. A female fruit fly is heterozygous for the following traits, cinnabar eyes, lo
ID: 34057 • Letter: #
Question
. A female fruit fly is heterozygous for the following traits, cinnabar eyes, long body, and winged. She is crossed with a homozygous recessive male that has brown eyes, dwarf body, and wingless. They produced the following offspring.
Cinnabar eyes, long body, winged
489
Cinnabar eyes, long body, wingless
202
Cinnabar eyes, dwarf body, winged (double cross-over)
10
Cinnabar eyes, dwarf body, wingless
74
Brown eyes, long body, winged
68
Brown eyes, long body, wingless (double cross-over)
12
Brown eyes, dwarf body, winged
198
Brown eyes, dwarf body, wingless
511
Total
1564
What are the parental phenotypes and what are the recombinant phenotypes? (3 pts)
Parental:
Recombinant:
Calculate the distances between the 3 traits in map units. (7 pts)
Draw a genetic map of these 3 traits that shows their orientation and distance from each other. (5 pts)
Cinnabar eyes, long body, winged
489
Cinnabar eyes, long body, wingless
202
Cinnabar eyes, dwarf body, winged (double cross-over)
10
Cinnabar eyes, dwarf body, wingless
74
Brown eyes, long body, winged
68
Brown eyes, long body, wingless (double cross-over)
12
Brown eyes, dwarf body, winged
198
Brown eyes, dwarf body, wingless
511
Total
1564
Explanation / Answer
It is very easy to find the parental and recombinant genotypes from such questions. The maximum showing phenotypes are PARENTAL i.e. [Cinnabar eyes, long body, winged] & [Brown eyes, dwarf body, wingless]. And for RECOMBINANTS is the least frequency offsprings i.e. [Cinnabar eyes, dwarf body, winged] & [Brown eyes, long body, wingless].
To find the distance between the traits see the Single Cross overs, Double Cross overs for two traits at a time.
Like first for Length and wings- [202+198] + [10+12] Now map distance between length and wing is= 422/1564 = = 26.98 MU
For Eyes and Length- [74+68] + [10+12] = 164 Map distance between Eyes and Length= 164/1564= 10.4 MU
The Genetic Map is- Wing- Length- Eye
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