In Genetics lab, you decide to further investigate Drosophila eye color. In addition to studying the...
- In Genetics lab, you decide to further investigate Drosophila eye color. In addition to studying the sepia gene (se), you decide to study the gene mahogany eyes (mo). Flies that are mo-/mo- have brown eyes. You wonder if a fly that is a double mutant for sepia and mahogany eyes would have very dark brown (almost black) eyes. Because se and mo are both on the third chromosome, you have to do a recombination.
You look through a former labmate’s notebook and see that she has previously done this experiment! Once she generated the se- mo-/se- mo- fly, she even performed a cross between the homozygous double mutant and a heterozygous fly to map the distance between the two alleles. Her resulting progeny were as follows:
1,339 wild-type flies
1,195 black-eyed flies
151 sepia-eyed flies
154 brown-eyed flies
- Which phenotypes are the parental phenotypes? How many progeny show each parental phenotype?
- Calculate the map distance between the sepia and mahogany eyes alleles.
- Sadly, the double mutant line that your labmate created died off, and you need to create it again. You generate 60 flies that you think may possible contain the recombined chromosome and create individual stocks. How many of those 60 stocks would you estimate to contain the desired recombinant outcome?
Homework Answers
Please look in the image for solution:
The double mutant type of 60 are=( 5.371/100) ×60=3.22 = 3( approx)
We found 3 double mutant progeny out of 60 .
Add Answer to:
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