Answer:
two types of moths are given,
rewb and re+wb+ and they crossed.
in F1 generation they will produce,
rewb × re+wb+ and will give, normal eyed and normal winged offsprings.
offsprings of F1 are crossed with red eyed and white-banded winged moth in a testcross. The progenies will be - (given in the question)
- the genes of the red eyed and white-banded winged are located on different chromosomes,
then their phenotypic proportions will be ,
25% of wild-type eyed (re+), and wild-type wings (wb+);
25% of red eyed (re), and wild-type wings (wb+);
25% of wild-type eyed (re+), white-banded wings (wb);
25% of red eyed (re), white-banded wings (wb).
- the recombination percentage between the red eyed and white-banded winged will be,
The recombinant progeny are the 19 with red eyes, wild-type wings and 16 with wild-type eyes, white banded wings.
Recombination Frequency = no. of recombinants/total progeny × 100%
= (19 +16)/879 × 100%
= 4.0%
The map distance between genes is 4 map units or 4 m.u or 4 centi Morgan.
