| 11/09/07 - 09:19 PM  
 
   
 
|   #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.
___________________ FORUM RULES-- Those who believe in telekinesis, raise my hand. I get enough exercise just by pushing my luck --P4U World.." The pure and simple truth is rarely pure and never simple."
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| Ujk Forum Elite

Topics: 36 Posts: 398
| | 11/11/07 - 05:38 AM  
 
   
 
|   #5 |

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| ShweHope99.9998 Forum Fanatic
Topics: 12 Posts: 2,436
| | 11/11/07 - 07:14 PM  
 
   
 
|   #6 |
Thanks for the Q [ allison], & answer [ doc 179 ] 
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| DRFP Forum Newbie

Topics: 3 Posts: 41
| | 11/12/07 - 06:05 PM  
 
   
 
|   #7 |
Thanks I thought maybe I was wrong
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| u3846158 Forum Newbie
Topics: 0 Posts: 7
| | 11/19/07 - 10:11 AM  
 
   
 
|   #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
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