Phenotypically wild-type F1 female Drosophila, whose mothers had light eyes (lt) and fathers had straw (stw)...
Phenotypically wild-type F1 female Drosophila, whose mothers had light eyes (lt) and fathers had straw (stw) bristles, produced the following offspring when crossed to homozygous light-straw males:
| Phenotype | Number |
| light-straw | 22 |
| wild-type | 18 |
| light | 970 |
| straw | 970 |
| Total: 2000 |
Compute the map distance between the light and straw loci.
| 2 map units |
| 8 map units |
| 98 map units |
| 4 map units |
| 12 map units |
Homework Answers
The correct answer is 2 map units:
When parents with light eyes (lt) and staw (stw) are crossed:
Total 2000 offsprings were produced, out of which two recombinants other than parentals are produced => light-straw 22, and wild 18.
% of recombinants = [(total number of recombinants)/(total offsprings)] x 100 = [(22+18)/2000] x 100 = (40x100)/2000 = 2%.
Since 2% recombination = 2 m.u.,
Map distance between the light and straw loci.= 2 map units.
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