EVOLUTION
HARDY WEINBERG EQUILIBRIUM
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Question
[CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
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0.7
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0.21
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0.42
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0.49
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0.91
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Detailed explanation-1: -We know that the frequency of the recessive homozygote genotype is q2 and equal to 0.09. Thus, p = 1 – q ∴ p = 1 – 0.30 = 0.70 ∴The homozygote dominants are represented by p2 = (0.70)2 = 0.49 or 49% Page 2 4.
Detailed explanation-2: -In a population that is in Hardy-Weinberg equilibrium, the frequency of the recessive homozygous genotype of a certain trait is 0.16.
Detailed explanation-3: -In a population that is in Hardy Weinberg equilibrium, the frequency of a recessive allele for a certain hereditary trait is 0.20.
Detailed explanation-4: -The Hardy-Weinberg Equation q = the frequency of the recessive allele in a population. 2 p q 2pq 2pq = the frequency of the heterozygous dominant genotype. p p2 = the frequency of homozygous dominant genotype.q2 = the frequency of homozygous recessive genotype.