Question

True-breeding red-eyed Drosophila flies with plain thoraxes were crossed with pink-eyed flies with striped thoraxes.
Red eye plain thorax x Pink eye striped thorax
The $F_1$ flies were then test crossed against the double recessive.
The details of the $F_2$ generation resulting from the cross is given below:
$\begin{array}{cccc}\text{80}& \text{20}& \text{8}& \text{92}\\ \text{Red eye}& \text{Red eye}& \text{Pink eye}& \text{Pink eye}\\ \text{Plain thorax}& \text{Striped thorax}& \text{Plain thorax}& \text{Striped thorax}\end{array}$
Mention the percentage of recombinant offsprings resulting from the test cross.

• A. 12
• B. 14
• C. 16
• D. 28

Answer 1

The correct option is B 14

Parental phenotypes include red eyed flies with plain thorax and pink eyed flies with striped thorax.
Recombinant phenotypes include red eyed flies with striped thorax and pink eyed flies with plain thorax.
Total number of progeny flies after test cross $=80+20+8+92=200$
Total number of progenies with recombinant phenotypes = 20 (red eyed, striped thorax) + 8 (pink eyed, plain thorax) = 28
Therefore, percentage of recombinants
= (No. of flies with recombinant phenotype)/ (Total no. of flies in F2 generation) $\times 100\%$
$=\left(28/200\right)\times 100$
=14%
Hence, 14% of the F2 offsprings were of recombinant phenotype.