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PrQ4C_key - Biology 352 Quiz#4 In problems 1-3 fitness was...

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Biology 352 Quiz #4 1 of 3 In problems 1-3, fitness was measured in a wild population of seagulls. 90% of individuals with genotypes AA and Aa survive to adulthood. AA and Aa gulls that survive have (on average) 3.2 offspring. 90% of individuals with genotype aa survive to adulthood, but aa gulls that survive have (on average) only 2.1 offspring. 1. What is the absolute fitness of genotype AA? a. 3.2 b. 2.9 c. 1.5 d. 1.0 e. 0.90 3.2*0.9 = 2.9 2. What kind of selection model fits this population? a. complete dominance b. partial dominance c. underdominance d. overdominance e. genotypic dominance The heterozygote has the same fitness as AA . This is the definition of complete dominance. 3. If a population of gulls has 2 of each genotype, what is w ? a. 1.00 b. 0.900 c. 0.885 d. 0.667 e. 0.656 The average fitness is obtained as the sum of {each genotype’s frequency times its relative fitness}. It is basically a weighted average of the fitnesses in the population. The relative fitness of AA and Aa is (0.90)(3.2) / (0.90)(3.2) = 2.88 / 2.88 = 1.0 for each. The relative fitness of aa is (0.90)(2.1) / 2.88 = 0.65625 The frequency of each genotype is 0.33. So w -bar is (2/6)(1) + (2/6)(1) + (2/6)(0.65625) = 0.885 You could also get this answer by taking the average of {1.0, 1.0, 1.0, 1.0, 0.65625, 0.65625}. The equations w p w pqw q w AA Aa aa =+ + 22 2 and ws q =− 1 2 are appropriate only if you have p and q and assume random mating. (The question above is much simpler. It simply asks what is the average fitness for the six individuals.) To summarize: the first method is appropriate if you are asking “What is the average fitness of these individuals?” The second method is appropriate if you ask “What is the average fitness of this population assuming random mating?”

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PrQ4C_key - Biology 352 Quiz#4 In problems 1-3 fitness was...

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