Question

Suppose a birth defect has a recessive form of inheritance. In a study population, the recessive gene (a) initially has a prevalence of 25%. A subject has the birth defect if both maternal and paternal genes are of type a.
a.In the general population, what is the probability that an individual will have the birth defect, assuming that maternal and paternal genes are inherited independently.

A further study finds that after 10 generations (≈ 200 years) a lot of inbreeding has taken place in the population. Two subpopulations (populations A and B), consisting of 30% and 70% of the general population, respectively, have formed. Within population A, prevalence of the recessive gene is 40%, whereas in population B it is 10%.

b.Suppose that in 25% of marriages both people are from population A, in 65% both are from population B, and in 10% there is one partner from population A and one from population B.

What is the probability of a birth defect in the next generation?

c.Suppose that a baby is born with a birth defect, but the baby’s ancestry is unknown. What is the posterior probability that the baby will have

1. both parents from population A?

2. both parents from population B?

3. mixed ancestry (i.e. one parent from population A and one from population B)?

Answer 1

Answer:

Step-by-step explanation:

a) Maternal gene = 0.25; Paternal gene = 0.25

Probability having a birth defect = 0.25 X 0.25 = 0.625

b) P(A) = 0.30; P(B) = 0.70

P(A)*P(B) = 0.30 X 0.70 = 0.021

For marriage:

25%: P(A)*P(A)*0.25 = 0.09 X 0.25 = 0.0225

65%: P(B)*P(B)*0.65 = 0.049 X 0.65 = 0.03185

10%: P(A)*0.10 = 0.03 and P(B)*0.10 = 0.07

P(A)*P(B) = 0.03 X 0.07 = 0.0021

Probability of a defect birth in the next generation = 0.0225 + 0.03185 + 0.0021 = 0.05645

c) 1. P(A)*P(A) = 0.09

2. P(B)*P(B) = 0.049

3. P(A)*P(B) = 0.021