Ok thanks. Also for this question
Cystic fibrosis is an autosomal recessive trait that affects many parts of the body particularly the lungs and other organs. Parents who show none of the characteristics of cystic fibrosis have an affected child. The chance that there next child will be phenotypically normal is
A three in four
B one in four
C one in two
D zero
Why is the answer for this question A
Before I answered this I answered this in a rush but since I’m not in a rush let me actually give you a concise answer.
So CF is an autosomal recessive disease; thus, an individual must express both recessive genes in order to express the diseased phenotype. Individuals who do not express both diseased alleles will be phenotypically normal.
Let F = normal gene and f = CF gene. Individuals who express FF or Ff genotypes will not have CF, but they will have CF if they have the ff genotype because the gene is autosomal recessive.
Father genotype = Ff
Mother genotype = Ff
The parents must have the Ff genotype because the question stem indicates that they do not show the characteristics of CF but do have an affected child. This finding indicates that the parents must be carriers of the CF gene, as carriers are phenotypically normal but are capable of producing children with the diseased phenotype.
Next, create a Punnett square to determine the potential genotypes of the offspring:
Ff × Ff = FF + 2Ff + ff
Thus, there is a 1/4 = 25% probability that the parents will have a child with CF; the Punnett square shows that 1/4 of the children have the ff genotype. In contrast, there is a 3/4 = 75% chance that the parents will have a phenotypically normal child (3/4 do NOT have the ff genotype).