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#1




Applying Hardy Weinberg Equation to carrier state?
A carrier with cystic fibrosis asks about the probability of having a child with the disease.
disease prevalence 1/2500?? A 1/25 B 1/50 C 1/100 D 1/1000 
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aktorque (03022011) 
#2




C. 1/100
The answer should be 1/100
The prevalence of the AR disease is 1/2500 That means that under Hardy Weinberg: p^2 = 1/2500 so p = root(1/2500) = 1/50 q = 1p = 49/50 (effectively 1) 2pq, which is the probablity that a random mate will be a heterozygous carrier of CF, is 2(1*1/50) = 1/25 A carrier has a 1/2 probability of passing on the bad allele, so the probability that a random partner will pass on a CF allele is 1/2*1/25 = 1/50. Your patient is a carrier, so the probability that your patient will pass on the CF allele is 1/2. The probability that both will pass on the CF allele is 1/50 * 1/2 = 1/100 
#3




WOW Mondoshawan, excellent explanation.



#4




Quote:
the question not passing one allele but passing the disease mean they have affected boy ( aa ) so two carrier have probability having a disease child is 1/4 child with carrier allele 1/2 if you have simple explanation 
#5




Quote:
(if it's too small, click on the image to go to the large version) 
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#6




LOL....A.W.E.S.O.M.E

#7




Quote:
very funny looking hardy and weinberg... keep it up. good luck. 
#8




is it possible that the carrier in this question gets married to affected homozygous(aa)???

#9




Not unknowingly....

#10




I think it's possible.
Some time ago I visited a patient who had undiagnosed cystic fibrosis until she was 23. Actually, she had comparatively few symptoms, and may still not have her diagnosis, because her sweat test was an "unconclusive borderline". In fact, she is probably waiting for DNA results. Well, she is probably double positive for the mutation, and would be one of these cases. Anyway, nobody is interested in this degree of detail. A question answer would never involve these veryveryvery rare possibilities. But that was a great opportunity to realize that patients don't read the books. And that life is not a picnic. 
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Mondoshawan (02022011) 
#11




Quote:
Can u please tell me if im wrong 
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Mondoshawan (03022011) 
#12




Quote:

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adorisz (03022011) 
#13




Quote:
Is is 1? 
#14




Quote:
If you were asking what the probability of passing on a CF allele if one of the parents were affected, then yes it's 1. All of their children will have at least one allele, because one of the parents was homozygous recessive and will pass on one of the chromosomes. 
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rapiddo (04132012) 
#15




I have a different way of looking at this questions, if anyone still checking this post , I would appreciate if they correct me if I was wrong.
The person in the question is carrier, chance of marrying carrier/normal partner is 1/2, we got q from above as 1/50, probability of their affected offspring is 1/25, and multiplying all probabilities will give us 1/100. Isn't it right that way too??
__________________
IT IS GETTING INTENSE !! 
#16




Well, CF patients have an increased risk of infertility, in women mucus builds up in the fallopian tubes and prevents conception (or causes ectopic pregnancy), in men there is usually agenesis of the vas deferens. So even if the carrier marries an homozygous the chance of having children is very low i guess

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BiostatisticsEpidemiology, Genetics, Step1Questions 
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