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6. A true-breeding male fruit fly with a gray body was crossed with a true-breeding ebony-bodied female. All their offspring
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6.

Parental Cross: True-breeding Grey male x True breeding Ebony female

F1: Grey Bodies (Grey is dominant to Ebony)

When F1 is selfed, each F1 has one Grey allele and one Ebony allele. Since Ebony is recessive to Grey, only the progeny that would receive both Ebony alleles from either F1 parent will be ebony. This has a probability of 1/2 * 1/2 = 1/4

Thus, 25% of the F2 will have an Ebony body.

7.

Similar to the previous question, here both parents are heterozygous for the Cystic Fibrosis causing allele. Therefore, the chance of each parent passing on the CF causing allele to their child is 1/2.

Thus, the probability that the child of this couple has Cystic Fibrosis is 1/4.

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