IceAge
Forum Senior
Topics: 21
Posts: 159
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I came across a question which read as follows:
Father has an X-linked dominant disease. Mother is normal. Penetrance is 80%. What is the chance of disease in daughters?
The correct ans was given as 80%, BUT I think it should be 40% because:
every daughter will have father's X-chr; however there is a 50% chance that it will become a Barrbody. Therefore the probability should be half of 80% i.e. 40%
Please give your feedback ........ thanks!!
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| alexaE
Forum Elite
Topics: 23
Posts: 334
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all daughters will have the mutated X chromosome, so all daughters will be diseased. if penetrance is 80% it means they have 80% chances to express it
nonetheless, all will carry the X mutated chromosome from their father
we cant tell which of the two chromosomes will become a barr body so we cannot assume the X chromosome from their father will be inactivated in every cell
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veni vidi...vincam
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| blueocean
Forum Senior
Topics: 15
Posts: 195
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The father will transmit the X chromosome to ALL his daughters; if he has the disease the risk for the daughter is 100%.If the penetrance is incomplete (80%) than the risk is 80%.
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| blueocean
Forum Senior
Topics: 15
Posts: 195
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The chance to become a Barr body is randomly distributed and this is the problem with Xdominant disease - the phenotype is expressed in heterozygous females as well as in hemizygous males.
http://www.uic.edu/classes/bms/bms655/lesson6.htm...
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