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Old 02-12-2013
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Genetics Complicated CF Genetics Pedigree

How do I do this?

Joan and Claude come to you seeking genetic counseling. Claude was married before, and he and his first wife had a child with cystic fibrosis, an autosomal recessive condition. A brother of Joan’s died of cystic fibrosis and Joan has never been tested for the gene. If they marry, what is the probability that Joan and Claude will have a son that WILL NOT be a carrier for or have cystic fibrosis?
A. 1/2
B. 1/4
C. 1/12
D. 1/6
E. 1/8
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b. 1/8????

Oh God I am bad at this.:sorry:

Last edited by Doc4Step1; 02-12-2013 at 10:35 AM.
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Originally Posted by Doc4Step1 View Post
b. 1/8????

Oh God I am bad at this.:sorry:
I guess thats the right one
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How did you get 1/8
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Quote:
Originally Posted by ashanair70 View Post
How did you get 1/8
Quote:
Originally Posted by ashanair70 View Post
How do I do this?

Joan and Claude come to you seeking genetic counseling. Claude was married before, and he and his first wife had a child with cystic fibrosis, an autosomal recessive condition. A brother of Joan’s died of cystic fibrosis and Joan has never been tested for the gene. If they marry, what is the probability that Joan and Claude will have a son that WILL NOT be a carrier for or have cystic fibrosis?
A. 1/2
B. 1/4
C. 1/12
D. 1/6
E. 1/8
Claude is carier of the gene cause otherwise he wouldnt have son with disease
He is either contributing or not the diseased gene to generation = 50%diseased/50%healthy gene goes to offspring
So we take 50%=1/2 for healthy gene

Now for female partner we dont know if she is a carrier or not so we take the same chance 50/50% so 1/2 chance that she is healthy and in case if she is carrier-50% chance that she is transferring healthy gene to child which totaly equals 1/2x1/2x1/2=1/8

but not sure if thats the way
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Old 02-13-2013
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Answer says it is 1/4. Can some one help.
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Originally Posted by ashanair70 View Post
Answer says it is 1/4. Can some one help.
Can u post the explanation of that question
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" a son that WILL NOT be a carrier for or have cystic fibrosis"

I was waiting on somebody to say 1/4 cause that what i thought the answer was.

My reason: cf is autosomal recessive. needing 2 parents which are carriers to produce a child with cf at 25% of the time.

based on the q stem, we know that both parents are carriers. we know that because Claude already had a CF child in his past marriage. Joan brother died from CF so we can assume she's a carrier. The questions is asking about what are the chances of producing a Normal child?

CF child = 25%
CF carrier = 50%
Normal child= 25%

thats why i picked 1/4.
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Answers 1/4.

I think it's because u guys forgot to take into account that both of Joan's parents must be carriers in order for her brother to be affected. Which mean's that there's 1/2 chance of joan being a carrier and 1/4 chance of her being homozygous for the normal alleles.

If you calculate the probability of joan having a homozygous normal child with her carrier husband you'll get two probabilities. One if joan was a carrier (1/2) and the other using joans probability of being homozygous normal (1/4). If you add both probabilities, you'll get 1/8 + 1/8 = 1/4
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Btw where'd u get this question from??
Cuz I think CF has complete penetrance and the answer should be 1/3. Does the book ur using have an erratum?
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This question is definitely pain in the ass
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dont think about it that way. I use to always freak out when i see these kind of answer choices.

But now i just relax and think it through with some deductive reasoning. You have all the information you need on the stem.
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It is an old practice exam paper (not from a text book).
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Quote:
based on the q stem, we know that both parents are carriers. we know that because Claude already had a CF child in his past marriage. Joan brother died from CF so we can assume she's a carrier. The questions is asking about what are the chances of producing a Normal child?

CF child = 25%
CF carrier = 50%
Normal child= 25%

thats why i picked 1/4.
You're assuming claude is a carrier and ignoring the fact that there's a 25% chance that she's homozygous normal. You can't simply ignore such a clear possibility.


Quote:
It is an old practice exam paper (not from a text book).
I think they made a mistake on this paper. :P
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Originally Posted by slowpoke View Post
[based on the q stem, we know that both parents are carriers. we know that because Claude already had a CF child in his past marriage. Joan brother died from CF so we can assume she's a carrier. The questions is asking about what are the chances of producing a Normal child?

CF child = 25%
CF carrier = 50%
Normal child= 25%

thats why i picked 1/4.]

You're assuming claude is a carrier and ignoring the fact that there's a 25% chance that she's homozygous normal. You can't simply ignore such a clear possibility.
Can u post the way how should it be solved
Thanks in advance
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Quote:
Joan and Claude come to you seeking genetic counseling. Claude was married before, and he and his first wife had a child with cystic fibrosis, an autosomal recessive condition. A brother of Joan’s died of cystic fibrosis and Joan has never been tested for the gene. If they marry, what is the probability that Joan and Claude will have a son that WILL NOT be a carrier for or have cystic fibrosis?
I'll try to solve it the way I learned it.. Hope it makes sense

Since Claude had a kid with CF, we know for sure that he's a carrier.
Since Joan had a brother with CF, we know her parents must've both been carriers. This means that Joan can either be homozygous normal (CC), or a carrier (Cc), but she can't be homozygous affected (cc) since she obviously doesn't have CF. Knowing that Joan's parents were both affected, we can calculate the probability of Joan being Normal/(CC) or a Carrier/(Cc).

If we do a punnet square with Joan's parents (two carriers) we get the values (CC) = 1/3 and (Cc) = 2/3 for Joan (the punnet square would only have 3 possibilities since we already know Joan's not affected and therefore the probability of cc is nonexistant)

Now that we have both of Joan's genotypic probabilities, we can calculate the probability of Joan's child with Claude. We'd do this in 2 steps.

1) If Joan were a carrier (probability of 2/3), the probability her child would be homozygous normal (CC) would be: 2/3 * 1/4 = 1/6

2) If Joan were homozygous normal (probability of 1/3), the probability her child would be normal as well: 1/3*1/2= 1/6

Adding both results gives us the true probability of Joan and Claude having a homozygous normal child: 1/6 + 1/6 = 1/3.
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Last edited by slowpoke; 02-13-2013 at 10:41 AM.
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Originally Posted by slowpoke View Post
I'll try to solve it the way I learned it.. Hope it makes sense

Since Claude had a kid with CF, we know for sure that he's a carrier.
Since Joan had a brother with CF, we know her parents must've both been carriers. This means that Joan can either be homozygous normal (CC), or a carrier (Cc), but she can't be homozygous affected (cc) since she obviously doesn't have CF. Knowing that Joan's parents were both affected, we can calculate the probability of Joan being Normal/(CC) or a Carrier/(Cc).

If we do a punnet square with Joan's parents (two carriers) we get the values (CC) = 1/3 and (Cc) = 2/3 for Joan (the punnet square would only have 3 possibilities since we already know Joan's not affected and therefore the probability of cc is nonexistant)

Now that we have both of Joan's genotypic probabilities, we can calculate the probability of Joan's child with Claude. We'd do this in 2 steps.

1) If Joan were a carrier (probability of 2/3), the probability her child would be homozygous normal (CC) would be: 2/3 * 1/4 = 1/6

2) If Joan were homozygous normal (probability of 1/3), the probability her child would be normal as well: 1/3*1/2= 1/6

Adding both results gives us the true probability of Joan and Claude having a homozygous normal child: 1/6 + 1/6 = 1/3.
I understand everything except
1) If Joan were a carrier (probability of 2/3), the probability her child would be homozygous normal (CC) would be: 2/3 * 1/4 = 1/6
The chance of her giving diseased/normal allele would be 1/2
Which should equal 2/6=1/3
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Quote:
The chance of her giving diseased/normal allele would be 1/2
Which should equal 2/6=1/3
Yes, the chance she'll pass the normal allele if she were a carrier is 1/2. So you're right upto that point, 1/2 * 2/3= 1/3. But ur forgetting about claude, who's also a definite carrier and he has 1/2 chance of passing on the normal allele as well. So to complete the childs probability: 1/3*1/2= 1/6.
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Originally Posted by slowpoke View Post
Yes, the chance she'll pass the normal allele if she were a carrier is 1/2. So you're right upto that point, 1/2 * 2/3= 1/3. But ur forgetting about claude, who's also a definite carrier and he has 1/2 chance of passing on the normal allele as well. So to complete the childs probability: 1/3*1/2= 1/6.
Why not add up probabilities of female giving healthy allele and multiply it by probability from male carrier ? It is more simple to understand i guess
1/2x(1/3+1/3)=1/3

Much thanks ur post was very helpful
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claude is a carrier
Joan has 2 probability
1-carrier 1/2
2-normal allel 1/4

claude(carrier) +Joan(carrier) = 1/4 chance for normal not carrier child
Claude(carrier)+ Joan(normal) = 1/2 chance for normal not carrier child

it is not separated event so we have to multiply
1/4 *1/2=1/8
This is what I think. correct me
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Quote:
Originally Posted by sami trailer View Post
claude is a carrier
Joan has 2 probability
1-carrier 1/2
2-normal allel 1/4

claude(carrier) +Joan(carrier) = 1/4 chance for normal not carrier child
Claude(carrier)+ Joan(normal) = 1/2 chance for normal not carrier child

it is not separated event so we have to multiply
1/4 *1/2=1/8
This is what I think. correct me
We had brain drilling discussion on this case in above posts
Its not as simple as it appears
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Old 02-20-2013
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Default what's the correct answer?

so...what's the consensus guys?
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