Women possess two X chromosomes. However, during development, when the embryo has about 32 - 64 cells, one of these chromosomes is randomly inactivated (in each cell) by an lncRNA named Xist. As a consequence, females are mosaics.

I was interested if this might affect phenotypes of X-linked genetic diseases, for example...

1) Is it possible for a woman who is heterozygote to color vision deficiency be "partially" color blind, considering that some of her cones will only possess the mutated allele?

2) Is it possible that a heterozygote female who carries a dominant allele for a X-linked disease be asymptomatic?

Any references on these issues are welcome.


Short answer
Skewed X-inactivation may lead to phenotypic manifestation of X-linked color blindness.

What an interesting question!

In a study by Jorgensen et al. (1992) two female identical twins that were heterozygotes for X-linked deuteranomaly were investigated. Phenotypically, deuteranomaly refers to anomalous trichromacy, i.e., reduced perception of a color. Note that deuteranopes are totally color blind. Deuteronomaly in the twins was associated with a defective gene derived from their father. Interestingly, while monozygotic twins are genetically identical, one of the twins was phenotypically deuteranomalous, while the other had normal color vision.

Analysis of skin cells of the color vision-defective twin revealed that almost all skin cells had the paternal abnormal X chromosome active, explaining her color-vision defect. In contrast, skin cells from her sister with normal color vision had predominantly the maternal X chromosome active. Hence, deuteranomaly in one of the twins could be explained by an extremely skewed X inactivation.

As pointed out by @mdperry, extreme skewed X-inactivation is rare, and most heterzygous females for color blindness (carriers) are indeed asymptomatic (Jordan & Mollon, 1993). Regarding your second question, all gene mutations leading to color blindness are recessive (NIH).

- Jordan & Mollon, Vis Res (1993); 33(1): 1495-1508
- Jorgensen et al. Am J Hum Genet (1992); 51: 291-8


Your theoretical understanding of the ramifications of X-chromosome inactivation is excellent (e.g., the two situations you described). However, it is important to consider the size of the resulting clones. At that early stage of embryogenesis only ~ half (or less) of those cells will go on to contribute to the fetus, the rest will form the placenta. Also at this stage each cell is likely totipotent, and capable of contributing to all of the different tissues, but eventually the daughters of these cells will become committed, or determined in their developmental lineage. Depending on how much mixing of cells occurs prior to determination, I think it unlikely that within one retina the cone cells would originate from more than a single clone--so I would think that either they all are + or the all are mutant. Since there is L-R symmetry between the two optic cups that will give rise to the developing eyes, I think it is more likely that one female could have one eye that is mutant and one eye that is +. Given that the prevalence of red-green color-blindness in males approaches 7 % of the population, we can infer that heterozygous females are very common, yet to the best of my knowledge there are not millions of females with partial color-blindness--they either have it (m/m), or they don't (+/+ & m/+). So I think a mosaic female with one eye color-blind and one eye normal must be exceedingly rare. Re: autosomal dominant diseases, it all depends on the tissue(s) and the affected developmental stage. The OMIM, Mendelian inheritance in Man database may have some discussion of this interesting point(?).

  • 1
    $\begingroup$ Although a nicely approached answer, it seems to be build from assumptions rather than facts. -1 With some references it would be an excellent answer $\endgroup$
    – AliceD
    Mar 26 '15 at 13:32

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