Nursing Elites

1. In assessing cancer risk, which woman is at greatest risk of developing breast cancer?

• A. A 35-year-old multipara who never breastfed.
• B. A 50-year-old whose mother had unilateral breast cancer.
• C. A 55-year-old whose mother-in-law had bilateral breast cancer.
• D. A 20-year-old whose menarche occurred at age 9.

Correct answer: B

Rationale: The correct answer is B because family history of breast cancer, specifically in the mother, is a significant risk factor for developing breast cancer. The age of 50 is also a risk factor for breast cancer. Choice A is less likely as breastfeeding can actually reduce the risk of breast cancer. Choice C is less relevant since the risk is higher with a direct family member. Choice D, although early menarche is a risk factor, the age of the individual is much lower compared to the other age-related risk factors.

2. A client who has undergone pleural biopsy is being monitored by a nurse. Which finding indicates a potential complication for the client?

• A. Warm, dry skin
• B. Mild pain at the biopsy site
• C. Complaints of shortness of breath
• D. Capillary refill time of less than 3 seconds

Correct answer: C

Rationale: Complaints of shortness of breath are a concerning finding post-pleural biopsy, as they may indicate a complication such as a pneumothorax or hemothorax. Shortness of breath can be a sign of respiratory distress that requires immediate attention. Warm, dry skin, mild pain at the biopsy site, and a capillary refill time of less than 3 seconds are not typically associated with immediate complications following a pleural biopsy. Warm, dry skin may be a normal finding, mild pain can be expected at the biopsy site, and a capillary refill time of less than 3 seconds is within normal limits.

3. A client who experienced partial-thickness burns involving over 50% body surface area (BSA) 2 weeks ago has several open wounds and develops watery diarrhea. The client's blood pressure is 82/40 mmHg, and temperature is 96°F (36.6°C). Which action is most important for the nurse to take?

• A. Increase the room temperature.
• B. Assess the oxygen saturation.
• C. Continue to monitor vital signs.
• D. Notify the rapid response team.

Correct answer: D

Rationale: In this scenario, the client is presenting with signs of sepsis, such as hypotension, hypothermia, and a recent history of partial-thickness burns with open wounds. The development of watery diarrhea further raises suspicion for sepsis. With a blood pressure of 82/40 mmHg and a low temperature of 96°F (36.6°C), the nurse should recognize the potential for septic shock. Notifying the rapid response team is crucial in this situation as the client requires immediate intervention and management to prevent deterioration and address the underlying septic process. Increasing the room temperature (Choice A) is not the priority as the low body temperature is likely due to systemic vasodilation and not environmental factors. While assessing oxygen saturation (Choice B) is important, the client's hypotension and hypothermia take precedence. Continuing to monitor vital signs (Choice C) alone is insufficient given the critical condition of the client and the need for prompt action to address the sepsis and potential septic shock.

4. What do crackles heard on lung auscultation indicate?

• A. Cyanosis.
• B. Bronchospasm.
• C. Airway narrowing.
• D. Fluid-filled alveoli.

Correct answer: D

Rationale: Crackles heard on lung auscultation are caused by the popping open of small airways that are filled with fluid. This is commonly associated with conditions such as pulmonary edema, pneumonia, or heart failure. Cyanosis (Choice A) is a bluish discoloration of the skin and mucous membranes due to low oxygen levels in the blood, not directly related to crackles. Bronchospasm (Choice B) refers to the constriction of the airway smooth muscle, causing difficulty in breathing but does not typically produce crackles. Airway narrowing (Choice C) can lead to wheezing but is not directly linked to crackles heard on auscultation.

5. What is the most common cause of urinary tract infections (UTIs)?

• A. Escherichia coli infection
• B. Staphylococcus aureus infection
• C. Pseudomonas aeruginosa infection
• D. Klebsiella pneumoniae infection

Correct answer: A

Rationale: Escherichia coli is the most common cause of urinary tract infections (UTIs). It is responsible for the majority of UTIs, especially in women. E. coli is a normal inhabitant of the bowel and can enter the urinary tract through the urethra, leading to infection. Staphylococcus aureus, Pseudomonas aeruginosa, and Klebsiella pneumoniae are less common causes of UTIs compared to E. coli. Staphylococcus aureus typically causes skin and soft tissue infections, Pseudomonas aeruginosa is more commonly associated with healthcare-associated infections, and Klebsiella pneumoniae is known for causing pneumonia and other respiratory infections.

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