a. for Alltriplody, when there is 2n from species I and n from speci.pdf

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a. for Alltriplody, when there is 2n from species I and n from species II; 3n=15. when there is n from species I and 2n from species II; 3n= 21 b. Autotetraploidy in species II will be 4n= 28 c. Trisomy in species I 2n + 1= 8 + 1= 9 d. Monosomy is species II 2n- 1= 14- 1= 13 e. Tetrasomy of species I 2n + 2 = 8 + 2 = 10 f. Allotetraploid of species I and II There should be chromosomes from both species with a total of 4n. The possible combinations for allotetraploids are : 2n from both species = 8+ 14= 22 3n from species I + 1n from species II = 3(4)+ 1(7)= 12 + 7 = 19 1n from species I + 3n from species II =1(4)+ 3(7)= 4 + 21 = 25 Solution a. for Alltriplody, when there is 2n from species I and n from species II; 3n=15. when there is n from species I and 2n from species II; 3n= 21 b. Autotetraploidy in species II will be 4n= 28 c. Trisomy in species I 2n + 1= 8 + 1= 9 d. Monosomy is species II 2n- 1= 14- 1= 13 e. Tetrasomy of species I 2n + 2 = 8 + 2 = 10 f. Allotetraploid of species I and II There should be chromosomes from both species with a total of 4n. The possible combinations for allotetraploids are : 2n from both species = 8+ 14= 22 3n from species I + 1n from species II = 3(4)+ 1(7)= 12 + 7 = 19 1n from species I + 3n from species II =1(4)+ 3(7)= 4 + 21 = 25.

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