Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
A couple are both carriers of a recessive allele that causes a disorder. A carrier has one copy of the allele and does not show the disorder. The couple plan to have children, and each child is an independent event.
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. Both parents are Aa. Three children, each independent of the others.
Plan. Find the chance for one child. Then use the product rule for three children. At least one affected is the rest.
Step 1. (a) Aa × Aa gives aa in one box of four. The chance is .
Step 2. (b) The chance that one child is not affected is . For three children: .
Step 3. (c) At least one affected is everything except none affected: .
Answer: ; ; .
Check. The two chances in (b) and (c) add up to .
Watch out. Having had one affected child does not change the chance for the next one. Every child starts again at .
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In a monohybrid cross with complete dominance, the F1 of TT × tt is all Tt and tall. The F2 has the genotypic ratio 1 TT : 2 Tt : 1 tt and the phenotypic ratio 3 tall : 1 dwarf. A test cross uses a homozygous recessive partner (tt). A Tt parent gives T and t with equal chance, so each child of two carriers has a chance of of showing the recessive trait.
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