Dentinogenesis imperfecta is a tooth disorder involving the production of dentin sialophosphoprotein, a bone-like component of the protective middle layer of teeth. The trait is inherited as an autosomal dominant allele located on chromosome 4 in humans and occurs in about 1 in 6000 to 8000 people. Assume that a man with dentinogenesis imperfecta, whose father had the disease but whose mother had normal teeth, married a woman with normal teeth. They have six children. What is the probability that their first child will be a male with dentinogenesis imperfecta? What is the probability that three of their six chil- dren will have the disease?

The given problem states that dentinogenesis imperfecta is inherited as an autosomal dominant allele. This means that an individual only needs one copy of the affected allele, paired with a normal allele, to express the condition. Here, the father has dentinogenesis imperfecta, so his genotype will be Dd (D represents the dominant allele for dentinogenesis imperfecta and d represents the normal, recessive allele). The mother has normal teeth, implying her genotype is dd.

To get a male child with dentinogenesis imperfecta, we first need to find out the probability of having a male child, which is 1/2. Then, we need to calculate the probability of the male child having dentinogenesis imperfecta by creating a Punnett square to show the possible genotypes for their children: D d d \|/ \|/ /dd Dd dd The probability of having a child with dentinogenesis imperfecta (Dd genotype) is 1/2. Combining these probabilities, the chance of having a male child with dentinogenesis imperfecta is: 1/2 (probability of a male child) * 1/2 (probability of having dentinogenesis imperfecta) = 1/4

The binomial probability formula is: P(k) = C(n, k) * p^k * (1-p)^{n-k} Where: - n is the total number of children - k is the number of children with dentinogenesis imperfecta - p is the probability of each child having dentinogenesis imperfecta - C(n, k) is the binomial coefficient, or the number of ways to choose k successes from n trials In this scenario, n = 6, k = 3, and p = 1/2 (since each child has a 1/2 chance of inheriting the Dd genotype from their parents). C(6,3) = 6! / (3! * (6-3)!) = 20 Plugging these values into the binomial probability formula, we get: P(3) = 20 * (1/2)^3 * (1/2)^{6-3} = 20 * (1/8) * (1/8) = 20/64 = 5/16 So, the probability of three out of their six children having dentinogenesis imperfecta is 5/16.