Pop.+Gen+Questions+2+Answers - Answers to Study Questions...

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-1- Answers to Study Questions #2 1. f(aa) = p 2 = .2 x .2 = .04 f(Aa) = 2pq = .2 x .8 x 2 = .32 f(AA) = q 2 = .8 x .8 = .64 The observed frequency of aa homozygotes is much higher than predicted by a Hardy-Weinberg model. When combined with information from the lectures and reading, these results imply that the frequency of the a allele is increasing, that the frequency of gray morphs is increasing, and that pollution is decreasing around Manchester (e.g. the lichen is recovering on the trees in that area). 2. Frequency of alleles in the parental generation: blue birds = yy. In this case, the green birds are all homozygotes, YY. Number of y alleles in yy birds = 40 x 2 = 80. Number of Y alleles in YY birds = 60 x 2 = 120. Number of y or Y alleles in Yy birds = 0 (because there aren't any heterozygotes in the parental generation) Check: total number of alleles in a population of N individuals should be 2N. In this case, there are 100 birds, and 80 + 120 alleles = 200 alleles. Frequency of y allele = 80/200 = 0.4 Frequency of Y allele = 120/200 = 0.6 Check: allele frequencies at a given locus must always add up to one. .4 + .6 = 1.00. Strict assortative mating means that every bird chooses its mate based on the phenotypic trait of interest. In this example, strict positive assortative mating means that: All green (YY) birds mate with other green (YY) birds, and all blue (yy) birds mate with other blue (yy) birds. If each mating produces the same number of surviving offspring (e.g. four offspring per pair), then the number of genotypes in the next generation is: Pairs consisting of 2 YY birds = 30 pairs, producing 120 YY offspring. Pairs consisting of 2 yy birds = 20 pairs, producing 80 yy offspring. We calculate the allele frequencies from the numbers of each genotype present in the offspring generation, using the same procedure used for the parents.
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-2- Total number of Y alleles in YY offspring = 2 x 120 = 240. Total number of y alleles in yy offspring = 2 x 80 = 160. Frequency of Y in the offspring generation = 240/400 = .6 Frequency of y in the offspring generation = 160/400 = .4 In contrast, if the birds had satisfied all of the assumptions of Hardy-Weinberg, including random mating, the genotypic frequencies in the offspring generation should have been as follows: Frequency of yy = p 2 = .4 x .4 = .16 Frequency of Yy = 2pq = .24 x 2 = .48 Frequency of YY = q 2 = .6 x .6 = .36 And in the offspring generation, allele frequencies: f(y) = .4, f(Y) = .6. Point : positive assortative mating affects genotypic frequencies, but not allele frequencies. In this extreme example of strict positive assortative mating, we are missing the heterozygote genotype and phenotype entirely. Hence the frequency of this genotype is zero, not 0.48. The allele frequencies did not change. 3. Green birds would have higher reproductive success than blue birds. In this case, you would expect a decrease in the frequency of the y allele across generations (compare to question 2).
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This note was uploaded on 10/08/2009 for the course BIS 2 taught by Professor Schwartzandkeen during the Spring '09 term at UC Davis.

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Pop.+Gen+Questions+2+Answers - Answers to Study Questions...

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