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  #1  
Old 05-28-2011
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Genetics Genetic Counseling of CF Parents

Cystic fibrosis shows autosomal recessive inheritance. Parents of a newly diagnosed affected child are referred for genetic counseling. It would be correct to tell them that:

a) the probability that their next child will be affected is 1 in 2.
b) the probability that the older unaffected sister of the affected child is a carrier is 1 in 2.
c) the fact that their last child was affected means that their next three children will not be affected.
d) the probability that each parent is a carrier is 1.
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  #2  
Old 05-28-2011
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the ans is D
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  #3  
Old 05-28-2011
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d) the probability that each parent is a carrier is 1
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  #4  
Old 05-28-2011
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This one is easy, Its D.
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  #5  
Old 05-28-2011
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Quote:
Originally Posted by eesfee View Post
This one is easy, Its D.
Well, all the questions are easy, if you know how
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  #6  
Old 05-28-2011
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Why not B?
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  #7  
Old 05-28-2011
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Quote:
Originally Posted by usluipek View Post
Why not B?
answer B) "the probability that the older unaffected sister of the affected child is a carrier is 1 in 2."

Since this is an AR disease, both parent are carriers... you will have:
two Aa
one AA
one aa

Then, among the "unaffected", the probability of being a carrier is 2/3.
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  #8  
Old 05-28-2011
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Thanks, bebix,
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  #9  
Old 05-29-2011
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please explain me whole concept of dominant and recessive disease?
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  #10  
Old 05-29-2011
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Quote:
Originally Posted by amresh kumar View Post
please explain me whole concept of dominant and recessive disease?
Dominant means you only need one copy of the gene to be affected, recessive means both copies of the gene need to be mutated for someone to be affected. Some key features:
Dominant:
1. There are no carriers because if you have one copy by definition you have the disease.
2. Only 1 parent need to have the disease for the child to be at risk, and if one parent has it the probability of the child getting it is 1/2

Recessive:
1. If only one parent is a carrier or has the disease, there is 0 probability that the child will get the disease
2. If both parents are carriers or have the disease, you need to calculate probabilities based on the genotypes you are given.
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Old 05-29-2011
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The answer should be ''B''.
@bebix nice and short explanation. good job.
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Old 05-29-2011
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thanks buddy
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  #13  
Old 05-30-2011
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Quote:
Originally Posted by dr_ram View Post
The answer should be ''B''.
@bebix nice and short explanation. good job.
answer is d i guess
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  #14  
Old 05-30-2011
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Quote:
Originally Posted by moonjane11 View Post
answer is d i guess
Yes, the correct answer is D.
I think @dr_ram meant to say why canīt be B)

"Since this is an AR disease, both parent are carriers... you will have:
two Aa
one AA
one aa

Then, among the "unaffected", the probability of being a carrier is 2/3."
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  #15  
Old 05-31-2011
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Default oops

Quote:
Originally Posted by moonjane11 View Post
answer is d i guess
i am sorry for the disastrous error in typing. the correct answer is D. my mind was pre-occupied with something else while typing. my apologies for that.
thank you for the correction moonjane.
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  #16  
Old 06-08-2011
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D is correct. The child is "aa" and must have received each gene from each of his parents. Since CF is AR condition and the parents are phenotypically healthy, they should be carrier for the disease, Thus they surely are "Aa".
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Old 07-24-2011
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Quote:
Originally Posted by bebix View Post
answer B) "the probability that the older unaffected sister of the affected child is a carrier is 1 in 2."

Since this is an AR disease, both parent are carriers... you will have:
two Aa
one AA
one aa

Then, among the "unaffected", the probability of being a carrier is 2/3.
I agree with u but for the probability of been a carrier is 2/3. I think the probability of been a carrier is 2/4. The "unaffectd" is just an adjective for the sister, not related to the probability ( I mean not probability of been a carrier from chances of unaffected.
Note: genetics is about chances of the individual fertilized ovum or person, not the chances of the whole family.-
That is a fertilized ovum has chances of( autosomal recessive )
1 chance to be AA= normal
2 " ". ". Aa= carrier
1. ". ". aa= disease manifest
So the probability of been normal is 1:4
" " Carrier is 2:4
" " " disease manifest 1:4
Now considering the family
If the family (God's will) all the children could b normal, all may manifest the disease, or mixture depending on thief individual outcome from thief chance

For autosomal dormant( the presence of only one gene manifest the disease)
1 chance of AA= the only normal chance
2 " " Aa= disease manifest
1 " " aa= disease manifest
Probability of having the disease is 3:4
". Been normal is 1:4
This is for both parent with gene Aa
I hope this willhelp someone

Last edited by Bibi; 07-24-2011 at 05:40 AM. Reason: typing error
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  #18  
Old 07-24-2011
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Quote:
Originally Posted by Bibi View Post
I agree with u but for the probability of been a carrier is 2/3. I think the probability of been a carrier is 2/4. The "unaffectd" is just an adjective for the sister, not related to the probability ( I mean not probability of been a carrier from chances of unaffected.
Note: genetics is about chances of the individual fertilized ovum or person, not the chances of the whole family.-
That is a fertilized ovum has chances of( autosomal recessive )
1 chance to be AA= normal
2 " ". ". Aa= carrier
1. ". ". aa= disease manifest
So the probability been normal is 1:4
" " Carrier is 2:4
" " " disease manifest 1:4
Now considering the family
If the family (God's will) all the children could b normal, all may manifest the disease, or mixture depending on thief individual outcome from thief chance

For autosomal dormant( the presence of only one gene manifest the disease)
1 chance of AA= the only normal chance
2 " " Aa= disease manifest
1 " " aa= disease manifest
Probability of having the disease is 3:4
". Been normal is 1:4
This is for both parent with gene Aa
I hope this willhelp someone





note
typing error earlier
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  #19  
Old 07-24-2011
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d) the probability that each parent is a carrier is 1
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  #20  
Old 07-24-2011
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Quote:
Originally Posted by Bibi View Post
I agree with u but for the probability of been a carrier is 2/3. I think the probability of been a carrier is 2/4. The "unaffectd" is just an adjective for the sister, not related to the probability ( I mean not probability of been a carrier from chances of unaffected.
Note: genetics is about chances of the individual fertilized ovum or person, not the chances of the whole family.-
That is a fertilized ovum has chances of( autosomal recessive )
1 chance to be AA= normal
2 " ". ". AA= carrier
1. ". ". AA= disease manifest
So the probability been normal is 1:4
" " Carrier is 2:4
" " " disease manifest 1:4
Now considering the family
If the family (God's will) all the children could b normal, all may manifest the disease, or mixture depending on thief individual outcome from thief chance

For autosomal dormant( the presence of only one gene manifest the disease)
1 chance of AA= the only normal chance
2 " " Aa= disease manifest
1 " " aa= disease manifest
Probability of having the disease is 3:4
". Been normal is 1:4
This is for both parent with gene Aa
I hope this willhelp someone


AA= normal
A

note
AR disease - carrier (Aa)
If they ask among the unaffected, is 2/3 = conditional probability
If they ask "just" the probability is 1/2
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