Genetic testing- it is not uncommon for the husband and wife couple to carry the gene for cystic fibrosis without heaving the disease themselves. suppose we select one of these couples at random according to the law of genetics, the probability that any child will develop cystic fibrosis is 0.25. (a) explain what this probability means

(b) if researchers randomly select 4 such couples, is one of these couples guaranteed to have a first child who develops cystic fibrosis? Explain your 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, =

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).

Step-by-step explanation:

the discriminant is 64 + 4*1*12 = 112

this is prime.

step-by-step explanation:

srry idk

step-by-step explanation: