Question

Individuals III-3 and III-4 are expecting their first child when they become aware that they both have a family history of this recessive condition. As their genetic counselor, you can calculate the probability that they are carriers and that their child will be affected with the condition.

1.) The probability that III- 3 is a carrier (Rr) =

2.) The probability that III - 4 is a carrier (Rr) =

3.) The probability that IV - 1 will be affected (rr) =


Options are 1/4, 1/2, 1/16, 1/3, 3/4, 2/3, 1/12, 1/6

Answer 1

The genealogy tree which is missing from the question is in the attachment.

Answer: 1) P1 =
`(1)/(2)`; 2) P2 =
`(1)/(2)`; 3)P3 =
`(1)/(16)`

Step-by-step explanation: 1) According to the genealogy tree, the individual III-3 has a P1 = 1/2, because each of the parents are heterozygous for the condition, since one of the ofspring is affected.

2) For individual III-4, one of the parents is heterozygous for the condition, so the probability of being a carrier is P2 = 1/2.

3) Now, for individual IV-1 be affected, the parents has to be carrier. So, this probability, depends on the probability of each parent being heterozygous and that the individual will be carrying the condition, which means:

P3 = 1/2 · 1/2 · 1/4 = 1/16.

The probability of having the condition is 1/16 or 6.25%.