In a family where one parent is heterozygous for Huntington's disease (an autosomal dominant disorder) and the other parent is unaffected, what is the percent probability that exactly four out of seven children will inherit Huntington's disease?

• A. 50.0%
• B. 42.9%
• C. 27.1%
• D. 12.5%

Answer: (C)

Verified step by step guidance

1. Identify the probability that a single child inherits Huntington's disease from a heterozygous affected parent. Since Huntington's is autosomal dominant, the affected parent has one dominant allele (H) and one recessive allele (h), so the chance of passing the disease allele to a child is 50%, or \(p = 0.5\).
2. Recognize that the problem involves a binomial probability scenario, where each child independently has a probability \(p = 0.5\) of inheriting the disease, and \(q = 1 - p = 0.5\) of not inheriting it.
3. Set the parameters for the binomial distribution: number of trials \(n = 7\) (children), number of successes \(k = 4\) (children with the disease), probability of success \(p = 0.5\).
4. Write the binomial probability formula to find the probability of exactly \(k\) successes in \(n\) trials: \(P(X = k) = \binom{n}{k} \times p^{k} \times q^{n-k}\) where \(\binom{n}{k}\) is the binomial coefficient calculated as \(\frac{n!}{k!(n-k)!}\).
5. Calculate the binomial coefficient \(\binom{7}{4}\), then compute \(p^{4}\) and \(q^{3}\), and multiply all these values together to find the probability that exactly four out of seven children inherit Huntington's disease.