Question

Over 1,800 mutations have been described within the CFTR gene. These mutations can be readily detected using PCR and DNA sequencing technologies. Prenatal genetic testing is available to identify 'carriers', and thus detect the risk of CFTR-related disorders. If a pregnant mother was tested and tested negative for any known disease-causing CFTR mutations, can it be assumed that the child born will be wild-type with respect to the CFTR phenotype? Explain.
a. No. If the father is a carrier and passes along a mutant allele to the child, the child will develop cystic fibrosis.
b. It depends on the genotype of the father.
c. Yes. Since this is a recessive disorder, even if the father is a carrier, the mother will contribute a wt allele, ensuring the child is wt.

Answer 1

Answer:

The correct answer is - option C.

Explanation:

CFTR is a short form of cystic fibrosis transmembrane conductance regulator protein which is caused by a recessive mutation in this protein. This mutation results in the accumulation of mucus in secretory organs.

As the disease is a recessive disease, for the mutant phenotype to expressed both copies of the allele required to be recessive.

The mother is negative for the CFTR. Hence, she will not transfer the mutant allele to her child So, the child would have a wild-type phenotype even if the father having two copies of recessive allele.

Thus, the correct answer is - option C.