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. 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.

3. 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.

4. 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.

5. Chromosomes contain thousands of segments called:

• A. nuclei.
• B. nodes.
• C. capillaries.
• D. genes.

Correct answer: D

Rationale: Chromosomes are structures composed of DNA and genes. Genes are the functional segments within chromosomes that encode specific traits and characteristics. The other choices ('nuclei,' 'nodes,' 'capillaries') do not accurately describe the segments found within chromosomes and are unrelated to their structure or function.

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