Grey eyes (b) are recessive to brown eyes (B) in rabbits. In an ideal rabbit population exhibiting Hardy–Weinberg equilibrium, if allele b has a frequency of 0.23, what percentage of the population is heterozygous for this trait?
Round your answer to the nearest whole number.
Answer%
3 years ago
Answered By Emily D
They're sneaking math into biology, how dare they!
We have two equations thanks to Hardy-Weinberg:
$1=p+q$1=p+q this one is for the frequency of alleles (how common is b and B?)
p is the dominant allele frequency (B in this case)
q is the recessive allele frequency (b)
$1=p^2+2pq+q^2$1=p2+2pq+q2 this one is for frequency of genotypes (how common is bb, Bb, and BB?)
p2 is the frequency of homozygous dominant (BB)
2pq is the frequency of heterozygous (Bb)
q2 is the frequency of homozygous recessive (bb)
Now that we have those equations, let's think about the information we're given: "if the allele b has a frequency of 0.23" - which equation uses alleles?
"what percentage of the population is heterozygous for this trait" - which equation will tell us about genotypes?
$1=p^2+2p\left(0.23\right)+\left(0.23\right)^2$1=p2+2p(0.23)+(0.23)2 - we're looking for the middle term, 2pq, which is the frequency of the heterozygous genotype. To get our answer, we need to figure out what p is. Thankfully, we have an easier equation for that so we don't need to deal with polynomials!
3 years ago
Answered By Emily D
They're sneaking math into biology, how dare they!
We have two equations thanks to Hardy-Weinberg:
$1=p+q$1=p+q this one is for the frequency of alleles (how common is b and B?)
p is the dominant allele frequency (B in this case)
q is the recessive allele frequency (b)
$1=p^2+2pq+q^2$1=p2+2pq+q2 this one is for frequency of genotypes (how common is bb, Bb, and BB?)
p2 is the frequency of homozygous dominant (BB)
2pq is the frequency of heterozygous (Bb)
q2 is the frequency of homozygous recessive (bb)
Now that we have those equations, let's think about the information we're given: "if the allele b has a frequency of 0.23" - which equation uses alleles?
"what percentage of the population is heterozygous for this trait" - which equation will tell us about genotypes?
$1=p^2+2p\left(0.23\right)+\left(0.23\right)^2$1=p2+2p(0.23)+(0.23)2 - we're looking for the middle term, 2pq, which is the frequency of the heterozygous genotype. To get our answer, we need to figure out what p is. Thankfully, we have an easier equation for that so we don't need to deal with polynomials!
$1=p+0.23$1=p+0.23
We can solve for p as 0.77
Now that we have both p and q...
$2pq=2\left(0.77\right)\left(0.23\right)$2pq=2(0.77)(0.23)
$2pq=0.3542$2pq=0.3542
This number gives us the frequency, to get it into percentage, we just have to multiply by 100%
So our answer is 35.42%