anatomy
Forum Guru
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A man whose brother has cystic fibrosis wants to know his risk of having an affected child . The prevalance of CF is 1 in 1600.. The risk in this case is-
1/8
1/16
1/60
1/120
1/256
pls explain also.
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| almina
Forum Senior
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Posts: 99
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q2 = 1/1600, q=1/40
q+p=1
2pq = 1/20 - carrier rate in population
the chace of 2 carriers to have having a child with disease is 1/4
Assuming that the man is a carrier himself, his risk of having a child with cystic fibrosis is 1/20 x 1/4 = 1/80
But there is no such answer...so where am I wrong?
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| anatomy
Forum Guru
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pls some one else?
explain it pls.
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| mjl1717
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Hardy Weinberg Law baby! I think me, you and the others will have to fight our way thru it! :shock:
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Smell the coffee! "Is That an Osler move??"
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| blinx420
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A man whose brother has cystic fibrosis wants to know his risk of having an affected child . The prevalance of CF is 1 in 1600
The mans brother is aa so the parents must be Aa x Aa. The man is either Aa or AA. The probability that he is Aa is thus 2/3.
The prevalance of CF is 1/1600. q2 = 1/1600 so q = 1/40 and p ~ 1. therefore, 2pq = 1/20. Thus, the probability that the man marries a woman who is Aa is 1/20.
The probability that the man (Aa) and this woman (Aa) have an aa child is 1/4
Therefore, the probability all this happens is: 2/3 * 1/20 * 1/4 = 1/120
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| Moctopod
Forum Elite
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Good explanation. Thanks!
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