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 wt × Tr1;Tr2 will have a wild-type chromosomal complement?

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In the cross between a wild-type (wt) Arabidopsis (with a normal diploid chromosome number of 2n = 10) and a plant trisomic for chromosomes 1 and 2 (Tr1;Tr2), the trisomic plant has one extra copy of chromosomes 1 and 2, making its chromosome number 2n + 1 + 1 = 12. During gamete formation in the trisomic plant, various combinations of chromosomes can be produced, including those with an extra chromosome 1, an extra chromosome 2, both extra chromosomes, or neither. For progeny to have a wild-type chromosomal complement (2n = 10), they must receive gametes from the trisomic parent that do not contain any extra chromosomes. Given that all possible pairing configurations are possible at meiosis, including nonpairing, the chance of forming a gamete without any extra chromosomes (thus leading to a wild-type progeny when fertilized by a wild-type gamete) is significantly reduced. Considering the multiple possible disjunctions and the need for the specific outcome where no extra chromosomes are included in the gamete, the probability is 1/9, reflecting the complexity of achieving a specific gamete composition from a trisomic parent.

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