A sweet pea plant of genotype A/A.B/B is crossed to one that is a/a.b/b to produ

Question

A sweet pea plant of genotype A/A.B/B is crossed to one that is a/a.b/b to produce a dihybrid F1.

One cell of one F1 individual (genotype A/a . B/b) goes through meiosis and produces four gametes of the following genotypes: A.b, A.b, a.B and a.B.

What can be concluded regarding the linkagerelationship between the two genes?

Answer

Genes A and B are linked; the result is indicative of linkage as the gametes are recombinants.

Genes A and B are linked; the result is indicative of linkage as the gametes are the products of crossovers.

Genes A and B could be linked or unlinked; the result is not indicative of either as in both cases it's possible to obtain 100%recombinant gametes form one meiosis.

Genes A and B could be linked or unlinked; the result is unusual as we should never obtain more than 50% recombinants from onemeiosis.

Genes A and B are unlinked; the result is indicative of independent assortment as only two genotypes are represented among thegametes"

The answer is marked C but I am not sure if this is correct.

Thanks

Explanation / Answer

Correct answer

Genes A and B could be linked or unlinked; the result is not indicative of either as in both cases it's possible to obtain 100%recombinant gametes form one meiosis.

Note;

If the loci are not linked, there are equal proportions 25% of each offspring haplotype. Two of the haplotypes are nonrecombinants (total probability 50%), and two are recombinants (total probability 50%)

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