Answer:
233 snakes are heterozygous for the banding allele
Explanation:
According to Hardy-Weinberg, the allelic frequencies in a locus are represented as p and q, referring to the alleles. The genotypic frequencies after one generation are p² (Homozygous for allele p), 2pq (Heterozygous), q² (Homozygous for the allele q). Populations in H-W equilibrium will get the same allelic frequencies generation after generation. The sum of these allelic frequencies equals 1, this is p + q = 1.
In the exposed example,
- The banding phenotype is autosomal recessive, bb
- The frequency of banded snakes on the island is 0.4
- There are 500 total snakes on the island
How many snakes are heterozygous for the banding allele?
The frequency of banded snakes refers to the genotypic frequency for the trait, which is bb= q2= 0.4.
If q2= 0.4, then q = √0.4 = 0.63
The allelic frequency for b is 0.63.
This means that the allelic frequency for B or p is 0.37, which we deduce by clearing the equation p + q = 1
p + 0.63 = 1
p = 1 - 0.63
p = 0.37
The allelic frequency of B is 0,37, and the allelic frequency for b is 0,63. The population heterozygote frequency for this allele is 2 x p x q = 2 x 0,37 x 0,63 = 0.466. The percentage of the population that is heterozygous for this allele is 46%.
As the population size is 500 individuals, then we can calculate how many of these snakes are heterozygous. This is: 0.466 x 500 = 233
