Wild-type Arabidopsis has 5 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?

Question

Quizplus · Accepted Answer

The correct answer is D because the probability of getting a wild-type chromosomal complement from the cross wt × Tr1;Tr2 is 1/9. This is because the wild-type parent will contribute 5 chromosomes, and the Tr1;Tr2 parent will contribute 6 chromosomes (3 from each trisomic chromosome). The probability of getting a wild-type chromosomal complement is therefore (1/2) ^ 6 = 1/64. However, since there are 9 possible chromosomal complements that can be produced from this cross, the probability of getting a wild-type chromosomal complement is 1/64 * 9 = 1/9.

Wild-Type Arabidopsis Has 5 Chromosomes (2n = 10)