Genetic risk calculation - USMLE Forums
 USMLE Forums         Your Reliable USMLE Online Community     Members     Posts
 Home USMLE Articles USMLE News USMLE Polls USMLE Books USMLE Apps
 USMLE Forums Genetic risk calculation
 Register FAQs Members List Search Today's Posts Mark Forums Read

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

#1
05-04-2011
 USMLE Forums Guru Steps History: --- Posts: 332 Threads: 48 Thanked 168 Times in 107 Posts
Genetic risk calculation

I saw this question somewhere else and thought I will share with you all to discuss: Please provide explanation with answer.

A woman has a genetic disease with autosomal dominance transmission. Her husband is unaffected. They plan to have a family with three children. What is the probability that one of the three children will be affected?
A. 1/8
B. 1/4
C. 3/8
D. 1/3
E. 7/8
 The above post was thanked by: heights (05-04-2011), Sparco (05-05-2011), Taiwan_Guy (05-15-2011)

#2
05-04-2011
 USMLE Forums Addict Steps History: 1+CK+CS Posts: 189 Threads: 20 Thanked 506 Times in 99 Posts Reputation: 516
I think the answer is A

Since no one has answered till now let me try..I may be wrong as i am not good in genetics

Considering the female is heterozygous for the trait..
(Homozygous would lead to all children being affected)
At each child birth there would be 1/2 chance of the child being affected.
**Therefore the chance of not being affected is also 1/2 (1-1/2=1/2 )

As asked in the question we want only one child to be affected.
So 3 events should take place..
That is one event with child having the disease i.e 1/2 chance
The 2nd event with child not having the disease i.e 1/2 chance(from ** above)
The third event with child not having the disease i.e 1/2 (again from **)
If we consider all the three events occurring together 1/2*1/2*1/2=1/8
so the probability of having one child affected out of three should be 1/8
This corresponds to option a)
AGAIN I SAY. I MAY BE WRONG, PLEASE CORRECT ME IF YOU THINK I AM WRONG
 The above post was thanked by: usluipek (05-04-2011)
#3
05-04-2011
 USMLE Forums Guru Steps History: --- Posts: 332 Threads: 48 Thanked 168 Times in 107 Posts

Other people pls join in, we want to reach a concensus.
#4
05-04-2011
 USMLE Forums Master Steps History: 1+CK+CS Posts: 590 Threads: 31 Thanked 1,227 Times in 410 Posts Reputation: 1251

I think mayursn started off fine, but I would have said that the probability that no child would be affected is 1/2*1/2*1/2=1/8, therefore the probability that at least one child would be affected is 7/8, so I would have picked E.
 The above post was thanked by: usluipek (05-04-2011)
#5
05-04-2011
 USMLE Forums Addict Steps History: 1+CK+CS Posts: 189 Threads: 20 Thanked 506 Times in 99 Posts Reputation: 516

I think I got the question wrong..The question says only one child out of three or at least one out of three?
because i think then the answer would change
#6
05-04-2011
 USMLE Forums Master Steps History: Not yet Posts: 557 Threads: 69 Thanked 1,038 Times in 351 Posts Reputation: 1058

Yeah, I think I'd probably go with e) also, because the more kids you have, the greater the probability that one will have the disease.
 The above post was thanked by: usluipek (05-04-2011)
#7
05-04-2011
 USMLE Forums Addict Steps History: 1+CK+CS Posts: 143 Threads: 4 Thanked 178 Times in 59 Posts Reputation: 188

Probability of each child should be taken as an individual event..
for every child probability of having disease is 1/2

so for 3 children by individual probability
1/2*1/2*1/2 = 1/8

but if u consider child not having disease for each child, apply same logic
we will arrive at 1/2*1/2*1/2 = 1/8
so child probability of child with disease is 1-1/8 = 7/8

Where am I going wrong?
#8
05-04-2011
 USMLE Forums Master Steps History: 1+CK+CS Posts: 590 Threads: 31 Thanked 1,227 Times in 410 Posts Reputation: 1251

Quote:
 Originally Posted by JaJeeK Probability of each child should be taken as an individual event.. for every child probability of having disease is 1/2 so for 3 children by individual probability 1/2*1/2*1/2 = 1/8 but if u consider child not having disease for each child, apply same logic we will arrive at 1/2*1/2*1/2 = 1/8 so child probability of child with disease is 1-1/8 = 7/8 Where am I going wrong?
Your first calculation is the probability that all three children will have the disease.

Your second calculation is the probability that at least one child will have the disease.

To answer mayursn39's idea that maybe they could be asking about one and only one child: that would be the combined probability of any one of the three fixed outcomes that fit the profile ∈ {(A+,B-,C-),(A-,B+,C-),(A-,B-,C+)}, right? So that would be 1/8+1/8+1/8 = 3/8. However, I think that the question is looking for 7/8...
 The above post was thanked by: mayursn39 (05-04-2011)
#9
05-05-2011
 USMLE Forums Scout Steps History: 1+CK+CS Posts: 88 Threads: 25 Thanked 17 Times in 10 Posts Reputation: 27

I would say, in most autosomal dominant cases, the diseased is... Aa
So mating Aa into aa produces 1/2 diseases if there are 4 children...
so if you have to calculate out of 3, the probability will increase from 1/2....so I think it will b 7/8....
I might be wrong though ...
#10
06-07-2011
 USMLE Forums Addict Steps History: CK Only Posts: 178 Threads: 61 Thanked 250 Times in 105 Posts Reputation: 260

If the question was "What is the probability that ONLY one of the three children will be affected?" then the answer would be 1/8.

If the question was "What is the probability that AT LEAST one of the three children will be affected?" then the answer would be 7/8.

I think the question is not clear. How do you feel about it?
#11
11-29-2011
 USMLE Forums Veteran Steps History: Step 1 Only Posts: 277 Threads: 36 Thanked 158 Times in 95 Posts Reputation: 168

How can we justify the 1/2?? could the mom not be AA and thus 100% of her offsprings will recieve it?

I HATE PROBABILITY! - anyone has a quick fix?
__________________
Set your mind to it, and you will be there.
To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
#12
11-29-2011
 USMLE Forums Guru Steps History: 1 + CS Posts: 300 Threads: 22 Thanked 97 Times in 62 Posts Reputation: 107

c
its 1/8+1/8+1/8 =3/8

 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

Human Verification

In order to verify that you are a human and not a spam bot, please enter the answer into the following box below based on the instructions contained in the graphic.