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Suppose you have a population of flour beetles with 1,000 individuals. Normally the beetles are red; however, this population is polymorphic for a mutant autosomal body color, black, designated by b/b. Red is dominant to black, so B/B and B/b genotypes are red. Assume the population is in Hardy-Weinberg equilibrium, with equal frequencies of the two alleles.
-What would be the allele frequencies (red, followed by black) if a population bottleneck occurred and only four individuals survived: one female red heterozygote and three black males?
A) 0.875; 0.125
B) 0.125; 0.875
C) 0.25; 0.75
D) 0.75; 0.25
E) 0.5; 0.5

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Quizplus · Accepted Answer

After the bottleneck event, the new population only has four individuals, consisting of one red heterozygous female (carrying the alleles B and b) and three black males (carrying the allele b). The allele frequency of B can be determined by adding together the frequency of the homozygous dominant (B/B) and the heterozygous (B/b) individuals in the original population: frequency of B = (2B/B + B/b) / (2 x 1,000) = 0.751. The frequency of b can be determined in the same way, using the homozygous recessive (b/b) and heterozygous (B/b) individuals: frequency of b = (2b/b + B/b) / (2 x 1,000) = 0.249. Therefore, after the bottleneck event, the allele frequencies are 0.125 for B and 0.875 for b.

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