Question

Huntington's disease is an inherited autosomal dominant disorder that can affect both men and women. Imagine a couple has had seven children, and later in life, the husband develops Huntington's disease. He is tested and it is discovered he is heterozygous for the disease allele, Hh. The wife is also genetically tested for the Huntington's disease allele, and her test results show she is unaffected, hh.

What is the percent probability that three of the seven children will have Huntington's disease?

Answer 1

The probability that three of the seven children will inherit Huntington's disease from a heterozygous affected father (Hh) and an unaffected mother (hh) is approximately 27.34%.

Step-by-step explanation:

The question of what is the percent probability that three of the seven children will have Huntington's disease involves calculating the probability of an autosomal dominant disorder being passed on from a heterozygous parent (Hh) to his offspring, given that the other parent is homozygous recessive (hh) for that trait. This scenario can be analyzed using a basic Punnett square, which shows that each child has a 50% chance of inheriting the disease allele (H) and therefore developing the disease. To solve this probability question, we use the binomial probability formula, which in this case is P(X=k) = C(n, k) * p^k * (1-p)^(n-k), where C(n, k) is the binomial coefficient representing the number of ways to choose k successes out of n trials, p is the probability of success on an individual trial, and n is the number of trials.

The formula becomes P(3) = C(7, 3) * (0.5)^3 * (0.5)^(7-3) to find the specific probability of three children inheriting Huntington's disease. Calculating this gives us P(3) = 35 * (0.5)^3 * (0.5)^4 = 35 * 1/8 * 1/16 = 35/128. Therefore, the percentage probability that three of the seven children will have Huntington's disease is (35/128) * 100%, which approximates to 27.34%.