A patient diagnosed with HIV-II is admitted to the hospital presenting symptoms of fever, night sweats, and a severe cough.
The laboratory results show a CD4+ cell count of 180/mm and a negative tuberculosis (TB) skin test conducted 4 days prior.
What is the first action the nurse should take?
Inform the primary health care provider about the CD4+ results.
Implement Airborne Precautions for the patient.
Initiate Droplet Precautions for the patient.
Provide care using Standard Precautions.
The Correct Answer is D
Rationale for Choice A:
While it's important for the primary healthcare provider to be informed about the CD4+ results, it's not the first action the nurse should take. The priority is to implement appropriate infection control measures to protect the patient, other patients, and healthcare staff.
CD4+ cell count is a crucial indicator of the patient's immune status. A count of 180/mm is significantly low, suggesting a weakened immune system and increased vulnerability to infections. However, informing the provider alone doesn't directly address the immediate need for infection control.
Rationale for Choice B:
Airborne Precautions are specifically used for patients with known or suspected airborne infections, such as tuberculosis, measles, or varicella. These precautions involve the use of negative pressure rooms and N95 respirators.
In this case, the patient's TB skin test was negative, indicating no evidence of active tuberculosis infection. Implementing Airborne Precautions unnecessarily could lead to excessive resource utilization and potential stigmatization of the patient.
Rationale for Choice C:
Droplet Precautions are used for patients with infections that can be spread through large respiratory droplets, such as influenza, pertussis, or meningococcal meningitis. These precautions involve the use of masks and eye protection.
While the patient's symptoms of fever, night sweats, and severe cough could be consistent with a droplet-spread infection, there's no definitive evidence to support this at the present time. Initiating Droplet Precautions without a clear indication could also lead to unnecessary resource use and potential anxiety for the patient.
Rationale for Choice D:
Standard Precautions are the foundation of infection control and should be used for all patients, regardless of their known or suspected infection status. These precautions include hand hygiene, use of personal protective equipment (PPE) when indicated, and safe handling of sharps and bodily fluids.
By implementing Standard Precautions, the nurse can effectively minimize the risk of transmission of pathogens, protecting both the patient and other individuals in the healthcare setting. This is the most appropriate first action to ensure a safe and appropriate level of care.
Nursing Test Bank
Naxlex Comprehensive Predictor Exams
Related Questions
Correct Answer is B
Explanation
Choice B rationale:
Benign tumors typically grow in the wrong place or at the wrong time. This is a key distinguishing feature of benign tumors compared to normal cells. Normal cells have precise mechanisms that control their growth, ensuring they divide and multiply only when and where they are needed. Benign tumors, however, have disruptions in these regulatory mechanisms, leading to abnormal growth patterns. This means they may grow in locations where they don't belong or continue to grow even when they are no longer needed.
Here's a detailed explanation of why the other choices are incorrect:
Choice A:
Benign tumors have not lost their cellular regulation from contact inhibition. Contact inhibition is a process that stops normal cells from growing when they come into contact with other cells. Benign tumors still maintain this ability, which helps to limit their growth and prevent them from spreading to other tissues.
Choice C:
Benign tumors do not grow through invasion of other tissue. Invasion is a hallmark feature of malignant tumors (cancers). Benign tumors, on the other hand, typically grow as encapsulated masses, meaning they are surrounded by a well-defined border that separates them from surrounding tissues. They do not infiltrate or invade surrounding structures.
Choice D:
Anaplasia refers to the loss of differentiation and resemblance to the parent cells, which is a characteristic of malignant tumors, not benign tumors. Benign tumors still maintain a degree of differentiation, meaning they retain some of the characteristics of the normal cells from which they originated.
Correct Answer is C
Explanation
Choice A rationale:
Uncompensated Respiratory Acidosis is characterized by a low pH (less than 7.35) and a high pCO2 (greater than 45 mmHg). In this case, the pH is slightly elevated (7.46), making this option less likely.
While the pCO2 is elevated (46 mmHg), the body has begun to compensate, as evidenced by the elevated HCO3 (29 mEq/L). This partial compensation does not align with an uncompensated respiratory acidosis.
Choice B rationale:
Compensated Metabolic Acidosis would present with a normal pH (7.35-7.45) due to full compensation by the respiratory system. In this case, the pH is slightly elevated (7.46), which is not consistent with full compensation.
Additionally, the HCO3 is elevated (29 mEq/L), which is characteristic of metabolic alkalosis, not acidosis.
Choice C rationale:
Partially Compensated Metabolic Alkalosis is the most likely interpretation based on the ABG results. The pH is elevated (7.46), indicating alkalosis.
The HCO3 is also elevated (29 mEq/L), which is the primary cause of metabolic alkalosis.
The pCO2 is elevated (46 mmHg), which is a compensatory mechanism to try to normalize the pH. However, the compensation is not complete, as the pH is still slightly elevated.
This partial compensation is consistent with partially compensated metabolic alkalosis.
Choice D rationale:
Partially Compensated Respiratory Acidosis would present with a low pH (less than 7.35) and an elevated pCO2 (greater than 45 mmHg).
The HCO3 would also be elevated, but to a lesser degree than in metabolic alkalosis, as it's a secondary compensatory mechanism.
In this case, the pH is slightly elevated (7.46), making respiratory acidosis less likely.
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