Nursing Elites

1. A multiparous woman has been in labor for 8 hours. Her membranes have just ruptured. What is the nurse’s highest priority in this situation?

• A. Prepare the woman for imminent birth.
• B. Notify the woman’s primary healthcare provider.
• C. Document the characteristics of the fluid.
• D. Assess the fetal heart rate (FHR) and pattern.

Correct answer: D

Rationale: The correct answer is to assess the fetal heart rate (FHR) and pattern (Choice D). When a multiparous woman's membranes rupture after 8 hours of labor, the nurse's priority is to assess the fetal well-being. Rupture of membranes can lead to potential complications such as umbilical cord prolapse. Monitoring the fetal heart rate and pattern immediately after the rupture of membranes is crucial to ensure the fetus is not in distress. This assessment helps in determining the need for immediate interventions to safeguard the fetus. Documenting the characteristics of the fluid (Choice C) may be necessary but is of lower priority compared to assessing fetal well-being. While preparing the woman for imminent birth (Choice A) is important, assessing the fetal heart rate takes precedence to ensure the fetus is not compromised. Notifying the woman's primary healthcare provider (Choice B) is also important but not the highest priority at this moment.

2. Which of the following statements is true of sickle-cell anemia?

• A. It is typically managed with treatments such as pain relief medications.
• B. It is caused by a mutation in the beta-globin gene.
• C. It leads to the obstruction of small blood vessels and decreased oxygen delivery.
• D. It is more prevalent in individuals of African, Mediterranean, Middle Eastern, and Indian descent.

Correct answer: C

Rationale: The correct answer is C. Sickle-cell anemia results from a mutation in the beta-globin gene, causing red blood cells to become sickle-shaped. These misshapen cells can obstruct small blood vessels, leading to reduced oxygen delivery to tissues. Choices A, B, and D are incorrect because sickle-cell anemia is typically managed with treatments such as pain relief medications, hydration, and in severe cases, blood transfusions. It is caused by a specific mutation in the beta-globin gene, not by the inability to metabolize phenylalanine. Additionally, sickle-cell anemia is more prevalent in individuals of African, Mediterranean, Middle Eastern, and Indian descent, not exclusive to any specific gender.

3. A client has bacterial vaginosis. Which of the following medications should the nurse expect to administer?

• A. Metronidazole
• B. Fluconazole
• C. Acyclovir
• D. Clindamycin

Correct answer: A

Rationale: Metronidazole is the correct choice for treating bacterial vaginosis as it is the first-line medication recommended for this condition. Metronidazole works by disrupting the DNA structure of bacteria, making it an effective treatment. Choice B, Fluconazole, is an antifungal medication primarily used for treating fungal infections, not bacterial vaginosis. Choice C, Acyclovir, is an antiviral medication used to treat viral infections, not bacterial vaginosis. Choice D, Clindamycin, is also used to treat bacterial infections but is not the first-line treatment for bacterial vaginosis, making it an incorrect choice in this scenario.

4. _________ are problems that stem from the interaction of heredity and environmental factors.

• A. Multifactorial problems
• B. Cognitive problems
• C. Horizon problems
• D. Coronal problems

Correct answer: A

Rationale: Multifactorial problems are conditions that result from the interplay of genetic and environmental factors. These conditions, such as diabetes or heart disease, are not solely determined by genetics or environment but are influenced by a combination of both factors. Choice B, cognitive problems, refers to difficulties related to thinking, learning, and memory and are not specifically linked to genetic and environmental interactions. Choices C and D, horizon problems and coronal problems, are nonsensical terms and do not relate to the interaction of heredity and environmental factors.

5. Before meiosis, a sperm cell:

• A. contains 46 chromosomes.
• B. contains two X chromosomes.
• C. is significantly larger than an egg cell.
• D. contains both an X and a Y chromosome.

Correct answer: A

Rationale: Before meiosis, a sperm cell contains 46 chromosomes. This is because sperm cells, like other somatic cells, have a diploid number of chromosomes. During meiosis, the number of chromosomes is halved to 23 to combine with an egg cell during fertilization. Choice B is incorrect because a sperm cell carries either an X or a Y chromosome, not both (Choice D). Choice C is incorrect as sperm cells are generally smaller than egg cells, which is an adaptation that aids in motility and penetration of the egg during fertilization.

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