1. Cystic fibrosis is a recessive disorder. This means an individual must inherit two copies of the recessive allele (be homozygous recessive) to have the condition.
2. The probability of inheriting one CF allele is 1 in 50 (or 1/50).
3. To have the disorder, a person must inherit the CF allele from their mother AND from their father.
4. The probability of this happening is the product of the individual probabilities: (1/50) × (1/50) = 1/2500.
Therefore, 1 in 2500 people are expected to have cystic fibrosis.
This question involves understanding both recessive inheritance and probability.
1. **Recessive Inheritance:** Cystic fibrosis is a recessive genetic disorder. This is a crucial piece of information. It means that an individual must have two copies of the faulty CFTR allele to have the disease. Their genotype would be, for example, 'ff'. Individuals with one faulty allele and one normal allele ('Ff') are carriers but do not have the disease themselves.
2. **Allele Frequency:** The question states that the frequency of the cystic fibrosis allele ('f') in the population is 1 in 50, or a probability of 1/50.
3. **Probability Calculation:** For a child to be born with cystic fibrosis, they must inherit one CF allele from their mother and one CF allele from their father. These are independent events. The probability of inheriting the allele from one parent is 1/50. The probability of inheriting it from the other parent is also 1/50. To find the probability of both events happening, we multiply their individual probabilities:
P(child has CF) = P(inherit from mother) × P(inherit from father)
P(child has CF) = (1/50) × (1/50)
P(child has CF) = 1/2500
This calculation, known as the Hardy-Weinberg principle for homozygous recessive individuals (q²), explains why the frequency of the disease (1 in 2500) is much lower than the frequency of the allele (1 in 50).
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