7T and 9T genetic carrier?

Question

Dear expert team,

I received a normal result from a gene test. The assessment states “heterozygous for 7T and 9T (normal alleles).” Does this mean I am genetically healthy or a genetic carrier for CF?

Many thanks for your help.

Answer

Dear questioner,

Short answer:

Yes, this is definitely a normal result (i.e., both CFTR genes are genetically healthy, a non-CF result) – the CFTR allele associated with CF in this context is called “5T,” which you do not carry on either of your two CFTR genes.

Long answer:

The actual protein CFTR is read from the messenger molecule “mRNA,” which contains more information than is necessary to put together the actual protein building blocks.

This information exists a fragmented way, roughly like this: informationforproteinCFTRnumber1—connector—informationforproteinCFTRnumber2—etc. For the CFTR protein, there is a total of 27 such information fragments. In one of these connectors (more precisely: in the 8th connector), there is a number of T building blocks back-to-back (T stands for “thymidine” here), which can be 9T (i.e., TTTTTTTTT), 7T (i.e., TTTTTTT) or 5T (i.e., TTTTT) long in humans.

The number of T building blocks determines how well the cell can process the information from the messenger molecule: the fewer Ts, the less functional CFTR will emerge from the gene – it is more difficult for the cell to correctly produce connectors if there are fewer Ts. Therefore, 5T is considered a risk gene that can involve CF; 7T and 9T, on the other hand, are completely normal for the CFTR gene and are present in all healthy people as well.

“Heterozygous 7T/9T” means that one of your two CFTR genes (i.e. for instance the one inherited from your father) was found with 7T and the other one (i.e. for instance the one inherited from your mother) with 9T. However, both are completely normal – nothing is pointing to CF here.

Kind regards,
Frauke Stanke

27.11.2013