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mario62 [17]
4 years ago
12

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?
Biology
1 answer:
fomenos4 years ago
3 0

Answer:

Explanation:

The homozygous recessive individual can only produce 1 type of gamete (<em>aabb</em>).

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.

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Answer:

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  • In case of Island B, each of the B1 and B2 sub-populations that got split from each other developed certain mutations that were necessary for them to adapt to the particular diverse environment each of them were exposed to, through the period of 50,000 years. These mutations were so varied that reproductive isolation was generated between them that resulted in each of them to develop into different species.Hence, speciation happens here and B1 and B2 are incapable of inter-breeding.
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