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Sagot :
Assuming Hardy-Weinberg equilbrium, we have
p + q = 1
where p = frequency of dominant allele and q = frequency of recessive allele. So
p = 1 - q = 1 - 0.04 = 0.96
Consequently we have
• p² = 0.9216 = frequency of hom ozygous domininant genotype
• pq = 0.0768 = frequency of heterozygous genotype
• q² = 0.0016 = frequency of hom ozygous recessive genotype
since
p² + 2pq + q² = 1
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