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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 expected frequencies of the homozygous dominant, heterozygous, and homozygous recessive after 100 generations?
A) 0.75; 0.20; 0.05
B) 0.25; 0.5; 0.25
C) All red, because it is the natural color.
D) All black, because all red alleles would mutate to black.
E) None of the above

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Quizplus · Accepted Answer

Since the population is in Hardy-Weinberg equilibrium, the genotype frequencies can be calculated using the Hardy-Weinberg equation:
p2 + 2pq + q2 = 1
where p is the frequency of the dominant allele (B) and q is the frequency of the recessive allele (b).
Since the population has equal frequencies of the two alleles, p = q = 0.5. Thus:
- Homozygous dominant (B/B): p2 = (0.5)2 = 0.25
- Heterozygous (B/b or b/B): 2pq = 2(0.5)(0.5) = 0.5
- Homozygous recessive (b/b): q2 = (0.5)2 = 0.25
Therefore, the expected frequencies after 100 generations would be 0.25, 0.5, and 0.25 for B/B, B/b or b/B, and b/b, respectively. 
Thus, the correct answer is B.

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