Phenotypically wild-type F1 female Drosophila, whose mothers had purple eyes (pr) and fathers had curved wings (cw), produced the following offspring when crossed to homozygous curved-wings and purple eyed males:

Phenotype                         Number

Purple-curved wing          220

wild-type                           200

purple                               770

curved wing                      810

                             Total: 2000

Compute the map distance between the purple and curved wing loci.

Question

Phenotypically wild-type F1 female Drosophila, whose mothers had purple eyes (pr) and fathers had curved wings (cw), produced the following offspring when crossed to homozygous curved-wings and purple eyed males:

Phenotype                         Number

Purple-curved wing          220

wild-type                           200

purple                               770

curved wing                      810

Total: 2000

Compute the map distance between the purple and curved wing loci.

BlackTom AI · Accepted Answer

To tackle this, I start by identifying the classes of progeny in the test cross and which phenotypes are parental versus recombinant.
Option-by-option analysis:
Option A: 2 map units. This would imply a very tight linkage with only about 2% recombination. Looking at the data, the two most frequent classes are purple and curved-wing alone (770 and 810), which are far more common than the double mutant (220) or wild-type (200). Such a small distance does not fit these observed frequencies.
Option B: 77 map units. A distance that large would approach independent assortment, producing roughly equal numbers of most phenotypes. However, the observed counts show a clear excess of the single-mutant progeny over the double-mutant and wild-type classes, indicating much tighter linkage than 77 cM.
Option C: 4 map units. This would predict mostly parental classes with very few recombinants. But the data show substantial recombinant-class counts (double mutant and wild-type together are only 420 of 2000, whereas the single-mutant classes total 1580), so 4 cM is inconsistent with these numbers.
Option D: 21 map units. Here, the two most frequent classes are single-mutant phenotypes (purple only and curved wing only) totaling 770 + 810 = 1580, while the double mutant (purple-curved) and wild-type classes total 220 + 200 = 420. This pattern matches a parental configuration where single mutants are parental and the double mutant plus wild-type are recombinants, giving recombination frequency 420/2000 = 0.21, i.e., 21 map units.
Option E: 11 map units. With such a small distance, recombinants would be rarer, and the distribution would skew even more toward the two parental classes. The observed data show a substantial number of recombinants (420) relative to total, which contradicts 11 cM.
Synthesis: The classification of progeny as parental vs recombinant points to the cross producing two main parental classes (purple only and curved-wing only) and two rarer classes (purple-curved and wild-type). Counting recombinants as the two rarer classes yields 220 + 200 = 420 recombinants out of 2000 offspring, which equals 21%. Therefore, the map distance is 21 map units.
Conclusion note: The only option compatible with the observed distribution of phenotypes and the standard method for calculating map distance from a test cross is 21 map units.

MOLGEN 4500E AU2025 (25000) Exam 4- Final Exam- Requires Respondus LockDown Browser

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