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ikadub [295]
3 years ago
6

Cystic fibrosis in humans is caused by recessive mutations in a single gene located in an autosome and follows mendelian laws. a

normal couple has two children. the first child has cystic fibrosis, and the second child is unaffected. what is the probability that the second child is a carrier (heterozygous) for the mutation that causes the disease? (hint: we already know that the second child is not homozygous recessive)
Biology
1 answer:
lara [203]3 years ago
4 0
I think the probability would be 50% bc there are four options FF Ff fF and ff. Half of those are heterozygous (the child would be a carrier). So I think it's 50%. However, if we already know that the child will not have the ff combination, then we could say that there are only three other options (so maybe it's 60%?) but that's probably just me being a smartass.
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