Dr. Disney has been raising exotic fruit flies for decades. Recently, he discovered a strain of fruit flies that in the recessive condition have baby blue eyes that he designates as bb. He also has another strain of fruit flies that in the recessive condition have pink wings that are designated as pw. He is able to establish flies that are homozygous for both mutant traits.He mates these two strains with each other. Dr. Disney then takes phenotypically wild- type females from this cross and mates them with double recessive males. In the resulting testcross progeny, he observes 500 flies that are of the following makeup:41 with baby blue eyes and pink wings207 with baby blue eyes only210 with pink wings only42 with wild-type phenotype14. Assuming the wild-type alleles for these two genes are b+ and pw+, what is the correct testcross of the F1 flies?A) b+ pw+/b pw ⋅ b pw/b pwB) b+ pw+/b pw ⋅ b pw+/b+ pwC) b+ pw/b pw+ ⋅ b pw/b pwD) b+ pw/b pw+ ⋅ b+ pw+/b pwE) b+ pw+/b pw ⋅ b+ pw/b pw+The answer is C but I need some help figuring out why. Thanks!

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Asked Jan 14, 2019
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Dr. Disney has been raising exotic fruit flies for decades. Recently, he discovered a strain of fruit flies that in the recessive condition have baby blue eyes that he designates as bb. He also has another strain of fruit flies that in the recessive condition have pink wings that are designated as pw. He is able to establish flies that are homozygous for both mutant traits.
He mates these two strains with each other. Dr. Disney then takes phenotypically wild- type females from this cross and mates them with double recessive males. In the resulting testcross progeny, he observes 500 flies that are of the following makeup:

41 with baby blue eyes and pink wings
207 with baby blue eyes only
210 with pink wings only
42 with wild-type phenotype

14. Assuming the wild-type alleles for these two genes are b+ and pw+, what is the correct testcross of the F1 flies?
A) b+ pw+/b pw ⋅ b pw/b pw
B) b+ pw+/b pw ⋅ b pw+/b+ pw
C) b+ pw/b pw+ ⋅ b pw/b pw
D) b+ pw/b pw+ ⋅ b+ pw+/b pw
E) b+ pw+/b pw ⋅ b+ pw/b pw+

The answer is C but I need some help figuring out why. Thanks!

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Expert Answer

Step 1

 

The given question refers to that the resulting progeny of 500 fruit flies consists of:

  1. 41 having baby blue eyes and pink wings
  2. 207 having baby blue eyes only
  3. 210 having pink wings only
  4. 42 having wild-type phenotype

This progeny shows variations due to a testcross between b+ pw/b pw+. b pw/b pw. The reason behind this is explained as follows.

 

Step 2

Homozygous refers to a specific gene which has same alleles present on both homologous chromosomes. It is denoted by two (XX) capital letters for some dominant character, and two (xx) small letters for some recessive trait.

A test cross in genetics refers to the breeding of an organism with a phenotypically recessive organism, in demand to govern the zygosity of the earlier by examining extents of progenies phenotypes. Zygosity could be homozygous or heterozygous.

Step 3

Baby blue eyes (bb) and pink wings (pw) are recessive traits.

Phenotypically wild types females are (b+ pw/b pw+) which are crossed with recessive male flies (b pw/b pw). And wild type allele is dominant over the other.

...

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