Nokea
The following pathological features is true regarding non-Hodgkin lymphoma (NHL)
- A. Classical Reed-Sternberg (RS) cells are neoplastic cells
- B. B cell phenotype is more common
- C. Contiguous group of lymph nodes are affected
- D. It is not associated with leukemic phase of disease
Correct Answer: B
Rationale: Step 1: Non-Hodgkin lymphoma (NHL) can arise from either B cells or T cells. B cell NHL is more common than T cell NHL.
Step 2: The B cell phenotype is predominant in NHL, making choice B correct.
Step 3: Classical Reed-Sternberg cells are characteristic of Hodgkin lymphoma, not NHL, making choice A incorrect.
Step 4: NHL can involve non-contiguous lymph nodes, ruling out choice C.
Step 5: NHL can present with leukemic involvement, refuting choice D.
Overall, the B cell phenotype being more common in NHL makes choice B the correct answer.
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A 4-year-old male child presents to the emergency department with his fourth invasive Staph infection. CBC consistently identifies moderate neutropenia. Sophisticated lab testing identifies lack of Toll-like receptor responses. The patient undergoes whole exome sequencing and is found to have pathogenic variants in IRAK4. What does 'IRAK4' stand for?
- A. Interferon gamma receptor-associated kinase 4
- B. Inducible RAS activating kinase 4
- C. Interleukin-1 receptor-associated kinase 4
- D. Immune response activating kinase 4
Correct Answer: C
Rationale: Step 1: Identify the function of IRAK4.
IRAK4 is involved in the signaling pathway of the immune system, particularly in response to interleukin-1 (IL-1) receptor activation.
Step 2: Break down the acronym IRAK4.
IRAK4 stands for Interleukin-1 Receptor-Associated Kinase 4.
Step 3: Link the information in the question to the correct answer.
Given that the patient has a lack of Toll-like receptor responses and pathogenic variants in IRAK4, it indicates a problem with the interleukin-1 signaling pathway, making choice C (Interleukin-1 receptor-associated kinase 4) the correct answer.
Summary:
A: Incorrect - Interferon gamma receptor-associated kinase does not match the function of IRAK4.
B: Incorrect - Inducible RAS activating kinase does not match the function of IRAK4.
C: Correct - Matches the function of IRAK4
After seven days of treatment with sulfonamides, a patient's hemoglobin had decreased from 14.7 gm/100ml to 10gm/100ml. The most likely cause of hemolysis in this patient is
- A. Sickle cell disease
- B. Thalassemia minor
- C. Hereditary spherocytosis
- D. Glucose 6-phosphate dehydrogenase deficiency (G6PD)
Correct Answer: D
Rationale: The correct answer is D: Glucose 6-phosphate dehydrogenase deficiency (G6PD). Sulfonamides can trigger hemolysis in patients with G6PD deficiency due to oxidative stress on red blood cells. G6PD enzyme deficiency impairs the ability of red blood cells to combat oxidative damage, leading to hemolysis. In this case, the patient's hemoglobin decreased significantly after sulfonamide treatment, indicating red blood cell destruction. The other choices (A: Sickle cell disease, B: Thalassemia minor, C: Hereditary spherocytosis) are not directly associated with sulfonamide-induced hemolysis and would not explain the observed decrease in hemoglobin levels after treatment.
You have a new 7-year-old female patient with a WBC count of 6,000/mm3, hemoglobin of 7.2 g/dL, and platelet count of 30,000/mm3. A bone marrow aspirate reveals 14% blasts with a monocytic morphologic appearance that are surface marker positive for CD33. You receive a call from the fluorescence in situ hybridization (FISH) lab that the bone marrow is positive for KMT2A rearrangement in 68% of cells. Your staff asks whether this represents a diagnosis of acute leukemia in the current classification scheme for this type of hematologic malignancy. What would you say?
- A. No, because for a diagnosis of acute leukemia you must have 30% or more blasts in the marrow.
- B. No, because for a diagnosis of acute leukemia you must have 20% or more blasts in the marrow.
- C. No, because the cytogenetics do not include +21, monosomy 7, or trisomy 8.
- D. Yes, because the FISH is positive for KMT2A rearrangement.
Correct Answer: D
Rationale: The correct answer is D: Yes, because the FISH is positive for KMT2A rearrangement.
Rationale:
1. KMT2A rearrangement is a genetic abnormality commonly associated with acute leukemia.
2. Presence of blasts (14%) with monocytic appearance and positive for CD33 also supports the diagnosis.
3. The percentage of blasts (14%) is not below the threshold for acute leukemia diagnosis.
4. The specific cytogenetic findings mentioned in choice C are not absolute requirements for diagnosing acute leukemia.
In summary, the presence of KMT2A rearrangement, along with morphologic and flow cytometry findings, supports the diagnosis of acute leukemia in this case, making choice D the correct answer.
A 15-year-old female presents with 1 month of fatigue and 3 days of chest pain and shortness of breath. Her physical exam is unremarkable. A chest x-ray shows a large mediastinal mass that is greater than 33% of the diameter of her chest cavity. A biopsy shows nodular sclerosing, classic Hodgkin lymphoma (cHL). Metastatic workup at diagnosis, including CT scan of neck, chest, abdomen, and pelvis and PET scan, shows no other site of disease. According to the Ann Arbor staging system, the patient has which stage of cHL?
- A. Stage I
- B. Stage II
- C. Stage III
- D. Stage IV
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 or a single extralymphatic organ or site. In this case, the patient's disease is limited to the mediastinum without involvement of other lymph nodes or organs. The large mediastinal mass is considered as a single site of disease. The absence of disease involvement in other areas based on the metastatic workup supports the classification as Stage I. The other choices (B, C, D) are incorrect because they represent more extensive disease involving multiple lymph node regions or extralymphatic organs, which is not the case in this patient.
In the laboratory diagnosis of leishmaniasis, the expected parasite stage in vitro is
- A. Amastigote
- B. Promastigote
- C. Trypomastigote
- D. Schizont
Correct Answer: B
Rationale: The correct answer is B: Promastigote. In the laboratory diagnosis of leishmaniasis, the expected parasite stage in vitro is the promastigote form. This is because the promastigote form is the stage of the parasite that lives in the sandfly vector and is transmitted to humans, causing infection. In vitro, the promastigote form can be cultured and studied for diagnostic purposes.
Incorrect choices:
A: Amastigote - This is the intracellular form of the parasite found in mammalian hosts and not typically used for laboratory diagnosis.
C: Trypomastigote - This form is typically associated with Trypanosoma species, not Leishmania.
D: Schizont - This term is used for malaria parasites, not Leishmania parasites.
In summary, the promastigote form is the most relevant stage for laboratory diagnosis of leishmaniasis due to its presence in the sandfly vector and its ability to