A female with a flattened face, small head, short neck, protruding tongue, small ears, and a poor muscle tone (hypotonia). She probably has a genetic disorder that's caused by?

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Pediatric Genetic Disorders Questions

Question 1 of 5

A female with a flattened face, small head, short neck, protruding tongue, small ears, and a poor muscle tone (hypotonia). She probably has a genetic disorder that's caused by?

Correct Answer: A

Rationale: Trisomy 21 (Down syndrome, A) matches these features. Rationale: Flat face, short neck, hypotonia, etc., are classic Down syndrome signs. Monosomy X (Turner) and Trisomy X have different phenotypes (e.g., short stature vs. subtle effects).

Question 2 of 5

A person whose karyotype is 45, XX, t(15q;21q). If she mates with a 46, XY normal individual. What is the possibility that they will have a zygote which will develop into Down syndrome? ('t' is translocation):

Correct Answer: A

Rationale: 1/6 (A) is correct. Rationale: Robertsonian t(15;21) carrier (45, XX) produces 6 gamete types: normal 15+21, t(15;21), 15, 21, null 15, null 21. With normal sperm (15+21+Y), only 21+t(15;21) yields 47, +21 (Down). 1/6 gametes result in trisomy 21.

Question 3 of 5

Chromosome 22 is:

Correct Answer: A

Rationale: Chromosome 22 is acrocentric, with a very short p-arm and centromere near one end. 'A' is correct. Rationale: Human acrocentric chromosomes (13, 14, 15, 21, 22) have nucleolar organizer regions on their p-arms; 22 fits this pattern, unlike metacentric (central centromere) or telocentric (end centromere).

Question 4 of 5

Which banding technique stains heterochromatin (dark) and euchromatin (light)?

Correct Answer: C

Rationale: G-banding (C) stains heterochromatin dark (G-positive) and euchromatin light. Rationale: Giemsa stains AT-rich, gene-poor heterochromatin dark, while GC-rich, gene-active euchromatin stays light, standard for karyotyping.

Question 5 of 5

In which phase of gametogenesis nondisjunction will produce all the cells aneuploidy?

Correct Answer: C

Rationale: Nondisjunction in meiosis I (MI, C) affects all daughter cells. Rationale: MI separates homologs; failure means all gametes get abnormal chromosome numbers (e.g., 24 or 22), unlike MII (sister chromatids, half affected).

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