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#1




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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MEERA@MD (4 Weeks Ago) 
#2




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 
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MEERA@MD (4 Weeks Ago), pradeeprox (05162012) 
#3




Quote:
I go for E as well.. 


#4




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




Quote:
2pq = carrier q^2 = 1/10,000Now, 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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DocSikorski (05162012), pradeeprox (05162012) 
#6




correct answer is E
correct answer is E
thnx a lot for that exhaustive explanation 
#7




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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pradeeprox (05162012) 
#8




Quote:
He asked for an explanation...
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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 
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pradeeprox (05162012) 
#9




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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#10




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. 
#11




Quote:
Explanation it's straight answer fact given fact asked, Why we calc ?
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#12




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