Question
In a population of jaguars, a gene with two alleles encodes the fur color. Allele B causes melanism (dark fur) and is dominant over allele b, which results in light colored fur. Suppose that there is a migration event, producing a population with 60% BB, 20% Bb and 20% bb individuals. If we assume Hardy-Weinberg equilibrium for future generations, what would the genotypes frequencies be after 5 generations?
So the answer is 49% BB, 42% Bb, 9% bb.
But I'm confused. So BB which is p^2 = .6. Square rooting .6 is rougly .775 so P = .775. this means that Q = .225
but the answer has P = .7 and Q = .3
I know that you get the answer by finding P + 1/2pq for allele frequency of P, but how come it isn't the same as the other method of finding the answer?
So the answer is 49% BB, 42% Bb, 9% bb.
But I'm confused. So BB which is p^2 = .6. Square rooting .6 is rougly .775 so P = .775. this means that Q = .225
but the answer has P = .7 and Q = .3
I know that you get the answer by finding P + 1/2pq for allele frequency of P, but how come it isn't the same as the other method of finding the answer?
Answers
Devron
The number of generations shouldn't affect the the genotype frequencies. the first thing to do is calculate allele frequencies, which is 0.7 for B and .3 for b.
B+b=1=allele frequencies
Since in the parental generation there are 49 individuals with the genotype BB, 42 individuals with the genotype Bb, and 9 individuals with the genotype bb, the frequency of B alleles will be the number of B alleles/total number of alleles, which is equal to [2(49)+42]/200=0.7. Solving for b using the equation above (b=1-B), we get b=0.3.
B=0.7
b=0.3
The genotype frequency is equal to the following:
(B+b)^2=B^2 +2Bb+b^2= 0.49 +0.42 + 0.09
Multiplying you values for B^2, 2Bb, and b^2 by 100
BB=49%
Bb=42%
bb=9%
B+b=1=allele frequencies
Since in the parental generation there are 49 individuals with the genotype BB, 42 individuals with the genotype Bb, and 9 individuals with the genotype bb, the frequency of B alleles will be the number of B alleles/total number of alleles, which is equal to [2(49)+42]/200=0.7. Solving for b using the equation above (b=1-B), we get b=0.3.
B=0.7
b=0.3
The genotype frequency is equal to the following:
(B+b)^2=B^2 +2Bb+b^2= 0.49 +0.42 + 0.09
Multiplying you values for B^2, 2Bb, and b^2 by 100
BB=49%
Bb=42%
bb=9%