Question

A young couple went to see a genetic counselor because each had a sibling affected with cystic fibrosis. (Cystic fibrosis is a recessive disease, and neither member of the couple nor any of their four parents is affected.) a. What is the probability that the female of this couple is a carrier? b. What are the chances that their child will be affected with cystic fibrosis? c. What is the probability that their child will be a carrier of the cystic fibrosis mutation?

Answer 1

Answer:

a) The correct answer is 1/2 or 0.5

Cystic fibrosis is a recessive disorder and her sibling has a disease but none of the parents had it. Thus, each of the parents must be the carrier of the disease, that is, both of them must be heterzygous.

So, if we cross Cc x Cc, it results in the genotype ratio of 1 (CC) : 2 (Cc) : 1 (cc)

Thus, 2 out of 4 would be the carrier of the disease.

b) 1/16

The probability of female to the carrier is 1/2

Same is the case with male partner due to the probability that he would be the carrier is 1/2

If both are carrier then, there is only 1/4 chances that the offspring will have the disease (as per cross shown above).

Now, the probability that the offspring will have the disease = product of all the above probabilities. (The probability of occurrence of two independent events is equal to the product of their individual probabilities).

So, $\frac{1}{2}x\frac{1}{2}x\frac{1}{4}$ = $\frac{1}{16}$

c) 1/8

The probability of female to be the carrier is 1/2

The probability male to be the carrier is 1/2

If both are carrier then, there is only 2/4 chances that the offspring will be the carrier of the disease (as per cross shown above).

Now, the probability that the offspring will be the carrier = product of all the above probabilities. (The probability of occurrence of two independent events is equal to the product of their individual probabilities).

$\frac{1}{2}x\frac{1}{2}x\frac{2}{4}$ = $\frac{1}{8}$