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In Drosophila, the genes crossveinless-c and Stubble are linked, about 8 map units apart on chromosome 3. cv-c is a recessive mutant allele of crossveinless-c (cv-c+ is wild type), while Sb is a dominant mutant allele of Stubble (Sb+ is wild type). A dihybrid female Drosophila with genotype cv-c Sb+/cv-c+Sb is testcrossed. The proportion of phenotypically wild-type individuals in the progeny of the testcross will be.

So I know the answer is .46. If 8% are recombinant then the remaining 92% are parental. This means that .46 are parental genotypes cv-c+ sb and .46 are cv-c and sb+. My question is wouldnt the wild type progeny be one of the recombinants (cv-c+ sb+)? I know this can't be because otherwise the answer would be .04 I just dont know why. My only guess is because Sb is dominant it will be the allele that is expressed more than the wild type Sb+...

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Bunny Greenfelder
Bunny GreenfelderLv2
29 Sep 2019
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