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This problem utilizes three variations in Drosophila (fruit flies)—all linked on the X chromosome.
 Red eyes (wild type) vs white eyes (mutant)
Normal length wings (wild type) vs cut wings (mutant)
Normal wing veins (wild type) vs. crossveinless wings (mutant)
Normal symbolism:
- Eye Color:
- wild type: w+(often abbreviated +)
- white: w
- Wing length:
- Wild type: cut+ (or +)
- Cut wings: cut
- Wing veins:
- Wild type: cv+ (or +)
- Crossveinless: cv
Parents:
- Female is heterozygous for all three genes: w+ w / cut+ cut / cv+ cv
- (or + w / + cut / + cv)
- Male is white, cut, crossveinless (w cut cv). Since these are sex linked genes, he has only one allele for each.
Offspring Phenotypes:
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| w cut cv: 35 |
w + +: 5 |
| w cut +: 64 |
+ cut +: 410 |
| w + cv: 395 |
+ + cv: 60 |
| + cut cv: 6 |
+ + +: 25 |
| Total: 1000 |
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Parentals: w + cv and + cut +
Double Crossovers: w + + and + cut cv
So the middle gene is cv.
Offspring phenotypes in correct order:
-
| w cv cut: 35 |
w + +: 5 |
| w + cut: 64 |
+ + cut: 410 |
| w cv +: 395 |
+ cv +: 60 |
| + cv cut: 6 |
+ + +: 25 |
| Total: 1000 |
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Single crossover classes:
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| w/cv: w + cut (64) and + cv + (60) |
| cv/cut: w cv cut (35) and + + + (25) |
Calculation of map distances:
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| w-cv: [(64+60+6+5)/1000]x100=13.5 LMU |
| cv-cut: [(35+25+6+5)/1000]x100=7.1 LMU |
| w-cut: 13.5 + 7.1 = 20.6 LMU |
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