Question

.  Eye color of the Oriental fruit fly (Bactrocera dorsalis) is determined by two genes that interact.  A...

.  Eye color of the Oriental fruit fly (Bactrocera dorsalis) is determined by two genes that interact.  A true-breeding fly having wild-type eyes is crossed with a true-breeding fly having yellow eyes.  All of the F1 flies from this cross have wild-type eyes.  When the F1 flies are interbred, 394 of the F2 progeny have wild-type eyes, 131 have amethyst (a bright, sparkling blue color) eyes and 175 have yellow eyes.

a).  Based on the number of progeny produced in the cross between F1 flies, describe how the two genes interact to determine eye color.  Establish a key which shows the alleles for each gene and how they interact.  Your answer should also include a list of the different genotypes which lead to each of the three phenotypes.  

A =                                                                  B =

a =                                                                   b =

Wild type eyes:

Yellow eyes:

Amethyst eyes:

b).  If two flies heterozygous at each loci mate and produce three offspring, what is the probability that the offspring will all have wild-type eyes or all have amethyst eyes or all have yellow eyes?  Use expected frequencies and not observed frequencies in your calculations.  

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Answer #1

The give phenotypic classes corerspond to 9:3:4 phenotypic ratio.
i.e. the given phenotype is an example of Recessive epistasis.
A B
Yellow ------> Amethyst --------> WT

Parental cross: AABB X aabb
F1 progeny: AaBb
F1 selfing: AaBb X AaBb
F2 progeny:
Phenotypic ratio: 9:3:4
9 = A_B_ = WT
3 = A_bb = Amethyst
3 = aaB_ = Yellow
1 = aabb = Yellow

The probability for all the progeny to have WT eyes = 9/16 X 9/16 X 9/16 = 0.1779
= 17.79%

The probability for all the progeny to have Amethyst eyes = 3/16 X 3/16 X 3/16 = 0.0065
= 0.65%

The probability for all the progeny to have yellow eyes = 4/16 X 4/16 X 4/16 = 0.0156
= 1.56%

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