Nursing Elites

1. A newborn with a respiratory rate of 40 breaths per minute at one minute after birth is demonstrating cyanosis of the hands and feet. What action should a nurse take?

• A. Assess bowel sounds.
• B. Continue to monitor.
• C. Assist with intubation.
• D. Rub the infant's back.

Correct answer: B

Rationale: Cyanosis of the hands and feet, known as acrocyanosis, is common in newborns shortly after birth and usually resolves on its own. It is not indicative of a need for immediate intervention. Therefore, the appropriate action is to continue monitoring the newborn's condition. Assessing bowel sounds (Choice A) is not relevant to the presenting issue of cyanosis and respiratory rate. Assisting with intubation (Choice C) is an invasive procedure that is not warranted based on the information provided. Rubbing the infant's back (Choice D) is not necessary for acrocyanosis and could potentially disturb the newborn.

2. 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.

3. 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.

4. In the prenatal record, the nurse should record for the pregnant client who has a 3-year-old child at home, a term birth, a miscarriage at 10 weeks’ gestation, and a set of twins who died within 24 hours:

• A. Gravida 2, para 1.
• B. Gravida 3, para 3.
• C. Gravida 4, para 2.
• D. Gravida 5, para 4.

Correct answer: C

Rationale: The correct answer is C: 'Gravida 4, para 2.' Gravida refers to the total number of pregnancies, including the current one. In this case, the client has been pregnant a total of 4 times, so gravida is 4. Para is the number of pregnancies that have reached viability, which is 2 in this case. The client has had a term birth and a set of twins who died within 24 hours, totaling 2 pregnancies that reached viability. Choices A, B, and D are incorrect because they do not accurately reflect the client's obstetric history based on the information provided.

5. A client at 20 weeks of gestation has trichomoniasis. Which of the following findings should the nurse expect?

• A. Thick, White Vaginal Discharge
• B. Urinary Frequency
• C. Vulvar Lesions
• D. Malodorous Discharge

Correct answer: D

Rationale: Malodorous discharge is a common symptom of trichomoniasis caused by the Trichomonas vaginalis parasite. It is typically described as frothy, greenish-yellow, and malodorous. Choices A, B, and C are incorrect findings associated with other conditions. Thick, white vaginal discharge is more characteristic of a yeast infection; urinary frequency may be seen in urinary tract infections; and vulvar lesions are commonly seen in herpes simplex virus infections.

Similar Questions