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#1
09-07-2011
 USMLE Forums Addict Steps History: CK Only Posts: 105 Threads: 19 Thanked 69 Times in 32 Posts Reputation: 79
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
 The above post was thanked by: dr_lizard (09-07-2011)

#2
09-07-2011
 USMLE Forums Scout Steps History: 1+CK+CS Posts: 69 Threads: 4 Thanked 76 Times in 21 Posts

It would be 75% chance of getting at least one diseased allele, then that times 80% penetrance = 60% disease risk, then C
#3
09-07-2011
 USMLE Forums Addict Steps History: Not yet Posts: 189 Threads: 17 Thanked 95 Times in 58 Posts Reputation: 105

Aa - Aa ==> AA(1/4x20%x20%)+Aa(1/2x20%)+aa(1/4)=36%
is it right?
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#4
09-07-2011
 USMLE Forums Master Steps History: Step 1 Only Posts: 566 Threads: 25 Thanked 522 Times in 263 Posts Reputation: 532

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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 The above post was thanked by: dr_lizard (09-07-2011)
#5
09-07-2011
 USMLE Forums Addict Steps History: Not yet Posts: 189 Threads: 17 Thanked 95 Times in 58 Posts Reputation: 105

Quote:
 Originally Posted by rulz 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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nothing is impossible!
#6
09-07-2011
 USMLE Forums Master Steps History: Step 1 Only Posts: 566 Threads: 25 Thanked 522 Times in 263 Posts Reputation: 532

Quote:
 Originally Posted by dr_lizard 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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#7
09-07-2011
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yes you r right, the answer is 40%
thank you for the explanation
#8
09-07-2011
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 623 Threads: 111 Thanked 424 Times in 264 Posts Reputation: 434

Quote:
 Originally Posted by laurier 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.
#9
09-07-2011
 USMLE Forums Master Steps History: Step 1 Only Posts: 566 Threads: 25 Thanked 522 Times in 263 Posts Reputation: 532

Quote:
 Originally Posted by laurier yes you r right, the answer is 40% thank you for the explanation
You welcome =)
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Nothing is Impossible, the Word Itself says "I'm Possible"
#10
09-07-2011
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 623 Threads: 111 Thanked 424 Times in 264 Posts Reputation: 434

Quote:
 Originally Posted by rulz 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
#11
09-07-2011
 USMLE Forums Master Steps History: Step 1 Only Posts: 566 Threads: 25 Thanked 522 Times in 263 Posts Reputation: 532

Quote:
 Originally Posted by aknz 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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Nothing is Impossible, the Word Itself says "I'm Possible"
 The above post was thanked by: aknz (09-07-2011)
#12
09-07-2011
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 623 Threads: 111 Thanked 424 Times in 264 Posts Reputation: 434

Quote:
 Originally Posted by rulz 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)
#13
09-07-2011
 USMLE Forums Master Steps History: Step 1 Only Posts: 566 Threads: 25 Thanked 522 Times in 263 Posts Reputation: 532

Quote:
 Originally Posted by aknz 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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Nothing is Impossible, the Word Itself says "I'm Possible"
 The above post was thanked by: aknz (09-07-2011)
#14
09-07-2011
 USMLE Forums Scout Steps History: 1+CK+CS Posts: 69 Threads: 4 Thanked 76 Times in 21 Posts

Quote:
 Originally Posted by dr_lizard 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
 The above post was thanked by: dr_lizard (09-07-2011)

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