Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
Two healthy parents have a child II-1 with an autosomal recessive disorder. Their healthy daughter II-2 grows up and wants to know the risk for her own children. She marries a healthy man who is known to be a carrier.
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Given. Her parents are Aa × Aa, because they have an affected child. Her husband is Aa.
Plan. Find the chance that she is a carrier. Then multiply by the chance of an affected child from two carriers.
Step 1. (a) She is healthy, so she is AA, Aa or Aa out of the four boxes. Two of these three are carriers: .
Step 2. (b) Both parents must be carriers, and then the child has a chance of . So .
Step 3. (c) The affected child shows that she is a carrier. Now both parents are known to be Aa, and the chance for the next child is .
Answer: ; ; .
Check. The chance in (c) is higher than in (b), because we now know that she is a carrier.
Watch out. In (a), do not use . The affected box aa is not possible for a healthy person, so it is removed from the count.
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In an autosomal recessive disorder, two carriers (Aa × Aa) have children in the ratio 1 AA : 2 Aa : 1 aa. The chance of an affected child is , of a carrier , and each child is an independent event. Thalassaemia is a problem of the amount of globin, and sickle-cell anaemia is a problem of one amino acid of the globin.
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