Multiple Choice
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 linkage relationship between the two genes?
A) Genes A and B are linked;the result is indicative of linkage as the gametes are recombinants.
B) Genes A and B are linked;the result is indicative of linkage as the gametes are the products of crossovers.
C) Genes A and B could be linked or unlinked;the result is not indicative of either,because in both cases it's possible to obtain 100% recombinant gametes from one meiosis.
D) Genes A and B could be linked or unlinked;the result is unusual as we should never obtain more than 50% recombinants from one meiosis.
E) Genes A and B are unlinked;the result is indicative of independent assortment as only two genotypes are represented among the gametes.
Correct Answer:

Verified
Correct Answer:
Verified
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