1. (5 pts) Peregrine falcons are one of the several species of raptors whose pop
ID: 177792 • Letter: 1
Question
1. (5 pts) Peregrine falcons are one of the several species of raptors whose populations declined dramatically in the mid-60's after exposure to pesticides that caused them to lay very thin-shelled eggs. In fact, the population in eastern North America disappeared entirely, but birds have been reintroduced. Even though use of DDT has been banned in the USA, it is still used elsewhere, and is manufactured in the USA, and so species such as peregrines are still exposed to some extent to these toxins. At a locus coding for an enzyme involved in laying down calcium in the eggshell, birds of genotype AA lay perfect eggs, while those that are Aa suffer 20% egg breakage (sAa = 0.20), and those that are aa suffer 50% egg breakage (saa = 0.50). The original populations starts with 16 individuals of genotype AA, 48 of genotype Aa, and 36 individuals of aa.
(a) What is the starting allele frequency for the (A) allele (in other words, what is p0)? What would the allele frequency for the (A) allele be in the adults after two generations (what is p2)?
(b) If the p2 generation of adults reproduced by random mating, how many chicks in the population would you expect to find with the aa genotype (the most fragile) if you counted a total of 100 baby falcons?
Explanation / Answer
The original populations starts with 16 individuals of genotype AA, 48 of genotype Aa, and 36 individuals of aa.
Aa suffer 20% egg breakage (sAa = 0.20), and those that are aa suffer 50% egg breakage (saa = 0.50). The original populations starts with 16 individuals of genotype AA, 48 of genotype Aa, and 36 individuals of a
sAa = 0.20
S=1-F
allele frequency for the (A) allele = D+H/2p^2 + 2pq/2
= AA+Aa
=0.16 +0.48/2
=0.4----------answer
Genotype
AA
Aa
aa
Genotype Frequencies
p2
2pq
q2
Selection coefficient
0
sAa = 0.20
saa = 0.50
0.4 x 0.4=0.16
2x 0.4 x 0.6=0.48
0.6x 0.6=0.0.36
Genotypic Fitness:
F=1
F=1-S=0.80
F=1-S=0.50
Genotype Frequencies after selection
0.16
0.384
=0.18
allele frequency for the (A) allele = D+H/2=
=0.16 +0.384/2=0.412
allele frequency for the (a) allele = R+H/2
=0.18 +0.384/2=0.37
Genotype
AA
Aa
aa
Genotype Frequencies
p2
2pq
q2
Selection coefficient
0
sAa = 0.20
saa = 0.50
0.4 x 0.4=0.16
2x 0.4 x 0.6=0.48
0.6x 0.6=0.0.36
Genotypic Fitness:
F=1
F=1-S=0.80
F=1-S=0.50
Genotype Frequencies after selection
0.16
0.384
=0.18
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