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Answer: The frequency of brown beetles is 0.32.
Explanation: The frequency of A1 allele is 0.8. As p+q=1, or the sum of dominant and recessive frequencies equals 1 or 100%:
1 - 0.8 = 0.2
In Hardy-Weinberg principle,
2pq represents the frequency of heterozygote individuals, so:
genotype A1A2 = 2*0.8*0.2 = 0.32.
Thus, the frequency of brown beetles (A1A2) in the population is 0.32.
<h2>Frequency of allele </h2>
Explanation:
Hardy Weinberg Equilibrium is used to calculate the allelic as well as genotypic frequency
Allelic frequency of dominant and recessive allele is represented by p and q respectively whereas genotypic frequency of dominant genotype is represented by and
respectively
Given:
H allele (p) = hairy heffalump (dominant)
h allele (q) = hairless heffalump (recessive)
36% of heffalump population is hairless represents the % of recessive genotype, hh () =36%
Calculation of frequency of the h allele (q) :
Frequency of genotype hh () will be: 36/100=0.36 or 0.6*0.6
Frequency of h allele (q) will be 0.6