








USMLE Step 1 Forum USMLE Step 1 Discussion Forum: Let's talk about anything related to USMLE Step 1 exam 

Thread Tools  Search this Thread  Display Modes 
#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.
__________________
"inflammable means flammable!? What a country." 
#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...
__________________
Ambition is a dream with a V8 engine. EP 
#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%) 
The above post was thanked by:  
Hope2Pass (05052012) 
#9




Quote:
__________________
Ambition is a dream with a V8 engine. EP 
#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




__________________
"inflammable means flammable!? What a country." 
#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?
__________________
"inflammable means flammable!? What a country." 
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?
__________________
"inflammable means flammable!? What a country." 
#19




.
Quote:
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 (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!
__________________
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 
The above post was thanked by:  
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!

Tags 
Genetics, Step1Questions 
Thread Tools  Search this Thread 
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
AUTOSOMAL Dominant disorders mnemonic  usmleman2020  USMLE Step 1 Mnemonics  13  10062015 07:46 AM 
List of Autosomal Dominant Diseases  rasheed  USMLE Step 1 Bits & Pieces  11  05302013 05:34 PM 
Autosomal Dominant, Planning to have 3 children  laurier  USMLE Step 1 Forum  26  07252011 11:44 AM 
Autosomal dominant disorder and normal offspring  bebix  USMLE Step 1 Forum  12  07122011 11:53 AM 
Autosomal Dominant Inheritance Question  Claus_CU  USMLE Step 1 Forum  8  06072011 09:45 PM 
