Nursing Elites

1. 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.

2. Which of the following statements about Rh incompatibility is true?

• A. Rh incompatibility occurs most commonly during a woman’s first pregnancy.
• B. Rh incompatibility is an untreatable condition that leaves a woman infertile for the rest of her life.
• C. Rh incompatibility is an abnormality that is transmitted from generation to generation and carried by a sex chromosome.
• D. Rh incompatibility occurs due to antibodies transmitted to a fetus during subsequent deliveries causing brain damage or death.

Correct answer: D

Rationale: Rh incompatibility occurs when the mother's antibodies attack the fetus's red blood cells, leading to serious complications, usually in subsequent pregnancies. Choice A is incorrect because Rh incompatibility often occurs in subsequent pregnancies, not necessarily the first one. Choice B is incorrect as Rh incompatibility does not render a woman infertile but can lead to complications during pregnancies. Choice C is incorrect as Rh incompatibility is not carried by a sex chromosome but involves the Rh factor on red blood cells.

3. A mother spontaneously delivers a newborn infant in the taxicab while on the way to the hospital. The emergency room nurse reported the mother has active herpes (HSV II) lesions on the vulva. Which intervention should the nurse implement first when admitting the neonate to the nursery?

• A. Document the newborn's temperature on the flow sheet.
• B. Place the newborn in the isolation area of the nursery.
• C. Obtain a blood specimen for a serum glucose level.
• D. Administer the vitamin K injection.

Correct answer: B

Rationale: Newborns exposed to active herpes lesions are at high risk for neonatal herpes, which can be severe. Placing the newborn in isolation is crucial as it helps prevent the spread of the virus and allows for close monitoring. Documenting the newborn's temperature, obtaining a blood specimen for a serum glucose level, and administering the vitamin K injection are important interventions but are not the priority when dealing with a potential infectious risk like neonatal herpes.

4. A client comes to the clinic for her first prenatal visit and reports that July 10 was the first day of her last menstrual period. Using Nagele’s Rule, the nurse calculates the estimated date of birth for the client to be _________.

• A. 4/17.
• B. 4/10.
• C. 5/10.
• D. 5/17.

Correct answer: A

Rationale: Nagele's Rule is a common method used to estimate the due date. To calculate it, subtract 3 months and add 7 days to the first day of the last menstrual period. In this case, if the last menstrual period started on July 10, subtracting 3 months (April) and adding 7 days gives an estimated due date of April 17. This is the correct answer. Choices B, C, and D are incorrect because they do not follow the Nagele's Rule calculation method.

5. A client is experiencing preterm labor and has a prescription for 4 doses of dexamethasone 6 mg IM every 12 hours. The available concentration is dexamethasone 10 mg/mL. How many mL of dexamethasone should the nurse administer per dose? (Round the answer to the nearest tenth. Use a leading zero if it applies. Do not use trailing zero.)

• A. 0.6 mL
• B. ---
• C. ---
• D. ---

Correct answer: A

Rationale: Calculation: (6 mg/dose) / (10 mg/mL) = 0.6 mL per dose. The correct answer is 0.6 mL. This calculation is obtained by dividing the dose needed (6 mg) by the concentration available (10 mg/mL). The resulting value is 0.6 mL per dose. Choices B, C, and D are not applicable as the correct answer has been calculated accurately.

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