43. Red-blue colour blindness is a human X-linked recessive disorder. The two parents with normal colour vision have two sons. Son 1 has 47, XXY chromosome composition and is colour blind. Son 2 has 46. XY and is also colour blind. Assuming that no crossing over took place in prophase I of meiosis, Klinefelter syndrome in Son 1 resulted due to nondisjunction during which one of the following events?
(1) Female gamete formation in meiosis I
(2) Female gamete formation in meiosis II
(3) Male gamete formation in meiosis I
(4) Male gamete formation in meiosis II

Genetic logic

• Disorder: X‑linked recessive (red‑blue colour blindness).

• Parents: both have normal colour vision, so:

  • Father: X⁺Y (normal X, normal vision).

  • Mother: must be carrier X⁺Xᶜ (one normal, one mutant) because both sons are affected.

Sons:

• Son 2: 46,XY and colour blind → genotype XᶜY.

  • Received Xᶜ from mother, Y from father – consistent with a carrier mother.

• Son 1: 47,XXY (Klinefelter) and colour blind.

  • Genotypes possible:

    • X⁺XᶜY (one normal, one mutant X) – would still be colour blind?

      • For an X‑linked recessive, a male with any normal X has the recessive masked, so he would be normal for colour vision.

    • XᶜXᶜY (two mutant X’s) – must be colour blind.

Since Son 1 is colour blind, he must be XᶜXᶜY.

Thus, both X chromosomes in Son 1 are the same maternal mutant allele Xᶜ; the father provided only Y, not an additional X.

Why this means nondisjunction in maternal meiosis II

• No crossing over is assumed, so the two maternal X chromatids at meiosis II are identical within each chromatid pair.

• To give an egg with two mutant X’s (XᶜXᶜ):

  • Meiosis I must separate X⁺ and Xᶜ normally into two secondary oocytes.

  • In the Xᶜ‑carrying secondary oocyte, nondisjunction of sister chromatids in meiosis II can send both Xᶜ chromatids into the same egg, making an XX egg (XᶜXᶜ).

• Fertilization by a normal Y sperm → XᶜXᶜY (47,XXY) Klinefelter son, colour blind.

So the extra X and the duplicated mutant allele both come from the mother’s meiosis II error.

Why the other options are wrong

1. Female gamete formation in meiosis I

   • Meiosis I nondisjunction would put both homologous X’s (X⁺ and Xᶜ) into the same egg → X⁺XᶜY son.

   • With a normal X present, an X‑linked recessive phenotype would be masked, so Son 1 would not be colour blind.

2. Male gamete formation in meiosis I

   • Paternal meiosis I nondisjunction yields XY or 0 sex‑chromosome sperm.

   • XXY offspring from XY sperm + X egg would carry only one maternal Xᶜ; to be colour blind they’d need the paternal X to be mutant, which is impossible because father is phenotypically normal.

3. Male gamete formation in meiosis II

   • Meiosis II nondisjunction in the father can create XX or YY sperm.

   • An XXY child from an XX sperm + Xᶜ egg would receive at most one maternal Xᶜ, again giving either X⁺XᶜY or XᶜX⁺Y, not XᶜXᶜY.

Only maternal meiosis II nondisjunction correctly explains both the XXY karyotype and two identical mutant X chromosomes causing colour blindness in Son 1.