PS9_Key - BICD100 Genetics Fall 2005 Problem Set 9 Answers...

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BICD100 Genetics Fall 2005 Reinagel Problem Set 9 Answers 1. A population called the “founder generation”, consisting of 2000 AA individuals, 2000 Aa individuals, and 6000 aa individuals is established on a remote island. Mating within this population occurs at random, the three genotypes are selectively neutral, and mutations occur at a negligible rate. a) What are the frequencies of the alleles A and a in the founder generation? to calculate allele frequencies, count the total alleles represented in individuals with each genotype and divide by the total number of alleles number of individuals number of A alleles number of a alleles 2000 AA 4000 0 2000 Aa 2000 2000 6000 aa 0 12,000 Total 6000 14000 Frequency of the A allele = 6000/20,000 = 0.3 = p Frequency of the a allele = 14,000/20,000 = 0.7 = q b) Is the founder generation at Hardy-Weinberg Equilibrium? Given p and q above, the genotype frequencies should be Frequency of AA should be p 2 = 0.3 x 0.3 = 0.09 Frequency of Aa should be 2pq = 2 x 0.3 x 0.7 = 0.42 Frequency of aa should be q 2 = 0.7 x 0.7 = 0.49 For a population of 10,000 individuals, the number of individuals should therefore be 900 AA, 4200 Aa, and 4900 aa. The founder population is therefore not in equilibrium. c) What is the frequency of the A allele in the second generation (that is, the generation subsequent to the founder generation)? Given the condition of random mating, selectively neutral alleles, and no new mutations, allele frequencies do not change from one generation to the next; p=0.3 and q=0.7. (You can demonstrate this by writing out the punnett square). d) What are the frequencies of the AA, Aa, and aa genotypes in the second generation? The genotype frequencies would be the ones calculated in part b because in one generation the population will go to equilibrium. (You can demonstrate this by writing out the punnett square).
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