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Suppose that 75% of chickadees (a bird) have a head crest and must therefore have at least one copy of the dominant H allele. Assuming that this population is in Hardy Weinberg Equilibrium for this gene, what percentage of the population must be homozygous dominant for this trait?

Respuesta :

Answer:

25%

Explanation:

If p is frequency of dominant allele and q is frequency of recessive allele, according to HWE,

p + q = 1

p² + 2pq + q² = 1

where p² = frequency of homozygous dominant

q² = frequency of homozygous recessive

2pq = frequency of heterozygous

Here,

p² + 2pq = 75% or 0.75

q² = 0.25

q = √0.25 = 0.5

p = 1-q = 0.5

p² = (0.5)² = 0.25 or 25%

Hence percentage of the population homozygous dominant for this trait = 25%

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