Drosophila melanogaster
In female fruit flies (Drosophlia melanogaster), the oocyte is located between maternal nurse cells and follicle cells providing nutrients, proteins and mRNA crucial for the development of the embryo. In one of the genes whose mRNA is transported to the oocyte, a mutation X has been found that leads to deformed, non-viable embryos. If the mutation is recessive and two individuals heterozygous for X are crossed to produce the F1, what fraction of the F2 would be homozygous for X. (Approximate to 3 decimal places)
The answer is 0.167.
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1 solution
To be homozygous for the mutation, an individual needs to receive a mutant allele from each parent. In the F1 of a cross between heterozygotes, the genotypes are distributed 1:2:1. Males can give the mutant allele with a probability of 2 1 , however, the homozygous mutant female are sterile, therefore, only heterozygous females representing 3 2 of the fertile females can give the mutant allele to their offspring, this with a probability of 2 1 . The final probability of being homozygous in the F2 IS 2 1 * 3 2 * 2 1 = 6 1 . This is approximately equal to 0.167