cystic fibrosis is caused by a recessive allele. A child has cystic fibrosis, even though neither of his parents has cystic fibrosis. If this couple has another child, what is the probability he or she will NOT have cystic fibrosis ?

if both parents are normal, then they must be heterozygous for the cystic fibrosis allele. Use a Punnett Square.

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To determine the probability that the next child will not have cystic fibrosis, we need to analyze the inheritance pattern of the recessive allele causing the condition.

Cystic fibrosis is an autosomal recessive disorder, meaning that it is caused by the presence of two copies of the recessive allele—one inherited from each parent. In this case, since neither of the parents has cystic fibrosis, we can infer that they are both carriers of the recessive allele. Carriers have one copy of the recessive allele but do not exhibit the disorder because the presence of the dominant allele masks its effects.

Let's denote the dominant allele as "C," representing a normal gene, and the recessive allele as "c," representing the gene responsible for cystic fibrosis. Since both parents do not have cystic fibrosis but are carriers, we can represent their genotypes as:

Parent 1: Cc
Parent 2: Cc

Now, let's determine the possible genotypes of the children:

C c
------------------
C | CC Cc
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c | Cc cc

There are four possible combinations of alleles between the parents: CC, Cc, Cc, cc. However, only two of these combinations (Cc and Cc) would result in a child without cystic fibrosis, as they would have one dominant allele (C) that can mask the effects of the recessive allele (c).

Therefore, the probability that their next child will not have cystic fibrosis is 2 out of 4, which simplifies to a 50% chance or a 1 in 2 probability.

It is important to note that this calculation assumes that the parents' carrier status is accurate, and no other genetic factors are involved in the manifestation of cystic fibrosis. Genetic counseling and testing could provide more accurate probabilities.