Nursing Elites

1. A 17-year-old client gave birth 12 hours ago. She states that she doesn't know how to care for her baby. To promote parent-infant attachment behaviors, which intervention should the nurse implement?

• A. Ask if she has help to care for the baby at home
• B. Provide a video on newborn safety and care
• C. Explore the basis of fears with the client
• D. Encourage rooming in while in the hospital

Correct answer: D

Rationale: Encouraging rooming in while in the hospital is the most appropriate intervention to promote parent-infant attachment behaviors. Rooming in allows the mother to stay with her baby continuously, facilitating bonding and providing the opportunity for the mother to learn how to care for her baby with the nurse's support. Asking if she has help at home (Choice A) does not directly address promoting attachment behaviors. Providing a video on newborn safety and care (Choice B) may offer information but does not actively facilitate immediate bonding. Exploring the basis of fears (Choice C) is important but may not directly address promoting attachment behaviors as effectively as encouraging rooming in.

2. When should the low-risk patient, who is 16 weeks pregnant, be advised to return to the prenatal clinic?

• A. 1 week.
• B. 2 weeks.
• C. 3 weeks.
• D. 4 weeks.

Correct answer: D

Rationale: The correct answer is D: 4 weeks. Low-risk pregnant patients typically have prenatal visits every 4 weeks until 28 weeks of gestation. This frequency allows for adequate monitoring of the pregnancy without being overly burdensome on the patient. Choices A, B, and C are incorrect as they do not align with the standard prenatal care schedule for low-risk pregnancies. Visits that are too frequent may cause unnecessary anxiety for the patient, while visits that are too infrequent may miss important opportunities for monitoring and intervention.

3. Twins that derive from a single zygote that has split into two are called:

• A. monozygotic (MZ) twins.
• B. fraternal twins.
• C. non-identical twins.
• D. dizygotic (DZ) twins.

Correct answer: A

Rationale: The correct answer is A: monozygotic (MZ) twins. Monozygotic twins, also known as identical twins, occur when a single zygote splits into two embryos, leading to two genetically identical individuals. Choice B, fraternal twins, are twins that develop from two separate eggs fertilized by two different sperm cells, resulting in non-identical siblings. Choice C, non-identical twins, is not a common term used to describe this type of twinning. Choice D, dizygotic (DZ) twins, refer to twins that develop from two separate eggs fertilized by two different sperm cells, leading to non-identical twins.

4. A woman at 26 weeks of gestation is being assessed to determine whether she is experiencing preterm labor. Which finding indicates that preterm labor is occurring?

• A. Estriol is not found in maternal saliva.
• B. Irregular, mild uterine contractions occur every 12 to 15 minutes.
• C. Fetal fibronectin is present in vaginal secretions.
• D. The cervix is effacing and dilated to 2 cm.

Correct answer: D

Rationale: The correct answer is D. Cervical changes such as effacement and dilation to 2 cm are strong indicators of imminent preterm labor. These changes, combined with regular contractions, can signify labor at any gestation. Estriol can be detected in maternal plasma as early as 9 weeks of gestation. Levels of salivary estriol have been linked to preterm birth. Irregular, mild contractions occurring every 12 to 15 minutes without cervical change are generally not concerning. While the presence of fetal fibronectin in vaginal secretions between 24 and 36 weeks of gestation may predict preterm labor, its predictive value is limited (20%-40%). Therefore, cervical changes provide more reliable information regarding the risk of preterm labor.

5. 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.

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