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#1




Autosomal Dominant Disease with 80% Penetrance?
Here's an easy selfmade 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




B) 25% :d:d:d

#3




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




@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




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




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




@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




my answer
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%) 
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Hope2Pass (05052012) 
#9




Quote:
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#10




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. 10.6 = 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 (05052012), DocSikorski (05052012), Dr. Mexito (05052012), Farrukh Munir (05102012), islandcrazy (05052012), koolkiller88 (05052012), neelima208 (05062012), tulipa (05052012) 
#11




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#12




It's just saying that there's 20% chance that the offspring will not show the disease phenotypically.

#13




.
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 (05052012) 
#14




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 (05052012) 
#15




.
Quote:
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




.
Quote:
the term penetrance doesn't refer to single allele. it refers to individual with a set of alleles. 
#17




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




oh does it?
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#19




.
Quote:
penetrance refers only to those individuals who have the mutaded allele (here it's dominant so A): Aa and AA. 
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husseinalaa (05052012) 
#20




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




that makes very good sense, appreciate it casandra.

#22




.
"(...)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




My logical method also deserves a credit!
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Ambition is a dream with a V8 engine. EP 
#24




Yes, your logical method did deserve a credit...It got you to the right answer some how

#25




Quote:
lol I finished at your step 1 so I get 25% haha thanks man very helpful 
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Hope2Pass (05052012) 
#26




How do u know the parents are heterozygous?
The parents can be diseased due to homozygous as well, right? 
#27




.
Quote:
"They are both heterozygotes." 
#28




Quote:
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 selfmade)... 
#29




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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