Imagine that you are a genetic counselor, and a couple planning to start a family comes to you for information. Charles was married once before, and he and his first wife had a child with cystic fibrosis. The brother of his current wife, Elaine, died of cystic fibrosis. What is the probability that Charles and Elaine will have a baby with cystic fibrosis? (Neither Charles nor Elaine has cystic fibrosis.)

The probability of the couple having a child with cystic fibrosis is 1/4

Explanation:

There is a huge chance both Charles and his current wife have heterogenous genes for cystic fibrosis. This is because Charles and his first wife must have contributed an allele each for the child to have cystic fibrosis. Also, for one of the current wife's siblings to have cystic fibrosis, it shows there parents both had a alleles for cystic fibrosis which could have been passed to her also. Thus, assuming the heterogenous gene for cystic fibrosis is Aa, the probability of them having a baby with cystic fibrosis will be determined by the crossing below.

Aa     ×      Aa

AA  Aa   Aa  aa

Where "A" is the normal allele and "a" is the allele for cystic fibrosis, the probability of the couple having a child with cystic fibrosis is 1/4.

a, b, c - sides of a triangle

a + b > c

a + c > b

b + c > a

we have a = 5m and b = 11m. substitute:

5 + 11 > c

16 > c → c < 16

5 + c > 11         subtract 5 from both sides

c > 6

11 + c > 5         subtract 11 from both sides

c > -6

c < 16 and c > 6 and c > -6

thereofre your answer is 6m < c < 16m