A nursing student wants to know why clients with chronic obstructive pulmonary disease tend to be polycythemic. What response by the nurse instructor is best?

Correct Answer: D

Rationale: The correct answer is D. In chronic obstructive pulmonary disease (COPD), the lungs are unable to efficiently exchange oxygen and carbon dioxide, leading to tissue hypoxia. The body compensates for this by producing more red blood cells (polycythemia) to increase oxygen-carrying capacity. This helps deliver more oxygen to tissues. Choice A is incorrect because medications for bronchodilation do not directly cause polycythemia. Choice B is incorrect because overactive bone marrow is not the primary reason for polycythemia in COPD. Choice C is incorrect because polycythemia in COPD is not a response to combat anemia but rather to address tissue hypoxia.

Correct Answer: C

Rationale: Rationale: 1. Blinatumomab is designed to engage T cells to target CD19 antigen on B-lymphoblasts. 2. CD19 is a specific marker expressed on B cells, making it a target for Blinatumomab. 3. CD10 is not targeted by Blinatumomab and is commonly found on early B-cell precursors. 4. CD15 and CD20 are not targeted by Blinatumomab and have different functions unrelated to B-lymphoblasts. Summary: - CD19 is the correct answer as it is the specific target for Blinatumomab on B-lymphoblasts. - CD10, CD15, and CD20 are incorrect as they are not the targeted antigens for Blinatumomab and have different roles in immune function.

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.