Answer:
<h2>
Allele frequencies for B and b.
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"b"(q) allele frequency = 0.60
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"B"(p) allele frequency = 0.40
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Genotype frequencies;
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BB = 0.16
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Bb = 0.48
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Bb = 0.36
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Explanation:
Given
Non-baldness (B) is dominant on baldness(b), so B is dominant over b.
Homozygous pattern baldness male (bb) = 360,
Heterozygous non- baldness (Bb)= 480,
Homozygous non-baldness (BB)= 160.
So, we can also denote then by genotypes only,
BB= 160;
Bb= 480;
bb= 360;
Total= 1000
Allele frequency q² (bb) = 360/1000=0.36
allele fequency for q( b)= √o.36=0.60.
Allele frequency for p²(BB) = 160/1000=0.16
allele frequency p(B)= √0.16 = 0.4
Expected genotype frequencies;
BB = 160/1000 = 0.16
Bb = 480/1000= 0.48
Bb = 360/1000= 0.36
