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Author9 Posts
  #1

t-Haoosaminidasa in made alpha and beta subunits Tay-Sachs disease in due to a mutation in the alpha-subunit gene, and Sandhoff disease is due to a mutation in the beta subunit gene. If a carrier of Tay-Sachs disease marries a carrier of Sandhoff disease, what in the probability that their first child will have either disease?
C A) 0%
o B) 12.8%
o C) 25%
o D) 50%
o E) 100%


Please esplain.




  #2

25% ARraised eyebrow


  #3

Both the chromosomes should be affected by the same disease mutation to get expressed in an individual in autosomal recessive inheritance. Having one chromosome for Sandoff and the other for Tay-sach's , that person will become a carrier for both the diseases and neither of them will be expressed in him. Hence the answer is A, 0%.


  #4

doc179 wrote:
Both the chromosomes should be affected by the same disease mutation to get expressed in an individual in autosomal recessive inheritance. Having one chromosome for Sandoff and the other for Tay-sach's , that person will become a carrier for both the diseases and neither of them will be expressed in him. Hence the answer is A, 0%.

Once again beautifully explained.


  #5

nod


  #6

Thanks for the Q [ allison], & answer [ doc 179 ] nod


  #7

Thanks I thought maybe I was wrong


  #8

The explanation in biochemical terms are
hexosaminidase enzymes have more than one isoenzymes A, B, S
each isoenzymes comprise alpha and beta subunit
beta subunit is common and produce from same gene, so if beta subnuit is deficient (homozygotes) , the patients will lack of both hesoxaminidase A and hexosaminidase B lead to sever form Sandhoff disease.
alpha unit is different from isoenzyme A and B, when mutation occur on alpha subunit producing gene of isoenzyme 1, they will lack only hexosaminidase A isoenzyme and produce Tay-Sachs disease which are more variable in age group of presentation.
If we calculat emore precisely (you can ignore it for USMLE)
paternal carries tay-sachs gene, the couple will have heterozygote frquency about 1/30(Jewish), so risk of having children with tay-sachs is 1/2x1/2x1/30=1/120
maternal carries sandhoff gene, the couple will have heterozygote frquency about 1/300 (jewish), so risk of having children with Sandhoff is 1/2x1/2x1/300=1/1200
risk of having child with either disease so =1/120+1/1200 =estimate 0.01 or 1 % that are substantial high risk that cannot be neglected.
The result willchange if the anceester of the parents are variable, some studies show Scottish, English might have more frequent gene frquency.

Ok enough
http://clinicalgenetics.blogspot.com


  #9

smiling facesmiling face





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