Colour Blindness in the Next Child
A couple in Gaya seeks advice. The man is colour-blind; the ophthalmologist says the trait is X-linked recessive. The woman has normal colour vision, but her younger brother is colour-blind, and neither of her parents is colour-blind. The couple is expecting their first child. The man argues that because the defect came from the woman's side, the child is at risk only if it is a boy, and that 'daughters will be safe because they get a normal X from me'. The woman's mother-in-law says that because the husband is affected, all sons will be colour-blind and all daughters will be carriers. The couple wants to know the actual probabilities before the child is born and asks whether a karyotype or a DNA test can tell them the child's colour vision before birth.
1. Determine the woman's possible genotypes using her parents' phenotypes and her brother's condition. Calculate the probability that a son will be colour-blind and that a daughter will be colour-blind or a carrier.
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Let X^C be the normal allele and X^c the colour-blind allele. The woman's brother is X^cY. His father is normal, X^CY, so the brother must have received X^c from his mother. The mother is therefore a carrier, X^CX^c, even though she has normal vision.
The woman receives X^C from her father. From her mother she receives X^C with probability 1/2 or X^c with probability 1/2. So she is 1/2 X^CX^C and 1/2 X^CX^c.
For a son: the father gives Y, and the mother gives X. If the woman is a carrier (probability 1/2), half her sons receive X^c; if she is not a carrier, none do. Probability a son is colour-blind = 1/2 × 1/2 = 1/4.
For a daughter: the father always gives X^c. If the mother is X^CX^C (1/2), the daughter is X^CX^c, a carrier. If the mother is X^CX^c (1/2), the daughter receives X^C with probability 1/2 (carrier) or X^c with probability 1/2 (affected). So carrier probability = 1/2 + (1/2 × 1/2) = 3/4, and affected probability = 1/2 × 1/2 = 1/4. No daughter can be a non-carrier normal.
2. Evaluate the man's statement and the mother-in-law's statement. Which parts are correct or incorrect?
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The man is partly right that a son is at risk, but wrong to say only boys are at risk. A daughter can be colour-blind if the mother is a carrier and passes X^c, because the father passes X^c to every daughter. His claim that daughters get a normal X from him is false: he has only X^c to give.
The mother-in-law is also wrong on both counts. Not all sons will be colour-blind; the overall probability is 1/4, and if the woman is not a carrier no son will be affected. Not all daughters will be carriers; 3/4 are expected to be carriers, but 1/4 can be affected.
The correct statement is that the risk depends on whether the woman is a carrier. Her brother's colour blindness makes carrier status possible but not certain.
3. Can a karyotype or a DNA test detect colour blindness before birth? Explain what each can and cannot show, and give appropriate counselling.
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A karyotype shows the number and gross structure of chromosomes, such as an extra chromosome 21 or a missing X. It does not show a single-gene trait like red-green colour blindness, so it cannot answer this question.
A DNA test can detect known X-linked opsin gene variants, but red-green colour blindness often involves complex rearrangements of the opsin genes on the X chromosome rather than one simple mutation. Routine prenatal testing is not done for colour blindness because it is not a serious medical condition and the result would not change management.
Counselling should give the probabilities: 1/4 for a son to be affected, 3/4 for a daughter to be a carrier and 1/4 for a daughter to be affected. The couple should not be blamed, and they should know that affected children can lead normal lives with aids such as coloured filters and good lighting.