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The homologous pair of chromosome 18 failed to separate which is known is meiotic disjunction. Now there are too many chromosomes in one of the cells at the start of meiosis II which results in the production of all irregular haploid cells
How else can T18 occur?
The second option is that sister chromatids do NOT separate in ANAPHASE II, this results in 2 normal haploid cells, 1 haploid cell with one less chromosome 18, and 1 haploid cell with one extra chromosome 18
Thanks for all this information Doctor. Before we go underway with any testing, I want to go home and do a little more research on the topic.
In this case, nondisjunction occurs in meiosis II, specifically in anaphase when the sister chromatids fail to separate which results in 2 regular haploid cells and 2 irregular haploid cells.
Statistics around T18: •occurs in 1 in 6000 live births •median lifespan of babies living with the syndrome is 5 to 15 days •8% of infants live past 1 year •1% live past 10 years •the chances of birthing a T18 baby increase with maternal age •the disorder is not inherited, and occurs randomly
My research indicated that the physical features of those affected by T18 are deformations such as an abnormally small head (microcephaly), widely spaced out eyes, a very small jaw which often results in cleft lip or cleft palette. Clenched fists and overlapping fingers are also common features.
I discovered that there is no cure for Edward's Syndrome, due to the incredibly short lifespan of T18 infants. However, one treatment option involves the Xist Gene (x inactivation gene) being inserted into the extra 18th chromosome of a cell
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