Colour Blindness and Turner's Syndrome Genotype

This question describes a specific genetic scenario involving a father who is colour blind and a daughter who is also colour blind and has Turner's syndrome. To understand the daughter's genotype, we need to consider the inheritance pattern of colour blindness and the cause of Turner's syndrome.

Understanding Colour Blindness and Turner's Syndrome

• Colour Blindness: This is an X-linked recessive trait. It means the gene responsible is located on the X chromosome. For a female to be colour blind, she must inherit two copies of the recessive allele (one on each X chromosome). For a male to be colour blind, he only needs one copy of the recessive allele because he has only one X chromosome.
• Turner's Syndrome: This is a genetic condition in females resulting from the absence of one X chromosome. The typical genotype is 45, XO, meaning the individual has one X chromosome and no other sex chromosome.

Father's Genotype and Daughter's Genotype

• The father is colour blind. Since colour blindness is X-linked recessive, his genotype must be \(X^cY\), where \(X^c\) represents the X chromosome carrying the allele for colour blindness and Y is the Y chromosome.
• The daughter has Turner's syndrome, so her genotype is XO. She is also colour blind. For an XO female to be colour blind, the single X chromosome she possesses must carry the allele for colour blindness. Therefore, the daughter's genotype is \(X^cO\).

Origin of the \(X^cO\) Genotype

A daughter with genotype \(X^cO\) receives one \(X^c\) chromosome from one parent and no sex chromosome (represented by O) from the other parent. Let's examine how the father and mother contribute sex chromosomes:

• A father with genotype \(X^cY\) produces sperm with either an \(X^c\) chromosome or a Y chromosome during normal meiosis.
• A mother (assuming she has two X chromosomes, say \(X^AX^B\)) produces eggs with one X chromosome (either \(X^A\) or \(X^B\)) during normal meiosis.

For the daughter to be \(X^cO\), one parent must contribute the \(X^c\) and the other must contribute 'O' (no sex chromosome).

Non-disjunction Events

The contribution of 'O' (no sex chromosome) typically occurs due to a mistake during meiosis called non-disjunction. Non-disjunction is the failure of homologous chromosomes or sister chromatids to separate properly during cell division.

• Non-disjunction in the father: If non-disjunction occurs during sperm formation, the father could produce sperm with no sex chromosome (0). If such a sperm fertilizes a normal egg (contributing an X chromosome), the resulting zygote could be XO. For the daughter to be \(X^cO\), the mother must contribute the \(X^c\) chromosome in her egg, and the father must contribute the 0 sperm due to non-disjunction. This requires the mother to be at least a carrier (\(X^CX^c\)) or colour blind (\(X^cX^c\)).
• Non-disjunction in the mother: If non-disjunction occurs during egg formation, the mother could produce eggs with no sex chromosome (0). If such an egg is fertilized by a normal sperm (contributing a sex chromosome), the resulting zygote could be XO. For the daughter to be \(X^cO\), the father must contribute the \(X^c\) chromosome in his sperm, and the mother must contribute the 0 egg due to non-disjunction. An \(X^cY\) father normally produces \(X^c\) sperm.

Analyzing the Scenario

The father is colour blind, meaning he has the \(X^c\) chromosome. He can pass this \(X^c\) to his offspring. The daughter is \(X^cO\), meaning her single X chromosome is \(X^c\).

• If the \(X^c\) came from the father (which he can pass via normal sperm production), then the 'O' (absence of the second sex chromosome) must have come from the mother. This happens if the mother's egg had no sex chromosome due to non-disjunction during meiosis.
• If the \(X^c\) came from the mother, then the 'O' must have come from the father (via sperm lacking a sex chromosome due to non-disjunction). This would require the mother to carry the \(X^c\) allele and non-disjunction to occur in the father.

The question asks for the event that explains the daughter's genotype given the father is colour blind. The most direct explanation aligning the father's known genotype (\(X^cY\)) with the daughter's \(X^cO\) genotype is that the father contributed the \(X^c\) via a normal sperm, and the mother contributed an egg lacking a sex chromosome due to non-disjunction. This results in the combination \(X^c\) (from father) + 0 (from mother) = \(X^cO\).

Translocation events involve structural changes where part of one chromosome breaks off and attaches to another, which is not the cause of monosomy X (Turner's syndrome).

Conclusion

The genotype \(X^cO\) of the colour blind daughter with Turner's syndrome, given her colour blind father (\(X^cY\)), is best explained by the father contributing the \(X^c\) chromosome and the mother contributing an egg with no sex chromosome due to a non-disjunction event during her meiosis.