diagnosis exclusion

Question

Hello,
I would like to know if there is any test that would exclude for 100% the existence of the disease. For example, a genetic test done for one parent, if it is negative, it is sure that baby can not have the disease. If there is a test please tell me witch one it is. Thanks

Answer

Hello,
Cystic fibrosis or mucoviscidosis is an autosomal recessive genetic disease that occurs in a child born of parents each of them being carrier of a CF mutation. Only if both parents have a genetic mutation and if the child inherits from both parents the mutations, the child will be ill. Positive diagnosis requires the presence of typical clinical signs and a positive sweat test. There are also so-called atypical cases, who don’t have the typical clinical picture, and/or the sweat tests values are in normal/borderline range, or the genetic test is negative (any mutation detected). The genetic test can detect a limited number of mutations, for example 27, 35 or 43 mutations, depending on the laboratory kit used; until now more than 1800 mutations were discovered, many of them are very rare and can be detected in specialized centers. There is no test that definitely exclude the existence of the disease. If one parent does not have a specific mutation for cystic fibrosis, it is not certain that he does not have a rare mutation, although this is a rare possibility.
Prenatal diagnosis (genetic test of amniotic fluid obtained by amniocentesis) can detect the presence of disease from week 14-16 of pregnancy. Even if both parents are carriers there is only 25% likely chance for the child to be ill, and the same percentage chance that the child will be perfectly healthy and 50% chance to be only carrier of a mutation.
If you have any questions or concerns, please contact us.
Sincerely , dr.Ioana Ciuca

03.11.2011