Respuesta :
The reason this is so different is because those that are carriers (1 in 20) are heterozygous and to have a child with CF, they have to have a child with another carrier and then have the unlucky occurrence that their child will end up homozygous recessive.
So, 2 parents are Cc
C= normal CFTR (the gene at question here - Cystic Fibrosis Transmembrane Receptor)
c = mutated CFTR
Cross: Cc x Cc
You get: CC (1); Cc (2) and cc (1)....there is only a 0.25 chance that each time you have a child that it will be diseased (cc). Highest chance that the child is a carrier (Cc - 0.50) and similar low chance to no longer carry the recessive damaged allele (CC - 0.25)
So, 2 parents are Cc
C= normal CFTR (the gene at question here - Cystic Fibrosis Transmembrane Receptor)
c = mutated CFTR
Cross: Cc x Cc
You get: CC (1); Cc (2) and cc (1)....there is only a 0.25 chance that each time you have a child that it will be diseased (cc). Highest chance that the child is a carrier (Cc - 0.50) and similar low chance to no longer carry the recessive damaged allele (CC - 0.25)
