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  #1  
Old 05-15-2012
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Genetics 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
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Old 05-15-2012
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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.
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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
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  #3  
Old 05-15-2012
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Quote:
Originally Posted by Hope2Pass View Post
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..
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Old 05-15-2012
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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.
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  #5  
Old 05-16-2012
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Quote:
Originally Posted by pradeeprox View Post
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
dad carrier = .02
mom * dad = 0.0004

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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Default correct answer is E

correct answer is E

thnx a lot for that exhaustive explanation
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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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Quote:
Originally Posted by DocSikorski View Post
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.

He asked for an explanation...
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Quote:
Originally Posted by Dr. Mexito View Post
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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Default my answer

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.
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Old 05-17-2012
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Quote:
Originally Posted by Casandra View Post
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 ?
Explanation it's straight answer fact given fact asked,
Why we calc ?
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Old 05-17-2012
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thanks for these genetic Q
but from which Q bank are they
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