Nokea
Which of the following is false regarding Burkitt's lymphoma?
- A. Associated with translocation of c-myc gene on chromosome 18
- B. Usually presents with abdominal involvement in HIV patients
- C. Common in Africa children due to Epstein Barr infection after malaria exposure.
- D. Histologically characterized by a 'starry sky' appearance
Correct Answer: A
Rationale: The correct answer is A because Burkitt's lymphoma is associated with the translocation of the c-myc gene on chromosome 8, not chromosome 18. This translocation results in the overexpression of c-myc, leading to uncontrolled cell proliferation. Choice B is incorrect as Burkitt's lymphoma commonly presents with jaw or facial bone involvement in HIV patients. Choice C is incorrect as Burkitt's lymphoma is commonly seen in African children due to Epstein-Barr virus infection, not necessarily after malaria exposure. Choice D is incorrect as Burkitt's lymphoma is histologically characterized by a "starry sky" appearance due to the presence of numerous macrophages engulfing debris from apoptotic cells.
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Plummer Vinson syndrome is not associated with:
- A. angular stomatitis
- B. splenomegaly
- C. clubbing
- D. post cricoid web
Correct Answer: C
Rationale: The correct answer is C: clubbing. Plummer Vinson syndrome is characterized by the triad of iron deficiency anemia, dysphagia, and esophageal webs. Clubbing is not a typical feature of Plummer Vinson syndrome. Angular stomatitis (A), splenomegaly (B), and post cricoid web (D) are commonly associated with Plummer Vinson syndrome due to chronic iron deficiency anemia. Clubbing is more commonly seen in conditions such as chronic respiratory or cardiac diseases, not in Plummer Vinson syndrome.
The nurse examines the lymph nodes of a patient during a physical assessment. Which assessment finding would be of most concern to the nurse?
- A. A 2-cm nontender supraclavicular node
- B. A 1-cm mobile and nontender axillary node
- C. An inability to palpate any superficial lymph nodes
- D. Firm inguinal nodes in a patient with an infected foot
Correct Answer: A
Rationale: The correct answer is A: A 2-cm nontender supraclavicular node. This finding is concerning because supraclavicular nodes are not normally palpable. Enlarged supraclavicular nodes can indicate metastatic cancer. B is incorrect because mobile and nontender axillary nodes are usually benign. C is incorrect as the inability to palpate superficial nodes may be normal. D is incorrect as firm inguinal nodes could indicate a localized infection or inflammation.
During haemostasis, prostacyclin
- A. Cleaves prothrombin into thrombin
- B. Causes vasodilation
- C. Stimulates platelet aggregation
- D. Activates fibrinolysis
Correct Answer: B
Rationale: Prostacyclin is a vasodilator, which means it causes blood vessels to widen, leading to decreased blood pressure and increased blood flow. During haemostasis, this vasodilation helps in maintaining blood flow and preventing platelet aggregation in the damaged area. Option A is incorrect because prostacyclin does not cleave prothrombin into thrombin. Option C is incorrect as prostacyclin inhibits platelet aggregation. Option D is incorrect because prostacyclin does not directly activate fibrinolysis. In summary, the correct answer is B because prostacyclin's vasodilation function plays a crucial role in haemostasis by preventing excessive clot formation and maintaining blood flow.
A young child with consanguineous parents has developmental delay and a history of multiple recurrent bacterial infections and short stature. He presents to the emergency department following trauma and requires a blood transfusion. Blood work identifies leukocytosis, neutrophilia, and the Bombay blood group (absent H antigen as well as absent A and B antigens). What is this patient's diagnosis?
- A. Chediak-Higashi syndrome
- B. Leukocyte adhesion deficiency (LAD) Type II
- C. CD18 deficiency
- D. Griscelli syndrome
Correct Answer: B
Rationale: The correct answer is B: Leukocyte adhesion deficiency (LAD) Type II. This patient's symptoms of recurrent bacterial infections, leukocytosis, neutrophilia, short stature, and Bombay blood group (lack of H antigen) are characteristic of LAD Type II. In LAD Type II, there is a defect in fucose metabolism, leading to impaired leukocyte adhesion and migration, causing immune dysfunction. Chediak-Higashi syndrome (A) presents with oculocutaneous albinism, recurrent infections, and giant granules in leukocytes. CD18 deficiency (C) is a type of LAD characterized by mutations in the CD18 gene causing impaired integrin function. Griscelli syndrome (D) presents with silvery hair, immunodeficiency, and neurological deficits, not matching this patient's symptoms.
A study is designed to investigate the rates of central line–associated blood stream infections (CLABSI) among pediatric hematology/oncology patients. Investigators wish to compare the length of stay (LOS) between subjects receiving three common central line types (totally implanted catheter [port], peripherally inserted central catheter [PICC], and tunneled externalized catheter [TEC]). It is discovered that LOS is not normally distributed. What is the appropriate test for comparing the LOS between patients receiving the three central line types?
- A. Student's t test
- B. ANOVA
- C. Wilcoxon-Mann-Whitney test
- D. Kruskal-Wallis test
Correct Answer: D
Rationale: The correct answer is D: Kruskal-Wallis test. This non-parametric test is used to compare the medians of three or more independent groups when the data is not normally distributed, as in this case with LOS. It is an extension of the Wilcoxon-Mann-Whitney test, which compares two groups. The Kruskal-Wallis test assesses whether the medians of the groups are equal or different by ranking all observations, calculating the sum of ranks for each group, and then comparing these sums. It is appropriate here because it allows for comparison of central line types without assuming normality. The other choices (A: Student's t test and B: ANOVA) require the data to be normally distributed, which is not the case in this scenario. Choice C: Wilcoxon-Mann-Whitney test is for comparing two groups, not three or more.