Most cases of Turner syndrome are attributed to nondisjunction of one or more
of the sex chromosomes during gametogenesis, from either the male or female
parent. However, some females possess a rare form of Turner syndrome in which
some of the cells of the body (somatic cells) lack an \(X\) chromosome, while
other
cells have the normal two X chromosomes. Often detected in blood
and/or skin cells, such individuals with mosaic Turner syndrome may exhibit
relatively mild symptoms. An individual may be specified as \(45,
\mathrm{X}(20) / 46, \mathrm{XX}(80)\) if, for example, 20 percent of the cells
examined were \(\mathrm{X}\) monosomic. How might mitotic events cause such
mosaicism, and what parameter(s) would likely determine the percentages and
distributions of X0 cells?