Nokea
Haemolytic anemia is not produced by:
- A. penicillin
- B. lithium
- C. Quinidine
- D. methyldopa
Correct Answer: B
Rationale: The correct answer is B: lithium. Lithium does not cause hemolytic anemia because it does not affect red blood cell destruction. Penicillin (A), Quinidine (C), and Methyldopa (D) are known to cause hemolytic anemia by various mechanisms, such as immune-mediated destruction or oxidative damage to red blood cells. In contrast, lithium's mechanism of action does not involve direct harm to red blood cells, making it the correct choice in this scenario.
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A nurse is caring for four clients. After reviewing today's laboratory results, which client would the nurse assess first?
- A. Client with an international normalized ratio of 2.8
- B. Client with a platelet count of 128,000/mm3 (128 x 109/L).
- C. Client with a prothrombin time (PT) of 28 seconds
- D. Client with a red blood cell count of 5.1 million/mcl. (5.1 x 1012/L)
Correct Answer: C
Rationale: The correct answer is C because a prothrombin time (PT) of 28 seconds indicates potential issues with blood clotting and can be a sign of bleeding disorders or liver dysfunction, requiring 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's not an immediate concern. Choice B (platelet count of 128,000/mm3) is low but not critically low, so it doesn't require immediate assessment. Choice D (red blood cell count of 5.1 million/mcl) is within the normal range and doesn't indicate urgent issues.
In summary, the nurse should assess the client with a PT of 28 seconds first due to the potential risk of bleeding or clotting disorders, while the other choices are not as urgent.
A 14-year-old Syrian male with beta thalassemia major has relocated to your community as a refugee. He has been receiving chronic transfusion therapy in Turkey for the past 3 years. On his first visit, you notice that his height is below the fifth percentile. He has skin discoloration and hepatosplenomegaly. His mother reports they have not had regular access to chelation therapy. Laboratory testing shows a serum ferritin of 6,200 ng/mL. A cardiac MRI shows grossly normal cardiac function but a T2* value of 9 ms. What is the most likely cause of his short stature?
- A. Lack of regular blood transfusion causing growth failure
- B. Cirrhosis and liver failure
- C. Ineffective erythropoiesis and chronic anemia
- D. Growth hormone deficiency due to iron deposition in the pituitary
Correct Answer: D
Rationale: The correct answer is D: Growth hormone deficiency due to iron deposition in the pituitary. Iron overload in patients with beta thalassemia major can lead to iron deposition in various organs, including the pituitary gland, impairing its function. This can result in growth hormone deficiency, leading to short stature. In this case, the patient's history of chronic transfusions and high serum ferritin level indicate iron overload, which can affect the pituitary gland. Choices A, B, and C are incorrect because growth failure in beta thalassemia major is primarily attributed to endocrine complications such as growth hormone deficiency, rather than lack of transfusions, cirrhosis, or ineffective erythropoiesis.
Presence of an --jaundice --splenomegaly with increase MCH is seen in:
- A. liver cirrhosis
- B. th.major
- C. PNH
- D. herditary spherocytosis
Correct Answer: D
Rationale: The correct answer is D: hereditary spherocytosis. In this condition, there is splenomegaly due to hemolysis, leading to jaundice. The increased MCH (mean corpuscular hemoglobin) is a characteristic finding in hereditary spherocytosis. Liver cirrhosis (choice A) may cause jaundice but is not typically associated with splenomegaly and increased MCH. Thalassemia major (choice B) presents with microcytic anemia and not typically associated with increased MCH. Paroxysmal nocturnal hemoglobinuria (choice C) is known for hemolysis but does not commonly present with splenomegaly and increased MCH.
A nurse is caring for a client who is about to begin alteplase therapy to treat pulmonary embolism. Which of the following drugs should the nurse have available in the event of a severe adverse reaction?
- A. Vitamin K
- B. Aminocaproic acid
- C. Protamine
- D. Deferoxamine
Correct Answer: B
Rationale: Rationale: Aminocaproic acid is used to manage bleeding complications associated with thrombolytic therapy, like alteplase. In case of severe adverse reaction such as uncontrolled bleeding, aminocaproic acid can help by inhibiting fibrinolysis. Vitamin K (A) is not used for this purpose. Protamine (C) is used to reverse heparin anticoagulation, not for thrombolytic therapy. Deferoxamine (D) is used for iron toxicity, not related to thrombolytic therapy. Thus, having aminocaproic acid available is crucial for managing potential adverse reactions during alteplase therapy.
A study is designed to investigate the rates of central line–associated blood stream infections among pediatric hematology/oncology patients. Three common central line types (totally implanted catheter [port], peripherally inserted central catheter [PICC], and tunneled externalized catheter [TEC]) were included in the study. What data structure is central line type?
- A. Continuous
- B. Dichotomous
- C. Nominal
- D. Ordinal
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.