CF & heterozygous?

Question

Dear reader,

After reading your articles on CF (thank you for that) I still have one question.

I was diagnosed with F508del and R117H, heterozygous, polymorphism IVS8-Tn: 7T/9T.
I am confused about the part “homozygous and heterozygous” and in what grade it can cause illness. I understand that heterozygous means in my case that 1 gene is found with 2 mutations. According to the articles I have been reading, it means that I have no risk on illness. However, I have no vas deferens and moreover my sweat is very salty. In other words: that seems to me real disease. I started with my partner an IVF treatment but I have doubts. My partner has been tested and was considered “clean”. We do not want to have children with a serious disease. So my question is: knowing this background, is there a chance that our children will have a severe kind of this disease, and if yes, how big is that chance?

Answer

Everyone has 2 CFTR genes. One CFTR gene inherited from the father, the other CFTR gene inherited from the mother. If you only have a CF mutation on one single gene (1 CFTR mutant), and the other CFTR gene is normal, then you are heterozygous for this mutation and a CF carrier but not ill. If you have a CF mutation on both genes then you have CF. If you have on both CFTR genes the same mutation you are homozygous for this mutation (and CF). If you have on both CFTR genes different mutations then you are 'compound heterozygous' for those 2 mutations (and you have CF).

You are "compound heterozygous" for the F508del and R117H mutations. On one CFTR gene you carry the F508del mutation, and on the other CFTR gene you carry the R117H mutation. The severity of a mutation varies from mutation to mutation (and even from individual to individual). The R117H mutation is a special mutation. It can cause a mild kind of CF, some symptoms of CF (for example the fact that you believe you have a high salt level in your sweat may explain this). It can also lead to infertility in men, but it usually does not imply severe disease. The factors that determine whether R117H provokes the disease or not are not fully understood. In any case the T tract influences disease severity: mainly subjects who carry T5 on the R117H allele have a higher risk of developing lung disease. Since you were tested T7/T9, this explains why you have no lung symptoms.

What is the risk that you and your partner will have a CF child. A risk can never be excluded, and can only be reduced. The risk is also determined by the sensitivity of the genetic test, namely the type of test used to screen your partner. There are more than 1900 mutations known in the CFTR gene, of which about 30 are more frequent. The majority of genetic tests only analyze the presence or absence of these more frequent mutations, and their combined frequency detects approximately 90% of the mutant genes found in patients with CF.

If your partner was tested for the 30 most frequent mutations found in 90% of all mutant genes of CF patients, then a negative result not necessarily means that a CF mutation is not present. Your partner can either not be a CF carrier, or she can be a carrier of a CF mutation that was not present in the test. 90% sensitivity means that when the test is done on 100 CF carriers, 90% CF carriers will be identified and 10% not.

For your family the worst case would be the following. The chance that you will give your F508del CFTR gene to your child is 50% since you have 2 CFTR genes and you can only pass through 1 CFTR gene to your child. 1 out of 25 individuals in the general population are CF carriers. The chance that your partner still is a CF carrier even with a negative result is 1/25 x 10/100 = 1/250. In that case, there is 1 chance out of 2 that your partner will pass through her mutant CFTR gene to his child. The combined probability of a CF child for you and your partner is (½) x (½ x 1/250) = 1/1000.

To put this risk in perspective following comparison:

The probability of each couple to have a child with cystic fibrosis in the general population without tests is (½ x1/25) x (½ x1/25) = 1/2500. The reason that you have a higher risk is because you already are sure that you have a severe CF mutation.

If your partner was not tested, the risk would be: (½) x (½ x1/25) = 1/100. The test of your partner did decrease the risk with a factor of 10.

If your partner was tested positive for a mutation present in the test, than the risk of a CF child would have been ½ x ½ = ¼.

In order to reduce the risk, there are expensive CF tests on the market that can detect 99% of all mutant CFTR genes in CF patients. But a test with 100% certainty does not exist in medicine. The risk of a CF child would then be (½) x (½ x 1/25 x 1/100) = 1/10000.
We understand this is really complex information. So please also discuss it with your doctor to be sure you understand it correctly.

With Regards,

Harry Cuppens

24.01.2013