PROBLEM SET 1 - EVOLUTIONARY BIOLOGY - FALL 2016 KEY Hardy-Castle-Weinberg (20 pts total) 1. A population of infants from Musoma, Tunisia, was scored for the β-hemoglobin locus genotype (determined by a single autosomal locus with two alleles, A and S) as follows: Genotype AA AS SS Total Number 189 89 9 287 Characterize this population by its genotype frequencies. Characterize the gene pool by the allele frequencies for A and S. Using Hardy-Weinberg Law, predict the genotype frequencies based on the allele frequencies. (3 pts) Observed Genotype Frequencies: Freq(AA) = 189/287 = 0.66 Freq(AS) = 89/287 = 0.31 Freq(SS) = 9/287 = 0.03 Allele Frequencies: Freq(A) = p = 189/287 + ½(89/287) = 0.81 Freq(S) = q = 9/287 + ½(89/287) = 0.19 (Or use p + q = 1 to solve for q) Expected HWE genotype frequencies p2 = (0.81)2 = 0.66 2pq = 2(0.81)(0.19) = 0.3 q2 = (0.19)2 = 0.03 2. Among people of southern Italian and Sicilian ancestry living in Rochester, New York (N = 10,000), about one birth in 2,500 has thalassemia major (a type of anemia) and about one birth in 25 has a milder anemia known as thalassemia minor. Are these data compatible with a single locus hypothesis as a basis for the heredity of these anemic conditions? Show why or why not. (3 pts) To answer this question, construct a model with all the elements of the basic Hardy-Weinberg model and see if the data you have been given fit. Assume two alleles and one locus and arbitrarily name the alleles A and a. If you assume that thalassemia major is a homozygous genotype (AA) and that thalassemia minor is a heterozygous genotype (Aa) then you have the genotype frequencies from the question to work with: AA = 1/2500 = 0.0004; Aa = 1/25 = 0.04; and aa= 1-0.04-0.0004 = 0.9596 Solve for the allele frequencies: pA= 0.0004 +1/2(0.04) = 0.0204 qa= 0.9596 + 1/2(0.04) = 0.9796 Given these values for p and q, the expected genotype frequencies under H-W are p2 = 0.0004; 2pq = 0.04; q2 = 0.9596. These frequencies match the given values in the population. Therefore, this trait fits the H-W two-allele, one-locus model. 3. Two populations are examined for the same gene locus with the following results: Population 1: AA 162; Aa 36; aa 2 Population 2: AA 18; Aa 84; aa 98 Now suppose these two populations are combined to form a new population: Population 3: AA 180; Aa 120; aa 100 a. Can the differences in genotype frequencies in populations 1 and 2 be explained by differences in the mating systems with respect to this gene locus? (2 pts) Pop 1 Genotype frequencies AA = 162/200 = 0.81; Aa = 36/200 = 0.18; aa = 2/200 = 0.01 Allele frequencies pA = 0.81 +1/2(0.18) = 0.9; qa= 0.01 + 1/2(0.18) = 0.1 Expected H-W frequencies p2 = (0.9)2= 0.81; 2pq = 2(0.9)(0.1) = 0.18; q2 = (0.1)2 = 0.01 This population is in HW equilibrium Pop 2 Genotype frequencies AA = 18/200 = 0.09; Aa = 84/200 = 0.42; aa = 98/200 = 0.49 Allele frequencies pA= 0.09 +1/2(0.42) = 0.03; qa= 0.49 + 1/2(0.42) = 0.7 Expected H-W frequencies p2 = (0.3)2= 0.09; 2pq = 2(0.3)(0.7) = 0.42; q2 = (0.7)2 = 0.49 This population is in HW equilibrium Since both populations are in HW expected proportions they do not violate the assumption of random mating. There is no evidence that differences in the mating system are responsible for the differences in genotype frequencies. b. Is population 3 in H-W equilibrium? If not, indicate whether heterozygotes are in excess or in deficiency. (2 pts) Pop 3 Genotype frequencies AA = 180/400 = 0.45; Aa = 120/400 = 0.3; aa = 100/400 = 0.25 Allele frequencies pA= 0.45 +1/2(0.3) = 0.6; qa= 0.25 + 1/2(0.3) = 0.4 Expected H-W frequencies p2 = (0.6)2= 0.36; 2pq = 2(0.6)(0.4) = 0.0.48; q2 = (0.4)2= 0.16 The HW expected genotype frequencies do not match the observed genotype frequencies so this population is not in HW equilibrium. There is an excess of homozygotes and a deficit of heterozygotes. c. For how many generations will the effect of this single episode of admixture of populations be detectable in a population established from population 3 with respect to genotype frequencies if mating is random with respect to this locus? (2 pts) After one generation of random mating this populations will be in HW proportions and will stay in those proportions. You can only see the mixture in the first mixed generation. 4. The frequency of phenylketonuria (PKU, caused by an autosomal recessive allele) is 0.00004 at birth. a. Assuming H-W, what is the frequency of the PKU allele? (1 pts) q2 = 0.00004 so q = 0.006 and p = 1-0.006 = 0.994 b. What is the expected H-W ratio of PKU carriers (heterozygotes) to affected individuals (PKU homozygotes)? (1 pts) 2pq / q2 = 2(0.006)(0.994) / 0.00004 = 314; For every homozygous affected individual there will be more than 300 carriers. c. Selection tends to reduce the frequency of deleterious alleles in a population. As the frequency of the PKU allele goes down, what is the limit to the ratio of carriers to affected individuals? (1 pts) This quantity tends toward as q 0 d. Why is it difficult for selection to eliminate a completely recessive deleterious allele from a population? (1 pts) It is very difficult for natural selection to eliminate a completely recessive allele because at low frequencies most of the alleles will be in the heterozygous state where they will have no phenotypic effect. As the allele gets increasingly rare the frequency of homozygous genotypes becomes even lower giving selection little opportunity for selection to act against the deleterious allele. 5. A person has blood type O blood only if they are homozygous for the O allele at the ABO locus. The frequency of the O allele is 0.67 in the United States. A person’s blood is Rh+ if he or she is homozygous or heterozygous for the D allele at the Rh locus, which has an allele frequency of 0.6. Assuming no linkage disequilibrium between these two unlinked loci, what is the frequency of people with blood type O+, (assume a random-mating population with respect to these loci)? (2 pts) The probability of being O+ is = (the probability of being OO) * (the probability of being RH+) Since the frequency of the O allele is 0.67 the frequency of the OO genotype is (0.67)2 = 0.45. To estimate the probability of being RH+ first you need to figure out the frequency of the DD and Dd genotypes. You know that the D allele has a frequency of 0.6 so the frequency of the alternate allele must be 0.4. The frequency of the DD homozygotes + the frequency of the Dd heterozygotes is then 1-(0.4)2 = 0.84. So, The frequency of O+ blood types is = 0.45 * 0.84 = 0.378. 6. In some species of butterflies the presence of spots on the wings is controlled by a single locus and determined by an allele that is dominant in females and recessive in males. If 84% of females have wing spots, what proportion of males have wing spots? (2 pts) This problem is similar to the last one. If 84% of the females have the dominant phenotype then 16% have the recessive homozygous phenotype and the allele frequency for the non-spotted allele is the square root of 0.16 = 0.4 and the allele frequency of the spotted allele is 0.6. In males the spotted allele is recessive not dominant so only the homozygous genotype for the spotted allele have wing spots. In males there are (0.6)2 = 0.36 individuals with spots. .