Hardy-Weinberg Homework FALL 2014 ANSWER KEY Due in class on 9/24 (late submissions will not be accepted) You must work independently [don’t share answers] and show your work (each question is worth 4 pts). 1. An allozyme analysis of variation in a species of grasshopper produced the following results: 83 B1B1, 49 B1B2, and 19 B2B2. Calculate the genotype frequencies, allele frequencies, and the expected genotype frequencies for the next generation. N=151 f B1B1 = 83/151 = 0.55; fB1 =(83x2) + 49 /302 = 0.71 f B1B2 = 49/151 = 0.32 f B2B2 = 19/151 = 0.13; fB2 =(19x2) + 49 /302 = 0.29 expected genotype frequencies: p2 (B1B1) = 0.50; 2pq (B1B2) = 0.42; q2 (B2B2) = 0.08 2. In Drosophila melanogaster, cn+ vs. cn (red vs. cinnabar eyes), bw+ vs. bw (brown vs. black body, (and by+ vs. by (normal vs. blistery wing) are autosomal pairs of alleles. Samples of three large natural adult populations, each classified for a different pair of traits, are found to have the following genotypes: Population A Population B Population C 31 cncn 182 bw+bw+ 100 by+by+ 171 cn+cn 391 bw+bw 372 by+by 60 cn+cn+ 152 bwbw 40 byby Compare these distributions with those expected for a population at Hardy-Weinberg equilibrium. Propose a reasonable explanation to account for any differences. Population A: p = (60 + 85.5)/262 = 0.56; q = 0.44 Expected cn+cn+ = p2 = 0.31 x 262 = 82 Expected cn+cn = 2pq = 0.49 x 262 = 129 Expected cncn = q2 = 0.19 x 262 = 51 Population B: p = 0.52; q = 0.48 Expected bw+bw+ = 196 Observed = 182 + Expected bw bw = 362 Observed = 391 Expected bwbw = 167 Observed = 152 Population C: p = 0.56 ; q = 0.44 Expected by+by+ = 161 Observed = 100 Expected by+by = 252 Observed = 372 Expected byby = 99 Observed = 40 Observed 60 Observed 171 Observed 31 Without doing Chi-square analysis, all three populations show increased heterozygosity compared to the expected, as well as decreased homozygosity of both genotypes. There may be selection for the heterozygote (overdominance), and all three genes could be possibly linked. 3. Cystic fibrosis (CF), which affects 1/2000 Caucasians, is characterized by respiratory infections and incomplete digestion. “It is typically described as one of the most common lethal autosomal recessive 1 disorders in North America” (Grody, 2001) yet it persists in HW equilibrium in certain populations. Provide an explanation for this. The rare allele persists because there are so many heterozygous carriers… and although it is a single gene disorder, there are multiple mutations (some with more severe effects than others). 4. The fitness of three genotypes are w11 = 1.0, w12 = 1.0 and w22 = 0.7. If the population starts at the allele frequency p = 0.5, what is the value of p in the next generation? If the frequency for A1 continues to increase over time, what type of selection would this be? (include a graph to explain) 5. Consider the following conditions, the allele frequency of a dominant allele in generation t is 0.8; the conversion of this dominant allele to the recessive allele by mutation is 2.5 x 10-4 Calculate the allele frequency for the dominant allele in the next generation. 6. What is H and how is it affected by inbreeding? H is heterozygosity (freq. of heterozygotes in the population). H decreases as inbreeding increases due to the likelihood of alleles being identical by descent in the homozygous individual. 2 7. In a large experimental population of house mouse (Mus musculus), the relative fitness of a recessive phenotype is calculated to be 0.9, and the mutation rate to the recessive allele is 6.6 x 10-6. What is the equilibrium frequency? 8. Consider the following conditions, the allele frequency of a dominant allele in generation t is 0.8; the conversion of this dominant allele to the recessive allele by mutation is 10-5 Calculate the allele frequency for the dominant allele in the next generation. 9. In a large, randomly mating population, the frequency for the allele (s) for sickle-cell hemoglobin is 0.028. The results of studies have shown that people with the following genotypes at the beta-chain locus produce the average number of offspring given: Genotype SS Ss ss # of offspring produced 5 6 0 What is the frequency of the sickle-cell allele (s) in the next generation? 10. The following data was produced from an assessment of a fruit fly population for the ADH gene. 3 Genotype Mean # of offspring AdhF/AdhF 120 AdhF/AdhS 60 AdhS/AdhS 30 a. Calculate the relative fitness of each genotype. b. If a population of fruit flies has an initial frequency of AdhS equal to 0.2, what will the frequency be in the next generation when alcohol is present? a. wFF = 120/120 = 1; wFS = 60/120 = 0.5; wSS = 30/120 = 0.25 11. A deletion mutation in the CCR5 receptor (Δ32) confers protection against HIV infection. Calculate the allele frequencies for the following human T-Cell CCR5 receptor genotype data: a) CCR5 genotypes: CCR5-1/ CCR5-1, 60%; CCR5-1/Δ32, 35.1%; Δ32/Δ32, 4.9% fCCR5-1 = 0.78 fΔ32 = 0.22 b) If the mutation rate of CCR5-1 to Δ32 is 1 x 10-3. Using the estimate calculated in problem #2, for the next generation what will be the change (Δp) in frequency of this allele? p’ = p – µp Δp = p – p’ so, Δp = 0.78 – 0.77922 = 0.00078 12. Is mutation alone enough of a force to significantly increase the frequency of this protection in human populations (why or why not)? No. It would take hundreds of thousands of generations to significantly increase this allele by mutation rate alone. However, if it does confer a fitness advantage it would increase significantly. 13. The two graphs below show the change in allele frequency of p, over 50 generations. (Each color represents an independent trial.) What is the most likely difference in the conditions simulated in the two graphs? The difference is due to the greater impact genetic drift has on the top simulations, which have much lower population sizes than the bottom. 4 14. Calculate the effective population size (Ne) in a population of frogs with 46 males and 14 females. Ne = 4NmNf/(Nm+Nf) = 42.98 15. Briefly explain the outcome for allele frequencies with the following interactions: a) mutation and selection b) selection and drift a) b) most mutations are harmful and so a balance is reached when the mutation rate is equal to the loss of that harmful allele due to selection. Positive mutations will increase in frequency due to selection. It depends on the effective population size. In small populations harmful alleles can become fixed because drift is stronger. In large populations, selection is the stronger force and drift will work with it to increase more fit alleles. 5