Homework6Solution

# Homework6Solution - C respectively Then the diploid states...

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Biol 214/414 Homework #6 Due 10-2-09 Solution A variety of rye is found in a remote farming area of Asia. It is brought to the lab where it is studied and found to be fertile and to have a (chromosome) n number of 21. Further study of wild grasses growing in this region identify three species with n numbers of 5, 7 and 9. Diagram or describe how the rye species could have arisen from these three wild grasses by the mechanism of allopolyploidy, showing all steps, including formation of gametes and zygotes. It would seem likely that the variety of rye is the result of two events of fusion of gametes from different species, followed each time by a doubling of the chromosomes. To diagram how this might have taken place, let’s use the same shorthand notation used in Figure 8-10 of our textbook. So let’s denote each haploid set of chromosomes (those which would exist in gametes) in the species with 5, 7, and 9 chromosomes as A, B, and
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Unformatted text preview: C, respectively. Then the diploid states for these species are AA, BB, and CC, respectively. Let’s suppose that the first step in the formation of the rye species was fusion of gametes (remember, they are haploid) from the n=5 and n=7 species. The resulting zygote (and eventually the plant that grows from it) can be denoted as AB. This hybrid is sterile, but if there is a chromosome doubling to get to the AABB state, the hybrid will be fertile, forming AB gametes. Now suppose an AB gamete fuses with a C gamete; the resulting zygote and eventually the plant from the zygote would be ABC, and after a second chromosome doubling would be AABBCC and would be fertile, a useful crop species that could then produce viable seeds. Such a hybrid would have an n of 5+7+9=21, just as we see in the rye species. Note that other sequences of hybrid formation ( eg AC or BC formed as the first step) would also work....
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