Nursing Elites

1. After a mother was diagnosed with gonorrhea immediately after delivery, what is an important goal of the nurse when providing care for her baby?

• A. Prevent the development of ophthalmia neonatorum.
• B. Lubricate the eyes.
• C. Prevent the development of infection.
• D. Teach about the risks of breastfeeding with gonorrhea.

Correct answer: A

Rationale: The correct answer is A: Prevent the development of ophthalmia neonatorum. When a mother has gonorrhea, the baby can be infected during delivery, leading to ophthalmia neonatorum, which can cause permanent blindness. Therefore, it is crucial for the nurse to prevent this condition by treating the baby's eyes with an antibiotic prophylactically after birth. Choice B, lubricating the eyes, is not the primary goal in this situation as preventing infection takes precedence. Choice C, preventing the development of infection, is too broad and does not specifically address the potential complication of ophthalmia neonatorum. Choice D, teaching about the risks of breastfeeding with gonorrhea, is important but not the immediate goal in this scenario where preventing ophthalmia neonatorum and potential blindness is the priority.

2. Which of the following pairs of bases is present in the rungs of the ladder-like structure of deoxyribonucleic acid (DNA)?

• A. Cytosine with guanine.
• B. Rhodamine with biotin.
• C. Diaminopurine with ribozyme.
• D. Serine with tyrosine.

Correct answer: A

Rationale: The correct answer is A: Cytosine with guanine. In the DNA double helix, cytosine always pairs with guanine forming a base pair, and adenine pairs with thymine. These complementary base pairs form the rungs of the ladder-like structure of DNA. Choice B, Rhodamine with biotin, is incorrect as they are not base pairs found in DNA. Choice D, Serine with tyrosine, is incorrect as they are amino acids, not DNA bases. Choice C, Diaminopurine with ribozyme, is also incorrect as ribozyme is an enzyme, not a base, and diaminopurine is not one of the standard bases found in DNA.

3. A 16-year-old gravida 1 para 0 client has just been admitted to the hospital with a diagnosis of eclampsia. She's not presently convulsing. Which intervention should the nurse plan to include in this client's nursing care plan?

• A. Allow liberal family visitation
• B. Keep an airway at the bedside
• C. Assess temperature every hour
• D. Monitor blood pressure, pulse, and respiration every 4 hours

Correct answer: B

Rationale: Keeping an airway at the bedside is crucial for a client with eclampsia, as there is a high risk of seizures that can obstruct the airway. Allowing liberal family visitation (choice A) may not be a priority at this time and can be overwhelming for the client. Assessing temperature every hour (choice C) is not directly related to managing eclampsia. Monitoring blood pressure, pulse, and respiration every 4 hours (choice D) is important but not as immediate as ensuring airway patency.

4. Humans begin life as a single cell that divides repeatedly. This cell is known as a(n):

• A. zygote.
• B. gonadotrope.
• C. embryo.
• D. chromaffin.

Correct answer: A

Rationale: A zygote is the correct answer. It is the initial cell formed when a sperm cell fertilizes an egg cell, marking the beginning of human development. Choice B, gonadotrope, is incorrect as it refers to a type of hormone-secreting cell in the pituitary gland. Choice C, embryo, is incorrect as it is the stage of development after the zygote has implanted into the uterine wall and undergone initial cell divisions. Choice D, chromaffin, is incorrect as it refers to cells found in the adrenal medulla that produce and store catecholamines.

5. Dizygotic (DZ) twins share _________ percent of their genes.

• A. 100
• B. 75
• C. 50
• D. 25

Correct answer: C

Rationale: Dizygotic (DZ) twins share approximately 50% of their genes. This is because dizygotic twins, also known as fraternal twins, originate from two separate fertilized eggs and share similar genetic similarity to regular siblings. Choice A (100%) is incorrect because if twins shared 100% of their genes, they would be identical twins (monozygotic). Choice B (75%) is incorrect as it is not the typical genetic similarity seen in dizygotic twins. Choice D (25%) is incorrect as it represents a significantly lower genetic similarity than what is observed in dizygotic twins.

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