Nursing Elites

1. What causes cystic fibrosis?

• A. Sex-linked abnormality.
• B. Abnormality in the 21st pair of chromosomes.
• C. Recessive gene.
• D. Single segment found only on the Y chromosome.

Correct answer: C

Rationale: Cystic fibrosis is a genetic disorder caused by inheriting two copies of a recessive gene, one from each parent. This means that both parents must carry at least one copy of the faulty gene for a child to inherit the condition. Choice A is incorrect because cystic fibrosis is not linked to the sex chromosomes. Choice B is incorrect as cystic fibrosis is not caused by an abnormality in the 21st pair of chromosomes but by a specific gene mutation. Choice D is also incorrect as cystic fibrosis is not related to the Y chromosome, which is specific to males.

2. Rico is a man who has enlarged breasts and suffers from mild mental retardation. He has a problem learning languages, and his body produces less of the male sex hormone testosterone than normal males. Rico is most likely suffering from:

• A. Klinefelter syndrome.
• B. Tay-Sachs disease.
• C. Turner syndrome.
• D. Down syndrome.

Correct answer: A

Rationale: Rico's symptoms align with Klinefelter syndrome, which is characterized by an extra X chromosome in males (XXY). Enlarged breasts (gynecomastia), mild mental retardation, learning difficulties, and reduced testosterone production are common features of Klinefelter syndrome. Choice B, Tay-Sachs disease, is a genetic disorder that affects the nervous system and is not associated with the symptoms described. Choice C, Turner syndrome, occurs in females with a missing or partially missing X chromosome and does not fit Rico's profile. Choice D, Down syndrome, is caused by an extra copy of chromosome 21 and typically does not present with the symptoms mentioned for Rico.

3. Which of the following is a sexually transmitted infection that, in advanced stages, can attack major organ systems?

• A. Rubella
• B. Syphilis
• C. Cystic fibrosis
• D. Phenylketonuria

Correct answer: B

Rationale: Syphilis is the correct answer. Syphilis is a sexually transmitted infection caused by the bacterium Treponema pallidum. If left untreated, it can progress through various stages and potentially attack major organ systems, causing severe complications. Rubella, Cystic fibrosis, and Phenylketonuria are not sexually transmitted infections. Rubella is a viral infection, Cystic fibrosis is a genetic disorder affecting the lungs and digestive system, and Phenylketonuria is a genetic metabolic disorder. These conditions do not typically affect major organ systems in the same way as untreated syphilis.

4. A woman has experienced iron deficiency anemia during her pregnancy. She had been taking iron for 3 months before the birth. The client gave birth by cesarean 2 days earlier and has been having problems with constipation. After assisting her back to bed from the bathroom, the nurse notes that the woman’s stools are dark (greenish-black). What should the nurse’s initial action be?

• A. Perform a guaiac test and record the results.
• B. Recognize the finding as abnormal and report it to the primary health care provider.
• C. Recognize the finding as a normal result of iron therapy.
• D. Check the woman’s next stool to validate the observation.

Correct answer: C

Rationale: The nurse should recognize that dark stools are a common side effect in clients who are taking iron replacement therapy. Dark stools are a known, expected result of iron supplementation and are not indicative of a complication unless other symptoms of GI bleeding are present. A guaiac test would be necessary if there were concerns about gastrointestinal bleeding. Recognizing dark stools as a consequence of iron therapy is an essential nursing assessment skill and does not require immediate reporting. Checking the next stool to confirm the observation is unnecessary as the presence of dark stools in this context is already an expected outcome of iron supplementation.

5. A nurse on the labor and delivery unit is assessing four clients. Which of the following clients is a candidate for an induction of labor with misoprostol?

• A. A client who has active genital herpes
• B. A client who has gestational diabetes mellitus
• C. A client who has a previous uterine incision
• D. A client who has placenta previa

Correct answer: B

Rationale: Misoprostol can be used for induction in clients with gestational diabetes mellitus. Choice A, a client with active genital herpes, is not a candidate for misoprostol induction due to the risk of viral shedding and transmission. Choice C, a client with a previous uterine incision, may be at risk for uterine rupture with misoprostol use. Choice D, a client with placenta previa, is not an appropriate candidate for misoprostol induction due to the risk of increased bleeding associated with the condition.

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