Article Text
Abstract
Pseudoxanthoma elasticum (PXE) is an inherited systemic disease of connective tissue primarily affecting the skin, retina, and cardiovascular system. It is characterised pathologically by elastic fibre mineralisation and fragmentation (so called “elastorrhexia”), and clinically by high heterogeneity in age of onset and the extent and severity of organ system involvement. PXE was recently associated with mutations in the ABCC6 (ATP binding cassette subtype C number 6) gene. At least one ABCC6 mutation is found in about 80% of patients. These mutations are identifiable in most of the 31 ABCC6 exons and consist of missense, nonsense, frameshift mutations, or large deletions. No correlation between the nature or location of the mutations and phenotype severity has yet been established. Recent findings support exclusive recessive inheritance. The proposed prevalence of PXE is 1/25 000, but this is probably an underestimate. ABCC6 encodes the protein ABCC6 (also known as MRP6), a member of the large ATP dependent transmembrane transporter family that is expressed predominantly in the liver and kidneys, and only to a lesser extent in tissues affected by PXE. The physiological substrates of ABCC6 remain to be determined, but the current hypothesis is that PXE should be considered to be a metabolic disease with undetermined circulating molecules interacting with the synthesis, turnover, or maintenance of elastic fibres.
- ABCC6, ATP binding cassette, subfamily C, member 6
- ECM, extracellular matrix
- MRP6, multidrug resistance associated protein 6
- NBF, nucleotide binding fold
- PXE, pseudoxanthoma elasticum
- ADCC6
- MRP6
- PXE
- pseudoxanthoma elasticum
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Footnotes
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↵* Hardy–Weinberg law: p2+2pq+q2 = 1, with p and q being the frequency of the wild type and the mutated alleles respectively, p2 the frequency of homozygosity for the wild type allele, 2pq the frequency of heterozygosity, and q2 the frequency of the disease. Here, q2 = 1/25 000, q = √(1/25 000) = 0.0063; p = 1−q = 1−0.0063 = 0.9937 Thus the frequency of heterozygous carriers (2pq) can be calculated as 2×0.9937×0.0063 ∼ 0.0125 (1.25%).
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↵† Here the frequency of heterozygous carriers (2pq) = 0.03 (3%). Thus, considering p ∼ 1, q = 0.03/2 = 0.015, and the frequency of the disease (q2) = 0.0152 ∼ 2.25×10−4 ∼ 1/4450.
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↵* NC and LM contributed equally to this paper
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Published Online First 13 May 2005
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Competing interests: none declared