Cystic Fibrosis (CF) Is Caused by a Recessive Allele. a Child Has CF, Even Though Neither of His Parents.
Tuesday, June 1st, 2010 at
3:31 am
Nikki asked:
…….has CF. If this couple has another child, what is the probability he or she will NOT have CF?
a. 3/4
b. 2/3
c. 2/4
d. 1/3
e. 1/4
Tagged with: Cf • Cystic Fibrosis • Recessive Allele
Filed under: Cystic Fibrosis
Like this post? Subscribe to my RSS feed and get loads more!
a. 3/4
The parents are heterozygous for the CF allele – their genotypes will both be Cc (with ‘c’ denoting the recessive disease causing allele).
Genotypes Cc Cc
Gametes C c C c
Offspring CC Cc cC cc
Since the parents are not affected, but had an affected, child, then the parents are both carriers (Cfcf). So, their possible offspring are:
Cfcf x Cfcf
offspring
CfCf, normal, non-carrier
Cfcf, normal, carrier
Cfcf, normal, carrier
cfcf, cystic fibrosis
So, as stated by the others, the probability of them having a child that does not have cystic fibrosis is 3/4.
3/4
If two carriers of the mutated CF gene have children then there is:
• A one in four chance that their baby will have CF
• A one in four chance that their baby will not have CF or carry a CFTR mutated gene
• A two in four chance that the baby will not have CF but will carry one CFTR mutated gene
The risk to other relatives depend on who is a carrier eg the risk to the offspring of an unaffected sibling of an affected individual is (2/3 x 1/25 x 1/4 = 1/150). 2/3 is the risk of an unaffected sibling being a carrier, 1/25 is the population carrier risk and a ¼ chance of parental carriers both handing on their mutated CFTR gene.