EVOLUTION
HARDY WEINBERG EQUILIBRIUM
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Question
[CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
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0.16
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0.24
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0.48
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0.6
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Detailed explanation-1: -This expression would be written as 2pq = 2 (0.6) (0.4) = 0.48. Therefore, the frequency of heterozygotes in this population is 0.48 or 48%.
Detailed explanation-2: -The frequency of the B allele is 6/10 = 60%. A shortcut to counting the number of B alleles is to count how many BB genotypes there are, and multiply this by 2 (since each BB genotype has 2 B alleles), then add the number of Bb genotypes (since each Bb genotype has 1 B allele).
Detailed explanation-3: -c B-allele frequency (BAF) is calculated by dividing the signal intensities of minor (B) allele by those of major (A) plus minor (B) alleles. In normal sample, three values of 1, 0.5, and 0 are obtained. A gain region generates four values (e.g., 0, 0.33, 0.67, and1) and it is interpreted as allelic imbalance.
Detailed explanation-4: -The frequency of heterozygous individuals. Answer: The frequency of heterozygous individuals is equal to 2pq. In this case, 2pq equals 0.32, which means that the frequency of individuals heterozygous for this gene is equal to 32% (i.e. 2 (0.8)(0.2) = 0.32).