The key bit of knowledge required to answer this question is that colour blindness is an X chromosome linked condition. Say the allele for normal vision is (X^N) and the allele for colour blindness is (X^n). A female can be heterozygous for these two alleles and be a carrier. She will remain unaffected by the condition because the allele for normal vision is dominant to the recessive, defective colour blindness allele. She can however, pass her defective X chromosome on to her children. A male who is heterozygous for the allele will always be affected, as he does not have a second X chromosome to cancel out the effect of the defective X^n (colour blindness) allele.

In our question, the female is a carrier and must be (X^N, X^n). The male is unaffected and must be (X^N, Y). Using a simple Punnett grid, the possible outcomes are (X^N, X^n), (X^N, X^N), (X^N, Y), (X^n, Y). The only outcome that produces an affected child is (X^n, Y). The chance that a child will be affected is therefore 1/4. However, since the question asks the probability of having an affected SON, and only two of these outcomes result in a male child, the correct answer to this question is 1/2.

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