Which of the two X-chromosomes remains active in female individuals, is determined at the early stages of development. It was observed by Lyon in 1961, that each of the paternal and maternal X-chromosomes has a chance to become inactive. In other words, the inactivation of a X-chromosome is a random phenomenon. This could be shown by utilizing an individual which is heterozygous for coat colour in mice. It was shown that such heterozygous mice had a variegated phenotype, because while in some cells the chromosome carrying the normal allele was inactivated giving the mutant phenotype, in other cells the chromosome carrying the mutant allele was inactivated giving these cells the wild type phenotype. When the gene for coat colour was located on an autosome, such variegation was not observed suggesting that inactivation takes place only in X-chromosome and not in autosomes. Variegation was also absent in mice having XO genotype, since no inactivation takes place in such a case.
There is no general rule for the time at which the inactivation will take place leading to formation of facultative heterochromatin. In mammals, the inactivation usually takes place in early embryogenesis. In mouse, late replicating heterochromatic X-chromosomes can not be detected until late blastocyst. Similar situation has also been observed in certain other mammal cells. In mammals, it has also been observed that the condensed 'Barr-body', characteristic of somatic cells, is absent from the female pre-meiotic cells.
The process of inactivation, therefore, seems to be reversed in the germ cells, so that all female gametes or eggs will carry an active X chromosome. The fate of the X-chromosome carried in an egg will depend upon whether it goes to a male individual or to a female individual. In a female individual they may again have a chance of inactivation while in male they will have no chance of becoming inactive.
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