Question

Assume that genes A and B are 50 map units apart on the same chromosome. An animal heterozygous at both loci is crossed with one that is homozygous recessive at both loci. What percentage of the offspring will show recombinant phenotypes? Without knowing that these genes are on the same chromosome, how would you interpret the results of this cross?

Answer 1

Answer:

Explanation:

The homozygous recessive individual can only produce 1 type of gamete (aabb).

The heterozygous individual can produce 8 types of gametes, of which 2 are parental and the rest are recombinant.

Genetic distance (m.u.) = Frequency of Recombination (%)

If the distance between genes A and B is 50 m.u., 50% of the gametes produced by the heterozygous individual, and therefore the offspring, will have recombinant phenotypes.

Without knowing that the genes are located on the same chromosomes, I'd think they are on different chromosomes, because you would get the same result: 50% recombinant offspring.

Whenever the genes on the same chromosome are separated by at least 50 m.u., or they are in different chromosomes, crossing over between them can happen with no restrictions and they will behave as independent of one another.