Questions: Aneuploidy: Trisomy, Monosomy, and Non-Disjunction

Question 1

Non-disjunction occurs during meiosis I in an oocyte. How many of the four resulting gametes will have an abnormal chromosome number?

• A. One — only the gamete that received the extra chromosome is abnormal
• B. Two — one gamete gains a chromosome and one loses it
• C. All four — meiosis I non-disjunction affects the entire gamete pool because it occurs before meiosis II
• D. It depends on whether the non-disjunction affects a large or small chromosome

Answer: (C)

Meiosis I non-disjunction means both homologs of a chromosome pair move to the same pole rather than separating. After meiosis I, one secondary oocyte/spermatocyte has two copies of the chromosome and the other has zero. When each of these then undergoes meiosis II (which separates sister chromatids), the cell with two copies produces two gametes each with an extra chromosome, and the cell with zero copies produces two gametes each missing the chromosome. All four final gametes are abnormal. This contrasts with meiosis II non-disjunction, where only the one cell that fails to separate its sister chromatids produces two abnormal gametes — leaving the other two gametes normal.

Question 2

Why are sex chromosome aneuploidies (e.g., XXY, 45,X) generally much better tolerated than autosomal aneuploidies of comparable chromosome size?

• A. Because sex chromosomes are smaller than most autosomes and therefore carry fewer genes
• B. Because X-inactivation silences extra X chromosomes, minimizing gene dosage imbalance, while autosomes have no equivalent dosage compensation
• C. Because sex chromosomes are inherited only from one parent, reducing the conflict between paternal and maternal gene products
• D. Because the Y chromosome is largely non-functional and its presence or absence has minimal effect on protein dosage

Answer: (B)

X-inactivation is the key mechanism. In mammals, one X chromosome in each somatic cell is randomly inactivated to equalize expression between XX females and XY males. This pre-existing dosage compensation means that in an XXY individual, one X is inactivated just as in a typical female, so dosage of most X-linked genes is essentially normal. Similarly, a 45,X individual has one active X — the same as both typical males and typical females. Autosomes have no such inactivation mechanism, so trisomy of any autosome means roughly 50% overproduction of every protein encoded on that chromosome, disrupting protein stoichiometry and regulatory networks.

Question 3

Non-disjunction during meiosis I produces more unbalanced gametes than non-disjunction during meiosis II.

• T. True
• F. False

Answer: True

Meiosis I non-disjunction prevents homologous chromosomes from separating, so the chromosome imbalance is present in both daughter cells after meiosis I. Both of these then undergo meiosis II, producing four gametes that are all abnormal (two with an extra chromosome, two with none). Meiosis II non-disjunction fails to separate sister chromatids in only one of the two meiosis I products, so only two of the four final gametes are abnormal while the other two receive a normal haploid set. This is also why meiosis I errors are more clinically significant: they have twice the impact on the gamete pool.

Question 4

Trisomy 21 is compatible with live birth because chromosome 21 is gene-rich and has strong dosage compensation mechanisms similar to X-inactivation.

• T. True
• F. False

Answer: False

This is incorrect. Trisomy 21 is compatible with live birth precisely because chromosome 21 is one of the smallest and most gene-poor of the human autosomes. There is no autosomal dosage compensation equivalent to X-inactivation. The reason some trisomies survive while most do not is simply that the dosage imbalance from a smaller chromosome with fewer genes is less disruptive to overall cellular stoichiometry. Trisomies 13, 18, and 21 are the only autosomal trisomies compatible with live birth — all three involve relatively gene-poor chromosomes — and even trisomies 13 and 18 are usually fatal within the first year.

Question 5

Why are sex chromosome aneuploidies generally better tolerated than autosomal aneuploidies, even when involving the same number of extra chromosomes?

Model answer: Because X-inactivation provides pre-existing dosage compensation for X chromosomes. In mammals, one X chromosome per cell is inactivated to equalize X-linked gene expression between XX females and XY males. When an extra X is present (as in XXY or XXX), it is simply inactivated like the second X in a normal female, so the dosage of most X-linked genes remains normal. Autosomes have no equivalent inactivation mechanism — an extra autosome means roughly 50% more expression of every gene on that chromosome, disrupting protein complex stoichiometry and gene regulatory networks throughout the cell.

The tolerance for sex chromosome aneuploidy is a direct consequence of a mechanism that already exists to handle dosage differences between sexes. X-inactivation converts the 'problem' of extra X chromosomes into a state the cell already knows how to manage. Without this pre-existing mechanism (as with autosomes), the cell has no way to buffer the extra gene dosage, and the resulting imbalances in protein complexes and transcription factor networks are typically incompatible with normal development.