Wild-type Arabidopsis has five chromosomes (2n = 10).Trisomic plants are designated as Tr followed by the trisomic chromosome number-that is, Tr1 is trisomic for chromosome 1.Assuming that trisomy is fully viable and that all possible pairing configurations (including nonpairing)are possible at meiosis, what proportion of the progeny from the cross Tr1 × Tr1 will have a wild-type chromosomal complement?

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Quizplus · Accepted Answer

In a Tr1 × Tr1 cross, both parents have an extra copy of chromosome 1, making their genotype 2n+1 for that chromosome. During meiosis, the chromosomes can segregate in several ways, but for a progeny to have a wild-type chromosomal complement (2n = 10), one gamete must contribute the normal two copies of chromosome 1, and the other must contribute only one copy. The probability of producing a gamete with two copies from a trisomic parent (2n+1) is 1/3, as the three copies of chromosome 1 can segregate as (2:1) or (1:2), with two equally likely configurations out of six total possibilities resulting in a gamete with two chromosomes. Therefore, the probability that one parent produces a gamete with two copies and the other produces a gamete with one copy is (1/3) * (1/3) = 1/9.

Wild-Type Arabidopsis Has Five Chromosomes (2n = 10)