A question asked by decemberkitty02: If two parents are carriers of cystic fibrosis (and do not have the disorder), what is the probability?
If two parents are carriers of cystic fibrosis (and do not have the disorder), what is the probability that their first child will have CF and their second child will not?

I thought that it would 3/16s? Is this correct?

The No 1 answer:

Answer by Russ
Theyre both Aa, so its {AA,Aa,Aa,aa}, so P(having the disease) = 1/4, and P(not having the disease) = 3/4
Multiply the two and you get 3/16. So you are correct

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Technorati Tags: carriers, cystic, disorder, fibrosis, Parents, Probability

A question from Itri: If two people are carriers for cystic fibrosis, what is the probability that they will have a child with cy?
If two people are carriers for cystic fibrosis, what is the probability that they will have a child with cystic fibrosis?

The No 1 answer:

Answer by diendvo
This is a genetic question that requires a Punett square. Cystic fibrosis is a recessive trait, meaning you need 2 alleles for cystic fibrosis to display it phenotypically. So if you define “a” as the allele for cystic fibrosis, “A” is the normal allele. A carrier of cystic fibrosis will have the genotype: Aa. If you do a Punett square, you’ll discover that there’s a 1/4 chance the child will have the aa genotype and will have cystic fibrosis.

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Technorati Tags: carriers, child, cystic, fibrosis, People, Probability, they

A question from Ty: Probability help please!!!!!!!!!!!!!!!?
Genetic testing reveals that two newlyweds are both carriers of the recessive mutation that causes cystic fibrosis. As a consequence, any child produced by the marriage has a probablility of 1/4 of being afflicted with cystic fibrosis. If this couple has three children (none being identical siblings), what is the probablility that….
(a) none of the children will have cystic fibrosis?
(b) at least one child will have cystic fibrosis?
(c) one child will have cystic fibrosis and the other two will not?
(d) the last born child will have cystic fibrosis, given that you already know that the first two have cystic fibrosis?

The best answer:

Answer by Gerry
a) Since there is a (1/4) chance of a child getting cystic fibrosis, there is a (1 – (1/4)) = (3/4) chance of a child not getting it. If three in a row are not to get it the probability is:

(3/4)^3 = (27/64) < -------------------- Answer to (a)
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b) This is the opposite of none getting it (see (a) above) and that is (1 - (27/64)) = (37/64)
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c) (1/4) * (3/4) * (3/4) * 3 = (27/64)
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d) The first two do not affect the third, they are independent of each other and are called independent events.

(1/4) <----------------- Answer to (d)

.

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A question from Aurore: Probability problem having to do with cystic fibrosis?
It is estimated that 1 in 25 Americans is a cystic fibrosis carrier. Find the probability that a randomly selected American couple’s child will have cystic fibrosis, assuming that they’re unrelated?

How do I do this? Thanks for your help; I want to understand the process.

The best answer:

Answer by kelby7670
probability of father a carrier is 1/25
probability of mother a carrier is 1/25
probability of both is 1/25 * 1/25 = 1/625 which is th probability of a cf baby.

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Technorati Tags: cystic, fibrosis, having, Probability, problem

A question asked by Michelle: What is the probability that their child will be a carrier of the cystic fibrosis mutation?
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.)
The answer is 4/9. Please explain why.

Top answer:

Answer by JOHN H
To have siblings with cf, both parents of each of the couple must be carriers.

Each of the young couple, not being a cf sufferer has a 2/3 chance of being a carrier
(1 normal ; 2 carriers ; 1 cf – as they are not cf there remains a 2/3 chance of being a carrier and a 1/3 chance of being normal)
2/3 x 2/3 (to combine the probabilities) = 4/9

John H

If you know better then please let us know below.

Technorati Tags: Carrier, child, cystic, fibrosis, Mutation, Probability, their

by mcbill

Question posed by sam: genetics problem with probability?
Imagine that you are a genetic counselor and a couple planning to start a family come 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 answer is 1/6 but WHY is this the answer???

No 1 answer:

Answer by Rainbow6562005
If I remember my Punnett Squares correctly, Charles is heterozygous for the defective gene because he had to have had it to contribute to his first child’s death.

Elaine’s parents had to be heterozygous also, since each had to contribute the defective gene to the deceased brother. There is no way, with the information given, to determine Elaine’s inheritance. If she’s carrying the defective gene (heterozygous), there’s a 1/4 chance a particular child will have the disease. If she isn’t (homozygous for the normal gene), there’s no chance any child will develop the disease.

Agree or disagree? Leave your own thoughts below.

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Technorati Tags: Autosomal Trait, Cystic Fibrosis, Probability

Technorati Tags: Autosomal Recessive Gene, Cystic Fibrosis, Probability

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Technorati Tags: Cy, People Carriers, Probability