A chromosomal analysis is obtained on a young woman with mild signs of Patau syndrome and reveals a 46,XX/47,XX +13 mosaic karyotype. Nondisjunction is most likely to have occurred in:

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Question 1 of 5

A chromosomal analysis is obtained on a young woman with mild signs of Patau syndrome and reveals a 46,XX/47,XX +13 mosaic karyotype. Nondisjunction is most likely to have occurred in:

Correct Answer: B

Rationale: Mitosis post-fertilization (B) causes mosaicism. Rationale: 46, XX/47, XX, +13 arises when nondisjunction in an early mitotic division of a 46, XX zygote loses or gains a 13, creating two cell lines. Meiotic errors produce uniform aneuploidy.

Question 2 of 5

Which phase of the cycle is most likely to be interrupted for smaller abnormalities detection?

Correct Answer: B

Rationale: Prometaphase (B) allows detection of smaller abnormalities. Rationale: Chromosomes are less condensed than metaphase, showing finer details with high-resolution banding, unlike S (replication) or G (non-dividing) phases.

Question 3 of 5

SMOH gene located in pseudoautosomal in Y chromosome, do you expect that it will be located in 46 XX?

Correct Answer: A

Rationale: Yes (A), pseudoautosomal regions (PAR) are on both X and Y. Rationale: PAR genes (e.g., SHOX, not SMOH—a typo?) are homologous on X and Y, present in 46, XX females, unlike Y-specific genes.

Question 4 of 5

Tetraploidy results from:

Correct Answer: B

Rationale: Endomitosis post-fertilization (B) causes tetraploidy (4n=92). Rationale: Zygote duplicates chromosomes without division; dispermy yields triploidy, not 4n.

Question 5 of 5

Possibility of combinations for diploid cell with 10 chromosomes?

Correct Answer: D

Rationale: 2^n where n=5 (haploid from 2n=10) gives 32 (D). Rationale: Independent assortment of 5 pairs yields 2^5 = 32 gamete combinations.

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