A 16-year-old girl is referred for evaluation of tall stature (5 cm above mid-parental height), mild learning disability, and speech delay. Karyotype analysis reveals 47,XXX. On buccal smear cytology, the structure marked **A** (three X chromosomes) results in the formation of multiple inactivated X chromosomes. How many Barr bodies would you expect to observe on buccal smear in this patient?

Explanation

## Why Two Barr bodies is right In Triple X Syndrome (47,XXX), the three X chromosomes undergo X-inactivation (Lyonization) according to the n−1 rule, where n is the total number of X chromosomes. With three X chromosomes, two undergo inactivation to form two Barr bodies, while one X remains active. This is the classic cytogenetic finding that distinguishes 47,XXX from normal females (1 Barr body) and helps confirm the diagnosis during developmental delay or infertility workup. The presence of two Barr bodies on buccal smear is a high-yield diagnostic clue for Triple X Syndrome (Nelson Textbook of Pediatrics 21e, Chapter 98). ## Why each distractor is wrong - **Zero Barr bodies**: This is seen in normal males (45,XY) who have no X inactivation. Triple X patients have three X chromosomes, not one, so inactivation occurs. - **One Barr body**: This is the normal finding in 45,XX females (one X inactivated, one active). Triple X patients have an extra X chromosome, resulting in one additional Barr body. - **Three Barr bodies**: This would occur only in 48,XXXX (tetrasomy X), not in 47,XXX. The n−1 rule means three X chromosomes yield exactly two Barr bodies. **High-Yield:** Triple X Syndrome → two Barr bodies on buccal smear; remember the n−1 rule: number of Barr bodies = total X chromosomes minus 1. [cite: Nelson Textbook of Pediatrics 21e, Chapter 98 (Sex Chromosome Aneuploidies); Williams Gynecology 4e]