In snorks, there is a single gene that is responsible for determining skin color
ID: 213867 • Letter: I
Question
In snorks, there is a single gene that is responsible for determining skin color. There are two different alleles, B and b, for this gene. Homozygous B individuals (BB) are blue, heterozygous individuals (Bb) are green, and homozygous b individuals (bb) are yellow. Snorks live in blue water with a sandy yellow bottom meaning that blue and yellow individuals have areas where they are well camouflaged and green individuals are never well camouflaged. The relative fitness of blue individuals is 1.00, the relative fitness of green individuals is 0.20, and the relative fitness of yellow individuals is 0.70. The allele frequencies in the parental population are fr(B)=0.80 and fr(b)=0.20.
What are the expected allele frequencies in the F2 generation of this population?
What are the predicted non-zero/non-one equilibrium allele frequencies of this population given the fitness values presented? Round to the nearest hundredth.
Explanation / Answer
From the given data, we have two alleles, B and b. BB represents homozygous, Bb for heterozygous, and bb for again homozygous.
Fitness Frequency of BB = 1.00
Fitness Frequency of Bb = 0.20
Fitness frequency of bb = 0.70
Also, from parental population figurer fr(B) = 0.80 and fr(b) = 0.20, if we calculate the initial frequencies of these three genotype BB, Bb, and bb will be 0.64, 0.32, and 0.04
This can be achieved by generating a probability of meeting
B * B i.e 0.8*0.8 = 0.64
B * b i.e 0.8*0.2 = 0.16
b * B i.e 0.2*0.8 = 0.16
b * b i.e 0.2*0.2 = 0.04
Hence the outcome product of the frequencies of the genotype exceeds 1 which is not possible.
Genotype
BB
Bb
bb
Alleles
B
b
Initial frequency
0.64
0.32
0.04
P=0.8
Q=0.2
Fitness
1.0
0.2
0.7
product
0.64
0.064
0.028
= 0.64+0.064+0.028 = 0.732
Frequencies after selection
0.64/0.732 = 0.874
0.064/0.732 = 0.0874
0.028/0.732 = 0.0382
P’= 0.874+0.0874/2 = 0.9177
Q’= 0.0382+0.0874/2 = 0.0819
F2 generation allele frequencies
(P’)2 = 0.84
2P’Q’ = 0.15
(Q’)2 = 0.0067
0.9177
0.0819
The expected frequencies for F2 generation is
fr2(B) = 0.9177
fr2(b) = 0.0819
Genotype
BB
Bb
bb
Alleles
B
b
Initial frequency
0.64
0.32
0.04
P=0.8
Q=0.2
Fitness
1.0
0.2
0.7
product
0.64
0.064
0.028
= 0.64+0.064+0.028 = 0.732
Frequencies after selection
0.64/0.732 = 0.874
0.064/0.732 = 0.0874
0.028/0.732 = 0.0382
P’= 0.874+0.0874/2 = 0.9177
Q’= 0.0382+0.0874/2 = 0.0819
F2 generation allele frequencies
(P’)2 = 0.84
2P’Q’ = 0.15
(Q’)2 = 0.0067
0.9177
0.0819
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