) + .1(3) = -0.7 + 2.4 + 0.3 = 2.0 mg/dl d. (4 points) Calculate the additive genotypic deviations for each genotype. What are the units of measurement? Genotype: 2/2 2/3 2/4 3/3 3/4 4/4 -26.4 -11.8 -11.2 2.8 3.4 4.0 all in mg/dl e. (2 points) Indicate the additive genetic variance for the population, including units of measurement (answer may be left in polynomial form). .01(-26.4)2 + . 16(-11.8)2 + .02(-11.2)2 + .64(2.8)2+ .01(3.4)2+ .01(4.0)2 = 6.97 + 22.28 + 2.51 + 5.02 + 1.16 + 0.16 = 38.1 mg2/dl2 f. (2 points) What information not provided above is necessary for determining the broad-sense and narrow-sense heritabilities for cholesterol level in this population? Either total phenotypic variance or environmental variance 6 NAME_______________________________ EXAM#_______ 9. (10 points) For each of the following statements regarding the hemoglobin beta locus in humans, indicate whether the statement is true or false. For false statements, indicate why the statement is false. a. The hemoglobin A and S alleles form a balanced polymorphism in central African populations because the heterozygous genotype has a higher fitness than the alternative AA and SS genotypes in a Malarial environment. True b. In some middle-eastern populations, persistence of fetal hemoglobin interacts epistatically with the homozygous hemoglobin S genotype to alleviate the anemia associated with that genotype in other genetic backgrounds. True c. At a selective equilibrium in central African populations, the hemoglobin A and C alleles occur at frequencies of 0.89 and 0.11 respectively. False - This equilibrium pertains to the A and S alleles, not C. d. The expected coalescence time for all copies of homologous DNA for this locus in the human population would be 4N generations, where N is the number of living individuals. True e. Given the following values of average excess for hemoglobin A, C and S alleles in a particular population, only the A allele is expected to increase in frequency by natural selection in the next generation: aA = -0.001, aS = 0.081, aC = 0.015. False - The A allele should decrease slightly and the other two alleles should increase in frequency. 7