Nursing Elites

1. Nurse Joy is caring for a client with an internal radiation implant. When caring for the client, the nurse should observe which of the following principles?

• A. Limit the time with the client to 1 hour per shift
• B. Do not allow pregnant women into the client’s room
• C. Remove the dosimeter badge when entering the client’s room
• D. Individuals younger than 16 years old may be allowed to go in the room as long as they are 6 feet away from the client

Correct answer: B

Rationale: Clients with internal radiation implants (also known as brachytherapy) emit a small amount of radiation, which can pose a risk to others. Pregnant women are particularly vulnerable to the harmful effects of radiation because it can affect both the mother and the developing fetus. Radiation exposure can lead to birth defects, miscarriage, or other developmental issues, so pregnant women should avoid any exposure by not entering the client's room.

2. A client who has been receiving radiation therapy for bladder cancer tells the nurse that it feels as if she is voiding through the vagina. The nurse interprets that the client may be experiencing which condition?

• A. Rupture of the bladder
• B. The development of a vesicovaginal fistula
• C. Extreme stress caused by the diagnosis of cancer
• D. Altered perineal sensation as a side effect of radiation therapy

Correct answer: B

Rationale: The correct answer is B. A vesicovaginal fistula is an abnormal connection between the bladder and the vagina, leading to the passage of urine through the vagina. This condition can occur due to various reasons, including radiation therapy. Choice A, rupture of the bladder, is incorrect because a rupture would present with more severe symptoms and is not consistent with the client's description. Choice C, extreme stress, is incorrect as it does not explain the physical symptom of voiding through the vagina. Choice D, altered perineal sensation, is incorrect as it does not involve a direct connection between the bladder and the vagina.

3. A patient on the oncology unit is receiving carmustine, a chemotherapy agent, and the nurse is aware that a significant side effect of this medication is thrombocytopenia. Which symptom should the nurse assess for in patients at risk for thrombocytopenia?

• A. Interrupted sleep pattern
• B. Hot flashes
• C. Epistaxis (nose bleed)
• D. Increased weight

Correct answer: C

Rationale: Thrombocytopenia is a condition characterized by a low platelet count, which increases the risk of bleeding and hemorrhage. Patients receiving chemotherapy agents like carmustine may experience thrombocytopenia as a significant side effect. Epistaxis (nosebleeds) is a common symptom associated with thrombocytopenia, as the blood vessels can become more fragile, and even minor trauma or spontaneous bleeding can occur. Therefore, assessing for signs of bleeding, including epistaxis, is crucial in patients at risk for thrombocytopenia.

4. A nursing student is caring for a patient with acute myeloid leukemia who is preparing to undergo induction therapy. In preparing a plan of care for this patient, the student should assign the highest priority to which nursing diagnosis?

• A. Activity Intolerance
• B. Risk for Infection
• C. Acute Confusion
• D. Risk for Spiritual Distress

Correct answer: B

Rationale: The correct answer is B: Risk for Infection. Induction therapy for acute myeloid leukemia suppresses the immune system, making the patient highly susceptible to infections due to neutropenia. Preventing infections is crucial in these patients to avoid complications. Activity Intolerance (Choice A) may be a concern, but infection prevention is of higher priority. Acute Confusion (Choice C) and Risk for Spiritual Distress (Choice D) are not the immediate priorities in this situation.

5. The cells of a normal individual can replicate in a specified rate. If the rate of replication becomes uncontrollable, which of the following is lacking from the patient?

• A. Apoptosis
• B. Contact inhibition
• C. Stable cells
• D. Labile cells

Correct answer: B

Rationale: Contact inhibition is a regulatory mechanism that prevents cells from proliferating once they reach a certain density. Normally, when cells grow and touch each other (such as in a monolayer), they stop dividing, maintaining tissue integrity and structure. When contact inhibition is lacking, as in many cancerous cells, cells continue to grow and divide uncontrollably, leading to tumor formation. This loss of regulation is a hallmark of cancerous growth.

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