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#1
05-05-2012
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Autosomal Dominant Disease with 80% Penetrance?

Here's an easy self-made question -
---------------------------------------------------------------

A couple is affected by an Autosomal Dominant disorder which has a penetrance of 80% in affected individuals. They are both heterozygotes. What is the probability of producing a phenotypically normal offspring ?

A) 20%
B) 25%
C) 60%
D) 40%
E) 80%

#2
05-05-2012
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B) 25% :d:d:d
#3
05-05-2012
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Aa x Aa = 25% chance of aa (normal)

the remaining 75% is affected. 50% of which is Aa, of which there is an 20% chance of being normal so 10% of the 50% is normal.

so now we're up to 35%.

the 25% that is AA has 0.20 x 0.20 chance of being normal; 25 x 0.04 = 1% of the 25%.

so now we're up to 36%.

which isn't an option so i fcuked up lol.
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#4
05-05-2012
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@tarek: Nice try
@drNick: You're kind of on the right path but don't confuse yourself with so many calculations! It's easier than it looks
#5
05-05-2012
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i would go with 40%
=============

in 100% penetrance there would be 25% chance to become homozygous normal and heterozygotes would be affected.

but since there is 80% penetrance, heterozygotes have 20% chance to be phenotypically normal, so overal chances should be higher than just 25...
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#6
05-05-2012
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Yeah I got 35% so I guess I'm missing something:

Parents are Aa.
1/4 can be aa= 0.25
2/4 of childern are Aa. There is a 0.2 those Aa childern are normal= 0.1
=35%

Can you explain what we are doing wrong?
#7
05-05-2012
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@DocSikorski: Do you have any way of calculating to prove your answer? Not saying if you're right or not, but how did you reach at 40% ?

@islandcrazy: I will explain the answer in a few minutes. I just want to see if anyone gets the right answer. A similar question was in the Qbank and only 37% of the people got it right (I was one of them) ... I'll post the explanation soon!
#8
05-05-2012
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I think it's 40% (D)

AA = 1/4 x 20% = 0.05

There will be 2 heterozygotes (Aa) and for each of them I would calculate the probability separately, so:

Aa = 1/4 x 20% = 0.05
Aa = 1/4 x 20% = 0.05

and 1 homozygous
aa = 0.25

so all together 0.40 (40%)
 The above post was thanked by: Hope2Pass (05-05-2012)
#9
05-05-2012
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,353 Threads: 94 Thanked 877 Times in 507 Posts Reputation: 890

Quote:
 Originally Posted by Hope2Pass @DocSikorski: Do you have any way of calculating to prove your answer? Not saying if you're right or not, but how did you reach at 40% ? @islandcrazy: I will explain the answer in a few minutes. I just want to see if anyone gets the right answer. A similar question was in the Qbank and only 37% of the people got it right (I was one of them) ... I'll post the explanation soon!
Well, just logically it seemed to me that having 2 parents with lets say, neurofibromatosis i would say that there is definitely less then 50% chance to look ok
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#10
05-05-2012
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,684 Threads: 213 Thanked 1,600 Times in 660 Posts Reputation: 1610

The correct answer is D) 40%

Here's how I worked it out:

Since its AD, there is 75% chance of producing an affected offspring (Aa, Aa, AA) and 25% chance of producing a normal offspring (aa), because the couple in question is heterozygous.

Since the penetrance of the disease is 80%, this means that out of the 75% offsprings produced, only 80% of the 75% will show the phenotype.

0.8 * 0.75 = 0.6

This means that the couple has a 60 % chance of producing an affected offspring that will have complete peneterance or show the disease phenotypically.

1-0.6 = 0.4 = 40%

Therefore, there is a 40% chance of producing a phenotypically normal offspring.

Also, the other way to calculate this is the way Casandra described it! Good job!
 The above post was thanked by: Casandra (05-05-2012), DocSikorski (05-05-2012), Dr. Mexito (05-05-2012), Farrukh Munir (05-10-2012), islandcrazy (05-05-2012), koolkiller88 (05-05-2012), neelima208 (05-06-2012), tulipa (05-05-2012)
#11
05-05-2012
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Quote:
 Originally Posted by Casandra AA = 1/4 x 20% = 0.05
i dont get that part
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#12
05-05-2012
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Quote:
 Originally Posted by Dr.NickRiviera i dont get that part
It's just saying that there's 20% chance that the offspring will not show the disease phenotypically.
#13
05-05-2012
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.

Quote:
 Originally Posted by Dr.NickRiviera i dont get that part
parents are both heterozygous so there's a 1/4 chance to get a AA homozygous child. And this homozygous has a 20% chance of being healthy.
So a chance for his couple of having AA healthy child is: 1/4 x 20% = 0.05
 The above post was thanked by: Hope2Pass (05-05-2012)
#14
05-05-2012
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Quote:
 Originally Posted by Casandra parents are both heterozygous so there's a 1/4 chance to get a AA homozygous child. And this homozygous has a 20% chance of being healthy. So a chance for his couple of having AA healthy child is: 1/4 x 20% = 0.05
isnt it 20% x 20% since you have two dominant alleles, and its 80% penetrance per allele?
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 The above post was thanked by: islandcrazy (05-05-2012)
#15
05-05-2012
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.

Quote:
 Originally Posted by Dr.NickRiviera isnt it 20% x 20% since you have two dominant alleles, and its 80% penetrance per allele?
but we're not calculating probability for the alleles but for an individual child (so 2 alleles .

Two heterozygous parents have:
- 1/4 chance of having homozygous AA (that's what i used in calculations)
- 1/2 chance of having heterozygots Aa
- 1/4 chance of having homozygous aa

80% is not a chance per allele but per individual (so per 2 alleles .

I used 20% because I was calculating a chance of having a healthy child (1 - 80% = 20%).

#16
05-05-2012
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.

Quote:
 Originally Posted by Dr.NickRiviera isnt it 20% x 20% since you have two dominant alleles, and its 80% penetrance per allele?

the term penetrance doesn't refer to single allele. it refers to individual with a set of alleles.
#17
05-05-2012
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I have a question for you guys,

why wouldn't be in this way;

diseased allel is A, so we have AA, Aa, Aa, aa.
aa is the normal with 25% chance, so it will be 0.8*0.25= 20%

is that wrong? and why?
#18
05-05-2012
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Quote:
 Originally Posted by Casandra the term penetrance doesn't refer to single allele. it refers to individual with a set of alleles.
oh does it?
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#19
05-05-2012
 USMLE Forums Master Steps History: 1+CK+CS Posts: 1,039 Threads: 189 Thanked 570 Times in 323 Posts Reputation: 580
.

Quote:
 Originally Posted by husseinalaa I have a question for you guys, why wouldn't be in this way; diseased allel is A, so we have AA, Aa, Aa, aa. aa is the normal with 25% chance, so it will be 0.8*0.25= 20% is that wrong? and why?
we know for sure that aa homozygous is healthy so there will be no penetrance issue here.
penetrance refers only to those individuals who have the mutaded allele (here it's dominant so A): Aa and AA.
 The above post was thanked by: husseinalaa (05-05-2012)
#20
05-05-2012
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,684 Threads: 213 Thanked 1,600 Times in 660 Posts Reputation: 1610

Hey guys, just realized that Bebix has posted this question before, back in July of last year! I got it wrong that time lol

Check out the discussion here:

Autosomal dominant disorder and normal offspring

Although, I just went over it and bebix explained one of the methods to be this:

b)
0.25 (aa) * [0.75 * 0.2]
where, [0.75 * 0.2] = phenotypically normal
= 40%

I dont know how that comes out to 40% because 0.75*0.2*0.25 = 0.0375

Anyways, you can read the discussion
#21
05-05-2012
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Quote:
 Originally Posted by Casandra we know for sure that aa homozygous is healthy so there will be no penetrance issue here. penetrance refers only to those individuals who have the mutaded allele (here it's dominant so A): Aa and AA.
that makes very good sense, appreciate it casandra.
#22
05-05-2012
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.

Quote:
 Originally Posted by Dr.NickRiviera oh does it?
"(...)disorder which has a penetrance of 80% in affected individuals."

in the stem that say about disorder penetrance so for a set of alleles.

of course if they had mentioned it for individual allele we would have calculated it differently
#23
05-05-2012
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Quote:
 Originally Posted by Hope2Pass The correct answer is D) 40% Also, the other way to calculate this is the way Casandra described it! Good job!
My logical method also deserves a credit!
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Ambition is a dream with a V8 engine. EP
#24
05-05-2012
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,684 Threads: 213 Thanked 1,600 Times in 660 Posts Reputation: 1610

Quote:
 Originally Posted by DocSikorski My logical method also deserves a credit!
Yes, your logical method did deserve a credit...It got you to the right answer some how
#25
05-05-2012
 USMLE Forums Scout Steps History: Not yet Posts: 16 Threads: 5 Thanked 3 Times in 3 Posts Reputation: 13

Quote:
 Originally Posted by Hope2Pass The correct answer is D) 40% Here's how I worked it out: Since its AD, there is 75% chance of producing an affected offspring (Aa, Aa, AA) and 25% chance of producing a normal offspring (aa), because the couple in question is heterozygous. Since the penetrance of the disease is 80%, this means that out of the 75% offsprings produced, only 80% of the 75% will show the phenotype. 0.8 * 0.75 = 0.6 This means that the couple has a 60 % chance of producing an affected offspring that will have complete peneterance or show the disease phenotypically. 1-0.6 = 0.4 = 40% Therefore, there is a 40% chance of producing a phenotypically normal offspring. Also, the other way to calculate this is the way Casandra described it! Good job!

lol I finished at your step 1 so I get 25% haha thanks man very helpful
 The above post was thanked by: Hope2Pass (05-05-2012)

#26
05-06-2012
 USMLE Forums Master Steps History: 1 + CS Posts: 675 Threads: 84 Thanked 428 Times in 207 Posts Reputation: 438

How do u know the parents are heterozygous?

The parents can be diseased due to homozygous as well, right?
#27
05-06-2012
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.

Quote:
 Originally Posted by Renaissance How do u know the parents are heterozygous? The parents can be diseased due to homozygous as well, right?
It says so in the stem

"They are both heterozygotes."
#28
05-06-2012
 USMLE Forums Master Steps History: CS Only Posts: 557 Threads: 29 Thanked 296 Times in 186 Posts Reputation: 316

Quote:
 Originally Posted by Hope2Pass Here's an easy self-made question - --------------------------------------------------------------- A couple is affected by an Autosomal Dominant disorder which has a penetrance of 80% in affected individuals. They are both heterozygotes. What is the probability of producing a phenotypically normal offspring ? A) 20% B) 25% C) 60% D) 40% E) 80%
This question seemed pretty easy to me, so I will try to explain is as good as possible. I know it has been explained already, maybe I can help clear up some last questions...

So 25% of the offspring are not affected at all, they have a normal genotype and therefore a normal phenotype.
The rest (75%) are affected - 25% are homozygotes, 50% heterozygotes. Their genotype is not normal. But since the disease only has 80% penetrance, 20% of those 75% have a NORMAL PHENOTYPE, which would be another 15% overall.
So, 25% of the offspring have a normal genotype and phenotype, and another 15% have an affected genotype but a normal phenotype -> 40% are phenotypically normal.

Hope this helps!

Btw, did you make this question yourself? (You said it is self-made)...
#29
05-06-2012
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,684 Threads: 213 Thanked 1,600 Times in 660 Posts Reputation: 1610

Quote:
 Originally Posted by val7 Btw, did you make this question yourself? (You said it is self-made)...
I came across the question in Kaplan Qbank but reworded it to make it seem like I made the question although I wanted to use the same concept! It's too small of a question to reword though lol Anyways, the thread bebix started is the same question as it appears in Qbank so I guess it doesn't matter if I reworded the question since its already been shared! Good concept behind the question though. Thanks for your explanation!

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