Salman23
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A couple presents to you. Husband had two brothers with PKU, wife has no family history of PKU. If the prevalence of PKU in that specific community is 1/400, what is the chance of their offspring getting PKU??
A. 1/50
B. 1/40
C. 1/30
D. 1/20
E. 1/10
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| rida
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is it E...
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| Idiopathic
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Random carrier rate is 1/10, since PKU is AR defect. Man MUST be a carrier, if you assume that with two brothere with PKU then one parent had to be homozygous and the other heterozygous.
So, 1/2 times 1/10 = 1/20
edit: if you assume both the husbands parents were heteroygous, then his risk of being a carrier is 2/3, and the anser 2/30 isnt listed 
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| rida
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oh i get it...thanx
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| anne
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i am sorry if i sound stupid,and do correct me if i am wrong.but when i did this Q,i got ans=B.because the q asks the about the chance of the offspring getting PKU,meaning the chance of the offspring being homozygous recessive for this condition.
the random carrier rate in 1/10th,and with the husband(as well as the wife) being heterozygous,the child has 1/4th chance of having PKU.so 1/10X1/4=1/40
do correct me,please.
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| Idiopathic
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You actually are correct...the husband has to be a carrier, so his rate is 1/1, the wife is 1/10, and their chance of having a PKU child is 1/4, which works out to 1/40. I am sorry, it was late :oops:
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| Salman23
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I dont know how you got the random carrier rate of 1/10. If you use hardy weinberg, 1/400 and take an underroot, gives u 1/20 for the p gene, q gene is 19/20 and 2xpxq =19/200....isnt that the randon carrier rate???
If you could please tell me how you got the random carrier rate??
Also, how can we assume that the husband is a carrier, he has a 2/3 chance of being a carrier???
and yes, I dont have the answer to this question :oops:
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| Idiopathic
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When given the pneotypic prevalence of an AR disease = 1/400...you must account for the 1/4 chance that two heterozygotes will have an affected child. That puts the chance that two heterozygotes will 'hook up' at 1/100 (or 1/400 divided by 1/4). That puts the heterozygote/allele frequency at 1/10.
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| Idiopathic
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"Salman23" wrote:
Also, how can we assume that the husband is a carrier, he has a 2/3 chance of being a carrier???
If we are told that the husband had two affected brothers, and no mention is made of any unaffected siblings, we have to assume that one of his parents was homozygous (and therefore affected). That puts his chance at being heterozygous at 100%. I did mention that his carrier rate, given two heterozygous parents, would be 2/3, but the answer that applies to that also isnt available, and we know that he has two brothers with the disease....Either he has 5 or 6 unaffected siblings, that we dont know about, or one parent is homozygous for the allelle.
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| DiaHuq
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I understand what salman did but i'm not understanding anything else....
help
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| sk
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helP??
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| mash
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ans is 1/60
PKU is AR
q^2 = 1/400 => q = 1/20
carrier rate = 2*p*q = 2*1*1/20 =1/10
bcoz his 2 sibs have PKU => both his parents r carriers
his chances of being a carrier r 2/3
for their offspring to have PKU both parents must be carriers
So, chances of their offspring having PKU r 1/4*2/3*1/10 = 1/60
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| mash
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and idiopathic i ve a ques
u said if we assume that with two brothers ve PKU then one parent had to be homozygous and the other heterozygous.
So, 1/2 times 1/10 = 1/20
i dint get this. can u plz explain?
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| namf
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HELP! I just read this thread and it is G-d Almighty confusing!
Can somebody please lay out in step-by-step and clear points how to go through this question? As I am very poor in these kind of qs, I would really appreciate the help. What's more, I hear a lot of people get these type of qs on their STEP 1 tests!! I will post my (suspectedly incorrect) reasonings and doubts/questions if no one will do me the honor of posting something more authorative (yes, that is a threat!  )
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| Ahab
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I got 1/60 using the following reasoning
You cannot assume that one parent is homozygous for PKU. The chance for two heterozygous to pass on a disease is 1/4 always. For two offspring the chances are 1/4*1/4, which is 1/16, rare but not impossible. I think that if one parent was homozygous it would be included in the question. Therefore the chance that the two heterozygous parents pass on a normal unaffected (does not have defective gene) child is 1/4. We are looking for the chance that he is heterozygous for a gene which is 2/4. However because he does not have PKU the chances of him having a gene is 2/3.
The chances of his wife being heterozygous for the gene is 1/10. The lack of family history would theoretically rule out her being homozygous for PKU.
Therefore the odds of this man being heterozygous and meeting up with a heterozygous female is 2/3*1/10=1/15
The odds that they give birth to a homozygous individual is 1/15*1/4 which is 1/60.
Any serious flaw in my arguement?
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| rida
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very good explanation!!!
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| Quinn
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I agree with Ahab....got the same answer. But, why is that not an answer choice? If we assume that the father is heterozygous, then the answer is
1/40....but, nothing in the question indicates that is true. :icon_scratch:
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| black_tulip
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If husband is a carrier, the anwser is 1/40.
Here it didn't mention the genotype of the husband, if he is phenotype normal, his chance of carrier is 2/3, so the answer should be 2/3 x 1/40 = 1/60
My question: is this husband phenotype normal in this question?
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| Malaysian
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I'm getting the answer as 1/60 too but its not in the option......so I guess its wrong!!
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