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Genetics, healthy carrier information

Hello, I ask myself this question: is it possible to know with the guthrie test if a baby is a healthy carrier of a gene causing later cystic fibrosis for his children and in this case if the families are warned. Thank you for your reply.
Best regards.

The Guthrie test is used for routine neonatal screening (DNS) of phenylketonuria. By extension this name is used (improperly) for all the neonatal screening tests of a genetic disease carried out on a drop of blood of the newborn.

In France, and in many other countries, the test used for the neonatal screening of cystic fibrosis (CF) involves 2 steps. The first step is the assay of immunoreactive trypsinogen (TIR). The second step is performed only if the TIR is greater than the threshold value: it consists of a search for a mutation of the cystic fibrosis gene (the CFTR gene) with a kit testing the most frequent mutations of this gene. A decree of law (2013-527) made the information of the parents compulsory before the genetic test was carried out:

"The person concerned is informed before the examination of his genetic characteristics of the obligation, if a serious genetic anomaly is diagnosed, to inform the members of his / her family who are potentially affected as soon as prevention or treatment measures can be proposed.”

This information is contained in the consent form for genetic testing that is presented to parents for signature when a newborn blood sample is collected for cystic fibrosis screening. You will find more information on the obligation and the information methods of the members of the family by clicking on the following link:

Https:// Inform-my-family [link in French]

Cystic fibrosis is considered as a serious genetic disorder even though its prognosis has improved considerably in recent decades. The so-called autosomal recessive transmission appears only when the child is the carrier of two CF causing mutations, one transmitted by the father and the other by the mother. In the French population, about one person in 33 has a single mutation in the CF gene (the CFTR gene) transmitted by one of its two parents: the person is said to be "heterozygous" for the CFTR gene or more simply "healthy carrier" but does not suffer from any signs of CF. These healthy adult carriers can however transmit CF to their offspring if their spouse is also a healthy carrier of a CFTR gene mutation. The frequency of these couples at risk of transmitting CF in the French population is of the order of 1 in 1000 (1/33 * 1/33 = 1/1089). For these couples, the risk that each parent transmits their CFTR gene mutation to their child is, at each pregnancy, of 1/4 (ie a frequency thus calculated of the disease of 1/1000 * 1/4 = 1/4000 births).

This has led some regions of the world where the disease is particularly common in offering voluntary screening for healthy carriers of reproductive age.

Hoping to have answered your question.
Best regards,
Gilles RAULT, CRCM of Roscoff
Marie-Piere AUDREZET, Molecular Genetics, Brest University Hospital