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Answer:
The closest to the number of flies expected to be heterozygous for the amylase mutation assuming all conditions of Hardy-Weinberg equilibrium are met is 1600 flies.
Explanation:
According to the Hardy-Weinberg equation : p² + 2pq + q² = 1
Also p + q = 1
Where p² is the frequency of individual flies with the similar alleles for the dominant trait;
2pq is the frequency of individuals with the heterozygous allele for breaking down starch, while q² is the frequency of the individuals having similar allele formthenrecessive trait.
q² = 410/10000 = 0.041
q = 0.2
p = 1 - 0.2
p = 0.8 p² = 0.8²
p² = 0.64
2pq = 1 - 0.2 - 0.64 = 0.16
2pq = 0.16 × 10000
2pq = 1600
Therefore, the closest to the number of flies expected to be heterozygous for the amylase mutation assuming all conditions of Hardy-Weinberg equilibrium are met is 1600 flies.