A nurse is caring for four clients. After reviewing today's laboratory results, which client should the nurse see first?

Correct Answer: C

Rationale: The correct answer is C because a prothrombin time (PT) of 28 seconds indicates a prolonged clotting time, putting the client at risk for bleeding. This requires immediate assessment and intervention to prevent complications. Choice A (INR of 2.8) is within the therapeutic range for clients on anticoagulant therapy, so it does not require immediate attention. Choice B (platelet count of 128000/mm³) is low but not critically low enough to warrant immediate action unless there are signs of active bleeding. Choice D (red blood cell count of 5.1 million/L) is within normal limits and does not indicate an urgent issue.

Correct Answer: C

Rationale: The correct answer is C: Nominal. The central line types in this study (port, PICC, TEC) are categorical and do not have a natural order or ranking. They are simply names or labels representing different types of central lines. This makes them fall under the nominal data structure category. Continuous data (choice A) would involve measurements with infinite possible values. Dichotomous data (choice B) would have only two categories. Ordinal data (choice D) would imply a natural ranking or order among the categories, which is not applicable in this context.

Correct Answer: C

Rationale: The correct answer is C because the infant's red cell findings of ovalocytes and polikilocytosis are consistent with hereditary elliptocytosis, which is caused by mutations in genes encoding spectrin. In this scenario, the inheritance pattern aligns with the infant receiving an alpha-spectrin mutation from both parents, leading to the observed erythrocyte abnormalities. This condition may improve over time due to compensatory mechanisms. Choice A (autosomal dominant ankyrin mutation causing hereditary spherocytosis) is incorrect because the clinical presentation and red cell morphology do not support a diagnosis of hereditary spherocytosis. Choice B (inherited band 3 variants and need for splenectomy) is incorrect as the infant's condition does not match the characteristics of hereditary spherocytosis requiring splenectomy. Choice D (PKLR variant and pyruvate kinase deficiency) is incorrect as the infant's red cell morphology is not indicative

Correct Answer: A

Rationale: The correct answer is A: Stage I. In the Ann Arbor staging system for Hodgkin lymphoma, Stage I indicates involvement of a single lymph node region (I) or a single extralymphatic site (IE) without systemic symptoms. In this case, the patient has a large mediastinal mass but no other sites of disease on metastatic workup. This corresponds to Stage I disease. Choice B: Stage II would indicate involvement of two or more lymph node regions on the same side of the diaphragm. Choice C: Stage III would indicate involvement of lymph node regions on both sides of the diaphragm. Choice D: Stage IV would indicate disseminated involvement of one or more extralymphatic organs or tissues. Therefore, based on the information provided, the correct stage for this patient is Stage I as per the Ann Arbor staging system.

Correct Answer: D

Rationale: Step-by-step rationale: 1. Lethargy, flank pain, nausea, vomiting in leukemia can suggest tumor lysis syndrome (TLS). 2. TLS can cause hyperkalemia, hyperphosphatemia, hyperuricemia, hypocalcemia, and elevated BUN. 3. Choice D has the highest potassium, phosphorus, uric acid, and BUN levels, and the lowest calcium level. 4. Therefore, choice D is the most consistent with the expected lab findings in tumor lysis syndrome. Summary: - Choice A has normal potassium, phosphorus, uric acid, calcium levels, and lower BUN. - Choice B has high potassium but normal phosphorus, uric acid, calcium, and slightly elevated BUN. - Choice C has low potassium, normal phosphorus, and elevated uric acid, calcium, and BUN. - Choice D aligns most closely with the expected lab findings in tumor lysis syndrome due to the pattern of