I have read this post but am still slightly confused about this. Do tissues in the human body not all develop from the same cell(s) in the embryo? If so, I do not see how the cell 'mosaic' would be grainy enough to mask the fact that half of the female's cells are not working properly. For example, all liver cells come from the same cell in the embryo which has an inactivated X. In that case, there is a 50% chance that the females liver is dysfunctional. So I would predict that some females would exhibit X-linked disorers even if they are heterozygous, but I haven't heard of this being the case...

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    $\begingroup$ think of a given organ in your body (if you're a female) like a calico cat's pattern $\endgroup$
    – dblyons
    Feb 3 '17 at 23:49

Barr bodies (X-chromosome inactivation) don't form in the initial fertilized embryo — it's not that one X-chromosome is inactivated, and then that same inactivation is passed down to daughter cells. Rather, X-chromosome inactivation occurs on a cell-by-cell basis in differentiated cells. Note how the accepted answer to the question you linked mentions that different cell lineages will have different X-chromosome inactivation patterns — in other words, inactivation occurs later in the differentiation and proliferation processes than I think you might be assuming.

To use your example: the liver cells do not all derive from a pluripotent (undifferentiated) cell with one inactivated X-chromosome, but instead derive from said pluripotent cell, and then undergo X-chromosome inactivation.

Although the jury is out, scientifically speaking, on the mechanisms of selection of the X-chromosome to be inactivated, it's been postulated that some sort of decision-making may occur on this cellular level, resulting in a higher likelihood of inactivating the damaged or deleterious chromosome, if one is present.

Regardless, as the accepted answer to the question you linked mentions, the 50% of functional cells that would result from chance or random X-chromosome inactivation tends to be sufficient for the body's needs.

  • $\begingroup$ Let me know if this answers your question! $\endgroup$
    – Asher F.
    Feb 10 '17 at 16:51

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