Genetics Question - USMLE Forums
 USMLE Forums         Your Reliable USMLE Online Community     Members     Posts
 Home USMLE Articles USMLE News USMLE Polls USMLE Books USMLE Apps
 USMLE Forums Genetics Question

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

#1
11-08-2009
 USMLE Forums Newbie Steps History: Not yet Posts: 8 Threads: 2 Thanked 11 Times in 5 Posts Reputation: 21
Genetics Question

A couple of Jews are about to marry. Both of them have siblings who died from a Tay Sach's disease. The frequency of heterozygotes in the Ashkenazi Jews is 0.1. The couple are asking what is the probability that they have a child affected by the disease?

a- 0.11
b- 0.25
c- 0.0025
d- 0.0625
e- 0.33
 The above post was thanked by: ashishkabir (04-30-2011), smile135 (11-08-2009), Taiwan_Guy (05-19-2011)

#2
11-08-2009
 USMLE Forums Scout Steps History: 1 + CK Posts: 33 Threads: 3 Thanked 35 Times in 14 Posts Reputation: 45

0.1 X 0.1 X 0.25 = 0.0025 answer D
#3
11-08-2009
 USMLE Forums Newbie Steps History: Not yet Posts: 8 Threads: 2 Thanked 11 Times in 5 Posts Reputation: 21

Quote:
 Originally Posted by indrag 0.1 X 0.1 X 0.25 = 0.0025 answer D

#4
11-08-2009
 USMLE Forums Scout Steps History: Not yet Posts: 18 Threads: 1 Thanked 14 Times in 5 Posts Reputation: 24

Quote:
What is the correct answer then?
#5
11-08-2009
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 3,316 Threads: 170 Thanked 5,039 Times in 1,820 Posts Reputation: 5092

I think that each parent has 2/3 chance of being a carrier (since they have an affected sibling) so
2/3 X 2/3 X 1/4 = 4/36 = 0.11 (answer A)

 The above post was thanked by: doc_study (04-29-2011), hugepuppy (05-02-2011), kirparo (01-03-2014), smile135 (11-08-2009), Taiwan_Guy (05-19-2011), vikash (10-12-2011)
#6
11-08-2009
 USMLE Forums Newbie Steps History: Not yet Posts: 8 Threads: 2 Thanked 11 Times in 5 Posts Reputation: 21

Quote:
 Originally Posted by Sabio I think that each parent has 2/3 chance of being a carrier (since they have an affected sibling) so 2/3 X 2/3 X 1/4 = 4/36 = 0.11 (answer A)
 The above post was thanked by: AshNiffler (02-20-2015)
#7
04-29-2011
 USMLE Forums Addict Steps History: 1+CK+CS Posts: 189 Threads: 20 Thanked 511 Times in 102 Posts Reputation: 521

I still havent figured it out..Could you help me.
My line of thinking was such
.
Now the frequency of heterozygosity is 0.1
If we use hardy Weinberg equation
2q=0.1 (Considering p being very large and almost equal to 1)
therefore q= 0.05
Now for the disease to manifest as homozygous we need q^2
So 0.05*0.05=0.0025

This is what i thougt..Correct me where i have gone wrong..
Or is it you cant apply this equation here?
 The above post was thanked by: doc_study (04-29-2011), noothan (04-29-2011)
#8
04-29-2011
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 3,316 Threads: 170 Thanked 5,039 Times in 1,820 Posts Reputation: 5092
There's a trick here

Quote:
 Originally Posted by mayursn39 I still havent figured it out..Could you help me. My line of thinking was such . Now the frequency of heterozygosity is 0.1 If we use hardy Weinberg equation 2q=0.1 (Considering p being very large and almost equal to 1) therefore q= 0.05 Now for the disease to manifest as homozygous we need q^2 So 0.05*0.05=0.0025 This is what i thougt..Correct me where i have gone wrong.. Or is it you cant apply this equation here?
The trick here, is that the probability of the parents being carriers is not similar to the population at large, it's because they had siblings who died from the disease, so their chances of being a carrier is 2/3 [One died, One Carrier, One Carrier, One Normal] i.e. the chance of being a carrier in each of the parents is 2 out of 3.
So in that case you don't need to apply the Hardy Weinberg equation, as you have more specific probabilities for these mating parents.

-
-

 The above post was thanked by: bebix (04-30-2011), Breex (04-29-2011), doc_study (04-29-2011), kirparo (01-03-2014), luisguaman (05-04-2011), Mondoshawan (04-30-2011), Taiwan_Guy (05-19-2011), usluipek (05-01-2011)
#9
04-30-2011
 USMLE Forums Addict Steps History: 1+CK+CS Posts: 189 Threads: 20 Thanked 511 Times in 102 Posts Reputation: 521

Thank you, now i get it.

 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 doctorF USMLE Step 1 Forum 5 02-14-2011 07:33 PM good_boy_1234 USMLE Step 1 Forum 3 06-08-2010 12:05 PM Dr. N Elham USMLE Step 1 Forum 3 12-07-2009 06:45 PM vascular USMLE Step 1 Forum 18 12-02-2009 03:20 PM