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In a linkage map, two genes A and B, are 70 cM apart. If individuals heterozygous for both the genes are test crossed number of progeny with parental phenotype will be equal to the number of progeny with recombinant phenotype.
Equal to the number of progeny with recombinant phenotype.
More than the number of progeny with recombinant phenotype.
Less than the number of progeny with recombinant phenotype.
Could be more or less than the number of progeny with recombinant phenotype depending on whether the genes are linked in cis or trans, respectively.
Answers (1)
In a test cross involving two genes that are 70 cM apart on a linkage map, the number of progeny with the parental phenotype (non-recombinants) is expected to be equal to the number of progeny with the recombinant phenotype. This is because a recombination frequency of 50% (or 0.5) is generally used as a threshold to determine whether two genes are linked or unlinked. In this case, the 70 cM distance between genes A and B indicates that there is a 50% chance of recombination occurring between them during gamete formation. As a result, the number of progeny with the parental and recombinant phenotypes would be expected to be equal. Hence, the correct answer is option 1.
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