Posted by pseudo from IP 71.64.122.31 on April 02, 2006 at 01:53:53:
In Reply to: Q bank question..please explain posted by adk on March 30, 2006 at 13:59:42:
Suppose AD parents have these genes....aA aA
So their kids would have .... AA aA aA aa.
All except the aa would have the phenotype. Only aa is normal.
The chance of phenotype in their kids is 75%.
Now since the gene has 80% penetrance, the risk becomes 80% X 75% = 60% for showing phenotype of the disease.
Since USMLE questions are never single step problems, they asked us what the change of a phenotypically normal kid would be.
That would be 100% - 60% = 40%