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#1




genetic Q
A man ans women are both affectedby an autosomal dominant disorder that has 80 % penetrance. They both heterozygotes for the diseasecausing mutation. What is the probabiity that they will produce phenotypically normal offspring?
a. 20% b.25% c.40% d.60% e.80% f.non of the above 
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dr_lizard (09072011) 
#2




It would be 75% chance of getting at least one diseased allele, then that times 80% penetrance = 60% disease risk, then C

#3




Aa  Aa ==> AA(1/4x20%x20%)+Aa(1/2x20%)+aa(1/4)=36%
is it right?
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#4




Interesting, lets do this ...
* A a *A AA Aa *a Aa aa assuming that AA = Normal Aa = Heterozigous aa = Affected Remember here, we just need 1 gene to be affected due the fact that is AD. so the Normal Calculation is that "75%" of this Disease will be afected the calculation is like this 75 Affected by 80% penetrance is 60 affected so we sustract 60100 = 40 ! In a normal 100% penetrance 75 would be affected, but this got 80 %, so we have to do the math for that. 75 Affected MINUS () 80 = 60 % will be affected in 80% of penetrance... so doing the subtraction is 40 for NORMAL. So... Answer is C. Corresponding the NORMAL Child Answer D correspond the Affected Childrens... Answer E correspond to the penetrance of the disease. Answer B and A wrong calculations... Hope it helps...
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dr_lizard (09072011) 
#5




Quote:
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nothing is impossible! 
#6




I didnt understand your answer. Could you please explain it to me, what did you do on it ? but im pretty sure my answer is the right one, if you have any hard time understand it let me know, i help you out ... =) That's the right way to do that calculation...
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#7




yes you r right, the answer is 40%
thank you for the explanation 
#8




Quote:

#9




You welcome =)
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#10




Please explain it

#11




First, read the answer above and let me know which part u dont understand... pls. =) I detailed the answer...
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Nothing is Impossible, the Word Itself says "I'm Possible" 
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aknz (09072011) 
#12




Quote:
AD disease,for 100% penetrance, 75% will be affected. For 80% penetrance,60% will be affected (did calculation for that) 
#14




I guess you are overthinking this. You are assuming that the child has 80% penetrance for each diseased allele he/she carries, in other words that a child with AA genotype has a higher than 80% chance of expressing the phenotype. That is why in your calculation you used 1/4x20%x20%. If you simply assume that the risk is still 80% even with the AA genotype, then the math does come to 40% = .25 x .2 + .5 x .2 + .25

The above post was thanked by:  
dr_lizard (09072011) 
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