roopashri
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The parents of a girl with Tay-Sachs disease decide to pursue bone marrow transplantation in an attempt to provide a source for the missing lysosomal enzyme. Preliminary testing of the girl's normal siblings is performed to assess their carrier status and their human leukocyte antigen (HLA) locus compatibility with their affected sister. What is the chance that one of the three siblings is homozygous normal (i.e., has a good supply of enzyme) and HLA-compatible.
A. 1/2
B. 1/3
C. 1/4
D. 1/6
E. 1/12
pls show ur calculation<?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" />
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| Doc2378
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Good question Roopashri!
All right...lemme try..
I'd go with D. 1/16
Probability of being homozygous normal (AA) is 1/4
Probability of HLA compatibility in the siblings is 1/4 (since the haplotype is passed from each parent)
Combined probability of 2 independent events= 1/4 *1/4= 1/16
Well??
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| Geroo
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I say 1/12.
1/3 probability of being homozygous normal and 1/4 of being HLA compatible.
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| msyamp
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hey probability of being homogygous normal is 1/3?
i think its 1/4
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| Doc2378
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hmmm...but wouldn't you have to consider the 4th child here too to calculate the overall probability of getting AA out of Aa x Aa ???
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Courage does not always ROAR. Sometimes courage is the quiet voice at the end of the day saying, "I will try again tomorrow" - Mary Anne Radmacher
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| Geroo
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as the girl is phenotypically normal that eliminate the probability aa,that leaves us with three probabilities(AA,Aa,Aa) so AA is one out of three.that what I think may be i'm wrong.
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| satyaking
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Suppose TT is homozygous normal,tt is homozygous affected,Tt is heterozygus
hence there is a cross between Tt X Tt, then TT is homozygous normal,we have
2 siblings out of three without disease(heterozygous or homozygous)hence probability of these siblings of being homozygous normal is 2/3 x 1/4 = 1/6 so it seems D is the correct answer
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| Doc2378
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Good work ya' all .....I think you guys are right...
it should be (AA probability) 1/3 x (HLA compatibility) 1/4 = 1/12
Hmmm...anyone wants to weigh in on this??
Edited by Doc2378 on 12/25/05 - 12:27 PM
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| drk1980
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the risk of recurrence is INDEPENDENT for each sibling, is it not?wch means each daughter born or to be born will hv the same recurrence risk statistics (i dont know if i hv worded this properly)
so i think, acc to the Q, the homozygous normal probability for one of the other siblings of the patient is 1/4....the same as it is for ANY autosomal recessive disorder. correct me plz.
also, cud u plz explain how u all derive HLA compatability chance amongst family? thanks!
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| roopashri
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Well the answer given is..1/4 and the explanation is...
tay sachs is AR , if the girl has tay sachs it means both parents r carriers(Aa and Aa)
now 4 possibilities wud be AA, Aa, Aa, aa. girl's normal siblings wud either be carriers or homozygous normal, so, we r left with 3 options Either Aa or Aa or AA. out of these 3, AA is homozygous normal, so chances of her sibling being homozygous normal is 1/3
now coming to 2nd part of the q , 2 sibs wud be HLA compatible if they inherit same chromosomes . if the pt got allele with A1-B8-DR2 frm the father , the chances of other sibling getting the same chr from father are 50%(1/2)
AND if pt has got ch with A9-B5-DR3 from mother then the chance of that sib getting same ch frm mother is 50% , the chances of inheriting same parental ch r 1/2*1/2 = 1/4
so, the chances of being homozygous normal and HLA compatible r 1/3*1/4 = 1/12
chances of one of the 3 siblings being homozygous normal and HLA compatible r 1/12 +1/12+1/12 =1/4
Can someone clarify the second part of this qn pls...<?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" />
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| Geroo
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good question roopashri.where is it from?
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| jonathon
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great explanation. In your experience do the genetic questions in the real usmle exam come this difficult
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| roopashri
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geroo, this qn is from this forum itself, i got it while i was searching for something else
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