Nursing Elites

1. Which information regarding the care of antepartum women with cardiac conditions is most important for the nurse to understand?

• A. Stress on the heart is greatest in the first trimester and the last 2 weeks before labor.
• B. Women with class II cardiac disease should avoid heavy exertion and any activity that causes even minor symptoms.
• C. Women with class III cardiac disease should get 8 to 10 hours of sleep every day and limit housework, shopping, and exercise.
• D. Women with class I cardiac disease need bed rest through most of the pregnancy and face the possibility of hospitalization near term.

Correct answer: B

Rationale: Class II cardiac disease is symptomatic with ordinary activity. Women in this category need to avoid heavy exertion and limit regular activities as symptoms dictate. Stress is greatest between weeks 28 and 32 of gestation, when hemodynamic changes reach their maximum. Class III cardiac disease is symptomatic with less-than-ordinary activity. These women need bed rest most of the day and face the possibility of hospitalization near term. Class I cardiac disease is asymptomatic at normal levels of activity. These women can perform limited normal activities with discretion, although they still need a good amount of sleep.

2. Twenty-year-old Jack is extremely tall and has very thick facial hair. Most of his male secondary sex characteristics are also more pronounced than men of his age. In this scenario, Jack is most likely:

• A. an XYY male.
• B. diagnosed with Klinefelter syndrome.
• C. an XXY male.
• D. diagnosed with Down syndrome.

Correct answer: A

Rationale: The correct answer is A: an XYY male. Individuals with XYY syndrome often exhibit increased height and more pronounced secondary male characteristics, such as thick facial hair. Choice B, Klinefelter syndrome (XXY), typically presents with less prominent male secondary sex characteristics due to the presence of an extra X chromosome. Choice C, XXY male, refers to Klinefelter syndrome, which does not align with the description of Jack having more pronounced male secondary sex characteristics. Choice D, Down syndrome, is caused by a trisomy of chromosome 21 and is not associated with the physical characteristics described in the scenario.

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. After meiosis, each new cell nucleus contains _____ chromosomes.

• A. 46
• B. 35
• C. 23
• D. 12

Correct answer: C

Rationale: After meiosis, each resulting cell contains 23 chromosomes. Meiosis is a process that involves two sequential divisions resulting in four daughter cells, each with half the number of chromosomes as the parent cell. In humans, the parent cell has 46 chromosomes (diploid), and after meiosis, the resulting cells (sperm or ova) have 23 chromosomes (haploid). Choice A (46 chromosomes) is incorrect because this is the number of chromosomes in a human diploid cell before meiosis. Choices B (35 chromosomes) and D (12 chromosomes) are incorrect as they do not represent the correct number of chromosomes after meiosis in human cells.

5. When reviewing the electronic medical record of a postpartum client, which of the following factors places the client at risk for infection?

• A. Meconium-stained amniotic fluid
• B. Placenta previa
• C. Midline episiotomy
• D. Gestational hypertension

Correct answer: C

Rationale: The correct answer is C: Midline episiotomy. An episiotomy is a surgical incision made during childbirth to enlarge the vaginal opening. This procedure increases the risk of infection in the postpartum period due to the incision site being a potential entry point for pathogens. Meconium-stained amniotic fluid (choice A) is a risk factor for fetal distress but does not directly increase the mother's risk of infection. Placenta previa (choice B) is a condition where the placenta partially or completely covers the cervix, leading to potential bleeding issues but not necessarily an increased risk of infection. Gestational hypertension (choice D) is a hypertensive disorder that affects some pregnant women but is not directly associated with an increased risk of infection in the postpartum period.

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