Cystic fibrosis is an autosomal recessive genetic disorder. What are the chances that the child would have the disease if any one of the parent (either mother or father)is a carrier of the faulty cystic fibrosis gene (Cc)?

The correct option is 0 per cent.

To understand why the chance of the child having cystic fibrosis is 0%, we can analyze the genetics of the disease step-by-step:

1. Understanding Autosomal Recessive Inheritance
Cystic fibrosis is an autosomal recessive disorder. This means that a person must inherit two copies of the faulty gene (one from each parent) to actually develop the disease. Let us denote:
- $C$ as the normal allele (dominant)
- $c$ as the faulty allele (recessive)

For a child to have cystic fibrosis, their genotype must be homozygous recessive ($cc$).

2. Analyzing the Parents' Genotypes
The question specifies that only one of the parents (either the mother or the father) is a carrier of the faulty gene. A carrier is heterozygous, meaning they have one normal allele and one faulty allele:
- Carrier Parent Genotype: $Cc$
Because the other parent is not mentioned as a carrier or affected, they are assumed to have two normal alleles (homozygous dominant):
- Non-carrier Parent Genotype: $CC$

3. Determining the Offspring Genotypes Using a Punnett Square
When we cross a carrier parent ($Cc$) with a normal parent ($CC$), the possible genetic combinations for their children are:
- Parent 1 ($Cc$) can contribute either a $C$ or a $c$ allele.
- Parent 2 ($CC$) can only contribute a $C$ allele.

This results in the following possible offspring genotypes:
- 50% chance of inheriting $CC$ (completely normal, non-carrier)
- 50% chance of inheriting $Cc$ (unaffected carrier)

Conclusion
Since none of the offspring can inherit two copies of the faulty allele ($cc$), there is a 0% chance that the child will have the disease. Therefore, the chance that the child would have the disease is 0 per cent.