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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.
The distance that shows the separation of two loci and their positions compared to mapped loci is called a map distance between loci and is measured in map unit or cM. It defines the chromosome distances.
The map distance between the red eyes and white-banded wings is:
Option B. 4 map units
This can be estimated as:
The moths having alleles for red eyes and white-banded wings (re wb) and wild-type eyes and wild-type wings (re+wb+) are crossed and F1 produced: rewb × re+wb+ (normal eyed and normal winged offspring).
F1 is crossed with red-eyed and white-banded winged moth by a test cross and the progenies produced was:
- Wild-type eyed (re+), and wild-type wings (wb+) = 25%
- Red eyed (re), and wild-type wings (wb+) = 25%
- Wild-type eyed (re+), white-banded wings (wb) = 25%
- Red eyed (re), white-banded wings (wb) = 25%
The phenotypes of the non-recombinant progeny are present in the highest numbers as F1 has (re wb) from one parent and (re+ wb+) from the other parent.
The recombinants are:
- Red eyes, wild-type wings = 19
- Wild-type eyes, white-banded wings = 16
[tex]\begin{aligned} \rm RF &= \dfrac{\rm Recombinants}{\rm Total \;progeny} \times 100\% \\\\&= \dfrac{(19 + 16)}{879} \times 100\% \\\\&= 4.0\% \end{aligned}[/tex]
RF = 4.0 mu
Therefore, distance between the genes is four map units.
To learn more about map distance follow the link:
https://brainly.com/question/9926973
