asmi
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Phenylketonuria is an autosomal recessive disease. What is the probability that the asymptomatic brother or sister of a child with the disease is a heterozygous carrier?
A
) 0
B
) 1/4
C
) 1/2
D
) 2/3
E
) ¾
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| mjl1717
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answer-1/2
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| asmi
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can you explain it mjl ?
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| rida
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If its autosomal recessive, both parents have the trait, so they both are aa* (a* being the allele carring the trait), if you make a punnet square of both parents have aa*, 1/4 kids would be normal with aa, 2/4 or 1/2 would be aa*, which is carrying the trait hence heterzygous, and only 1/4 would actually have the disease, as he would be a*a*. Hope this helps!
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| piter
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i think it's D) 2/3 becouse you mast calculate without a child with the disease.
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| rida
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oh yea, thats rite i agree
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| asmi
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agree with 2/3..thanks to all
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| Idiopathic
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I dont like this, because one parent could be homozygous recessive, in which case it would be 1/2
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| piter
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don't agree Idiopathic, becouse even if one parent is homozygous recessive and other is hetero- the chanses is 2 homozygous and 2 heterozygous, but again - for probability you have to calculate without a child, who is already with the disease. So it will be 2/3 for hetero
its very classic usmle question, it can't be with too probable answer
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| Idiopathic
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no, you need more info. We need to know the parental status. Something like, "nobody else in the family is affected."
If there is a 50% chance of having a homozygote recessive and a 50% chance of having a heterozygote, then obviously the chance of having a heterozygote child (knowing that he isnt affected) is 100%...i screwed it up the first time. Since 1 isnt an option, we have to assume that both parents are heterozygotes. 2/3 is the answer
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| mjl1717
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I stand corrected -You guys are right if PKU testing is mandatory in most states-EXCLUDE an affected child in autosomal recessive disease we have
left 2 carrier states and 1 normal state. The carrier state probability is 2 out of 3 :|
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| Idiopathic
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Even though I dont think there is enough info. It had to be 2/3.
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| mjl1717
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On the serious side Id say usmle would give a little more info because of too many unknowns. [Actually thought I knew PKU well,guess not]
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| santaclara
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hi rida.
can u explain me this q again plsssssssssssssssss.
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| rida
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Hey santaclara, to solve this problem you need to make a punnet square, and you have to assume that both parents are heterzygous, because i dont know where, but i remember one of my teachers in med school saying that when its not told, assume parents are heterozygous. Neways, so we will assume that they have aa* so they both have the trait, not the disease as to have the disease you need to have a*a*. So if you make a punnet square with both parents having aa*, you find that one kid would be aa (normal), one kid would be a*a* (would have the disease) and two kids would be aa* (have only the trait). The question asks for the ones only with the trait, which would be 2/3, not four since one child is normal and we don't wanna count him. Hope i didn't confuse you!
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