A colour-blind man marries a woman who has normal vision and whose family has no colour-blind…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
A colour-blind man marries a woman who has normal vision and whose family has no colour-blind allele. Their daughter marries a man with normal vision.
- What per cent of their daughters are carriers?
- What per cent of their sons are colour blind?
- The daughter has sons with her husband. What per cent of her sons are colour blind?
Need a reminder? The key ideaShow
An X-linked recessive allele (h) shows in every man who has it, and in a woman only if she has two copies. A carrier woman is . Carrier mother × normal father: one quarter of all the children are haemophilic sons. Haemophilic father × normal mother: all daughters are carriers and all sons are normal.
The worked solution
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. First couple: and . Second couple: the carrier daughter and a normal man.
Plan. Find the genotypes of the children. Then repeat with the daughter.
Step 1. (a) Every daughter gets the of her father and the of her mother, so all are : carriers.
Step 2. (b) A son gets the Y of his father and the of his mother, so no son is colour blind: .
Step 3. (c) The daughter crossed with gives sons and in equal numbers. So of her sons are colour blind.
Answer: ; ; .
Check. The allele went from the grandfather through his daughter to a grandson. This is criss-cross inheritance.
Watch out. The colour-blind man has no colour-blind sons here, because he gave them only his Y.
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