How did they discover that the child was ill ?
The story began when a 6 week-old boy developed an abnormal blistering sunburn after a short time outside (45 minutes during a 2-day period in early April without any sunscreen or hat).
The baby at the age of 6 weeks
His uncle at the same age
His parents immediately thought that he could be ill with XP. Indeed, they were aware of this illness because the mother’s brother was diagnosed with XP after a similar problem in infancy. This is why the mother knew what to do and began to protect her son from UV, using notably a UV meter to measure the level of UV in the child environment.
It was almost the same case with the baby's uncle, but his parents were not aware of the existence of this illness, so he was less protected in his first year. XP was suspected to be the cause at that time because of his extreme sun sensibility, his abnormal freckles, and a developmental delay.
At the age of 7 months, he has no skin damages.
At the age of 17 weeks, the baby was brought to the Children’s Memorial Hospital of Chicago but the findings of his skin and gross neurologic examinations at that time were unremarkable. The baby was diagnosed with XP later thanks to DNA testing, but the patient was sent to NIH (National Institute of Health) for further evaluation. Cultured fibroblasts from the patient were hypersensitive to the killing effects of UV.
Furthermore, host cell reactivation studies showed that his XPA gene was mutated, just like his uncle's. They were both heterozygous, with the same two mutations of the same gene (XPA) : a deletion of a T at position 288 (inherited from the mother) and another deletion of 5 bases (CTTAT) in the exon 3 (inherited from the father). These mutations can be detected thanks to PCR-based RFLP. If you have already read the previous pages of this blog, you may be wondering: "Why are they ill, if they are heterozygous? Don't they have to be homozygous to have XP?" You're right. But here's a rare case called compound heterozygosity. You're a compound heterozygote when you have two recessive alleles for the same gene, but with these two alleles being different from each other (for example, both alleles can be mutated but at different locations). And this particular case of heterozygosity might lead to a genetic illness, here XP.
Symptoms, and how to live with the illness?
The symptoms of these patients were quite usual among XP patients: they suffered from severe blistering sunburns after short exposures to sunlight.
The major difference was that these symptoms appeared earlier than they would have on most patients. Indeed, children usually begin to show xp symptoms after the age of 2 years, whereas the two patients studied here were sunburnt when they weren’t even two months old. Both also suffered from severe neurologic deterioration. For instance, the uncle, at the age of 20, was profoundly deaf, unable to speak, confined to a wheelchair, incontinent, and fed by a gastrostomy tube. As far as his nephew is concerned, he was only able to speak a few words at the age of 3, in order to communicate his needs.
Fortunately for the patient, his parents were very careful after the diagnosis. They quickly learned to cope with the disease and to live with it every day. As a matter of fact, they covered their windows at home and their cars with sun blocking film, and always checked the levels of UV in the environment with a UV meter. Moreover, when the child had to go outside, they applied sun blocking skin creams on him and made sure he was wearing long-sleeved shirts, long pants, and gloves. To protect his face, neck and ears, his mother designed a hood with a protective face shield made out of UV-blocking materials. With this caution, he could almost live as a non-XP child.
Prenatal diagnosis of xeroderma pigmentosum
There is roughly one person per million that suffers from XP. The rarity of the disease means that a diagnosis is not frequently required. Hence, robust diagnosis methods have not been developed yet. The potential market size is small, which inhibits expensive investments in new technological approaches. Furthermore, the diagnosis takes a long time (around 5 to 6 weeks).
Sequencing the wanted gene is the ultimate standard for positive identification of an XP patient, but this approach can only be applied effectively once the gene in question has been identified. Without this identification, it would require the sequencing of up to a total of 25 kb of coding sequence. It is possible, but it requires multiplex polymerase chain reaction amplification, independent cloning, sequencing and a lot of time. Also, the methods to identify the mutated gene require a panel of characterized XP cells to compare to a chosen cell line by cell-fusion techniques.
However, two methods of prenatal diagnosis have been successful: the unscheduled DNA synthesis (UDS) and the single cell gel electrophoresis assay (also known as ''comet assay''). UDS is the classic method for diagnosis. It requires 4 to 5 weeks before being able to conclude. The use of the alkaline comet assay is proposed as a simple repair test for early prenatal diagnosis.
The comet assay and the UDS tests are usually performed in parallel. The first one is a simple method for measuring DNA strand breaks in eukaryotic cells, whereas the second one measures the ability of a cell to perform global genomic nucleotide excision repair (NER).
In repair-proficient cells, DNA strands break due to the incision of UV-induced DNA damage, resulting in an increased migration of high molecular weight DNA in the comet assay. In incision repair-deficient XP cells, after post-UV incubation, migration does not occur, so normal fetal cells demonstrate repair capacity while XP ones don’t.
Nonetheless, comet assay method isn’t unerring, due to slightly sensitive XP cells coming from patients who have a very rare mutation.
Comet assay and Unscheduled DNA synthesis offer advantages in view of prenatal diagnosis, including the reduced number of cells required and the absence of need for radioactivity.
It’s also possible to calculate the probability for a child to be born to be ill. For instance, with the example above, knowing that his uncle is ill, the risk is equal to M * F * B :
- M the probability of the mother being a healthy carrier of a mutated allele. Knowing that she is healthy, and she is the sister of an XP patient, this probability is 2/3.
- F the probability of the father being a healthy carrier of a mutated allele in the same gene (not the same mutation (as the child is a compound heterozygote) but the mutation has to be on the same gene). This probability can be calculated by the Hardy-Weinberg equilibrium equation, which is x² + 2xy + y² = 1, with x the frequency of the normal allele (so x² is the probability of having 2 normal alleles), y the frequency of the mutated allele (so y² is the probability of being ill) and 2xy is the calculated risk of being a heterozygote for the mutant allele (so 2xy is what we are looking for : 2xy=F). In the US and Europe, the frequency of XP is 1 in a million, so y² = 1/10-6, and y = 1/1000. It’s possible to estimate that x² is equal to 1 (it’s actually equal to 999999/1000000), which means that x=1. Eventually, F = 2xy = 2*1*(1/1000) = 1/500.
- B the probability of two carriers with the same mutated gene (again, not necessarily the same mutation) giving birth to an XP child. This probability is trivially 1/4.
Therefore, the risk is (2/3)*(1/500)*(1/4) = 1/3000.
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