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#1




Genetic risk calculation
I saw this question somewhere else and thought I will share with you all to discuss: Please provide explanation with answer.
A woman has a genetic disease with autosomal dominance transmission. Her husband is unaffected. They plan to have a family with three children. What is the probability that one of the three children will be affected? A. 1/8 B. 1/4 C. 3/8 D. 1/3 E. 7/8 
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#2




I think the answer is A
Since no one has answered till now let me try..I may be wrong as i am not good in genetics
Considering the female is heterozygous for the trait.. (Homozygous would lead to all children being affected) At each child birth there would be 1/2 chance of the child being affected. **Therefore the chance of not being affected is also 1/2 (11/2=1/2 ) As asked in the question we want only one child to be affected. So 3 events should take place.. That is one event with child having the disease i.e 1/2 chance The 2nd event with child not having the disease i.e 1/2 chance(from ** above) The third event with child not having the disease i.e 1/2 (again from **) If we consider all the three events occurring together 1/2*1/2*1/2=1/8 so the probability of having one child affected out of three should be 1/8 This corresponds to option a) AGAIN I SAY. I MAY BE WRONG, PLEASE CORRECT ME IF YOU THINK I AM WRONG 
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usluipek (05042011) 
#3




Other people pls join in, we want to reach a concensus.

#4




I think mayursn started off fine, but I would have said that the probability that no child would be affected is 1/2*1/2*1/2=1/8, therefore the probability that at least one child would be affected is 7/8, so I would have picked E.

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usluipek (05042011) 
#5




I think I got the question wrong..The question says only one child out of three or at least one out of three?
because i think then the answer would change 
#6




Yeah, I think I'd probably go with e) also, because the more kids you have, the greater the probability that one will have the disease.

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usluipek (05042011) 
#7




Probability of each child should be taken as an individual event..
for every child probability of having disease is 1/2 so for 3 children by individual probability 1/2*1/2*1/2 = 1/8 but if u consider child not having disease for each child, apply same logic we will arrive at 1/2*1/2*1/2 = 1/8 so child probability of child with disease is 11/8 = 7/8 Where am I going wrong? 
#8




Quote:
Your second calculation is the probability that at least one child will have the disease. To answer mayursn39's idea that maybe they could be asking about one and only one child: that would be the combined probability of any one of the three fixed outcomes that fit the profile ∈ {(A+,B,C),(A,B+,C),(A,B,C+)}, right? So that would be 1/8+1/8+1/8 = 3/8. However, I think that the question is looking for 7/8... 
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mayursn39 (05042011) 
#9




I would say, in most autosomal dominant cases, the diseased is... Aa
So mating Aa into aa produces 1/2 diseases if there are 4 children... so if you have to calculate out of 3, the probability will increase from 1/2....so I think it will b 7/8.... I might be wrong though ... 
#10




If the question was "What is the probability that ONLY one of the three children will be affected?" then the answer would be 1/8.
If the question was "What is the probability that AT LEAST one of the three children will be affected?" then the answer would be 7/8. I think the question is not clear. How do you feel about it? 
#11




How can we justify the 1/2?? could the mom not be AA and thus 100% of her offsprings will recieve it?
I HATE PROBABILITY!  anyone has a quick fix?
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#12




c
its 1/8+1/8+1/8 =3/8 
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