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img019 - of spontaneous abortions suggests that human...

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Unformatted text preview: Chapter 14 261 of spontaneous abortions suggests that human embryos tolerate the genetic imbalance for three copies of a gene much better than one copy. Also, one copy of a chromosome will be lethal if that copy carries any lethal mutations. Monosomies usually arrest zygotic development so early that a pregnancy is not recognized, and thus they are not seen in karyotypic analysis of spontaneous abortions. 14-36. Both types of Turner's mosaics could arise from chromosome loss or from mitotic nondisjunction early in zygotic development. Chromosome loss would involve the loss of one of the X chromosomes early in development in an XX embryo (producing a mosaic with both 46, XX and 45, X0 cells) or the loss of the Y chromosome in the XY embryo (yielding a mosaic with 46, XY and 45, X0 cells). Mitotic nondisjunction in a normal XX embryo should produce an XXX daughter cell in addition to an XO, while mitotic nondisjunction in an XY embryo yields an X0 and an XYY daughter cell. If the XXX or XYY daughter cells did not expand into large clones of cells during development, karyotype analysis would not be able to detect their presence. Note that for mitotic nondisjunction to have given rise to the described mosaic individuals, the nondisjunction event must have occurred after the first mitotic division so that there would be some XX or XY cells. 14-37. You have three marked fourth chromosomes: ci+ ey, ci ey+ and ci ey. Drosophila can survive with two or three copies of the fourth chromosome, but not with one copy or four copies. You are looking for mutations that are defective in meiosis and cause an elevated level of nondisjunction. a. Mate potential meiotic mutants that are ci+ ey / ci ey+ with ey ci / ey ci homozygotes. The normal segregants should be ci+ ey / ey ci (ey) and ci ey+ / ey ci (ci). Nondisjunction in MI will be seen as the rare ci+ ey / ci ey+ / ey ci (wild-type) progeny. Nullo-4 gametes without any copy of chromosome 4 would produce zygotes with only one copy of this chromosome that would not survive. b. The cross in part a will detect nondisjunction in MI, but it will not distinguish MII nondisjunction. c. Diagram the testcross: ci+ ey / ci ey+ / ey 61' x ey ci / ey cz' ——> ? Remember that in a trisomic individual, two of the three copies of the chromosome pair normally at metaphase I of meiosis, while the third copy assorts randomly to one pole or the other. There are three different ways to pair the fourth chromosomes in the trisomic individual. The first option is: 1/3 (ci+ ey segregating from 01' ey+ with ey ci assorting independently) = 1/6 probability of(l/2 ci+ ey / ey ci : 1/2 ci ey+) and 1/6 probability of(l/2 ci+ ey : 1/2 ci eyJr / ey ci). The second option is: 1/3 (Ci+ ey segregating from ey ci with ci ey+ assorting independently) ...
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