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#1




Third Generation Marriage Counseling!
Bob arrives in your office for genetic counseling. Bob's brother Tom died at a young age from TaySachs disease. Both he and his sister Sarah are unaffected. Bob's son Adam and Sarah's daughter Jennifer have recently married and are expecting their first child. What is the chance that the child would have TaySachs disease?
(A) 1/4 (B) 1/9 (C) 1/12 (D) 1/36 (E) 1/44 
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Larisa (05072010), Taiwan_Guy (05192011) 
#2




1/12 I guess



#4




I don't understand
Would you explain for us how you came to this conclusion. I really would like to understand this tricky question

The above post was thanked by:  
Larisa (05072010) 
#5




Okie this is a long one, but here's my calculation.
1st generation: When you grid the parents Bob's and Sarah's parents were: Mom (Aa) x Dad (Aa) so Bob has a 2/3 chance of being a carrier Sarah since she has the same parents also has a 2/3 chance of being a carrier We say 2/3 since their brother Tom was the aa 2nd generation : We're not given any info about Bob's wife or Sarah's husband so they have a 1/2 (50%) chance of being carriers as well. 3rd generation: Bob's son Adam (??) X Sarah's daughter Jennifer (??) The child has (1/2) chance of being a boy and (1/2) chance of being a girl. SO: (2/3)(2/3)(1/2)(1/2)(1/2)= 1/36 Hope that helps. 
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13monument (04132010), drgabroomunda (05192011), Larisa (05072010), leeusmle (04132010), Sabio (04132010), Taiwan_Guy (05192011) 
#6




yep it is 1/36
1. parents of Bob and Sarah are heterozygous Aa (that is why one of their sons died in past  he was with aa genotype); 2. Bob has possibility to be a heterozygous Aa 2/3, his sister Sarah has the same possibility 2/3, both of them hav 2/3*2/3=4/9 possibility to be Aa together, 3. We suppose that Bob and Sarah married people with AA genotype, so their chldren have possibility to be a geterozygous Aa 1/2 for Adam and 1/2 for Jennifer, their possibility together 1/2*1/2=1/4 4. Adam's and Jennifer's child has a possibility to have aa genotype and be sick 1/4 (because we suppose that parents Aa), 5. 4/9*1/4*1/4=1/36 
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drgabroomunda (05192011), Larisa (05072010), noothan (05192011), shahra (07152013), Taiwan_Guy (05192011) 
#7




0!
Hi,
It doesn't seem to be correct! The problem should gave us the frequencies of the allels in population and with that we can calculate the probability of being carrier for Bob's wife and Sarah's husband. I think for these types of question which we don't have the allel's frequencies, we should consider Bob's wife and Sarah's husband to be phenotipically and genotypically healthy 'cause this disease is considered to be UNCOMMON. if anyone is Consentaneous me, plz say! 
#8




hmmmmm is it really 1/36 ??

#9




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