In a population of deer mice, the allele for white hair is recessive and the allele for brown hair is dominant. If the population consists of 500 individuals and the frequency of homozygous brown mice is 49%, what is the frequency of the recessive allele

According to Hardy-Weinberg equilibrium, sum of both dominant allele frequency (p) and recessive allele frequency (q) should be equal to 1 (p+q = 1).

Brown hair = dominant (p)

white hair = recessive (q)

49% mice are brown hair

so dominant genotype frequency = 0.49

According to Hardy-Weinberg principle, square root of the genotype (homozygous) is equal to allele frequency.

√p =√0.49 = 0.70

The dominant allele frequency is 0.7

Now, by this value we can find the recessive allele frequency by

p + q = 1

1 - 0.7 = 0.3

So, the recessive allele (white hair mice) frequency is 0.30.

luisvaschetto luisvaschetto · Answer 2 · 2021-12-31T13:50:15+01:00

In this case, the frequency of the recessive allele is equal to 0.30.

The Hardy Weinberg (HW) equation is used in population genetics in order to estimate genotype and allele frequencies.

According to this principle, the sum of the frequency of the dominant (p) allele and the frequency of the recessive (q) allele must be equal to one (1) in HW equilibrium.

In this case, the frequency of the homo-zygous dominant phenotype (p²) is equal to 0.49.

In consequence, we have:

p²=0.49 >> p=√0.49 = 0.70
p + q= 1 >> 1 - 0.70 = q = 0.30

In conclusion, the frequency of the recessive allele is equal to 0.30.

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