A 38-year-old woman delivers a baby with trisomy 21. Genetic studies show that the extra chromosome 21 arose from nondisjunction. In the vast majority of such cases of free trisomy 21, the error occurs during:
- A Maternal meiosis I, due to failure of disjunction of homologous chromosomes ✓
- B Maternal meiosis II, due to failure of separation of sister chromatids
- C Paternal meiosis II, due to failure of separation of sister chromatids
- D Mitotic division of the early zygote, producing mosaicism
Correct answer: A. Maternal meiosis I, due to failure of disjunction of homologous chromosomes
Explanation
About 95 percent of cases of free trisomy 21 result from meiotic nondisjunction of chromosome 21 in the ovum, and most of those errors occur in maternal meiosis I, where homologous chromosomes fail to separate. The incidence rises sharply with advancing maternal age because oocytes remain arrested in prophase of meiosis I from fetal life onward. Paternal nondisjunction accounts for only about 4 percent of cases.
Reference: Robbins and Cotran Pathologic Basis of Disease, 10th ed.
High-yield for: NEET PGINI-CETNExTFMGEUSMLEPLABMRCP
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