Which of the following will be present in chronic bacterial infection?

Correct Answer: D

Rationale: The correct answer is D: Monocytosis. In chronic bacterial infections, monocytosis occurs due to the increased presence of monocytes in response to the persistent infection. Monocytes differentiate into macrophages to help eliminate the bacteria. Neutrophilia (A) is more common in acute bacterial infections. Basophilia (B) is associated with conditions like chronic myeloid leukemia, not bacterial infections. Lymphocytosis (C) is often seen in viral infections or chronic inflammatory conditions, not specifically in chronic bacterial infections.

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.

Correct Answer: D

Rationale: The correct answer is D: Myelodysplastic syndrome. The patient's symptoms of weakness, lassitude, and reduced erythropoiesis along with increased myeloblasts in bone marrow are indicative of a myelodysplastic syndrome. This condition is characterized by ineffective hematopoiesis leading to cytopenias and an increased risk of progression to acute myeloid leukemia. Acute myeloid leukemia (choice A) typically presents with more aggressive symptoms and higher blast count. Acute lymphoid leukemia (choice B) primarily affects lymphoid cells, not myeloid cells. Myelofibrosis (choice C) is characterized by fibrosis in the bone marrow, leading to extramedullary hematopoiesis, which is not seen in this case.

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.

Correct Answer: B

Rationale: The correct answer is B: Promastigote. In the laboratory diagnosis of leishmaniasis, the parasite stage typically observed in vitro is the promastigote stage. This is because promastigotes are the infective form of the parasite found in the sandfly vector. When a sample is taken from a patient, it is cultured in a suitable medium where promastigotes develop. Amastigotes (choice A) are the intracellular form found in the host, not typically observed in vitro. Trypomastigotes (choice C) are the stage found in Trypanosoma species, not Leishmania. Schizonts (choice D) are a stage in the life cycle of Plasmodium species causing malaria, not Leishmania. Promastigotes are the stage most commonly used for laboratory diagnosis and research in leishmaniasis.