44. A colour blind father has a daughter who is also colourblind and has Turner’s syndrome. The genotype of the daughter is due to:
(1) Translocation event in the father.
(2) Translocation event in the mother.
(3) Non-disjunction event in the mother.
(4) Non-disjunction event in the father.
Genotype reasoning
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Trait: red–green colour blindness, X‑linked recessive.
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Father is colour blind → genotype XᶜY (Xᶜ = mutant).
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Daughter is Turner (XO) and colour blind → genotype must be 45,Xᶜ (only one X, and it carries the mutant allele).
For any child of this father:
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Sperm types: Xᶜ or Y (normally 50:50).
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To get a Turner girl (XO), the zygote must receive:
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an Xᶜ from father, and
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no sex chromosome from mother (an egg of type “O”).
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Thus, the abnormal gamete is the maternal gamete (lacking an X), while the paternal gamete is just the usual Xᶜ sperm.
This arises from nondisjunction in maternal meiosis (loss of an X into a polar body), producing an egg with zero sex chromosomes; fertilization by an Xᶜ sperm → 45,Xᶜ Turner, colour blind.
Therefore the crucial event is a non‑disjunction in the mother.
Why the other options are incorrect
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Translocation event in the father
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Translocation rearranges parts between chromosomes but does not explain loss of an entire maternal X chromosome needed for Turner syndrome (XO).
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Translocation event in the mother
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Again, this could change structure, not the total number (monosomy X) as in classic Turner.
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Non-disjunction event in the mother – correct
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Explains production of an O egg, so that the child’s single X must come from the father (Xᶜ) → Turner (45,Xᶜ) and colour blind.
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Non-disjunction event in the father
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Paternal nondisjunction might give XX or YY sperm, but the daughter clearly has only one X, and that X is paternal. The missing chromosome must be maternal, so the error is not in the father.
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Hence, the daughter’s genotype (colour blind Turner female) is best explained by a non‑disjunction event in the mother (option 3).
