Colour Blindness in a Family Seeking Genetic Advice
Both parents in a family have normal colour vision. The mother's father - the children's maternal grandfather - was red-green colour blind. The father's family has no history of colour blindness at all. The couple already has two sons and one daughter, and all three children have normal colour vision. The mother is now pregnant again and the parents ask for advice. A relative tells them: 'The two boys are normal, so the colour-blindness gene has skipped this generation - the new baby is safe.' Meanwhile the daughter, now 25, plans to marry a man with normal colour vision and wants to know whether her own children could be affected. Use X^B for the normal allele and X^b for the colour-blind allele; colour blindness is a recessive trait carried on the X chromosome.
1. Write out the genotypes of the parents and use a genetic diagram to show the possible children. Then explain whether the two normal sons are evidence that the gene has 'skipped' the family.
Read the worked answer
Genotypes: the mother's father was X^b Y, so he could only pass X^b to his daughters. Since the mother has normal vision, she must also carry X^B - she is X^B X^b (a carrier). The father has normal vision, so he is X^B Y.
Genetic diagram:
| X^B (father) | Y (father) | |
|---|---|---|
| X^B (mother) | X^B X^B - daughter, normal | X^B Y - son, normal |
| X^b (mother) | X^B X^b - daughter, normal carrier | X^b Y - son, colour blind |
So each child has a 1 in 4 chance of being a colour-blind son, a 1 in 4 chance of being a carrier daughter, and a 1 in 2 chance of being a wholly normal child.
Is it 'skipped'? No. Each son has a 1/2 chance of being colour blind, so the chance that both existing sons are normal is 1/2 × 1/2 = 1/4 - an ordinary outcome, not evidence of anything. Each pregnancy is an independent event with the same 1 in 4 chance of a colour-blind boy. The relative is treating past results as if they change future probabilities.
2. State the probability that the new baby will be a colour-blind boy, and the probability that the baby will be colour blind given that it is a boy. Explain why the two answers differ.
Read the worked answer
Probability the baby is a colour-blind boy = 1/4 (25%). This is one of the four equally likely outcomes in the diagram: X^B X^B, X^B X^b, X^B Y, X^b Y.
Probability the baby is colour blind, given that it is a boy = 1/2 (50%). Among sons only, two outcomes are possible - X^B Y and X^b Y - and each is equally likely.
Why they differ: the first question asks about one specific combination out of all four outcomes, while the second has already excluded the two daughter outcomes. Restricting the sample space from four to two doubles the probability.
Extra reasoning worth stating: no daughter in this family can be colour blind, because the father contributes X^B to every daughter, and one normal dominant allele is enough for normal colour vision. A daughter could only be colour blind if her father were colour blind (X^b Y) and her mother carried X^b.
3. The daughter wants to know the risk to her own children. Explain what must be determined first and calculate the probability that each of her sons would be colour blind.
Read the worked answer
Step 1 - her genotype is uncertain. From the diagram she is either X^B X^B (probability 1/2) or X^B X^b (probability 1/2). Her normal vision does not tell us which, because a carrier is phenotypically normal. Only a genetic test (or the birth of a colour-blind son) would settle it.
Step 2 - her partner. He has normal vision, so he is X^B Y.
Step 3 - combine the probabilities:
- If she is X^B X^B (chance 1/2): every son is X^B Y - risk of a colour-blind son = 0.
- If she is X^B X^b (chance 1/2): each son has a 1/2 chance of inheriting X^b - risk = 1/2.
- Combined risk for each son = 1/2 × 1/2 = 1/4 (25%).
Additional reasoning: her daughters cannot be colour blind (father gives X^B), but each daughter has a 1/2 × 1/2 = 1/4 chance of being a carrier, so the allele can continue quietly through the family line. This is why counselling uses probabilities rather than a yes/no answer: the outcome for any one child is all-or-nothing, but the risk across many pregnancies is 1 in 4.