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Old 03-12-2015
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Default Genetics Question

A man whose brother has cystic fibrosis wants to know his risk of having an unaffected child. The prevalence of CF is 1 in 1600 individuals. The risk in this case is:

A. 1/8
B. 1/16
C. 1/60
D. 1/120
E. 1/256

The answer is D. Can anyone explain the reasoning? Thx

Last edited by samusmle33; 03-12-2015 at 07:13 AM.
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Old 03-12-2015
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The man in this Question has 2/3 chance of being a carrier and his wife has 1/20 chance of being a carrier.

How do v figure out the chance of wife being a carrier??

Last edited by samusmle33; 03-12-2015 at 07:11 AM.
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Old 03-12-2015
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Lol I think I just got it. The wifez probability of bring a heterozygote came from hardy Weinberg equation..in which v assume p = 1 for aut rec conditions. Hence 2q=1/20..

Is this right??
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