× H). Adding these two terms gives I = µ + ( × H). H is the chance a carrier inherits the mutation from a surviving, reproducing affected father (I × f) (where f is the fitness of hemophilia) plus the chance of a new paternal mutation (µ) plus the chance of a new maternal mutation (µ) plus the chance of inheriting it from a carrier mother ( × H). Adding these four terms gives H = (I × f) + µ + µ + ()H.
a. If hemophilia A has a fitness (f) of approximately 0.70, that is, hemophiliacs have approximately 70% as many offspring as do controls, then what is the incidence of affected males? of carrier females? (Answer in terms of multiples of the mutation rate.) If a woman has a son with an isolated case of hemophilia A, what is the risk that she is a carrier? What is the chance that her next son will be affected?
c. They have had three unaffected children and now wish to know their risk for having an affected child. Using Bayesian analysis to take into consideration that they have already had three unaffected children, calculate the chance that their next child will be affected.
6. A 30-year-old woman with myotonic dystrophy comes in for genetic counseling. Her son, aged 14 years, shows no symptoms, but she wishes to know whether he will be affected with this autosomal dominant condition later in life. Approximately half of individuals carrying the mutant gene are asymptomatic before the age of 14 years. What is the risk that the son will eventually develop myotonic dystrophy? Should you test the child for the expanded repeat in the gene for myotonic dystrophy?
7. A couple arrives in your clinic with their 7-month-old son, who has been moderately developmentally delayed from birth. The couple is contemplating having additional children, and you are asked whether this could be a genetic disorder.
8. You are addressing a Neurofibromatosis Association parents’ meeting. A severely affected woman, 32 years old, comments that she is not at risk for passing on the disorder because her parents are not affected, and her neurofibromatosis therefore is due to a new mutation. Comment.
9. The figure shows the family from Figure 16-6, but with additional information that the consultand III-2 has two unaffected sons. There are now seven possible scenarios to explain this pedigree. List the scenarios, and use them to calculate the carrier risk for individual III-2.