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Average: 79.3 Exam II 2006 80 70 60 50 40 30 20 10 0 A B C D F Question 21 had no answer and was thrown out. Denominator was 24 instead of 25. 1 Evolutionary Genetics of Sickle Cell Disease in US • • • Normal Allele = S; Disease Allele = s. SS Homozygotes have normal hemoglobin A and do NOT have SCD. ss Homozygotes have hemoglobin S and SCD. • Ss Heterozygotes have both hemoglobin A and S but do not have SCD. Thus, by Mendel’s Definitions: SCD is a Recessive Lethal Disease in the US 2 Evolutionary Genetics of SCD • At birth in West Africa, 1-2% of babies have SCD if they have SCD, they must be ss homozygotes. Therefore, using the Hardy-Weinberg Equilibrium Observe: Gss = 0.02 2 Infer: Gss = Ps , Ps = √0.02 = 0.14 PS = 1- Ps = 1 – 0.14 = 0.86 In a large human population that is in HWE at birth: GSS = 0.74 normal individuals GSs = 0.24 carrier individuals Gss = 0.02 SCD sufferers 3 Evolutionary Genetics of SCD in US Viability Fitness Phenotypes Genotypes Normal SS WSS = 1.0 Normal Ss WSs = 1.0 SCD Wss = 0.1 ss 4 How much will natural selection change Ps, the frequency of s, the sickle cell allele, in the US population? Population Before Selection (babies) Population After Selection (adults) Natural Selection {GSS, GSs, Gss} {PS, Ps} {wSS, wSs, wss } {G’SS, G’Ss, G’ss} {P’S, P’s} Evolution of a Recessive Disease: {WSS, WSs, Wss} = {1.0, 1.0, 0.1} {Genotype Viabilities} What is the Difference between {WSS, WSs, Wss} and {wSS, wSs, wss }? {Genotype Viabilities} and {Genotype RELATIVE Viabilities}? 5 What is the Difference between {WSS, WSs, Wss} and {wSS, wSs, wss }? {WSS, WSs, Wss} = Absolute Fitnesses, the actual Viability or Reproductive Success of Genotypes. {wSS, wSs, wss } = Relative Fitnesses, the fitness of a genotype relative to the population’s Average Absolute Fitness “To escape the lion, a zebra does not have to be faster than the lion. A zebra only needs to be faster than the AVERAGE zebra.” 6 How much will natural selection change q, the frequency of the sickle cell allele in the population? Population Before Selection Population After Selection Natural Selection {GSS, GSs, Gss} {wSS, wSs, wss } {G’SS, G’Ss, G’ss} = {wSSGSS, wSsGSs, wssGss} If w > 1, then that genotype will increase in frequency. If w < 1, then that genotype will decrease in frequency. 7 Calculating Relative Fitness Relative fitness is equal to the absolute fitness of a given phenotype (WSS, WSs, or Wss) divided by the average fitness of the population (W) wSS wSs wss WSS W WSs W Wss W Recessive Disease: {WSS, WSs, Wss} = {1.0, 1.0, 0.1} How to calculate average viability fitness (W)? Average Fitness: W = (GSS) WSS + (GSs) WSs + (Gss) Wss = (0.74)*1 + (0.24)*1 + (0.02)*0.1 = 0.982 Average Viability Fitness 8 How much will the frequency of the sickle cell allele change in the US population? babies GSS = 0.74 adults adults G’SS = wSSGSS = 1.02 x 0.74 = 0.755 GSs = 0.24 G’Ss = wSsGSs = 1.02 x 0.24 = 0.245 Gss = 0.02 G’ss = wssGss = 0.102 x 0.02 = 0.002 PS = 0.86, before Selection. PS’ = G’SS + ½ G’Ss = 0.755 + 0.1225 ≈ 0.88 after selection Ps = 0.14, before Selection. Ps’ = G’ss + ½ G’Ss = 0.002 + 0.1225 ≈ 0.12 after 9 How much will the frequency of the sickle cell allele in the US population change? Ps = 0.14, before Selection Ps’ ≈ 0.12, after Selection ΔPs = After – Before ΔPs = 0.12 – 0.14 = - 0.02 10 How Many Recessive Alleles are Hidden From Natural Selection in Heterozygotes? In a population of 1,000,000 that is in HWE at birth, we have 2 GSS = p = 0.74 GSs = 2pq = 0.24 Hidden s alleles 2 Gss = q = 0.02 Exposed s alleles Number of Exposed s alleles: 2(1,000,000)(0.02) = 40,000 s alleles Number of Hidden s alleles: 1(1,000,000)(0.24) = 240,000 s alleles 6 times as more alleles Hidden than Exposed! 11 The Geographic Distribution of ‘s’ allele frequencies across Africa % of ‘s’ allele in population 14+ 6-8 12-14 4-6 10-12 2-4 8-10 12 Why is SCD so much more prevalent in Central and West Africa than in the US? Frequency of the ‘s’ allele is 0.14 in some parts of Africa, which is very high for a strongly deleterious recessive allele. 13 Distribution of primary malarial parasite, Plasmodium falciparum 14 Distribution of Plasmodium falciparum Distribution of ‘s’ allele ‘s’ allele is an adaptation to malaria because Ss heterozygotes are resistant 15 The presence of malaria in the environment changes directional selection against SCD in US into balancing selection in Africa because heterozygotes (Ss) have a survival advantage over SS and ss, homozygotes. The red blood cells of Heterozygous Ss individuals’ sickle when invaded by the malarial parasite. These sickled cells are then filtered out of the blood by the spleen. This purges the parasitic infection from Ss. 16 Relative Genotypic Fitnesses differ between US and West Africa US: No malaria in environment West Africa: Malaria in environment {WSS, WSs, Wss} = {1.0, 1.0, 0.1} ‘s’ is a recessive, deleterious allele {WSS, WSs, Wss} = {0.6, 1.0, 0.1} ‘s’ is an allele with Heterozygous Advantage ~ balancing selection {wSS, wSs, wss} = {0.88, 1.4, 0.15} 17 Relative Genotypic Fitnesses in West Africa {WSS, WSs, Wss} = {0.6, 1.0, 0.1} ‘s’ is an allele with Heterozygous Advantage ~ balancing selection Relative Fitness of the Heterozygotes Is GREATER THAN the relative fitness of either Homozygote: wSs > wss, wss 1.4 > 0.88, 0.15 18 Heterozygous Advantage in Africa leads to a Balanced Genetic Polymorphism: 0 < P*s < 1.0 Because the Relative Fitness of Heterozygotes is GREATER THAN that of either Homozygote, individuals with the highest fitness have BOTH S and s alleles. This results in a permanent, stable equilibrium allele frequency, Ps* = 0.12 Natural Selection reaches a genetic equilibrium and maintains genetic variation in the population. 19