In human females one X chromosome is inactivated forming a Barr Body. Then how is it that haemophilia is dominant?

Suppose a female has one normal X chromosome and one chromosome with the haemophilia gene. Now suppose that the normal X chromosome is inactivated — will the female show haemophilia?


1 Answer 1


X-linked Hemophilia is caused by the lack of production of the clotting factors VIII or IX. It is not inherited dominantly, but instead inherited in an X-linked recessive manner.

These proteins are produced in multiple regions around the body. In a female heterozygous for the gene coding for Factor VIII or IX, X-inactivation would randomly prevent one of the chromosomes from expressing its allele. Therefore, the production of active Factor VIII or IX would only occur in half the cells which have the faulty allele inactivated.

However, half the production of the relevant factor results in a close to normal phenotype, as the hemophilic phenotype would not manifest unless the clotting factor activity levels are very low.

This Medscape article on hemophilia states that:

Normal values for FVIII assays are 50-150%. Values in hemophilia are as follows:

Mild: >5%

Moderate: 1-5%

Severe: < 1%

Therefore, despite X-inactivation reducing the production of clotting factors by 50%, the heterozygous female still displays the healthy, non-hemophilic phenotype.

While the clotting factor concentration can only be reduced to approximately 50% of normal, it is still possible for heterozygotes to suffer from mild hemophilia. This is because other genetic factors may also affect the hemophilia phenotype.

For example, the gene encoding FVIII may have point mutations which reduce the effectiveness of the protease, therefore reducing activity levels below the expected 50%. Mutations in regulatory elements or interacting proteins may also result in hemophilia due to the regulatory elements downregulating the expression or activity of the protein.

Approximately 40% of cases of severe FVIII deficiency arise from a large inversion that disrupts the FVIII gene. Deletions, insertions, and point mutations account for the remaining 50-60% of the F8C defects that cause hemophilia A.

Low FVIII levels may arise from defects outside the FVIII gene, as in type IIN von Willebrand disease, in which the molecular defect resides in the FVIII-binding domain of von Willebrand factor.

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    $\begingroup$ Note that X-inactivation is a random process. Therefore not all indivduals will have 50% cells lacking the healthy allele (as you write). This is a process known as skewing. One heterozygotic could have no cells with the faulty allele, all cells with the faulty, or any other intermediate form (this follows a Gaussian distribution in the population). Thus heterozygotes display asymptomatic - mild - severe phenotypes. $\endgroup$
    – Wolgast
    Apr 23, 2015 at 11:38
  • $\begingroup$ @Wolgast While this is true and the percentage of cells with inactivation of the faulty allele does in fact follow a Gaussian distribution, the law of large numbers makes it extremely unlikely that the deviation from 50% would be very large. Answer was updated to account for these factors. $\endgroup$
    – March Ho
    Apr 23, 2015 at 11:47
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    $\begingroup$ @Wolgast. Also not all genes on Barr bodies are inactivated. There are some regions where you will see a small amount of expression. As March Ho pointed out, hemophilia appears to be a dosage dependent disease and as long as enough Megakaryocytes are expressing the functional allele, then the disease will be masked in heterozygous females. $\endgroup$
    – AMR
    Sep 6, 2015 at 15:51

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