Question

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 10% egg breakage (s_Aa = 0.10), and those that are aa suffer 70% egg breakage (s_aa = 0.70).  The original populations starts with 450 individuals of genotype AA, 380 of genotype Aa, and 170 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 one generation (what is p₁)?

(b)   If the p₁ 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?

Answer 1

Question:

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 10% egg breakage (sAa = 0.10), and those that are aa suffer 70% egg breakage (saa = 0.70).  The original population starts with 450 individuals of genotype AA, 380 of genotype Aa, and 170 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 one generation (what is p1)?

(b)   If the p1 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?

Answer a:

The original population starts with 450 individuals of genotype AA, 380 of genotype Aa, and 170 individuals of aa.

Total genotypes = (AA+Aa+aa) = (450+380+170) = 1000

Frequency of any genotype = number of that particular genotype / number of total genotypes

Thus, Frequency of AA = 450/1000 = 0.45

Frequency of Aa = 380/1000 = 0.38

Frequency of aa = 170/1000 = 0.17

Now, Frequency of allele A = Frequency of AA + ½ of frequency of Aa (because there are half of the alleles are A in the genotype Aa and other half are a)

Frequency of allele A = [0.45 + 1/2(0.38)] = [0.45 + 0.19] = 0.64

Thus, the starting allele frequency for the (A) allele (in other words, what is p0) is 0.64

The selection co-efficient (s) for AA [sAA] is 0 because birds of genotype AA lay perfect eggs; the selection co-efficient (s) for Aa [sAa] is 0.1 because Aa suffer 10% egg breakage; and the selection co-efficient (s) for aa [saa] is .7 because aa suffer 70% egg breakage.

The fitness (w) = 1-s; thus, fitness for AA is (1-0) = 1, Aa is (1-0.1) = 0.9 and aa is (1-0.7) = 0.3

p0 generation	AA	Aa	aa	Total
Number	450	380	170	1000
Frequency	0.45	0.38	0.17	1.0
Fitness (w)	1.0	0.9	0.3
p1 generation	AA	Aa	aa	Total
Number	(450×1.0) = 450	(380×0.9) = 342	(170×0.3) = 51	(450+342+51) = 843
Relative Frequency	(0.45×1.0) = 0.45	(0.38×0.9) = 0.342	(0.17×0.3) = 0.051	(0.45+0.342+ 0.051) = 0.843
Frequency	0.45/0.843 = 0.5338	0.342/0.843 = 0.4057	0.051/0.843 = 0.0605	(0.5338+0.4057 +0.0605) = 1.0

Thus, genotype frequency in the adults after one generation (in p1 generation) is, AA = 0.5338, Aa = 0.4057 and aa = 0.0605.

Now, Frequency of allele A = Frequency of AA + ½ of frequency of Aa (because there are half of the alleles are A in the genotype Aa and other half are a)

Frequency of allele A = [0.5338+ 1/2(0.4057)] = [0.5338+ 0.20285] = 0.73665

Frequency of allele a = Frequency of aa + ½ of frequency of Aa (because there are half of the alleles are A in the genotype Aa and other half are a)

Frequency of allele a = [0.0605+ 1/2(0.4057)] = [0.5338+ 0.20285] = 0.26335

Thus, the allele frequency for the (A) allele be in the adults after one generation (what is p1) will be 0.73665

Answer b:

We assume that frequency of allele A = p and frequency of allele a = q

Thus, the allele frequency in the adults after one generation for the (A) allele will be 0.73665 and for (a) allele will be 0.26335

If the p1 generation of adults reproduced by random mating, the genotype frequency will be AA = p2, Aa = 2pq and aa = q2.(according to Hardy-Weinberg equilibrium)

So, AA = (0.73665)2 = 0.5265, Aa = (2×0.73665×0.26335) = 0.388 and aa = (0.26335)2 = 0.06935

Thus, if the p1 generation of adults reproduced by random mating, the number of 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 = (frequency of aa ×100) = (0.06935×100) = 6.935; as baby falcons cannot come in fraction number thus approximately 7 baby falcon will of aa genotype.