Understanding Red-Blue Colour Blindness and Klinefelter Syndrome

Red-blue colour blindness is an X-linked recessive disorder. This means the gene responsible is located on the X chromosome. For a male (XY) to be colour blind, he only needs one copy of the recessive allele on his single X chromosome. For a female (XX) to be colour blind, she needs two copies of the recessive allele, one on each X chromosome. A female with one recessive allele and one dominant allele is a carrier; she has normal vision but can pass the recessive allele to her children.

Klinefelter syndrome is a chromosomal disorder in males resulting from an extra X chromosome, leading to a 47, XXY karyotype instead of the typical 46, XY.

Parents' Genotypes and Son 2's Condition

The parents have normal colour vision. Their son (Son 2) with a normal 46, XY karyotype is colour blind. Since colour blindness is X-linked recessive, Son 2's genotype must be X^cY, where X^c represents the chromosome carrying the allele for colour blindness. Males inherit their X chromosome from their mother and their Y chromosome from their father.

For Son 2 to be X^cY:

• He must have inherited the X^c chromosome from his mother.
• He must have inherited the Y chromosome from his father.

Since the father has normal vision, his X chromosome must carry the dominant normal allele (X^C). So, the father's genotype is X^CY.

The mother has normal vision but produced a son with X^c. This means she carries the recessive allele X^c on one of her X chromosomes. Since she has normal vision, she must also have the dominant allele X^C on her other X chromosome. Therefore, the mother's genotype is X^CX^c (a carrier).

So, the parental genotypes are: Mother X^CX^c and Father X^CY.

Analyzing Son 1's Condition: XXY and Colour Blind

Son 1 has Klinefelter syndrome with a 47, XXY chromosome composition and is colour blind. His genotype must include the alleles explaining both the XXY condition and the colour blindness.

To get an XXY individual, nondisjunction must occur during gamete formation in either the mother or the father, resulting in a gamete with two sex chromosomes which then fuses with a normal gamete carrying one sex chromosome.

Since the individual is male (has a Y chromosome), the Y chromosome must come from the father in a normal gamete (Y) or through nondisjunction (XY or YY, etc.). Since the individual has two X chromosomes, these must come from either the mother (both Xs) or the father (both Xs, not possible as he has only one X) or one X from each parent.

Considering the parents' genotypes (Mother X^CX^c, Father X^CY) and the condition of Son 1 (XXY, colour blind), let's examine the possibilities for nondisjunction, assuming no crossing over.

Nondisjunction during Gamete Formation

Case 1: Nondisjunction in the Mother (X^CX^c)

• Meiosis I Nondisjunction: Homologous chromosomes (X^C and X^c) fail to separate. Gametes produced are (X^CX^c) and (null).
  • Fusion with Father's normal Y gamete: (X^CX^c) + Y → X^CX^cY (XXY). This individual has X^C, so would have normal vision. This does not match Son 1 being colour blind.
• Meiosis II Nondisjunction: Homologous chromosomes (X^C and X^c) separate normally in Meiosis I, resulting in cells with X^C and X^c. Then, sister chromatids fail to separate in Meiosis II.
  • Cell with X^C in Meiosis I undergoes Meiosis II nondisjunction: produces gametes (X^CX^C) and (null).
    • Fusion with Father's normal Y gamete: (X^CX^C) + Y → X^CX^CY (XXY). This individual has X^C, so would have normal vision. This does not match Son 1 being colour blind.
  • Cell with X^c in Meiosis I undergoes Meiosis II nondisjunction: produces gametes (X^cX^c) and (null).
    • Fusion with Father's normal Y gamete: (X^cX^c) + Y → X^cX^cY (XXY). This individual only has X^c alleles on the X chromosomes, so would be colour blind. This matches Son 1's condition.

Case 2: Nondisjunction in the Father (X^CY)

• Meiosis I Nondisjunction: Homologous chromosomes (X^C and Y) fail to separate. Gametes produced are (X^CY) and (null).
  • Fusion with Mother's normal X^C gamete: X^C + (X^CY) → X^CX^CY (XXY). This individual has X^C, so would have normal vision. This does not match Son 1 being colour blind.
  • Fusion with Mother's normal X^c gamete: X^c + (X^CY) → X^cX^CY (XXY). This individual has X^C, so would have normal vision. This does not match Son 1 being colour blind.
• Meiosis II Nondisjunction: Homologous chromosomes (X^C and Y) separate normally in Meiosis I, resulting in cells with X^C and Y. Then, sister chromatids fail to separate in Meiosis II.
  • Cell with X^C in Meiosis I undergoes Meiosis II nondisjunction: produces gametes (X^CX^C) and (null).
    • Fusion with Mother's normal X gamete (X^C or X^c): X + (X^CX^C) → XXX. This does not result in an XXY individual.
  • Cell with Y in Meiosis I undergoes Meiosis II nondisjunction: produces gametes (YY) and (null).
    • Fusion with Mother's normal X gamete (X^C or X^c): X + (YY) → XYY. This does not result in an XXY individual.

Let's summarize the possibilities for the origin of the XXY chromosomes in Son 1 (karyotype 47, XXY) and his colour blindness (genotype must be X^cX^cY for colour blindness from these parents without crossing over, or potentially X^cX*Y where X* doesn't have C, but with no crossing over, the mother's X's are X^C and X^c). The only scenario that results in the required X^cX^cY genotype is nondisjunction during the mother's meiosis II.

As shown in the analysis and table, the only event that explains the combination of XXY karyotype and colour blindness (X^cX^cY genotype) in Son 1 from parents X^CX^c and X^CY (assuming no crossing over) is nondisjunction of sister chromatids carrying the X^c allele during Meiosis II in the mother.

Conclusion: Nondisjunction Location

The Klinefelter syndrome (XXY) in Son 1, combined with his colour blind phenotype, indicates that the extra X chromosome carries the recessive colour blindness allele (X^c), and there is no dominant normal allele (X^C) to mask it among the two X chromosomes. From parents with genotypes X^CX^c and X^CY, this specific outcome (X^cX^cY) arises when the mother produces a gamete with two X^c chromosomes (X^cX^c) due to nondisjunction during meiosis II, which is then fertilized by a normal Y gamete from the father.

Therefore, Klinefelter syndrome in Son 1 resulted due to nondisjunction during female gamete formation in meiosis II.