Nursing Elites

1. Which of the following medications is typically used to treat asthma?

• A. Aspirin
• B. Metformin
• C. Albuterol
• D. Lisinopril

Correct answer: C

Rationale: The correct answer is C, Albuterol. Albuterol is a bronchodilator commonly used to treat asthma by relaxing the muscles around the airways, helping to relieve symptoms such as coughing, wheezing, shortness of breath, and chest tightness. Aspirin (Choice A) is not typically used to treat asthma and can actually trigger asthma symptoms in some individuals. Metformin (Choice B) is a medication for managing type 2 diabetes and is not indicated for asthma treatment. Lisinopril (Choice D) is an angiotensin-converting enzyme (ACE) inhibitor primarily used to treat high blood pressure and heart failure, not asthma.

2. A client with autosomal dominant polycystic kidney disease (ADPKD) asks, “Will my children develop this disease?” How should the nurse respond?

• A. No genetic link is known, so your children are not at increased risk.
• B. Your sons will develop this disease because it has a sex-linked gene.
• C. Only if both you and your spouse are carriers of this disease.
• D. Each of your children has a 50% risk of having ADPKD.

Correct answer: D

Rationale: Children whose parent has the autosomal dominant form of PKD have a 50% chance of inheriting the gene that causes the disease. ADPKD is transmitted as an autosomal dominant trait and therefore is not gender-specific. Both parents do not need to have this disorder. Choice A is incorrect because ADPKD has a known genetic link and a definitive mode of inheritance. Choice B is incorrect as ADPKD is not sex-linked but autosomal dominant. Choice C is incorrect because ADPKD follows an autosomal dominant inheritance pattern and does not require both parents to be carriers for the child to inherit the disease.

3. During spring break, a young adult presents to the urgent care clinic and reports a stiff neck, a fever for the past 6 hours, and a headache. Which intervention is most important for the nurse to implement first?

• A. Initiate isolation precautions
• B. Administer an antipyretic
• C. Draw blood cultures
• D. Prepare for lumbar puncture

Correct answer: A

Rationale: The correct answer is to initiate isolation precautions. This is the priority action because the patient presents with symptoms that could be indicative of meningitis, an infectious disease that requires isolation to prevent its spread. Administering an antipyretic (Choice B) may help manage the fever but does not address the need for isolation. Drawing blood cultures (Choice C) and preparing for a lumbar puncture (Choice D) are important steps in diagnosing meningitis but should come after initiating isolation precautions to prevent potential transmission of the infection to others.

4. A female patient will receive doxycycline to treat a sexually transmitted infection (STI). What information will the nurse include when teaching this patient about this medication?

• A. Nausea and vomiting are uncommon adverse effects.
• B. The drug may cause possible teratogenic effects.
• C. Increase intake of dairy products with each dose of this medication.
• D. Use a backup method of contraception if taking oral contraceptives.

Correct answer: D

Rationale: The correct answer is D. The desired action of oral contraceptives can be reduced when taken with tetracyclines like doxycycline. Therefore, patients on oral contraceptives should be advised to use a backup contraception method while taking doxycycline. Choice A is incorrect because nausea and vomiting are common adverse effects of doxycycline. Choice B is incorrect because doxycycline is not known for causing teratogenic effects. Choice C is incorrect because dairy products can interfere with the absorption of doxycycline, so they should be avoided when taking this medication.

5. The client is being educated by the nurse on home blood glucose monitoring. Which of the following blood glucose measurements indicates hypoglycemia?

• A. 59 mg/dL (3.3 mmol/L)
• B. 75 mg/dL (4.2 mmol/L)
• C. 108 mg/dL (6 mmol/L)
• D. 119 mg/dL (6.6 mmol/L)

Correct answer: A

Rationale: A blood glucose level of 59 mg/dL (3.3 mmol/L) is considered hypoglycemia, which is an abnormally low blood sugar level. This level requires immediate attention as it can lead to symptoms such as confusion, shakiness, and even loss of consciousness if left untreated. Choices B, C, and D have blood glucose levels within the normal range or slightly higher, indicating euglycemia or normal blood sugar levels, and not hypoglycemia.

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