Question

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)

Answer 1

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.