Question

Difficulty: HardGene Mutations and Chromosomal Aberrations

During gametogenesis, non-disjunction of chromosome 21 occurs specifically during Meiosis I. If all resulting gametes are subsequently fertilized by normal haploid gametes, what percentage of the produced zygotes will have trisomy 21?

  1. A
    100%
  2. B
    75%
  3. 50%Answer
  4. D
    25%

Answer

50% of the resulting zygotes will exhibit trisomy 21.
When non-disjunction of a chromosome pair takes place during Meiosis I, the pair fails to segregate. Consequently, one daughter cell receives both homologous chromosomes while the other receives none. Following normal chromatid separation in Meiosis II, half of the gametes carry an extra chromosome (n+1n+1) and half are deficient by one chromosome (n1n-1). Fertilization of the n+1n+1 gametes with normal haploid gametes (nn) produces trisomic (2n+12n+1) zygotes. Therefore, 50% of the zygotes will have trisomy 21.

Step-by-Step Solution

1
Analyze the meiotic stage where non-disjunction occurs
In Meiosis I, non-disjunction means homologous chromosomes fail to separate into distinct daughter cells.
Understanding the exact stage of meiotic failure determines the chromosome distribution in the daughter cells after Meiosis I.
2
Determine the chromosomal content of the resulting four gametes after Meiosis II
Meiosis I yields one cell with an extra chromosome (n+1n+1) and one cell missing that chromosome (n1n-1). Normal chromatid separation in Meiosis II creates two (n+1)(n+1) gametes and two (n1)(n-1) gametes.
All four gametes produced are genetically abnormal after a Meiosis I non-disjunction event.
3
Calculate the proportion of trisomic zygotes post-fertilization
Fertilization of the two (n+1)(n+1) gametes by normal haploid (nn) gametes gives (2n+1)(2n+1), which is trisomy 21. Two out of four total zygotes (50%) will be trisomic (and 50% will be monosomic).
Trisomy requires the combination of an (n+1)(n+1) gamete with a normal nn gamete.

Key Concept

Chromosomal Non-disjunction and Aneuploidy
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