Question

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
0 0
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Answer #1

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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