Question

Polydactyly is inherited as an autosomal dominant trait. Two individuals that are heterozygous for this trait marry and plan a family of five children. Assume a 1:1 sex ratio and that the gene is completely penetrant. What is the probability that the couple's five children will consist of at least 3 with polydactyly?

A. 0.395
B. 0.775
C. 0.237
D. 0.551
E. 0.895

Answer 1

When two heterozygous individuals with an autosomal dominant trait have children, there is a 25% chance that each child will be homozygous dominant (have polydactyly), a 50% chance that each child will be heterozygous (carrier), and a 25% chance that each child will be homozygous recessive (not have polydactyly). The probability of having at least 3 children with polydactyly can be calculated using the binomial probability formula, and in this case, it is approximately 0.237 (option C).

Step-by-step explanation:

In this case, both parents are heterozygous for the polydactyly trait, which is inherited as an autosomal dominant trait. When two heterozygous individuals with an autosomal dominant trait have children, there is a 25% chance that each child will be homozygous dominant (have polydactyly), a 50% chance that each child will be heterozygous (carrier), and a 25% chance that each child will be homozygous recessive (not have polydactyly).

Using the binomial probability formula, we can calculate the probability of having at least 3 children with polydactyly:

1. There are 5 children, so we add up the probabilities of having 3, 4, or 5 children with polydactyly: P(X ≥ 3) = P(X = 3) + P(X = 4) + P(X = 5)
2. For each case, we use the binomial probability formula: P(X = k) = (n choose k) × p^k × (1 - p)^(n-k), where n is the total number of children, k is the number of children with polydactyly, and p is the probability of a child having polydactyly (0.25).

After calculating these probabilities and adding them up, we find that the probability of having at least 3 children with polydactyly is approximately 0.237 (option C).