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  #1  
Old 09-07-2011
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Default genetic Q

A man ans women are both affectedby an autosomal dominant disorder that has 80 % penetrance. They both heterozygotes for the disease-causing 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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Old 09-07-2011
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It would be 75% chance of getting at least one diseased allele, then that times 80% penetrance = 60% disease risk, then C
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Aa - Aa ==> AA(1/4x20%x20%)+Aa(1/2x20%)+aa(1/4)=36%
is it right?
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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 60-100 = 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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Quote:
Originally Posted by rulz View Post
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 60-100 = 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...
tanx, but which part of my answer's logic is incorrect?
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Quote:
Originally Posted by dr_lizard View Post
tanx, but which part of my answer's logic is incorrect?
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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Old 09-07-2011
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yes you r right, the answer is 40%
thank you for the explanation
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Quote:
Originally Posted by laurier View Post
A man ans women are both affectedby an autosomal dominant disorder that has 80 % penetrance. They both heterozygotes for the disease-causing 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
its b,25% chance to produce normal offspring.
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Quote:
Originally Posted by laurier View Post
yes you r right, the answer is 40%
thank you for the explanation
You welcome =)
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Quote:
Originally Posted by rulz View Post
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...
Please explain it
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Quote:
Originally Posted by aknz View Post
Please explain it
First, read the answer above and let me know which part u dont understand... pls. =) I detailed the answer...
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  #12  
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Quote:
Originally Posted by rulz View Post
First, read the answer above and let me know which part u dont understand... pls. =) I detailed the answer...
Thanks I understand it

AD disease,for 100% penetrance, 75% will be affected.
For 80% penetrance,60% will be affected (did calculation for that)
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Quote:
Originally Posted by aknz View Post
Thanks I understand it

AD disease,for 100% penetrance, 75% will be affected.
For 80% penetrance,60% will be affected (did calculation for that)
Exactly!! you got it.. =)
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  #14  
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Originally Posted by dr_lizard View Post
tanx, but which part of my answer's logic is incorrect?
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
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