1. 1 Average: 79.3 Question 21 had no answer and was thrown out. Denominator was 24 instead of 25.

2. 2 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

3. 3 In a large human population that is in HWE at birth: G SS = 0.74 normal individuals G S s = 0.24 carrier individuals G ss = 0.02 SCD sufferers 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: G ss = 0.02  Infer: G ss = P s 2, P s = √0.02 = 0.14 P S = 1- P s = 1 – 0.14 = 0.86

4. 4 Evolutionary Genetics of SCD in US SS Ss ss PhenotypesGenotypes Viability Fitness W SS W ss W Ss = 1.0 = 0.1 Normal SCD

5. 5 Population Before Selection (babies) Population After Selection (adults) {G SS, G Ss, G ss } {G’ SS, G’ Ss, G’ ss } {P S, P s } Evolution of a Recessive Disease: {W SS, W Ss, W ss } = {1.0, 1.0, 0.1} {Genotype Viabilities} What is the Difference between {W SS, W Ss, W ss } and {w SS, w Ss, w ss }? {Genotype Viabilities} and {Genotype RELATIVE Viabilities}? How much will natural selection change P s, the frequency of s, the sickle cell allele, in the US population? Natural Selection { w SS, w Ss, w ss } {P’ S, P’ s }

6. 6 What is the Difference between {W SS, W Ss, W ss } and {w SS, w Ss, w ss }? {W SS, W Ss, W ss } = Absolute Fitnesses, the actual Viability or Reproductive Success of Genotypes. {w SS, w Ss, w ss } = 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.”

7. 7 Population Before Selection Population After Selection {G SS, G Ss, G ss } {G’ SS, G’ Ss, G’ ss } = { w SS G SS, w Ss G Ss, w ss G ss } How much will natural selection change q, the frequency of the sickle cell allele in the population? Natural Selection { w SS, w Ss, w ss } If w > 1, then that genotype will increase in frequency. If w < 1, then that genotype will decrease in frequency.

8. 8 Calculating Relative Fitness Relative fitness is equal to the absolute fitness of a given phenotype (W SS, W Ss, or W ss ) divided by the average fitness of the population (W) W W SS W W Ss W W ss How to calculate average viability fitness (W)? w SS w Ss w ss Recessive Disease: {W SS, W Ss, W ss } = {1.0, 1.0, 0.1} Average Fitness: W = (G SS ) W SS + (G Ss ) W Ss + (G ss ) W ss = (0.74)*1 + (0.24)*1 + (0.02)*0.1 = 0.982 Average Viability Fitness

9. 9 How much will the frequency of the sickle cell allele change in the US population? G SS =0.74 G Ss =0.24 G ss =0.02 G’ SS = w SS G SS = 1.02 x 0.74 = 0.755 G’ Ss = w Ss G Ss = 1.02 x 0.24 = 0.245 G’ ss = w ss G ss =0.102 x 0.02 = 0.002 P S = 0.86, before Selection. P S ’ = G’ SS + ½ G’ Ss = 0.755 + 0.1225 ≈ 0.88 after selection P s = 0.14, before Selection. P s ’ = G’ ss + ½ G’ Ss = 0.002 + 0.1225 ≈ 0.12 after babiesadults

10. 10 How much will the frequency of the sickle cell allele in the US population change? ΔP s = After – Before ΔP s = 0.12 – 0.14 = - 0.02 P s = 0.14, before Selection P s ’ ≈ 0.12, after Selection

11. 11 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  G SS = p 2 = 0.74  G Ss = 2pq = 0.24 Hidden s alleles  G ss = q 2 = 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!

12. 12 The Geographic Distribution of ‘s’ allele frequencies across Africa 14+ 12-14 10-12 8-10 6-8 4-6 2-4 % of ‘s’ allele in population

13. 13 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.

14. 14 Distribution of primary malarial parasite, Plasmodium falciparum

15. 15 Distribution of Plasmodium falciparum Distribution of ‘s’ allele ‘s’ allele is an adaptation to malaria because Ss heterozygotes are resistant

16. 16 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.

17. 17 US: No malaria in environment { w SS, w Ss, w ss } = {0.88, 1.4, 0.15} West Africa: Malaria in environment Relative Genotypic Fitnesses differ between US and West Africa {W SS, W Ss, W ss } = {1.0, 1.0, 0.1} ‘s’ is a recessive, deleterious allele {W SS, W Ss, W ss } = {0.6, 1.0, 0.1} ‘s’ is an allele with Heterozygous Advantage ~ balancing selection

18. 18 Relative Genotypic Fitnesses in West Africa {W SS, W Ss, W ss } = {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: w Ss > w ss, w ss 1.4 > 0.88, 0.15

19. 19 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, P s * = 0.12 Natural Selection reaches a genetic equilibrium and maintains genetic variation in the population.