Autosomal dominant disorder and normal offspring - USMLE Forums
 USMLE Forums         Your Reliable USMLE Online Community     Members     Posts
 Home USMLE Articles USMLE News USMLE Polls USMLE Books USMLE Apps
 USMLE Forums Autosomal dominant disorder and normal offspring

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

#1
07-12-2011
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,357 Threads: 194 Thanked 3,360 Times in 900 Posts Reputation: 3370
Autosomal dominant disorder and normal offspring

A man and woman are both affected by an autosomal dominant disorder that has 80% penetrance. They are both heterozygotes for the disease-causing mutation. What is the probability that they will produce phenotypically normal offspring?

A) 20%
B) 25%
C) 40%
D) 60%
E) 80%
 The above post was thanked by: pass7 (07-12-2011), usluipek (07-12-2011)

#2
07-12-2011
 USMLE Forums Addict Steps History: Not yet Posts: 140 Threads: 5 Thanked 120 Times in 68 Posts Reputation: 130

phenotypically- 40%?
 The above post was thanked by: bebix (07-12-2011)
#3
07-12-2011
 USMLE Forums Addict Steps History: 1+CK+CS Posts: 124 Threads: 5 Thanked 217 Times in 73 Posts Reputation: 227
Ans.

 The above post was thanked by: bebix (07-12-2011)

#4
07-12-2011
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,684 Threads: 213 Thanked 1,652 Times in 669 Posts Reputation: 1662

If you work out a punnett square then the chance of a normal child is 25%.
But Im confused about the 80% penetrance. Does that mean you subtract 20% (100-80) of 25% ? .. so 25% x 0.2 = 5%. Therefore the probability to have a normal child will be 20%.
 The above post was thanked by: bebix (07-12-2011)
#5
07-12-2011
 USMLE Forums Guru Steps History: --- Posts: 332 Threads: 48 Thanked 181 Times in 117 Posts

CCCCCCCCCC
Husband: Aa, Mom: Aa
mating results: AA, Aa Aa aa
offspings' chance of having disease = 75 % x 80% =0.6
chance of being normal = 1- 0.6 =0.4
 The above post was thanked by: bebix (07-12-2011)
#6
07-12-2011
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,357 Threads: 194 Thanked 3,360 Times in 900 Posts Reputation: 3370

Quote:
 Originally Posted by usluipek CCCCCCCCCC Husband: Aa, Mom: Aa mating results: AA, Aa Aa aa offspings' chance of having disease = 75 % x 80% =0.6 chance of being normal = 1- 0.6 =0.4
So, in this case the correct answer is C)
but offspings' chance of having disease = 75 %!!!...
phenotypically chance of having disease = 75 % x 80% =0.6
 The above post was thanked by: jahn77 (07-12-2011), usluipek (07-12-2011)
#7
07-12-2011
 USMLE Forums Guru Steps History: --- Posts: 332 Threads: 48 Thanked 181 Times in 117 Posts

chances of offspring to carry abnormal genes is 75%

if the abnormal gene does not express itself, then there is no disease manifestations.

so when you say disease, it depends on whether you mean genotype or phenotype
#8
07-12-2011
 USMLE Forums Master Steps History: 1+CK+CS Posts: 570 Threads: 26 Thanked 554 Times in 276 Posts Reputation: 564

80% x 1/4 ( which represent the normal phenotype ) = 20% ...
Im not good at this,, i just try...

----- A ------ a
A---A A ----- Aa
a---A a------ aa

This disease is DOMINANT which means just 1 alele is require to express the disease so 3/4 are Sicks 1/4 is normal...

Please if im wrong explain it... =)

So According to my plots Answer is A
#9
07-12-2011
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,357 Threads: 194 Thanked 3,360 Times in 900 Posts Reputation: 3370

Quote:
 Originally Posted by rulz 80% x 1/4 ( which represent the normal phenotype ) = 20% ... Im not good at this,, i just try... ----- A ------ a A---A A ----- Aa a---A a------ aa This disease is DOMINANT which means just 1 alele is require to express the disease so 3/4 are Sicks 1/4 is normal... Please if im wrong explain it... =) So According to my plots Answer is A
The correct answer is this case is C)
If both parent are heterozygotes, there is a 75% chance that their offspring will receive one or two copies of the disease-causing gene. With 80% penetrance, the probability that the offspring will be affected (phenotypically) is 0.75 * 0.8 = 60%. The probability that the offspring will be phenotypically normal is 1 - 0.6 = 0.4 or 40%
 The above post was thanked by: pass7 (07-12-2011), rulz (07-12-2011)
#10
07-12-2011
 USMLE Forums Master Steps History: 1+CK+CS Posts: 570 Threads: 26 Thanked 554 Times in 276 Posts Reputation: 564

Thanks Bebix....=)
#11
07-12-2011
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,684 Threads: 213 Thanked 1,652 Times in 669 Posts Reputation: 1662

Quote:
 Originally Posted by bebix The correct answer is this case is C) If both parent are heterozygotes, there is a 75% chance that their offspring will receive one or two copies of the disease-causing gene. With 80% penetrance, the probability that the offspring will be affected (phenotypically) is 0.75 * 0.8 = 60%. The probability that the offspring will be phenotypically normal is 1 - 0.6 = 0.4 or 40%
Thanks for the explanation! But the question asks for the probability of producing a phenotypically normal offspring. Yes there is a 75% chance of transferring the gene with 80% penetrance but the 25% chance of not transferring the gene will also result in a phenotypically normal offspring. Dont you have to take that 25% in consideration ?
 The above post was thanked by: drgsarunprasath (07-12-2011)
#12
07-12-2011
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,357 Threads: 194 Thanked 3,360 Times in 900 Posts Reputation: 3370

Quote:
 Originally Posted by Hope2Pass Thanks for the explanation! But the question asks for the probability of producing a phenotypically normal offspring. Yes there is a 75% chance of transferring the gene with 80% penetrance but the 25% chance of not transferring the gene will also result in a phenotypically normal offspring. Dont you have to take that 25% in consideration ?
well, actually you are taking the 75% in consideration (which is 1-.25)...
#13
07-12-2011
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,357 Threads: 194 Thanked 3,360 Times in 900 Posts Reputation: 3370

two ways to answer this question:
Aa x Aa = AA Aa Aa aa
80% penetrance

a)
0.75 * 0.80 = 60% affected (phenotypically)
100-60 = 40% "normal"

b)
0.25 (aa) * [0.75 * 0.2]
where, [0.75 * 0.2] = phenotypically normal
= 40%
 The above post was thanked by: drgsarunprasath (07-12-2011), drortho (11-25-2011), Hope2Pass (07-12-2011), monsalg2502 (08-24-2011)

 Tags Genetics-, Step-1-Questions

Message:
Options

Register Now

In order to be able to post messages on the USMLE Forums forums, you must first register.
User Name:
Medical School
Choose "---" if you don't want to tell. AMG for US & Canadian medical schools. IMG for all other medical schools.
 AMG IMG ---
USMLE Steps History
What steps finished! Example: 1+CK+CS+3 = Passed Step 1, Step 2 CK, Step 2 CS, and Step 3.

Choose "---" if you don't want to tell.

 Not yet Step 1 Only CK Only CS Only 1 + CK 1 + CS 1+CK+CS CK+CS 1+CK+CS+3 ---
Favorite USMLE Books
 What USMLE books you really think are useful. Leave blank if you don't want to tell.
Location
 Where you live. Leave blank if you don't want to tell.

Log-in

 Similar Threads Thread Thread Starter Forum Replies Last Post rasheed USMLE Step 1 Bits & Pieces 11 05-30-2013 05:34 PM Claus_CU USMLE Step 1 Forum 8 06-07-2011 09:45 PM Ayshee USMLE Step 2 CK Mnemonics 1 10-30-2010 12:33 PM Telsa-Farad USMLE Step 2 CK Forum 2 06-15-2010 04:08 PM obgynaim USMLE Step 1 Forum 1 12-20-2009 10:06 AM