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#1
05-15-2012
 USMLE Forums Scout Steps History: 1+CK+CS Posts: 65 Threads: 14 Thanked 23 Times in 15 Posts Reputation: 33
Population Genetics; Autosomal Recessive Disease; Probability of affected child?

A couple approach a physician concerned about the risk of having a child with autosomal recessive disease which has a frequency of 1/10,000 newborn. The couple is unrelated and both of them have no family history of the disease. Which represents the risk of having a child with the disease?

A. 1/100
B. 1/200
C. 1/400
D. 1/2,500
E. 1/10,000

Answer with explanation will be appreciated
 The above post was thanked by: MEERA@MD (4 Weeks Ago)

#2
05-15-2012
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,684 Threads: 213 Thanked 1,600 Times in 660 Posts Reputation: 1610

E.1/10,000

q squared = frequency of the AR disease = 1/10,000
q = 1/100 (take the square root)

assume p = 1

2q = frequency of heterozygous carriers in population = 2(1/100) = 1/50

Chance of father being a carrier = 1/50
Chance of mother being a carrier = 1/50
Chance of having an affected offspring if both are carriers = 1/4 (for AR diseases)

multiply all the probabilites =

1/50 * 1/50 * 1/4 = 1/10,000

I hope that's right.
---------------------------------------------------------------------

YOU HAVE TO ASSUME P = 1 in this question .. Don't try to solve it or else you'll end up with something like 9801/100,000,000
That was the first answer I got and I was like WTF
 The above post was thanked by: MEERA@MD (4 Weeks Ago), pradeeprox (05-16-2012)
#3
05-15-2012
 USMLE Forums Master Steps History: Not yet Posts: 590 Threads: 8 Thanked 239 Times in 192 Posts Reputation: 249

Quote:
 Originally Posted by Hope2Pass E.1/10,000 q squared = frequency of the AR disease = 1/10,000 q = 1/100 (take the square root) assume p = 1 2q = frequency of heterozygous carriers in population = 2(1/100) = 1/50 Chance of father being a carrier = 1/50 Chance of mother being a carrier = 1/50 Chance of having an affected offspring if both are carriers = 1/4 (for AR diseases) multiply all the probabilites = 1/50 * 1/50 * 1/4 = 1/10,000 I hope that's right. --------------------------------------------------------------------- YOU HAVE TO ASSUME P = 1 in this question .. Don't try to solve it or else you'll end up with something like 9801/100,000,000 That was the first answer I got and I was like WTF

I go for E as well..

#4
05-15-2012
 Guest Steps History: --- Posts: 197 Threads: 9 Thanked 98 Times in 64 Posts Reputation: 108

I'd go for E too. There's no need to do all the math here. Unless the question stem gives you some very vital info, such as a +ve family hx/genetic test result in the parents, you can just go ahead and blindly pick whatever is the current rate in the population.
#5
05-16-2012
 USMLE Forums Master Steps History: 1+CK+CS Posts: 662 Threads: 72 Thanked 370 Times in 204 Posts Reputation: 380

Quote:
 Originally Posted by pradeeprox a couple approach a physician concerned abt the risk of havin a child with autosomal recessive disease which has a freq of 1/10,000 newborn. The couple is unrelated and both of them have no family history of the disease . which represents the risk of having a child with the disease A.1/100 B.1/200 C.1/400 D.1/2,500 E.1/10,000 ans with explanation will be appreciated
This is what I think

2pq = carrier
q^2 = 1/10,000
q = 1/100 = 0.01 (This is the way I do it, other wise I get stumble)

p = 1-q (I know you can approximate this, but if I do so... my brain cannot handle that much complexity )
p = 0.99

2pq = 2(0.99*0.01)
2pq = 0.0198 = 0.02

Now, since mom and dad are not affected... we have to think that they are carriers:
mom carrier = .02

Now, in AR disorders, mom and dad have to pass the defective gene to the baby, and the probability of doing so is 1/25 = 25% = 0.25

So, now... we multiply the probability of mom * dad * probability of passing the disease =

0.0004 * 0.25 = 0.0001 = 1/10,000

I do not know... maybe I'm wrong... lets see what the rest of the people think about it.

(Don't be so hard on me if I made a fool out of myself here)
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"Disease is very old, and nothing about it has changed. It is we who change as we learn to recognize what was formerly imperceptible." JMC
 The above post was thanked by: DocSikorski (05-16-2012), pradeeprox (05-16-2012)
#6
05-16-2012
 USMLE Forums Scout Steps History: 1+CK+CS Posts: 65 Threads: 14 Thanked 23 Times in 15 Posts Reputation: 33

thnx a lot for that exhaustive explanation
#7
05-16-2012
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,353 Threads: 94 Thanked 877 Times in 507 Posts Reputation: 890

I think you do not even need to use calculation in here...

basically q2 shows the frequency of homozygote in the population...

So if 2 random individuals from this population reproduce the risk is q2.

- you just saved a minute for the next question.
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Ambition is a dream with a V8 engine. EP
 The above post was thanked by: pradeeprox (05-16-2012)
#8
05-16-2012
 USMLE Forums Master Steps History: 1+CK+CS Posts: 662 Threads: 72 Thanked 370 Times in 204 Posts Reputation: 380

Quote:
 Originally Posted by DocSikorski I think you do not even need to use calculation in here... basically q2 shows the frequency of homozygote in the population... So if 2 random individuals from this population reproduce the risk is q2. - you just saved a minute for the next question.

__________________
"Disease is very old, and nothing about it has changed. It is we who change as we learn to recognize what was formerly imperceptible." JMC
 The above post was thanked by: pradeeprox (05-16-2012)
#9
05-16-2012
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,353 Threads: 94 Thanked 877 Times in 507 Posts Reputation: 890

Quote:
 Originally Posted by Dr. Mexito He asked for an explanation...
Yes, and u did a good job with it.

but its a good way (from the point of view of the author of the question) to slow you down
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Ambition is a dream with a V8 engine. EP
#10
05-16-2012
 USMLE Forums Master Steps History: 1+CK+CS Posts: 1,039 Threads: 189 Thanked 570 Times in 323 Posts Reputation: 580

I'd go with E.

My explanation would be: there's no reason (no info in the stem) why a child of that couple would have a higher risk than the general population risk (given in the stem - 1/10,000). I wouldn't do any math here.
#11
05-17-2012
 USMLE Forums Guru Steps History: Not yet Posts: 414 Threads: 30 Thanked 87 Times in 70 Posts Reputation: 97

Quote:
 Originally Posted by Casandra I'd go with E. My explanation would be: there's no reason (no info in the stem) why a child of that couple would have a higher risk than the general population risk (given in the stem - 1/10,000). I wouldn't do any math here.
I agree why we even using the formula ?
Why we calc ?
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#12
05-17-2012
 USMLE Forums Veteran Steps History: Not yet Posts: 205 Threads: 58 Thanked 19 Times in 16 Posts Reputation: 29

thanks for these genetic Q
but from which Q bank are they

 Tags Genetics-, Step-1-Questions

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