Review Question - QID 107031

The proportion of fruit flies with white eyes in the environment is 99%.

If all the conditions of Hardy Weinberg equilibrium are met, then one can use the equation p^2 + 2pq + q^2 = 1 (where p is the white-eyed allele and q is the red-eyed allele) to determine the frequencies of the alleles in the population and the number of white-eyed flies. The frequency of the red-eyed allele is 0.1 (q). Since p + q = 1, p = 0.9. Since the white-eyed allele is the dominant allele, both heterozygotes and homozygotes with the white-eyed allele will have white eyes. Thus the proportion of homozygous white-eyed flies (p^2) is 0.81 and the proportion of heterozygotes (2pq) is 0.18. Therefore, 99% ([0.81 + 0.18] x 100) of the population of fruit flies will have white eyes.

Harris and Torres review Marfan syndrome, an autosomal dominant connective tissue disorder, which is due to mutations in the FBN1 gene. Mutations result in defects in the ocular, skeletal, and cardiovascular systems such as myopia, displacement of the lens, retinal detachment, bone overgrowth, joint laxity, scolioisis, dilatation of the aorta, and mitral valve prolapse. With proper management, individuals can have a normal life expectancy.

Griffiths et al. discuss the Hardy Weinberg equilibrium and state that sexual reproduction does not cause genetic change with each subsequent generation, but remains constant. This equilibrium is only true if the following conditions are met: random mating, no mutations occur, no migration into or out of the population, no natural selection is taking place, or no random drift.

Illustration A shows describes the relationship between the three possible genotypes in Hardy Weinberg equilibrium.

Incorrect Answers:
Answers 1-4: The proportion of fruit flies with white eyes is 99%. 18% of the population will be heterozygotes (2pq), 81% will be homozygotes (p^2) for the white-eyed allele, and only 1% will be homozygotes for the red-eyed allele (p^2).