spo 2. Three species in the genus Brassica; campestris, nigra and diploid and ha
ID: 280650 • Letter: S
Question
spo 2. Three species in the genus Brassica; campestris, nigra and diploid and have the genome constitutions AA, BB and CC, respectively. The rde Peri netic oleracea are known to be chromosome numbers of 3 other species and the number of bivalents an in Fi hybrids of various crosses among the 6 species are as follows published in Jpn. J. Bot. 7:389, 1935) d uni (based on wor Do Number of Species and/or Fi hybrid Chromosome number Bivalents univalents B. juncea B. carinata B. napus Fi from juncea X nigra Fi from napus X campestris F1 from carinata X oleracea Fi from juncea X oleracea F1 from carinata X campestris Fi from napus X nigra 34 38 19 29 27 27 27 27 27 What are the haploid chromosome numbers of the 3 diploid species? Explain. (15 ts a. b. Are juncea, carina constitutions of juncea, carinata, and napus (in terms of A, B, C as above). Explain your answer. (15 pts.) a, and napus autopolyploids or allopolyploids? Give the genomeExplanation / Answer
Answer to Qa)
The three diploid species given in the table provided are B. juncea, B. carinata and B. napus. These species have chromosomal no. of 36, 34 and 38 respectively.
Now, these three are diploids (as the chromosomal set contains bivalents only) but actually they are double diploids and also allopolyploids i.e. they have been formed by hybridization between two species and then the chromosomes have duplicated due to which they occur as bivalent sets. The chromosomes from each parent pairs up with other chromosome from same parent (because duplicate sets are present) and not with that of the other parent. This is because the parents do not belong to same species and so have different chromosomes. That is why these species are also called as Allotetraploids.
In case of B. juncea, the parents are B. campestris (2n = 20, designated as AA) and B. nigra (2n = 16, designated as BB). So, the cross produced hybrid AB (chromosome number, 10+8 = 18) which then undergoes duplication to form AABB with chromosome no. 36. The haploid number is 18 because there are 18 different types of chromosomes (10 from one parent and 8 from another).
In case of B. carinata, the parents are B. nigra (2n = 16, designated as BB) and B. oleracea (2n = 18, designated as CC). So, the cross produced hybrid BC (chromosome number, 8+9 = 17) which then undergoes duplication to form BBCC with chromosome no. 34. The haploid number is 17 because there are 17 different types of chromosomes (8 from one parent and 9 from another).
In case of B. napus, the parents are B. campestris (2n = 20, designated as AA) and B. oleracea (2n = 18, designated as CC). So, the cross produced hybrid AC (chromosome number, 10+9 = 19) which then undergoes duplication to form AACC with chromosome no. 38. The haploid number is 19 because there are 19 different types of chromosomes (10 from one parent and 9 from another).
In all these cases because of duplication the chromosome appear as bivalents and so no univalent exists.
Answer to Qb)
As already explained in part a) all three are allopolyploids and have following genetic constitution:
B. juncea = AABB
B. carinata = BBCC
B. napus = AACC
Further, the allopolyploidy is confirmed by the F1 generations given in the table:
B. juncea x B. nigra resulted in chromosome number = 26, out of which 18 are bivalents and 8 are univalent which can be explained as follows :
B. juncea (AABB, chromosome number = 36) will form a gamete as AB (chromosome number = 18) and B. nigra (BB, chromosome number = 16) will form a gamete as B (chromosome number = 8). The F1 offspring would be designated as ABB (chromosome number = 18+8 =26), Out of these 8 chromosomes (from each B) will appear as bivalents while 10 from A will remain univalent.
Similarly the bivalents and univalents can be explained in the F1 crosses between B. napus and B. campestris as well as between B. carinata and B. oleracea.
Further confirmation is by F1 cross between B. juncea (AABB) and B. oleracea (CC), the gametes will be (AB, chromosome number = 18) and (C, chromosome number = 9), the F1 offspring would be ABC (chromosome number = 27), with three separate chromosomal sets and that is why none appear as bivalents but all appear as univalent.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.