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Question 1
Correct
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The histological exam of a tuberculous granuloma shows a periphery of multinuclear giant cells, with a central area of:
Your Answer: Caseous necrosis
Explanation:Granulomas with necrosis tend to have an infectious cause. The chronic infective lesion in this case typically presents with a central area of caseous (cheese-like) necrosis. Foam cells are the fat-laden M2 macrophages seen in atherosclerosis
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This question is part of the following fields:
- Cell Injury & Wound Healing; Dermatology
- Pathology
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Question 2
Correct
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An electronic manufacturing engineer had abdominal distension and underwent a CT scan of the abdomen. Thereafter he was diagnosed with hepatic angiosarcoma. Exposure to what agent is responsible for the development of this neoplasm?
Your Answer: Arsenic
Explanation:Hepatic angiosarcomas are associated with particular carcinogens which includes: arsenic , thorotrast, and polyvinyl chloride. With exposure to this three agents, there is a very long latent period of many years between exposure and the development of tumours.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 3
Incorrect
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A 62-year-old male patient in the intensive care unit was found to have a low serum phosphate level. What is the serum level of phosphate which is considered as normal in adults?
Your Answer: 0.3-1.3 mmol/l
Correct Answer: 0.8–1.45 mmol/l
Explanation:After calcium, phosphorus is the most plentiful mineral in the human body. It is an important and vital element which our body needs to complete many physiologic processes , such as filtering waste and repairing cells. Phosphorus helps with bone growth and approximately 85% of phosphate in the body is contained in bone. Phosphate is involved in energy storage, and nerve and muscle production. A normal range of plasma phosphate in adults teenagers generally from 0.8 mmol/l to 1.45 mmol/l. The normal range varies depending on age. Infants and children have higher phosphorus levels because more of this mineral is needed for their normal growth and bone development.
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This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
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Question 4
Correct
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Driving pressure is considered to be a strong predictor of mortality in patients with ARDS. What is the normal mean intravascular driving pressure for the respiratory circulation?
Your Answer: 10 mmHg
Explanation:Driving pressure is the difference between inflow and outflow pressure. For the pulmonary circulation, this is the difference between pulmonary arterial (pa) and left atrial pressure (pLA). Normally, mean driving pressure is about 10 mmHg, computed by subtracting pLA (5 mmHg) from pA (15 mmHg). This is in contrast to a mean driving pressure of nearly 100 mmHg in the systemic circulation.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 5
Incorrect
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A 40-year old Caucasian male came to the hospital with complaints of fatigue and lethargy. On examination, he was found to have raised blood pressure. Urine examination showed >300 mg/dl proteinuria (4+) and 24-hour urine protein 3.5g. No glucose, blood, nitrites, urobilinogen or casts were present in urine. What is the most likely diagnosis?
Your Answer: Post-streptococcal glomerulonephritis
Correct Answer: Membranous glomerulonephritis
Explanation:Membranous glomerulonephritis or nephropathy, is a renal disorder with insidious course and usually affects people aged 30-50 years. 85% cases are primary (or idiopathic). The other 15% are secondary to autoimmune conditions like SLE, infections like malaria or hepatitis B, drugs like captopril and NSAIDs, or malignancies (particularly lung or colonic carcinoma). This disease is caused due to circulating immune complexes which are said to form by binding of antibodies to antigens in glomerular basement membrane. This antigens could be endogenous or derived from systemic circulation. This immune complex triggers the complement system, resulting in formation of membrane attack complex (MAC) on glomerular epithelial cells. This further results in release of proteases and oxidants which damage the capillaries making them ‘leaky’. Moreover, the epithelial cells also secrete a mediator to reduce nephron synthesis and distribution.
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This question is part of the following fields:
- Pathology
- Renal
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Question 6
Incorrect
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Acute respiratory distress syndrome (ARDS) is a medical condition that occurs in critically ill patients, and can be triggered by events such as trauma and sepsis. Which of the following variables is most likely to be lower than normal in a patient with ARDS?
Your Answer:
Correct Answer: Lung compliance
Explanation:Acute (or Adult) respiratory distress syndrome (ARDS) is a medical condition occurring in critically ill patients characterized by widespread inflammation in the lungs. The development of acute respiratory distress syndrome (ARDS) starts with damage to the alveolar epithelium and vascular endothelium, resulting in increased permeability to plasma and inflammatory cells. These cells pass into the interstitium and alveolar space, resulting in pulmonary oedema. Damage to the surfactant-producing type II cells and the presence of protein-rich fluid in the alveolar space disrupt the production and function of pulmonary surfactant, leading to micro atelectasis and impaired gas exchange. The pathophysiological consequences of lung oedema in ARDS include a decrease in lung volumes, compliance and large intrapulmonary shunts. ARDS may be seen in the setting of pneumonia, sepsis, following trauma, multiple blood transfusions, severe burns, severe pancreatitis, near-drowning, drug reactions, or inhalation injuries.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 7
Incorrect
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Out of the following options, which malignancy has the highest potential for multicentricity?
Your Answer:
Correct Answer: Transitional cell carcinoma
Explanation:Transitional cell carcinomas can arise anywhere in the urothelium lining the urinary tract; and hence are known to be multicentric and recur commonly. Prostatic adenocarcinoma most commonly involves the posterior lobe of the prostate gland. Although renal cell carcinomas occasionally show multicentricity, it is not common. Penile carcinomas are usually locally infiltrative lesions. Wilm’s tumours are usually solitary, but can be bilateral or multicentric in 10% cases. Small cell carcinoma of lung and teratomas are usually solitary.
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This question is part of the following fields:
- Pathology
- Renal
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Question 8
Incorrect
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The passage of leukocytes through the wall of the blood vessels is best described by which of the following terms?
Your Answer:
Correct Answer: Diapedesis
Explanation:The steps involved in leukocyte arrival and function are:
1. margination: cells migrate from the centre to the periphery of the vessel
2. rolling: selectins are upregulated on the vessel walls
3. adhesion: upregulation of the adhesion molecules ICAM and VCAM on the endothelium interact with integrins on the leukocytes, interaction of these results in adhesion
4. diapedesis and chemotaxis: diapedesis is the transmigration of the leukocyte across the endothelium of the capillary and towards a chemotactic product
5. phagocytosis: engulfing the offending substance/cell.
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This question is part of the following fields:
- Cell Injury & Wound Healing
- Pathology
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Question 9
Incorrect
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A 47-year-old male smoker, who had been self-medicating with oral steroids for the last two years due to persistent breathlessness presented to the doctor complaining of a productive cough, fever and chest pain. A chest X-ray revealed bilateral patchy opacities. He was diagnosed with bilateral bronchopneumonia. Which of these organisms is most probably causing these findings?
Your Answer:
Correct Answer: Nocardia asteroides
Explanation:Nocardia is a Gram-positive aerobic actinomycete. Several species have been identified but the most common human pathogen is Nocardia asteroides. The predominant clinical finding in the majority of patients affected by nocardiosis is pulmonary disease. Predisposing factors for pulmonary nocardiosis include leukaemia, human immunodeficiency virus (HIV) infection, organ transplantation, diabetes and receiving prolonged corticosteroids.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 10
Incorrect
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What is the normal duration of PR interval on an electrocardiogram of a healthy individual?
Your Answer:
Correct Answer: 0.12–0.20 s
Explanation:PR interval extends from the beginning of the P-wave until the beginning of the QRS complex. The normal duration of the PR interval is 0.12-0.20 s. It can be prolonged in first degree heart block, and reduced in Wolff-Parkinson-White syndrome.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 11
Incorrect
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Lung compliance is increased by:
Your Answer:
Correct Answer: Emphysema
Explanation:Lung compliance is increased by emphysema, acute asthma and increasing age and decreased by alveolar oedema, pulmonary hypertension, atelectasis and pulmonary fibrosis.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 12
Incorrect
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Hormones of the anterior pituitary include which of the following?
Your Answer:
Correct Answer: Prolactin
Explanation:The anterior pituitary gland (adenohypophysis or pars distalis) synthesizes and secretes:
1. FSH (follicle-stimulating hormone)
2. LH (luteinizing hormone)
3. Growth hormone
4. Prolactin
5. ACTH (adrenocorticotropic hormone)
6. TSH (thyroid-stimulating hormone).
The posterior pituitary gland (neurohypophysis) stores and secretes 2 hormones produced by the hypothalamus:
1. ADH (antidiuretic hormone or vasopressin)
2. Oxytocin
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This question is part of the following fields:
- Endocrine
- Physiology
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Question 13
Incorrect
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A 40-year old gentleman underwent a computed tomographic scan for the abdomen to rule out blunt abdominal trauma, after a vehicular accident. The scan revealed no abnormal finding except for a 1 cm-sized cortical mass in the left adrenal gland. The doctor-on-call decided to not intervene for this mass because it was likely to be a:
Your Answer:
Correct Answer: Non-functioning adrenal adenoma
Explanation:Adrenal adenomas are common, benign lesions which are asymptomatic and seen in 10% of population. Usually detected incidentally on Computed tomography (‘incidentaloma’), only around 1 in 10,000 are malignant (adenocarcinoma). Adrenal adenomas rarely need to be investigated, especially if they are homogenous and less than 3 cm in diameter. Follow-up imaging can be done after an interval of 3-6 months to assess any change in size. Some adenomas can secrete cortisol (leading to Cushing syndrome), or aldosterone (leads to Conn syndrome) or androgens (leading to hyperandrogenism).
Haematomas and simple cysts are not usually seen in adrenal gland. Infection due to Histoplasma capsulatum is usually bilateral and leads to multiple granulomas. Adrenal metastasis will usually demonstrate a lung primary and the adrenal lesions will be often multiple and larger than 1 cm.
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This question is part of the following fields:
- Endocrine
- Pathology
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Question 14
Incorrect
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The blood-brain barrier is a membrane that separates the circulating blood from the brain extracellular fluid in the central nervous system (CNS). Which of the following statements regarding the blood– brain barrier is CORRECT?
Your Answer:
Correct Answer: It breaks down in areas of brain that are infected
Explanation:The blood–brain barrier is a membrane that controls the passage of substances from the blood into the central nervous system. It is a physical barrier between the local blood vessels and most parts of the central nervous system and stops many substances from travelling across it. During meningitis, the blood–brain barrier may be disrupted. This disruption may increase the penetration of various substances (including either toxins or antibiotics) into the brain. A few regions in the brain, including the circumventricular organs, do not have a blood–brain barrier.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 15
Incorrect
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A 40-year old gentleman, known with a history of peptic ulcer disease, was brought to the clinic in a dehydrated state with persistent vomiting. His blood investigations revealed:
- sodium = 142 mmol/l
- potassium = 2.6 mmol/l
- chloride = 85 mmol/l
- pH = 7.55
- p(CO2) = 50 mmHg
- p(O2) = 107 mmHg
- standard bicarbonate = 40 mmol/l
Your Answer:
Correct Answer: Metabolic alkalosis
Explanation:High pH with high standard bicarbonate indicates metabolic alkalosis. The pa(CO2) was appropriately low in compensation. This is hypokalaemic hypochloraemic metabolic acidosis due to prolonged vomiting. Treatment includes treating the cause and intravenous sodium chloride with potassium.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 16
Incorrect
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A 57-year-old male smoker noted a lump on his inner lip. Upon physical examination the lump measured more than 2 cm but less than 4 cm in its greatest dimension. He is diagnosed with squamous cell carcinoma of the lip. What is the stage of the patient's cancer according to the TNM staging for head and neck cancers?
Your Answer:
Correct Answer: T2
Explanation:Head and neck cancer is a group of cancers that starts within the mouth, nose, throat, larynx, sinuses, or salivary glands. The TNM staging system used for head and neck cancers is a clinical staging system that allows physicians to compare results across patients, assess prognosis, and design appropriate treatment regimens. The staging is as follows; Primary tumour (T): Tis: pre-invasive cancer (carcinoma in situ), T0: no evidence of primary tumour, T1: tumour 2 cm or less in its greatest dimension, T2: tumour more than 2 cm but not more than 4 cm, T3: tumour larger than 4 cm, T4: tumour with extension to bone, muscle, skin, antrum, neck, etc and TX: minimum requirements to assess primary tumour cannot be met. Regional lymph node involvement (N): N0: no evidence of regional lymph node involvement, N1: evidence of involvement of movable homolateral regional lymph nodes, N2: evidence of involvement of movable contralateral or bilateral regional lymph nodes, N3: evidence of involvement of fixed regional lymph nodes and NX: Minimum requirements to assess the regional nodes cannot be met. Distant metastases (M): M0: no evidence of distant metastases, M1: evidence of distant metastases and MX: minimum requirements to assess the presence of distant metastases cannot be met. Staging: Stage I: T1 N0 M0, Stage II: T2 N0 M0, Stage III: T2NOMO and T3N1MO, Stage IV: T4N1M0, any TN2M0, any TN3M0, any T and any NM1. The depth of infiltration is predictive of the prognosis. With increasing depth of invasion of the primary tumour, the risk of nodal metastasis increases and survival decreases. The patient in this scenario therefore has a T2 tumour.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 17
Incorrect
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A 40 year old woman, who is under anaesthesia for an elective procedure, received an antibiotic injection. She immediately developed a rash and her airway constricted raising the airway pressure. Which mechanism is responsible for this reaction?
Your Answer:
Correct Answer: Type I hypersensitivity
Explanation:Type I hypersensitivity reaction is an immediate reaction that occurs due to binding of the antigen with antibodies attached to mast cells in a previously sensitized person. It has an immediate phase, which is characterised by vasodilation, leakage of plasma, smooth muscle spasm, or glandular secretions. This manifests in about 5-30 min and usually resolves within 60 mins. The delayed phase follows after 24 hours and can persist up to several days. It is due to infiltration of eosinophils, neutrophils, basophils and CD4+ cells and leads to tissue destruction. The nature of the reaction varies according to the site. It can take the form of skin allergy, hives, allergic rhinitis, conjunctivitis, bronchial asthma or food allergy.
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This question is part of the following fields:
- Inflammation & Immunology
- Pathology
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Question 18
Incorrect
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A 29-year-old pregnant woman develops severe hypoxaemia with petechial rash and confusion following a fracture to her left femur. Which is the most probable cause of these symptoms in this patient?
Your Answer:
Correct Answer: Fat embolism
Explanation:Fat embolism is a life-threatening form of embolism in which the embolus consists of fatty material or bone marrow particles that are introduced into the systemic venous system. It may be caused by long-bone fractures, orthopaedic procedures, sickle cell crisis, parenteral lipid infusion, burns and acute pancreatitis. Patients with fat embolism usually present with symptoms that include skin, brain, and lung dysfunction.
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This question is part of the following fields:
- Cardiovascular
- Pathology
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Question 19
Incorrect
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A 53 year old women with a history of atrial fibrillation developed an acute abdomen. On laparoscopic examination her bowels appeared to be dusky to red-purple in colour and her mesenteric veins appeared to be patent. Which of the following is most likely to occur in this situation?
Your Answer:
Correct Answer: Wet gangrene
Explanation:Infarction of the small bowel following a sudden and complete occlusion of the mesenteric artery can involve any portion of the bowel, whether small or a large. The splenic flexure is at most risk for infarction as it is the watershed area between the superior and inferior mesenteric vessels. Regardless of whether the arterial or the venous blood vessels are occluded, because of the blood reflow into the damaged portion, it will appear haemorrhagic. The bowel appearing congested at first and then becoming oedematous. If the artery is occluded then there will be a clear cut demarcation and in venous occlusion the dusky colour fades with the rest of the normal bowel. Wet gangrene is characteristic of ischaemic injury to the gut.
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This question is part of the following fields:
- Cell Injury & Wound Healing; Gastrointestinal
- Pathology
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Question 20
Incorrect
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A 30 year old lawyer presented with a history of pleural effusions, pericarditis, arthralgia without joint deformity and episodes of myalgia for the past 12 months. Blood tests reveal a normocytic anaemia. Which of the following tests should be performed to investigate this condition further?
Your Answer:
Correct Answer: Antinuclear antibody test
Explanation:These non specific findings are suggestive of autoimmune disease. Antinuclear antibody (ANA) is the first test that should be performed when autoimmune diseases are suspected and later on more specific tests should be performed. This tests aids in the diagnosis of SLE, scleroderma, Sjogren’s syndrome, Raynaud’s disease, juvenile chronic arthritis, rheumatoid arthritis and antiphospholipid antibody syndrome etc. To confirm the disease a history, physical examination along with specific tests are required.
CPK is more specific for acute conditions.
ESR is a non specific test for inflammation.
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This question is part of the following fields:
- Inflammation & Immunology
- Pathology
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Question 21
Incorrect
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Which of the following is the most likely cause of prolonged thrombin clotting time?
Your Answer:
Correct Answer: Hypofibrinogenemia
Explanation:Thrombin clotting time, also called thrombin time (TT), is test used for the investigation of possible bleeding or clotting disorders. TT reflects the conversion of fibrinogen to fibrin and it’s also very sensitive to the presence of the anticoagulant heparin. A prolonged thrombin time may indicate the presence of hypofibrinogenemia (decreased fibrinogen level ), dysfibrinogenaemia, disseminated intravascular coagulation (DIC), end stage liver disease or malnutrition.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 22
Incorrect
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A 59-year old gentleman admitted for elective cholecystectomy was found to have a haemoglobin 12.5 g/dl, haematocrit 37%, mean corpuscular volume 90 fl, platelet count 185 × 109/l, and white blood cell count 32 × 109/l; along with multiple, small mature lymphocytes on peripheral smear. The likely diagnosis is:
Your Answer:
Correct Answer: Chronic lymphocytic leukaemia
Explanation:CLL or chronic lymphocytic leukaemia is the most common leukaemia seen in the Western world. Twice more common in men than women, the incidence of CLL increases with age. About 75% cases are seen in patients aged more than 60 years. The blood, marrow, spleen and lymph nodes all undergo infiltration, eventually leading to haematopoiesis (anaemia, neutropenia, thrombocytopenia), hepatomegaly, splenomegaly and decreased production of immunoglobulin. In 98% cases, CD+5 B cells undergo malignant transformation.
Often diagnosed on blood tests while being evaluated for lymphadenopathy, CLL causes symptoms like fatigue, anorexia, weight loss, pallor, dyspnoea on exertion, abdominal fullness or distension. Findings include multiple lymphadenopathy with minimal-to- moderate hepatomegaly and splenomegaly. Increased susceptibility to infections is seen. Herpes Zoster is common. Diffuse or maculopapular skin infiltration can also be seen in T-cell CLL.
Diagnosis is by examination of peripheral blood smear and marrow: hallmark being a sustained, absolute leucocytosis (>5 ×109/l) and increased lymphocytes in the marrow (>30%). Other findings can include hypogammaglobulinemia (<15% of cases) and, rarely, raised lactate dehydrogenase (LDH). Only 10% cases demonstrate moderate anaemia and/or thrombocytopenia.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 23
Incorrect
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Which of the following mediators of inflammation requires arachidonic acid for synthesis?
Your Answer:
Correct Answer: Prostaglandins
Explanation:Arachidonic acid is normally present in the phospholipids that make up the cell membrane and is cleaved by phospholipase A2 from the phospholipid. Arachidonic acid is a precursor for the production of eicosanoids which include: 1) prostaglandins, prostacyclins and thromboxane, 2) leukotrienes and 3) anandamides. The production of these products along with their action on the body is called the arachidonic acid cascade.
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This question is part of the following fields:
- Inflammation & Immunology
- Pathology
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Question 24
Incorrect
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A 79-year-old has been bedridden for 2 months after suffering from a stroke. She suddenly developed shortness of breath and chest pain, and was diagnosed with a pulmonary embolism. Which of the following is most likely to increase in this case?
Your Answer:
Correct Answer: Ventilation/perfusion ratio
Explanation:Pulmonary embolism (PE) is a blockage of an artery in the lungs by an embolus that has travelled from elsewhere in the body through the bloodstream. The change in cardiopulmonary function is proportional to the extent of the obstruction, which varies with the size and number of emboli obstructing the pulmonary arteries. The resulting physiological changes may include pulmonary hypertension with right ventricular failure and shock, dyspnoea with tachypnoea and hyperventilation, arterial hypoxaemia and pulmonary infarction. Consequent alveolar hyperventilation is manifested by a lowered pa(CO2). After occlusion of the pulmonary artery, areas of the lung are ventilated but not perfused, resulting in wasted ventilation with an increased ventilation/perfusion ratio – the physiological hallmark of PE – contributing to a further hyperventilatory state. The risk of blood clots is increased by cancer, prolonged bed rest, smoking, stroke, certain genetic conditions, oestrogen-based medication, pregnancy, obesity, and post surgery.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 25
Incorrect
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In a cardiac cycle, what event does the opening of the atrioventricular (AV) valves coincide with?
Your Answer:
Correct Answer: Beginning of diastole
Explanation:Cardiac diastole refers to the time period when the heart is relaxed after contraction and is preparing to refill with blood. Both ventricular and atrial diastole are together known as complete cardiac diastole. At its beginning, the ventricles relax, causing a drop in the ventricular pressure. As soon as the left ventricular pressure drops below that in left atrium, the mitral valve opens and there is ventricular filling of blood. Similarly, the tricuspid valve opens filling the right atrium.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 26
Incorrect
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A 20-year old boy presented with low grade fever, night sweats and weakness over two months. On examination, he had multiple, non-tender, cervical, supraclavicular and axillary adenopathy. Microscopy of lymph node biopsy showed the presence of Reed-Sternberg cells. He is likely suffering from:
Your Answer:
Correct Answer: Hodgkin’s lymphoma
Explanation:Hodgkin’s lymphoma is a disease characterized by malignant proliferation of cells of the lymphoreticular system. It can be localized or disseminated, and can involve the nodes, spleen, liver and marrow. Symptoms of the disease include non-tender lymphadenopathy, fever, night sweats, weight loss, itching and hepatosplenomegaly. Histologically, the involved nodes show the presence of Reed-Sternberg cells, which are large, binucleated cells, in a heterogenous cellular infiltrate of histiocytes, lymphocytes, monocytes, plasma cells and eosinophils.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 27
Incorrect
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Ventricular filling follows a delay caused by?
Your Answer:
Correct Answer: AV node
Explanation:The AV node is a conducting tissue found between the atria and the ventricles of the heart. It conducts electrical signal from the atria to the ventricles and acts a delaying mechanism preventing the atria and the ventricles from contracting at the same time. This decremental conduction prevents premature ventricular contraction in cases such as atrial fibrillation. A delay in the AV node is the reason for the PR segment seen on the ECG. In certain types of supraventricular tachycardia, a person could have two AV nodes; this will cause a loop in electrical current and uncontrollably rapid heart beat. When this electricity catches up with itself, it will dissipate and return to a normal heart rate.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 28
Incorrect
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Which statement is correct regarding secretions from the adrenal glands?
Your Answer:
Correct Answer: Aldosterone is producd by the zona glomerulosa
Explanation:The secretions of the adrenal glands by zone are:
Zona glomerulosa – aldosterone
Zona fasciculata – cortisol and testosterone
Zona reticularis – oestradiol and progesterone
Adrenal medulia – adrenaline, noradrenaline and dopamine.
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This question is part of the following fields:
- Physiology
- Renal
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Question 29
Incorrect
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A 15-day old baby was brought to the emergency department with constipation for 4 days. On examination, the abdomen of the baby was found to be distended and tender all over. No bowel sounds were heard. A sigmoid colon biopsy was carried out, which showed absent ganglion cells. What is the diagnosis?
Your Answer:
Correct Answer: Hirschsprung’s disease
Explanation:Hirschsprung’s disease is characterized by congenital absence of the autonomic plexus (Meissner’s and Auerbach’s plexus) in the intestinal wall. Usually limited to the distal colon, it can occasionally involve the entire colon or even the small bowel. There is abnormal or absent peristalsis in the affected segment, resulting in continuous spasm of smooth muscle and partial/complete obstruction. This causes accumulation of intestinal contents and dilatation of proximal segment. Skip lesions are highly uncommon. This disease is seen early in life with 15% patients presenting in first month, 60% by 1 year of age and 85% by the age of 4 years. Symptoms include severe and complete constipation, abdominal distension and vomiting. Patients with involvement of ultra-short segments might have mild constipation with intervening diarrhoea. In older children, symptoms include failure to thrive, anorexia, and lack of an urge to defecate. On examination, an empty rectum is revealed with stool palpable high up in the colon. If not diagnosed in time, it can lead to Hirschsprung’s enterocolitis (toxic megacolon), which can be fulminant and lead to death. Diagnosis involves a barium enema or a rectal suction biopsy. Barium enema shows a transition in diameter between the dilated, normal colon proximal to the narrowed, affected distal segment. It is to be noted that barium enema should be done without prior preparation, which can dilate the abnormal segment, leading to a false-negative result. A 24-hour post-evacuation film can be obtained in the neonatal period – if the colon is still filled with barium, there is a high likelihood of Hirschsprung’s disease. Full-thickness rectal biopsy is diagnostic by showing the absence of ganglion cells. Acetylcholinesterase staining can be done to highlight the enlarged nerve trunks. Abnormal innervation can also be demonstrated by rectal manometry.
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This question is part of the following fields:
- Gastrointestinal; Hepatobiliary
- Pathology
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Question 30
Incorrect
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A 24-year old, lactating mother presents to the clinic with a tender, 1.5cm mass just below the right nipple, which shows multiple fissures. What finding is likely associated with her condition?
Your Answer:
Correct Answer: Staphylococcus aureus infection
Explanation:Breast abscess occur commonly in lactating mothers in the postpartum period due to cracking of the nipple. It is commonly caused due to Staphylococcus aureus infection. Fat necrosis usually results from trauma wherein an ill-defined mass is formed. Ductal carcinomas are malignant masses which are not tender usually, and rare in the young age group. Plasma cell mastitis affect women in an older age group. Sclerosing adenosis is a type of fibrocystic disease which can lead to a tender, cystic mass but no fissuring or cracks are seen in the nipple. Fibroadenoma and lipomas are non-tender, well-defined masses.
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This question is part of the following fields:
- Pathology
- Women's Health
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Question 31
Incorrect
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Calculate the stroke volume in a patient admitted for coronary bypass surgery, with the following parameters pre-operatively:
Oxygen consumption = 300 ml/min
Arterial oxygen content = 20 ml/100 ml blood
Pulmonary arterial oxygen content = 15 ml/100 ml blood and Heart rate = 100 beats/min.Your Answer:
Correct Answer: 60 ml
Explanation:By Fick’s principle, cardiac output can be calculated as follows: VO2 = CO × (CAO2– CVO2) where VO2= oxygen consumption, CO = cardiac output, CAO2 = arterial oxygen content and CvO2 = mixed venous oxygen content. Thus, in the given problem, 300 ml/min = CO × (20 – 15) ml/100 ml CO = 300 × 100/5 ml/min CO = 6000 ml/min. Also, cardiac output = stroke volume × heart rate. Thus, 6000 ml/min = stroke volume × 100 beats/min. Hence, stroke volume = 6000/100 ml/min which is 60 ml/min.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 32
Incorrect
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A medical student is asked to calculate the net pressure difference in a capillary wall, considering: Interstitial fluid hydrostatic pressure = –3 mmHg, Plasma colloid osmotic pressure = 28 mmHg, Capillary hydrostatic pressure = 17 mmHg, Interstitial fluid colloid osmotic pressure = 8 mmHg, and Filtration coefficient = 1. Which is the correct answer?
Your Answer:
Correct Answer: 0 mmHg
Explanation:The rate of filtration at any point along a capillary depends on a balance of forces sometimes called Starling’s forces after the physiologist who first described their operation in detail. The Starling principle of fluid exchange is key to understanding how plasma fluid (solvent) within the bloodstream (intravascular fluid) moves to the space outside the bloodstream (extravascular space). Fluid movement = k[(pc– pi)–(Πc– Πi)] where k = capillary filtration coefficient, pc = capillary hydrostatic pressure, pi= interstitial hydrostatic pressure, Πc = capillary colloid osmotic pressure, Πi = interstitial colloid osmotic pressure. Therefore: 1 × [capillary hydrostatic pressure (17) – interstitial fluid hydrostatic pressure (–3)] – [plasma colloid osmotic pressure (28) – interstitial fluid colloid osmotic pressure (8)] = 0 mmHg
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 33
Incorrect
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Where do the cells belonging to the mononuclear phagocyte system originate?
Your Answer:
Correct Answer: Bone marrow
Explanation:The macrophage originates from a committed bone marrow stem cell. It is called the pluripotent hematopoietic stem cell. This differentiates into a monoblast and then into a promonocyte and finally matures into a monocyte. When called upon they leave the bone marrow and enter into the circulation. Upon entering the tissue they transform into macrophages. Tissue macrophages include: Kupffer cells (liver), alveolar macrophages (lung), osteoclasts (bone), Langerhans cells (skin), microglial cells (central nervous system), and possibly the dendritic immunocytes of the dermis, spleen and lymph nodes.
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This question is part of the following fields:
- Inflammation & Immunology
- Pathology
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Question 34
Incorrect
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A 45-year old male, who was a chronic smoker presented to the clinic with backache and dry, incessant cough. On examination, he was found to have raised blood pressure, purplish striae on his abdomen, truncal obesity and tenderness over the lower thoracic spine. These findings are suggestive of which condition?
Your Answer:
Correct Answer: Small-cell anaplastic (oat cell) carcinoma
Explanation:The symptoms suggest Cushing syndrome due to increased glucocorticoid levels. One cause of Cushing syndrome is ectopic production of adrenocorticotrophic hormone from oat cell carcinoma. As oat cell carcinoma is known to be highly metastatic, the tenderness in lower back could represent metastatic involvement.
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This question is part of the following fields:
- Endocrine
- Pathology
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Question 35
Incorrect
-
A 30 year old female presented in the emergency with an irregular pulse. Her ECG showed absent P-waves with irregular RR interval. What is the most likely diagnosis?
Your Answer:
Correct Answer: Atrial fibrillation
Explanation:Atrial fibrillation is one of the most common cardiac arrhythmias. It is often asymptomatic but may present with symptoms of palpitations, fainting, chest pain and heart failure. Characteristic findings are: absence of P-waves, unorganised electrical activity in their place, irregularity of RR interval due to irregular conduction of impulses to the ventricles and if paroxysmal AF is suspected, episodes may be documented with the use of Holter monitoring
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 36
Incorrect
-
Diuretics that act on the ascending limb of the loop of Henle produce:
Your Answer:
Correct Answer: Reduced active transport of sodium
Explanation:The loop of Henlé leads from the proximal convoluted tubule to the distal convoluted tubule. Its primary function uses a counter current multiplier mechanism in the medulla to reabsorb water and ions from the urine. It can be divided into four parts:
1. Descending limb of loop of Henlé – low permeability to ions and urea, while being highly permeable to water
2. Thin ascending limb of loop of Henlé – not permeable to water, but it is permeable to ions
3. Medullary thick ascending limb of loop of Henlé – sodium (Na+), potassium (K+) and chloride (Cl–) ions are reabsorbed by active transport. K+ is passively transported along its concentration gradient through a K+ channel in the basolateral aspect of the cells, back into the lumen of the ascending limb.
4. The cortical thick ascending limb – the site of action where loop diuretics such as furosemide block the K+/Na+/2Cl− co-transporters = reduced active transport.
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This question is part of the following fields:
- Physiology
- Renal
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Question 37
Incorrect
-
A 36-year old gentleman with a history of cough for 4 weeks came to the hospital. Examination revealed multiple lymphadenopathy with splenomegaly. Investigations revealed haemoglobin 11 g/dl, haematocrit 32.4%, mean corpuscular volume (MCV) 93 fl, white blood cell count 63 × 109/l, and platelet count 39 × 109/l; along with characteristic Auer rods on peripheral blood smear. What is the likely diagnosis?
Your Answer:
Correct Answer: Acute myelogenous leukaemia (AML)
Explanation:AML, or acute myeloid leukaemia is the commonest acute leukaemia affecting adults. increasing in incidence with age. It is a malignancy of the myeloid line of white blood cells. It results in rapid proliferation of abnormal cells, which accumulate in the marrow. Interference with normal cell production leads to a drop in red blood cells, white blood cells and platelets. This causes symptoms such as tiredness, shortness of breath, tendency to bleed or bruise easily and recurrent infections. AML is known to progress quickly and can lead to death in weeks and months if not treated. Leukemic blasts of AML show presence of Auer rods. These are clumps of azurophilic granular material that form needles in the cytoplasm. Composed of fused lysosomes, these contain peroxidase, lysosomal enzymes and crystalline inclusions. Auer rods are classically present in myeloid blasts of M1, M2, M3 and M4 acute leukaemia. They also help to distinguish the preleukemia myelodysplastic syndromes.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 38
Incorrect
-
After a prolonged coronary artery bypass surgery, a 60-year old gentleman was transfused 3 units of fresh-frozen plasma and 2 units of packed red cells. Two days later, the nurse noticed that he was tachypnoeic and chest X-ray showed signs consistent with adult respiratory distress syndrome. Which of the following variables will be low in this patient?
Your Answer:
Correct Answer: Compliance of the lung
Explanation:Acute or adult respiratory distress syndrome (ARDS) is a reaction to several forms of lung injuries and is commonly associated with sepsis and SIRS (systemic inflammatory response syndrome), severe traumatic injury, severe head injury, narcotics overdose, drowning, pulmonary contusion, and multiple blood transfusions. There is an increase in risk due to pre-existing liver disease or coagulation abnormalities. It results due to indirect toxic effects of neutrophil-derived inflammatory mediators in the lungs. ARDS is defined by the 1994 American–European Consensus Committee as the acute onset of bilateral infiltrates on chest X-ray, a partial pressure of arterial oxygen (pa(O2)) to fraction of inspired oxygen Fi(O2) ratio of less than 200 mmHg and a pulmonary artery occlusion pressure of less than 18 or the absence of clinical evidence of left arterial hypertension. ARDS is basically pulmonary oedema in the absence of volume overload or poor left ventricular function. This is different from acute lung injury, which shows a pa(O2)/Fi(O2) ratio of less than 300 mmHg. Pathogenesis of ARDS starts from damage to alveolar epithelium and vascular endothelium, causing increased permeability. Damage to surfactant-producing type II cells disrupts the production and function of pulmonary surfactant, causing micro atelectasis and poor gas exchange. There is a decrease in lung compliance and increase in work of breathing. Eventually, there is resorption of alveolar oedema, regeneration of epithelial cells, proliferation and differentiation of type II alveolar cells and alveolar remodelling. Some show resolution and some progress to fibrosing alveolitis, which involves the deposition of collagen in alveolar, vascular and interstitial spaces.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 39
Incorrect
-
Evaluation of a 60-year old gentleman, who has been a coal miner all his life and is suspected to have pulmonary fibrosis reveals the following: FEV1 of 75% (normal > 65%), arterial oxygen saturation 92%, alveolar ventilation 6000 ml/min at a tidal volume of 600 ml and a breathing rate of 12 breaths/min. There are also pathological changes in lung compliance and residual volume. Calculate his anatomical dead space.
Your Answer:
Correct Answer: 100 ml
Explanation:Dead space refers to inhaled air that does not take part in gas exchange. Because of this dead space, taking deep breaths slowly is more effective for gas exchange than taking quick, shallow breaths where a large proportion is dead space. Use of a snorkel by a diver increases the dead space marginally. Anatomical dead space refers to the gas in conducting areas such as mouth and trachea, and is roughly 150 ml (2.2 ml/kg body weight). This corresponds to a third of the tidal volume (400-500 ml). It can be measured by Fowler’s method, a nitrogen wash-out technique. It is posture-dependent and increases with increase in tidal volume. Physiological dead space is equal to the anatomical dead space plus the alveolar dead space, where alveolar dead space is the area in the alveoli where no effective exchange takes place due to poor blood flow in capillaries. This physiological dead space is very small normally (< 5 ml) but can increase in lung diseases. Physiological dead space can be measured by Bohr’s method. Total ventilation per minute (minute ventilation) is given by the product of tidal volume and the breathing rate. Here, the total ventilation is 600 ml times 12 breaths/min = 7200 ml/min. The problem mentions alveolar ventilation to be 6000 ml/min. Thus, the difference between the alveolar ventilation and total ventilation is 7200 – 6000 ml/min = 1200 ml/min, or 100 ml per breath at 12 breaths per min. This 100 ml is the dead space volume.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 40
Incorrect
-
After a car accident, a 30-year-old woman is alert and only has minor, superficial injuries. 2 hours later, she becomes unconscious and a CT scan reveals a convex, lens-shaped haemorrhage over the right parietal region. The most likely diagnosis is:
Your Answer:
Correct Answer: Epidural haematoma
Explanation:Epidural haematomas are usually caused by arterial bleeding, classically due to damage to the middle meningeal artery by a temporal bone fracture. Symptoms develop within minutes to several hours after the injury and consist of increasing headache, decreased level of consciousness, hemiparesis and pupillary dilation with loss of light reactivity. Around 15–20% of epidural hematomas are fatal.
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This question is part of the following fields:
- Neurology
- Pathology
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Question 41
Incorrect
-
During a normal respiratory exhalation, what is the recoil alveolar pressure?
Your Answer:
Correct Answer: +10 cmH2O
Explanation:To determine compliance of the respiratory system, changes in transmural pressures (in and out) immediately across the lung or chest cage (or both) are measured simultaneously with changes in lung or thoracic cavity volume. Changes in lung or thoracic cage volume are determined using a spirometer with transmural pressures measured by pressure transducers. For the lung alone, transmural pressure is calculated as the difference between alveolar (pA; inside) and intrapleural (ppl; outside) pressure. To calculate chest cage compliance, transmural pressure is ppl (inside) minus atmospheric pressure (pB; outside). For the combined lung–chest cage, transmural pressure or transpulmonary pressure is computed as pA – pB. pA pressure is determined by having the subject deeply inhale a measured volume of air from a spirometer. Under physiological conditions the transpulmonary or recoil pressure is always positive; intrapleural pressure is always negative and relatively large, while alveolar pressure moves from slightly negative to slightly positive as a person breathes.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 42
Incorrect
-
A 25-year-old woman complains of generalised swelling and particularly puffiness around the eyes which is worst in the morning. Laboratory studies showed:
Blood urea nitrogen (BUN) = 30 mg/dl
Creatinine = 2. 8 mg/dl
Albumin = 2. 0 mg/dl
Alanine transaminase (ALT) = 25 U/l
Bilirubin = 1 mg/dl
Urine analysis shows 3+ albumin and no cells.
Which of the following is the most likely diagnosis?Your Answer:
Correct Answer: Nephrotic syndrome
Explanation:Nephrotic syndrome is a disorder in which the glomeruli have been damaged, characterized by:
– Proteinuria (>3.5 g per 1.73 m2 body surface area per day, or > 40 mg per square meter body surface area per hour in children)
– Hypoalbuminemia (< 2,5 g/dl) – Hyperlipidaemia, and oedema (generalized anasarca).
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This question is part of the following fields:
- Physiology
- Renal
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Question 43
Incorrect
-
In a cardiac cycle, what event does the closing of atrioventricular (AV) valves coincide with?
Your Answer:
Correct Answer: First heart sound
Explanation:In the cardiac cycle, the closing of the atrioventricular (AV) valves coincides with the onset of ventricular systole. This event marks the beginning of the isovolumetric contraction phase, where the ventricles begin to contract, but the volume of blood in the ventricles remains the same because both the AV valves and the semilunar valves (aortic and pulmonary valves) are closed. The closing of the AV valves produces the first heart sound, known as “S1” or “lub.”
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 44
Incorrect
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A woman that presented with dyspnoea, chest pain and cough was found to have a serous pleural effusion. This finding is most likely to be associated with which of the following conditions?
Your Answer:
Correct Answer: Congestive heart failure
Explanation:A pleural effusion is defined as an abnormal collection of fluid in the pleural space. Pleural effusion can result from excess fluid production or decreased absorption or both. Thoracentesis and laboratory testing help determine the origin of the accumulated fluid. Serous fluid accumulation in the pleural space indicates the presence of a hydrothorax and is most likely to develop secondary to congestive heart failure.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 45
Incorrect
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Question 46
Incorrect
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Calculate the total peripheral resistance for a patient with a blood pressure of 130/70 mm HG and cardiac output of 5 litres / min?
Your Answer:
Correct Answer: 18 mmHg × min/l
Explanation:Total peripheral resistance = Mean arterial pressure/Cardiac output. And the mean arterial pressure = Diastolic pressure + 1/3 (Systolic pressure – Diastolic pressure), i.e., 70 + 1/3 (130-70) = 90 mmHg. Therefore, total peripheral resistance = 90 mmHg/5 l per min = 18 mmHg × min/l.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 47
Incorrect
-
What is the normal duration of the ST segment?
Your Answer:
Correct Answer: 0.08 s
Explanation:The ST segment lies between the QRS complex and the T-wave. The normal duration of the ST segment is 0.08 s. ST-segment elevation or depression may indicate myocardial ischaemia or infarction.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 48
Incorrect
-
What is the 5 year survival rate of a patient who is diagnosed with stage III colon cancer, who underwent successful resection and completed the prescribed session of adjuvant chemotherapy?
Your Answer:
Correct Answer: 30%–60%
Explanation:In this patient who has stage III colon cancer, the survival rate is 30-60%. For stage I or Dukes’ stage A disease, the 5-year survival rate after surgical resection exceeds 90%. For stage II or Dukes’ stage B disease, the 5-year survival rate is 70%–85% after resection, with or without adjuvant therapy. For stage III or Dukes’ stage C disease, the 5-year survival rate is 30%– 60% after resection and adjuvant chemotherapy and for stage IV or Dukes’ stage D disease, the 5-year survival rate is poor (approximately 5%).
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 49
Incorrect
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A 73-year-old woman goes to the doctor complaining of pain and stiffness in her shoulders and hips for 4 months, which is worst in the mornings. She has also been suffering from fatigue, weight loss and depression. There were no abnormal findings on physical examination. The erythrocyte sedimentation rate was 110 mm/hour, and serum rheumatoid factor and antinuclear antibody assays were negative. Mild normochromic normocytic anaemia was also found. What is the most likely diagnosis in this case?
Your Answer:
Correct Answer: Polymyalgia rheumatica
Explanation:Polymyalgia rheumatica (PMR) affects older adults, with an acute or subacute onset. Symptoms include severe pain and stiffness of the neck and pectoral or pelvic girdles, which is worse in the morning or after a period of inactivity and is usually bilateral. Other symptoms can include fatigue, weight loss, depression and fever. The erythrocyte sedimentation rate is usually elevated, and normochromic normocytic anaemia can occur. Other tests are usually normal in this disease.
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This question is part of the following fields:
- Orthopaedics
- Pathology
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Question 50
Incorrect
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Which portion of the renal tubule absorbs amino acids and glucose?
Your Answer:
Correct Answer: Proximal convoluted tubule
Explanation:In relation to the morphology of the kidney as a whole, the convoluted segments of the proximal tubules are confined entirely to the renal cortex. Glucose, amino acids, inorganic phosphate and some other solutes are reabsorbed via secondary active transport in the proximal renal tubule through co-transport channels driven by the sodium gradient.
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This question is part of the following fields:
- Physiology
- Renal
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Question 51
Incorrect
-
An organ transplant patient may be at risk of developing which type of cancer?
Your Answer:
Correct Answer: Skin cancer
Explanation:The most common malignancies encountered in the post–solid organ transplant setting are non-melanoma skin cancers, post-transplant lymphoproliferative disorders and Kaposi’s sarcoma (KS). The pathogenesis of these tumours is likely related to the immunosuppressive drugs used post-transplantation and subsequent viral infection.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 52
Incorrect
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Whilst snorkelling, a 30-year old gentleman has the respiratory rate of 10/min, tidal volume of 550 ml and an effective anatomical dead space of 250 ml. What is his alveolar ventilation?
Your Answer:
Correct Answer: 3000 ml/min
Explanation:Alveolar ventilation is the amount of air reaching the alveoli per minute. Alveolar ventilation = respiratory rate × (tidal volume – anatomical dead space volume). Thus, alveolar ventilation = 10 × (550 − 250) = 3000 ml/min.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 53
Incorrect
-
A 25-year-old female had a painful abdomen and several episodes of vomiting. She was severely dehydrated when she was brought to the hospital. Her ABG showed a pH 7.7, p(O2) 75 mmHg, p(CO2) 46 mmHg and bicarbonate 48 mmol/l. The most likely interpretation of this ABG report would be:
Your Answer:
Correct Answer: Metabolic alkalosis
Explanation:Metabolic alkalosis is a primary increase in bicarbonate (HCO3−) with or without compensatory increase in carbon dioxide partial pressure (Pco2); pH may be high or nearly normal. Metabolic alkalosis occurs as a consequence of a loss of H+ from the body or a gain in HCO3 -. In its pure form, it manifests as alkalemia (pH >7.40). As a compensatory mechanism, metabolic alkalosis leads to alveolar hypoventilation with a rise in arterial carbon dioxide tension p(CO2), which diminishes the change in pH that would otherwise occur. Normally, arterial p(CO2) increases by 0.5–0.7 mmHg for every 1 mmol/l increase in plasma bicarbonate concentration, a compensatory response that occurs very rapidly. If the change in p(CO2) is not within this range, then a mixed acid–base disturbance occurs. Likewise, if the increase in p(CO2) is less than the expected change, then a primary respiratory alkalosis is also present. However an elevated serum bicarbonate concentration can also occur due to a compensatory response to primary respiratory acidosis. A bicarbonate concentration greater than 35 mmol/l is almost always caused by metabolic alkalosis (as is the case in this clinical scenario). Calculation of the serum anion gap can also help to differentiate between primary metabolic alkalosis and the metabolic compensation for respiratory acidosis. The anion gap is frequently elevated to a modest degree in metabolic alkalosis because of the increase in the negative charge of albumin and the enhanced production of lactate. However, the only definitive way to diagnose metabolic alkalosis is by performing a simultaneous blood gases analysis, which reveals elevation of both pH and arterial p(CO2) and increased calculated bicarbonate.
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This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
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Question 54
Incorrect
-
A middle aged man presented in OPD with a low grade fever and a persistent cough. His blood smear showed an increase in cells with large bi-lobed nuclei. Which of these cells represent the one seen on the smear?
Your Answer:
Correct Answer: Monocytes
Explanation:Monocytes are white cells that protect the body against harmful pathogens. They are mobile and are produced in the bone marrow, mature there and circulate in the blood for about 1-3 days, where they enter the tissues and transform into macrophages. They are characteristically identified by their large bi-lobed nuclei.
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This question is part of the following fields:
- Inflammation & Immunology
- Pathology
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Question 55
Incorrect
-
Which of the following is responsible for the maximum increase in total peripheral resistance on sympathetic stimulation?
Your Answer:
Correct Answer: Arterioles
Explanation:Arterioles are also known as the resistance vessels as they are responsible for approximately half the resistance of the entire systemic circulation. They are richly innervated by the autonomic nervous system and hence, will bring about the maximum increase in peripheral resistance on sympathetic stimulation.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 56
Incorrect
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A suspected recreational drug user was brought to the Emergency department in an unconscious state, and was found to be hypoventilating. Which of the following set of arterial blood gas analysis report is most consistent with hypoventilation as the primary cause? pH, pa(CO2) (mmHg), pa(O2) (mmHg).
Your Answer:
Correct Answer: 7.28, 55, 81
Explanation:Hypoventilation (or respiratory depression) causes an increase in carbon dioxide (hypercapnia) and respiratory acidosis. It can result due to drugs such as alcohol, benzodiazepines, barbiturates, opiates, mechanical conditions or holding ones breath. Strong opioids such as heroin and fentanyl are commonly implicated and can lead to respiratory arrest. In recreational drug overdose, acute respiratory acidosis occurs with an increase in p(CO2) over 45 mm Hg and acidaemia (pH < 7.35)
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 57
Incorrect
-
A 27-year-old woman has chronic low serum calcium levels. Which of the following conditions may be responsible for the hypocalcaemia in this patient?
Your Answer:
Correct Answer: Hypoparathyroidism
Explanation:Chronic hypocalcaemia is mostly seen in patients with hypoparathyroidism as a result of accidental removal or damage to parathyroid glands during thyroidectomy.
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This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
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Question 58
Incorrect
-
When at rest, which of the following will be higher in a marathon runner compared to an untrained individual?
Your Answer:
Correct Answer: Cardiac stroke volume
Explanation:Cardiac muscle hypertrophy is seen in trained athletes as compared to the normal population. This hypertrophy results in higher stroke volume at rest and increased cardiac reserve (maximum cardiac output during exercise). However, the cardiac output at rest is almost the same in both trained and untrained people. This is because in trained athletes, the heart rate is slower, even up to 40-50 beats/min. There is minimal affect of athletic training on oxygen consumption and respiratory rate.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 59
Incorrect
-
Causes of metabolic acidosis with a normal anion gap include:
Your Answer:
Correct Answer: Diarrhoea
Explanation:Excess acid intake and excess bicarbonate loss as in diarrhoea, are causes of metabolic acidosis with a normal anion gap. The other conditions all result in an increased anion gap.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 60
Incorrect
-
A 34-year-old woman is diagnosed with cerebral oedema after suffering a severe head trauma. Which of the following conditions is not likely to be associated with the extracellular oedema?
Your Answer:
Correct Answer: Increased plasma colloid osmotic pressure
Explanation:Cerebral oedema is extracellular fluid accumulation in the brain. Increased capillary permeability, increased capillary pressure, increased interstitial fluid colloid osmotic pressure and lymphatic blockage would increase fluid movement into the interstitial spaces. Increased plasma colloid osmotic pressure, however, would oppose fluid movement from the capillaries into the interstitial compartment.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 61
Incorrect
-
The ability of the bacteria to cause disease or its virulence is related to :
Your Answer:
Correct Answer: Toxin and enzyme production
Explanation:The pathogenicity of an organism or its ability to cause disease is determined by its virulence factors. Many bacteria produce virulence factors that inhibit the host’s immune system. The virulence factors of bacteria are typically proteins or other molecules that are synthesized by enzymes. These proteins are coded for by genes in chromosomal DNA, bacteriophage DNA or plasmids. The proteins made by the bacteria can poison the host cells and cause tissue damage.
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This question is part of the following fields:
- Microbiology
- Pathology
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Question 62
Incorrect
-
If a patient takes long-term corticosteroid therapy, which of the following diseases is most likely to develop?
Your Answer:
Correct Answer: Osteoporosis
Explanation:One of the complications of long-term intake of corticosteroids is osteoporosis. Some guidelines recommend prophylactic calcium and vitamin D supplementation in patients who take more than 30 mg hydrocortisone or 7.5 mg of prednisolone daily.
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This question is part of the following fields:
- Orthopaedics
- Pathology
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Question 63
Incorrect
-
Which of the following toxins most likely results in continuous cAMP production, which pumps H2O, sodium, potassium, chloride and bicarbonate into the lumen of the small intestine and results in rapid dehydration?
Your Answer:
Correct Answer: Cholera toxin
Explanation:The cholera toxin (CTX or CT) is an oligomeric complex made up of six protein subunits: a single copy of the A subunit (part A), and five copies of the B subunit (part B), connected by a disulphide bond. The five B subunits form a five-membered ring that binds to GM1 gangliosides on the surface of the intestinal epithelium cells. The A1 portion of the A subunit is an enzyme that ADP-ribosylates G proteins, while the A2 chain fits into the central pore of the B subunit ring. Upon binding, the complex is taken into the cell via receptor-mediated endocytosis. Once inside the cell, the disulphide bond is reduced, and the A1 subunit is freed to bind with a human partner protein called ADP-ribosylation factor 6 (Arf6). Binding exposes its active site, allowing it to permanently ribosylate the Gs alpha subunit of the heterotrimeric G protein. This results in constitutive cAMP production, which in turn leads to secretion of H2O, Na+, K+, Cl−, and HCO3− into the lumen of the small intestine and rapid dehydration. The gene encoding the cholera toxin was introduced into V. cholerae by horizontal gene transfer.
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This question is part of the following fields:
- Microbiology
- Pathology
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Question 64
Incorrect
-
A patient is suspected to have a chromosomal abnormality. Which tumour and chromosomal association is correct?
Your Answer:
Correct Answer: Neuroblastoma – chromosome 1
Explanation:Neuroblastoma is associated with a deletion on chromosome 1 and inactivation of a suppressor gene. Neurofibromas and osteogenic sarcoma are associated with an abnormality on chromosome 17. Retinoblastoma (Rb) is associated with an abnormality on chromosome 13. Wilms’ tumours of the kidney are associated with an abnormality on chromosome 11.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 65
Incorrect
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Paracentesis of ascetic fluid in a 45-year old woman revealed the following : clear, yellow fluid with protein 2.0 g/dl and a few mesothelial and mononuclear cells seen. No malignant cells seen. What is the likely diagnosis?
Your Answer:
Correct Answer: Cirrhosis
Explanation:Ascites develops either from:
-
- Increased accumulation
- Increased capillary permeability
- Increased venous pressure
- Decreased protein (oncotic pressure)
- Decreased clearance
- Increased lymphatic obstruction
- Increased accumulation
Cause
- Transudate (<30g/L protein) (Systemic disease)
- Liver (Cirrhosis)
- Cardiac e.g. RHF, CCF, SBE right heart valve disease and constrictive Pericarditis
- Renal failure
- Hypoalbuminaemia (nephrosis)
- Exudate (>30g/L protein) (Local disease)
- Malignancy
- Venous obstruction e.g. Budd-Chiari, Schistosomiasis
- Pancreatitis
- Lymphatic obstruction
- Infection (especially TB)
Cirrhosis is disease of the liver that is characterized by fibrosis leading to disorganization of the hepatic architecture. It shows the development of regenerative nodules surrounded by dense fibrotic tissue. Cirrhosis shows non-specific symptoms initially, which include fatigue, anorexia and weight loss. It can later progress to portal hypertension, ascites and liver failure.
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This question is part of the following fields:
- Gastrointestinal; Hepatobiliary
- Pathology
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Question 66
Incorrect
-
Renal function is an indication of the state of the kidney, measured by glomerular filtration rate (GFR). In a healthy person, GFR would be greatly increased by:
Your Answer:
Correct Answer: Substantial increases in renal blood flow
Explanation:An increase in the rate of renal blood flow (RBF) greatly increases the glomerular filtration rate (GFR). The more plasma available (from increased RBF), the more filtrate is formed. Glomerular filtration rate (GFR) is the volume of fluid filtered from the renal (kidney) glomerular capillaries into the Bowman’s capsule per unit time. Central to the physiologic maintenance of GFR is the differential basal tone of the afferent and efferent arterioles.
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This question is part of the following fields:
- Physiology
- Renal
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Question 67
Incorrect
-
T lymphocytes that express the MCH type II antigen are most likely to produce which of the following cytokines?
Your Answer:
Correct Answer: Gamma interferon
Explanation:Interferon gamma is a soluble cytokine previously known as the macrophage activating factor. It is the only member belonging to the type II class of the interferons. It is secreted by a number of cells taking part in the immune reaction including: T-helper cells (CD-4), cells with immunological memory (CD45PA), killer cells (CD8), dendrite cells (CD23,35), natural killer cells (CD16) and B lymphocytes (CD22,CD23). It has both a defending as well as a pathological effect. It induces differentiation in the myeloid cell in the bone marrow. If macrophages are infected by parasites it activates the macrophages to destroy them. IFN-γ strengthens the anti-tumour activities of the cytotoxic lymphocytes. Together with CD4 or CD8 toxins, produced by lymphocytes, it suppresses the growth of the tumour cells. along with these functions it increases the non specific response of the natural killer cells, causing changes in the cell membrane surface to prevent adhesion and penetration of a virus. It can either increase or decrease B cell response and it activates osteoclasts which increases bone resorption.
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This question is part of the following fields:
- Inflammation & Immunology
- Pathology
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Question 68
Incorrect
-
What is the 5-year survival rate of carcinoma of the pancreas?
Your Answer:
Correct Answer: 30 per cent
Explanation:Pancreatic cancer typically has a poor prognosis, partly because the cancer usually initially remains symptomless, leading to locally advanced or metastatic disease at the time of diagnosis. Median survival from diagnosis is around 3–6 months. Even in those suitable for resectional surgery, 5-year survival rates are still only 30 per cent.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 69
Incorrect
-
Which of the following is the most abundant WBC seen in a smear from a healthy person.
Your Answer:
Correct Answer: Neutrophils
Explanation:neutrophils are the most abundant cell type of the WBC. These phagocytes are found normally in the blood and increase in number are seen during an acute inflammation. These the percentages of WBC in blood Neutrophils: 40 to 60%
Lymphocytes: 20 to 40%
Monocytes: 2 to 8%
Eosinophils: 1 to 4%
Basophils: 0.5 to 1%
Band (young neutrophil): 0 to 3%. eosinophils, basophils, neutrophils are known as granulocytes and monocytes and lymphocytes as agranulocytes.
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This question is part of the following fields:
- Inflammation & Immunology
- Pathology
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Question 70
Incorrect
-
Whilst snorkelling, a 30-year old gentleman has the respiratory rate of 10/min, tidal volume of 550 ml and an effective anatomical dead space of 250 ml. Which of the following will bring about a maximum increase in his alveolar ventilation?
Your Answer:
Correct Answer: A 2x increase in tidal volume and a shorter snorkel
Explanation:Alveolar ventilation = respiratory rate × (tidal volume − anatomical dead space volume). Increase in respiratory rate simply causes movement of air in the anatomical dead space, with no contribution to the alveolar ventilation. By use of a shorter snorkel, the effective anatomical dead space will decrease and will cause a maximum rise in alveolar ventilation along with doubling of tidal volume.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 71
Incorrect
-
What is the normal glomerular filtration rate?
Your Answer:
Correct Answer: 125 mL/min
Explanation:The normal glomerular filtration rate (GFR) in humans is 125 mL/min. After the age of 40, GFR decreases progressively by about 0.4–1.2 mL/min per year.
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This question is part of the following fields:
- Physiology
- Renal
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Question 72
Incorrect
-
How much blood can the pulmonary vessels of a 45-year-old healthy man accommodate when he is at rest?
Your Answer:
Correct Answer: 500 ml
Explanation:Pulmonary circulation is the portion of the cardiovascular system which carries deoxygenated blood away from the heart, to the lungs, and returns oxygenated blood back to the heart. The vessels of the pulmonary circulation are very compliant (easily distensible) and so typically accommodate about 500 ml of blood in an adult man. This large lung blood volume can serve as a reservoir for the left ventricle, particularly during periods when left ventricular output momentarily exceeds venous return.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 73
Incorrect
-
A young lady visited a doctor with complaints of fever and a dull, continuous pain in the right lumbar region for 6 days. On, enquiry, she recalled passing an increasing number of stools with occasional blood in last few months. Lower gastrointestinal endoscopic biopsy was taken 5 cm proximal to ileocaecal valve which showed transmural inflammation with several granulomas. Tissue section showed the absence of acid-fast bacillus. She denies any history of travel and her stool cultures were negative. What is the likely diagnosis?
Your Answer:
Correct Answer: Crohn’s disease
Explanation:Crohn’s disease is a chronic, inflammatory disease that can affect any part of the gastrointestinal tract but is usually seen in the distal ileum and colon. It is transmural and symptoms include chronic diarrhoea, abdominal pain, fever, anorexia and weight loss. On examination, there is usually abdominal tenderness with a palpable mass or fullness seen occasionally. Rectal bleeding is uncommon (except in isolated colonic involvement) which manifests like ulcerative colitis. Differential diagnosis includes acute appendicitis or intestinal obstruction. 25%-33% patients also have perianal disease in the form of fissure or fistulas.
Extra intestinal manifestations predominate in children, and include: arthritis, pyrexia, anaemia or growth retardation. Histologically, the disease shows crypt inflammation and abscesses initially, which progress to aphthoid ulcers. These eventually develop into longitudinal and transverse ulcers with interspersed mucosal oedema, leading to the characteristic ‘cobblestoned appearance’. Transmural involvement leads to lymphoedema and thickening of bowel wall and mesentery, leading to extension of mesenteric fat on the serosal surface of bowel and enlargement of mesenteric nodes. There can also be hypertrophy of the muscularis mucosae, fibrosis and stricture formation, which can cause bowel obstruction.
Abscesses are common and the disease can also leas to development of fistulas with various other organs, anterior abdominal wall and adjacent muscles. Pathognomonic non-caseating granulomas are seen in 50% cases and they can occur in nodes, peritoneum, liver, and in all layers of the bowel wall. The clinical course does not depend on the presence of granulomas. There is sharp demarcation between the diseased and the normal bowel (skip areas).
35% cases show only the ileal involvement, whereas in 45% cases, both the ileum and colon are involved with a predilection for right side of colon. 20% cases show only colonic involvement, often sparing the rectum (unlike ulcerative colitis). In occasional cases, there is jejunoileitis – involvement of the entire small bowel. The stomach, duodenum and oesophagus are rarely involved, although there has been microscopic evidence of disease involving the gastric antrum in younger patients. The affected small bowel segments show increased rick of cancer. Moreover, patients with colonic disease show a long-term risk of cancer similar to that seen in ulcerative colitis.
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This question is part of the following fields:
- Gastrointestinal; Hepatobiliary
- Pathology
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Question 74
Incorrect
-
Purkinje fibres in the heart conduct action potentials at the rate of:
Your Answer:
Correct Answer: 1.5–4.0 m/s
Explanation:Purkinje fibres control the heart rate along with the sinoatrial node (SA node) and the atrioventricular node (AV node). The QRS complex is associated with the impulse passing through the Purkinje fibres. These fibres conduct action potential about six times faster than the velocity in normal cardiac muscle.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 75
Incorrect
-
A 76-year-old woman is diagnosed with diabetes mellitus after a urine test revealed she has glucosuria. Glucosuria may occur due to inadequate glucose reabsorption at:
Your Answer:
Correct Answer: Proximal convoluted tubule
Explanation:Glucose is reabsorbed almost 100% via sodium–glucose transport proteins (apical) and GLUT (basolateral) in the proximal convoluted tubule. Glycosuria or glucosuria is a condition of osmotic diuresis typical in those suffering from diabetes mellitus. Due to a lack of insulin, plasma glucose levels are above normal. This leads to saturation of receptors in the kidneys and glycosuria usually at plasma glucose levels above 11 mmol/l. Rarely, glycosuria is due to an intrinsic problem with glucose reabsorption within the kidneys (such as Fanconi syndrome), producing a condition termed renal glycosuria.
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This question is part of the following fields:
- Physiology
- Renal
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Question 76
Incorrect
-
Which part of the nephron would have to be damaged to stop the reabsorption of the majority of salt and water?
Your Answer:
Correct Answer: Proximal tubule
Explanation:The proximal tubule is the portion of the duct system of the nephron of the kidney which leads from Bowman’s capsule to the loop of Henle. It is conventionally divided into the proximal convoluted tubule (PCT) and the proximal straight tubule (PST). The proximal tubule reabsorbs the majority (about two-thirds) of filtered salt and water. This is done in an essentially iso-osmotic manner. Both the luminal salt concentration and the luminal osmolality remain constant (and equal to plasma values) along the entire length of the proximal tubule. Water and salt are reabsorbed proportionally because the water is dependent on and coupled with the active reabsorption of Na+. The water permeability of the proximal tubule is high and therefore a significant transepithelial osmotic gradient is not possible.
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This question is part of the following fields:
- Physiology
- Renal
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Question 77
Incorrect
-
A 45-year old lady underwent biopsy of a soft, fleshy mass involving her left breast. The biopsy showed lymphoid stroma with minimal fibrosis, surrounding sheets of large vesicular cells with frequent mitoses. Which condition is she most likely suffering from?
Your Answer:
Correct Answer: Medullary carcinoma of breast
Explanation:Medullary carcinoma is a malignant tumour of the breast with well-defined boundaries and accounts for 5% of all breast cancers. Other special features include a larger size of the neoplastic cells and presence of lymphoid cells at tumour edge. Differential diagnosis includes invasive ductal carcinoma. Prognosis is usually good.
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This question is part of the following fields:
- Pathology
- Women's Health
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Question 78
Incorrect
-
A 40-year old lady with a flail chest due to trauma was breathing with the help of a mechanical ventilator in the ICU, and was heavily sedated on muscle relaxants. Due to sudden power failure, a nurse began to hand-ventilate the patient with a Ambu bag. What change will occur in the following parameters: (Arterial p(CO2), pH) in the intervening period between power failure and hand ventilation?
Your Answer:
Correct Answer: Increase, Decrease
Explanation:Respiratory acidosis occurs due to alveolar hypoventilation which leads to increased arterial carbon dioxide concentration (p(CO2)). This in turn decreases the HCO3 –/p(CO2) and decreases pH. Respiratory acidosis can be acute or chronic. In acute respiratory acidosis, the p(CO2) is raised above the upper limit of normal (over 45 mm Hg) with low pH. However, in chronic cases, the raised p(CO2) is accompanied with a normal or near-normal pH due to renal compensation and an increased serum bicarbonate (HCO3 – > 30 mmHg). The given problem represents acute respiratory acidosis and thus, will show a increase in arterial p(CO2) and decrease in pH.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 79
Incorrect
-
What is the most likely cause of bitemporal hemianopia?
Your Answer:
Correct Answer: Prolactinoma
Explanation:Prolactinoma is the mot common pituitary adenoma; leading to hyperprolactinaemia. By virtue of their size, macroprolactinomas press on the adjacent structures leading to headaches and loss of vision due to the pressure effect on optic chiasm. Women notice a change in their menstrual cycle due to raised prolactin levels. In comparison, the problem goes unnoticed in men in the initial stages. Craniopharyngioma is a less common space-occupying lesions affecting children and young adults.
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This question is part of the following fields:
- Endocrine
- Pathology
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Question 80
Incorrect
-
Out of the given options, which of the following is the most likely diagnosis in a 25-year old gentleman presenting with a testicular germ cell tumour?
Your Answer:
Correct Answer: Embryonal carcinoma
Explanation:Embryonal carcinoma is a non-seminomatous germ cell tumour of the testis, accounting for 25% testicular tumours. Other germ cell tumours include seminoma, teratoma and choriocarcinoma. Embryonal carcinomas are known to occur in men aged 25-35 years, and occasionally in teens. They are rarely seen in ovaries of females. It can spread to the vas deferens and also to the aortic lymph nodes. Embryonal carcinomas are known to have elements of fetal origin such as cartilage. Usually, the main tumour is about 2.5cm long, with an extension of 8-9cm along the testicular cord. Contiguous spread to the testicle is seen in less than 1% cases.
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This question is part of the following fields:
- Pathology
- Urology
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Question 81
Incorrect
-
Pseudomonas aeruginosa is a multidrug resistant pathogen that causes hospital-acquired infections. It is usually treated with piperacillin or another antibiotic. Which of the following is the other antibiotic?
Your Answer:
Correct Answer: Azlocillin
Explanation:Azlocillin, like piperacillin, is an acylampicillin antibiotic with an extended spectrum of activity and greater in vitro potency than the carboxypenicillins. Azlocillin is similar to mezlocillin and piperacillin. It demonstrates antibacterial activity against a broad spectrum of bacteria, including Pseudomonas aeruginosa.
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This question is part of the following fields:
- Pathology
- Pharmacology
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Question 82
Incorrect
-
A 45-year-old-female is suspected to have a pulmonary mass. Supposing that she has a neoplasm, which of the following are most commonly found to involve the lung:
Your Answer:
Correct Answer: Pulmonary metastases
Explanation:Lung metastases occur when a cancer started in another part of the body (primary site) spreads to the lungs. The lungs are among the most common site where cancer can spread due to its rich systemic venous drainage, almost every type of cancer can spread to the lung. The most common types of cancer that spread to the lung are breast, colorectal, kidney, testicular, bladder, prostate, head and neck cancers.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 83
Incorrect
-
Some substances, such as Chromium-51 and Technetium-99, are freely filtered but not secreted or absorbed by the kidney. In these cases, their clearance rate is equal to:
Your Answer:
Correct Answer: Glomerular filtration rate
Explanation:If a substance passes through the glomerular membrane with perfect ease, the glomerular filtrate contains virtually the same concentration of the substance as does the plasma and if the substance is neither secreted nor reabsorbed by the tubules, all of the filtered substance continues on into the urine. Glomerular filtration rate (GFR) describes the flow rate of filtered fluid through the kidney.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 84
Incorrect
-
The most likely cause of prominent U waves on the electrocardiogram (ECG) of a patient is:
Your Answer:
Correct Answer: Hypokalaemia
Explanation:The U-wave, not always visible in ECGs, is thought to represent repolarisation of papillary muscles or Purkinje fibres. When seen, it is very small and occurs after the T-wave. Inverted U-waves indicate myocardial ischaemia or left ventricular volume overload. Prominent U-waves are most commonly seen in hypokalaemia. Other causes include hypercalcaemia, thyrotoxicosis, digitalis exposure, adrenaline and class 1A and 3 anti-arrhythmic agents. It can also be seen in congenital long-QT syndrome and in intracranial haemorrhage.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 85
Incorrect
-
Evaluation of a 60-year old gentleman, who has been a coal miner all his life and is suspected to have pulmonary fibrosis reveals the following: normal FEV1, arterial oxygen saturation 92%, alveolar ventilation 6000 ml/min at a tidal volume of 600 ml and a breathing rate of 12 breaths/min. There are also pathological changes in lung compliance and residual volume. Which of the following is most accurate about his residual volume?
Your Answer:
Correct Answer: Cannot be measured directly with a spirometer
Explanation:Residual volume is the air left in the lungs after maximal expiration is done. Thus, this is not a part of vital capacity and cannot be measured with a spirometer directly. It can be measured by the methods such as body plethysmography or inert gas dilution. Expiratory reserve volume is vital capacity minus inspiratory capacity. Resting volume of lungs is he sum of residual volume and expiratory reserve volume. Lungs recoil inward until the recoil pressure becomes zero, which corresponds to a volume significantly lower than residual volume.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 86
Incorrect
-
Which of the following is found to be elevated in a case of hepatocellular carcinoma?
Your Answer:
Correct Answer: AFP
Explanation:Alpha-fetoprotein (AFP) is a glycoprotein that is normally produced by the yolk sac of the embryo, and then the fetal liver. It is elevated in the new-born and thus, also in the pregnant women. Eventually, it decreases in the first year of life to reach the adult normal value of < 20 ng/ml by 1 year of age. Markedly elevated levels (>500 ng/ml) in a high-risk patient is considered diagnostic for primary hepatocellular carcinoma (HCC). Moreover, due to smaller tumours secreting less quantities of AFP, rising levels can be a better indication. However, not all hepatocellular carcinomas produce AFP. Also, the level of AFP is not a prognostic factor. Populations where hepatitis B and HCC are common (e.g.: sub-Saharan Africans, ethnic Chinese) can see AFP levels as high as 100,000 ng/ml, whereas levels are low (about 3000 ng/ml) in regions with lesser incidences of HCC.
AFP can also be elevated up to 500 ng/ml in conditions like embryonic teratocarcinomas, hepatoblastomas, fulminant hepatitis, hepatic metastases from gastrointestinal tract cancers, some cholangiocarcinomas). Lesser values are seen in acute and chronic hepatitis.
Overall, the sensitivity of AFP value ≥20 ng/ml is 39-64% and the specificity is 76%–91%. Value of 500 ng/ml is considered as the diagnostic cut-off level for HCC.
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This question is part of the following fields:
- Gastrointestinal; Hepatobiliary
- Pathology
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Question 87
Incorrect
-
A 27-yeaar-old woman is diagnosed with candidiasis and prescribed fluconazole. What is the mechanism of action of fluconazole?
Your Answer:
Correct Answer: Inhibits cytochrome P450
Explanation:Fluconazole is a triazole antifungal drug used in the treatment and prevention of superficial and systemic fungal infections. Like other imidazole- and triazole-class antifungals, fluconazole inhibits the fungal cytochrome P450 enzyme, 14-demethylase. It is used to treat candidiasis, blastomycosis, coccidioidomycosis, cryptococcosis, histoplasmosis, dermatophytosis, and pityriasis versicolor.
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This question is part of the following fields:
- Pathology
- Pharmacology
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Question 88
Incorrect
-
A 55 year old lady underwent an uneventful appendicectomy. Two hours later, her arterial blood gas analysis on room revealed pH: 7.30, p(CO2): 53 mmHg and p(O2): 79 mmHg. What is the most likely cause of these findings?
Your Answer:
Correct Answer: Alveolar hypoventilation
Explanation:In the given problem, there is respiratory acidosis due to hypercapnia from a low respiratory rate and/or volume (hypoventilation). Causes of hypoventilation include conditions impairing the central nervous system (CNS) respiratory drive, impaired neuromuscular transmission and other causes of muscular weakness (drugs and sedatives), along with obstructive, restrictive and parenchymal pulmonary disorders. Hypoventilation leads to hypoxia and hypercapnia reduces the arterial pH. Severe acidosis leads to pulmonary arteriolar vasoconstriction, systemic vascular dilatation, reduced myocardial contractility, hyperkalaemia, hypotension and cardiac irritability resulting in arrhythmias. Raised carbon dioxide concentration also causes cerebral vasodilatation and raised intracranial pressure. Over time, buffering and renal compensation occurs. However, this might not be seen in acute scenarios where the rise in p(CO2) occurs rapidly.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 89
Incorrect
-
Renin is secreted by pericytes in the vicinity of the afferent arterioles of the kidney from the juxtaglomerular cells. Plasma renin levels are decreased in patients with:
Your Answer:
Correct Answer: Primary aldosteronism
Explanation:Primary aldosteronism, also known as primary hyperaldosteronism or Conn’s syndrome, is excess production of the hormone aldosterone by the adrenal glands resulting in low renin levels. Most patients with primary aldosteronism (Conn’s syndrome) have an adrenal adenoma. The increased plasma aldosterone concentration leads to increased renal Na+ reabsorption, which results in plasma volume expansion. The increase in plasma volume suppresses renin release from the juxtaglomerular apparatus and these patients usually have low plasma renin levels. Salt restriction and upright posture decrease renal perfusion pressure and therefore increases renin release from the juxtaglomerular apparatus. Secondary aldosteronism is due to elevated renin levels and may be caused by heart failure or renal artery stenosis.
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This question is part of the following fields:
- Physiology
- Renal
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Question 90
Incorrect
-
A 25-year-old football player suffers severe head trauma during a game. Physical examination reveals a decerebrate posture and bilateral papilledema. A CT scan reveals marked diffuse cerebral oedema. This condition will be most severe in which component of the brain?
Your Answer:
Correct Answer: White matter
Explanation:The greatest amount of salt and water increase with cerebral oedema occurs within the white matter. Any swelling from oedema, haemorrhage or haematoma increases the intracranial pressure (ICP). As the ICP increases, the cerebral perfusion decreases and brain tissue can become ischaemic, even leading to brain death.
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This question is part of the following fields:
- Neurology
- Pathology
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Question 91
Incorrect
-
A 56-year old male with history of previous abdominal surgery presents to the emergency department with severe abdominal pain that is increasing in intensity. This pain is associated with abdominal distension and faint, high-pitched bowel sounds. An old scar is also noted on the abdomen. An erect abdominal X-ray shows multiple air-fluid levels in dilated bowel loops. No occult blood was found in stool sample. What is the most likely predisposing factor for his present condition?
Your Answer:
Correct Answer: Adhesions from previous surgery
Explanation:The described features suggest acute bowel obstruction. The scar described points toward previous surgery, which suggests development of peritoneal adhesions that could lead to obstruction. Hepatitis does not lead to dilated bowel loops. Amoebiasis could lead to inflammatory bowel disease, however, occult blood in stool is usually positive. Ileal adenocarcinoma is rare. Meckel’s diverticulum can possibly lea to obstruction but the findings described here are more consistent with obstruction due to peritoneal adhesions from a past surgery.
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This question is part of the following fields:
- Gastrointestinal; Hepatobiliary
- Pathology
-
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Question 92
Incorrect
-
Question 93
Incorrect
-
A 56-year-old man undergoes tests to determine his renal function. His results over a period of 24 hours were:
Urine flow rate: 2. 0 ml/min
Urine inulin: 1.0 mg/ml
Plasma inulin: 0.01 mg/ml
Urine urea: 260 mmol/l
Plasma urea: 7 mmol/l
What is the glomerular filtration rate?Your Answer:
Correct Answer: 200 ml/min
Explanation:Glomerular filtration rate (GFR) is the volume of fluid filtered from the renal (kidney) glomerular capillaries into the Bowman’s capsule per unit time. GFR is equal to the inulin clearance because inulin is freely filtered into Bowman’s capsule but is not reabsorbed or secreted. The clearance (C) of any substance can be calculated as follows: C = (U × V)/P, where U and P are the urine and plasma concentrations of the substance, respectively and V is the urine flow rate. Thus, glomerular filtration rate = (1.0 × 2. 0)/0.01 = 200 ml/min.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 94
Incorrect
-
A 60-year-old male is suspected to have pancreatic cancer. What particular tumour marker should be requested to aid in the confirmation of the diagnosis of pancreatic cancer?
Your Answer:
Correct Answer: Carcinoembryonic antigen (CEA)
Explanation:Carcinoembryonic antigen (CEA) is used as a tumour marker. CEA test measures the amount of this protein that may appear in the blood of some people who have certain types of cancers especially cancer of the colon and rectal cancer. It may also be present in the pancreas, breast, ovary or lung.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 95
Incorrect
-
A young 16 year old boy presented to the ENT clinic with a history of sore throat for the past 1 day. On examination there was a pharyngeal purulent discharge. Which of the following types of inflammation is seen in this boy?
Your Answer:
Correct Answer: Acute inflammation
Explanation:A 1 day history suggests the purulent discharge is due to acute inflammation. Acute inflammation has 3 features:
1) the affected area is occupied by a purulent discharge composed of proteins, fluids and cells from local blood vessels
2) the infective agent i.e. bacteria is present in the affected area
3) the damaged tissue can be liquified and the debris removed from the site.
If the inflammation lasts over weeks or months, then it is termed as chronic inflammation.
Granulomatous inflammation is characterised by the presence and formation of granulomas.
Exudate is not a feature of resolution or a complication of inflammation.
Abscess formation takes more than 1 day to form and is usually within a capsule/cavity.
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This question is part of the following fields:
- Inflammation & Immunology; Respiratory
- Pathology
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Question 96
Incorrect
-
A 48-year-old woman has a mass in her right breast and has right axillary node involvement. She underwent radical mastectomy of her right breast. The histopathology report described the tumour to be 4 cm in its maximum diameter with 3 axillary lymph nodes with evidence of tumour. The most likely stage of cancer in this patient is:
Your Answer:
Correct Answer: IIB
Explanation:Stage IIB describes invasive breast cancer in which: the tumour is larger than 2 centimetres but no larger than 5 centimetres; small groups of breast cancer cells — larger than 0.2 millimetre but not larger than 2 millimetres — are found in the lymph nodes OR the tumour is larger than 2 centimetres but no larger than 5 centimetres; cancer has spread to 1 to 3 axillary lymph nodes or to lymph nodes near the breastbone (found during a sentinel node biopsy) OR the tumour is larger than 5 centimetres but has not spread to the axillary lymph nodes.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 97
Incorrect
-
Which of these laboratory findings will indicate a fetal neural tube defect when done between 15 and 20 weeks of pregnancy?
Your Answer:
Correct Answer: Increased alpha-fetoprotein
Explanation:Maternal serum screening during the second trimester is a non-invasive way of identifying women at increased risk of having children with a neural tube defect and should be offered to all pregnant women. The results are most accurate when the sample is taken between 15 and 20 weeks of gestation. Elevated levels of alpha-fetoprotein suggest open spina bifida, anencephaly, risk of pregnancy complications, or multiple pregnancy.
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This question is part of the following fields:
- Neurology
- Pathology
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Question 98
Incorrect
-
Which of the following is likely to result in hematocolpos in a 12-year old girl?
Your Answer:
Correct Answer: Imperforate hymen
Explanation:Hematocolpos means accumulation of blood in vagina and hematometra is accumulation of blood in the uterus. These are most likely seen with an imperforate hymen; which is seen I 1 in 2000 females. If spontaneous resolution does not occur, treatment involves making a hole in the hymen to allow discharge of menstrual blood.
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This question is part of the following fields:
- Pathology
- Women's Health
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Question 99
Incorrect
-
Which of the following over-the-counter drugs can cause a prolonged bleeding time?
Your Answer:
Correct Answer: Acetylsalicylic acid
Explanation:Acetylsalicylic acid, or aspirin, is a nonsteroidal anti-inflammatory drug that is widely used as an analgesic and antipyretic. Aspirin is as a cyclo-oxygenase inhibitor that leads to decreased prostaglandin production. Decreased platelet aggregation is another effect of this drug, achieved by long-lasting use of aspirin.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 100
Incorrect
-
Which of the following variables are needed to calculate inspiratory reserve volume of a patient?
Your Answer:
Correct Answer: Tidal volume, vital capacity and expiratory reserve volume
Explanation:Vital capacity = inspiratory reserve volume + tidal volume + expiratory reserve volume. Thus, inspiratory reserve volume can be calculated if tidal volume, vital capacity and expiratory reserve volume are known.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 101
Incorrect
-
A medical student is told a substance is freely filtered but is not metabolised, secreted, or stored in the kidney. It has a plasma concentration of 1000 mg/l and its urine excretion rate is 25 mg/min, and the inulin clearance is 100 ml/min. What is the rate of tubular reabsorption of the substance?
Your Answer:
Correct Answer: 75 mg/min
Explanation:Reabsorption or tubular reabsorption is the process by which the nephron removes water and solutes from the tubular fluid (pre-urine) and returns them to the circulating blood. To calculate the reabsorption rate of substance Z we use the following equation: excretion = (filtration + secretion) – reabsorption. As this substance is freely filtered, its filtration rate is equal to that of inulin. So 25 = (100 + 0) – reabsorption. Reabsorption = 100 – 25 therefore reabsorption = 75 mg/min.
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This question is part of the following fields:
- Physiology
- Renal
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Question 102
Incorrect
-
During cardiac catheterisation in a 20-year old man, the following data is obtained: Pressure (mmHg), O2 saturation (%) Right atrium 7 (N = 5) 90 (N = 75), Right ventricle 35/7 (N = 25/5) 90 (N = 75), Pulmonary artery 35/8 (N = 25/15), 90 (N = 75), Left atrium 7 (N = 9) 95 (N = 95), Left ventricle 110/7 (N = 110/9) 95 (N = 95), Aorta 110/75 (N = 110/75) 95 (N = 95) where N = Normal value. What is the likely diagnosis?
Your Answer:
Correct Answer: Atrial septal defect
Explanation:A congenital heart disease, ASD or atrial septal defect leads to a communication between the right and left atria due to a defect in the interatrial septum. This leads to mixing of arterial and venous blood from the right and left side of the heart. The hemodynamic significance of this defect depends on the presence of shunting of blood. Normally, the left side of the heart has higher pressure than the right as the left side has to pump blood throughout the body. A large ASD (> 9 mm) will result in a clinically significant left-to-right shunt, causing volume overload of the right atrium and ventricle, eventually leading to heart failure. Cardiac catheterization would reveal very high oxygen saturation in the right atrium, right ventricle and pulmonary artery. Eventually, the left-to-right shunt will lead to pulmonary hypertension and increased afterload in the right ventricle, along with the increased preload due to the shunted blood. This will either cause right ventricular failure, or raise the pressure in the right side of the heart to equal or more than that in the left. Elevation of right atrial pressure to that of left atrial pressure would thus lead to diminishing or complete cessation of the shunt. If left uncorrected, there will be reversal of the shunt, known as Eisenmenger syndrome, resulting in clinical signs of cyanosis as the oxygen-poor blood form right side of the heart will mix with the blood in left side and reach the peripheral vascular system.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 103
Incorrect
-
What's the nodal stage of a testicular seminoma if several lymph nodes between 2cm and 5cm are found?
Your Answer:
Correct Answer: N2
Explanation:According to the American Joint Committee on Cancer (AJCC) 2002 guidelines, the nodal staging of testicular seminoma is the following:
N0: no regional lymph node metastases
N1: metastasis with lymph nodes 2 cm or less in their greatest dimension or multiple lymph nodes, none more than 2 cm
N2: metastasis with lymph nodes greater than 2 cm but not greater than 5 cm in their greatest dimension, or multiple lymph nodes, any one mass greater than 2 cm, but not more than 5 cm
N3: metastasis with lymph nodes greater than 5 cm in their greatest dimension.
The patient in this case has N2 testicular seminoma. This TNM staging is extremely important because treatment options are decided depending on this classification.
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This question is part of the following fields:
- Pathology
- Urology
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Question 104
Incorrect
-
The most likely cause of a low p(O2) in arterial blood is:
Your Answer:
Correct Answer: Hypoxic hypoxia
Explanation:Hypoxia is when the whole body or a region is deprived of adequate oxygen supply. Different types of hypoxia include the following:
– Hypoxic hypoxia, which occurs due to poor oxygen supply, as a result of low partial pressure of oxygen in arterial blood. This could be due to low partial pressure of atmospheric oxygen (e.g., at high altitude), sleep apnoea, poor ventilation because of chronic obstructive pulmonary disease or respiratory arrest, or shunts. The other types of hypoxia have a normal partial pressure of oxygen.
– Anaemic hypoxia occurs due to low total oxygen content of the blood, with a normal arterial oxygen pressure.
– Hyperaemic hypoxia occurs due to poor delivery of oxygen to target tissues, such as in carbon monoxide poisoning or methemoglobinemia.
– Histotoxic hypoxia results due to inability of the cells to use the delivered oxygen due to disabled oxidative phosphorylation enzymes.
– Ischaemic (or stagnant) hypoxia occurs due to local flow restriction of well-oxygenated blood, seen in cases like cerebral ischaemia, ischaemic heart disease and intrauterine hypoxia.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 105
Incorrect
-
The wound healing process is documented in patients undergoing laparoscopic procedures. The port incisions are sutured closed and the wounds observed every few weeks for re-epithelialisation and tensile strength. Which substance is mostly likely to be found at a cellular level involved in wound healing?
Your Answer:
Correct Answer: Tyrosine kinase
Explanation:Cell surface growth factor receptors require intercellular proteins such as tyrosine kinase which are necessary to initiate a series of events that eventually lead to cell division and growth. Tyrosine kinase is an enzyme that transfers a phosphate group to the tyrosine residue in a protein. This phosphorylation will lead to an up regulation of the enzyme activity.
Fibronectin acts in the extracellular matrix to bind macromolecules (such as proteoglycans) via integrin receptors to aid attachment and migration of cells.
Laminin is an extracellular matrix component that is abundant in basement membranes.
Hyaluronic acid is one of the proteoglycans in the extracellular matrix.
Collagen fibres are part of the extracellular matrix that gives strength and stability to connective tissues.
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This question is part of the following fields:
- Cell Injury & Wound Healing
- Pathology
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Question 106
Incorrect
-
Dicloxacillin is an antibiotic; the subclass that dicloxacillin belongs to is:
Your Answer:
Correct Answer: Penicillin
Explanation:Dicloxacillin is a narrow-spectrum beta-lactam antibiotic. It is used to treat infections caused by susceptible Gram-positive bacteria and most effective against beta-lactamase-producing organisms such as Staphylococcus aureus. To decrease the development of resistance, dicloxacillin is recommended to treat infections that are suspected or proven to be caused by beta-lactamase-producing bacteria.
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This question is part of the following fields:
- Pathology
- Pharmacology
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Question 107
Incorrect
-
During strenuous exercise, what else occurs besides tachycardia?
Your Answer:
Correct Answer: Increased stroke volume
Explanation:During strenuous exercise there is an increase in:
– Heart rate, stroke volume and therefore cardiac output. (CO = HR x SV)
– Respiratory rate (hyperventilation) which will lead to a reduction in Paco2.
– Oxygen demand of skeletal muscle, therefore leading to a reduction in mixed venous blood oxygen concentration.
Renal blood flow is autoregulated, so renal blood flow is preserved and will tend to remain the same. Mean arterial blood pressure is a function of cardiac output and total peripheral resistance and will increase with exercise, mainly as a result of the increase in cardiac output that occurs.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 108
Incorrect
-
A 35 year-old female developed food poisoning 24H after eating canned food. She complained of abdominal cramps, with nausea and vomiting. Shortly after she suddenly developed weakness, blurring of vision, difficulty in swallowing and breathing. Which of the following organisms is most likely associated with fatal food poisoning?
Your Answer:
Correct Answer: Clostridium botulinum
Explanation:C. botulinum is a Gram-positive, rod-shaped, spore-forming bacterium. It is an obligate anaerobe, meaning that oxygen is poisonous to the cells. Only botulinum toxin types A, B, E, and F cause disease in humans. Types A, B, and E are associated with foodborne illness. Botulism poisoning can occur due to preserved or home-canned, low-acid food that was not processed using correct preservation times and/or pressure. Signs and symptoms of foodborne botulism typically begin between 18 and 36 hours after the toxin gets into the body, but can range from a few hours to several days, depending on the amount of toxin ingested. Botulinum that is produced by Clostridium botulinum can cause respiratory and muscular paralysis.
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This question is part of the following fields:
- Microbiology
- Pathology
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Question 109
Incorrect
-
A 23-year-old woman decides to donate a kidney through a kidney chain. Which of the following indices would be expected to be decreased in the donor after full recovery from the operation?
Your Answer:
Correct Answer: Creatinine clearance
Explanation:Since medication to prevent rejection is so effective, donors do not need to be similar to their recipient. Most donated kidneys come from deceased donors; however, the utilisation of living donors is on the rise. Most problems encountered with live donation are associated with the donor. Firstly, there are the potentially harmful investigative procedures carried out in the assessment phase, the most hazardous being renal angiography, where there is cannulation of the artery and injection of a radio-opaque dye to determine the blood supply to the kidney. Secondly, there are the short-term risks of nephrectomy surgery. According to the literature, there is a mortality rate of between 1 in 1600 and 1 in 3000, but this is no more than is associated with any anaesthetic. In the initial postoperative period creatinine clearance may be decreased but this recovers fully over a few weeks to months. Long-term complications include prolonged wound pain.
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This question is part of the following fields:
- Physiology
- Renal
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Question 110
Incorrect
-
Which of the following physiological changes will you see in a young man who has been trekking in the Himalayas for 3 years?
Your Answer:
Correct Answer: Increased renal excretion of HCO3 –
Explanation:The atmospheric pressure is lower at high altitudes as compared with sea level. This leads to a decrease in the partial pressure of oxygen. Once 2100 m (7000 feet) of altitude is reached, there is a drop in saturation of oxyhaemoglobin. The oxygen saturation of haemoglobin determines the oxygen content in the blood. The body physiological tries to adapt to high altitude by acclimatization. Immediate effects include hyperventilation, fluid loss (due to a decreased thirst drive), increase in heart rate and slightly lowered stroke volume. Long term effects include lower lactate production, compensatory alkali loss in urine, decrease in plasma volume, increased erythropoietin release and red cell mass, increased haematocrit, higher concentration of capillaries in striated muscle tissue, increase in myoglobin, increase in mitochondria, increase in aerobic enzyme concentration such as 2,3-DPG and pulmonary vasoconstriction.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 111
Incorrect
-
Which of the following diseases can cause paraesthesia along the distribution of the median nerve of the hand, especially after activities which require flexion and extension of the wrist?
Your Answer:
Correct Answer: Carpal tunnel syndrome
Explanation:Carpal tunnel syndrome tends to occur in women between the ages of 30-50. There are many risk factors, including diabetes, hypothyroidism, obesity, pregnancy, and repetitive wrist work. Symptoms include pain in the hand and wrist, tingling, and numbness distributed along the median nerve (the palmar side of the thumb, the index and middle fingers, and the radial half of the ring finger), which worsens at night.
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This question is part of the following fields:
- Orthopaedics
- Pathology
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Question 112
Incorrect
-
What class of drugs does buspirone belong to?
Your Answer:
Correct Answer: Anxiolytic
Explanation:Buspirone is an anxiolytic agent and a serotonin-receptor agonist that belongs to the azaspirodecanedione class of compounds. It shows no potential for addiction compared with other drugs commonly prescribed for anxiety, especially the benzodiazepines. The development of tolerance has not been noted. It is primarily used to treat generalized anxiety disorders. It is also commonly used to augment antidepressants in the treatment of major depressive disorder.
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This question is part of the following fields:
- Pathology
- Pharmacology
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Question 113
Incorrect
-
After total thyroidectomy, which of the following investigations is recommended in the immediate post-operative period?
Your Answer:
Correct Answer: Serum calcium
Explanation:Total thyroidectomy might sometimes result in inadvertent excision or damage of parathyroid glands, leading to hypoparathyroidism. Monitoring serum calcium levels in the post-operative period to detect hypocalcaemia is essential to diagnose and prevent this condition.
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This question is part of the following fields:
- Endocrine
- Pathology
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Question 114
Incorrect
-
Different substances have different renal clearance rates. Which of the following substances should have the lowest renal clearance rate in a healthy patient?
Your Answer:
Correct Answer: Glucose
Explanation:Under normal conditions the renal clearance of glucose is zero, since glucose is completely reabsorbed in the renal tubules and not excreted. Glycosuria – the excretion of glucose into the urine- is nearly always caused by elevated blood glucose levels, most commonly due to untreated diabetes mellitus.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 115
Incorrect
-
During cardiac catheterisation, if the blood sample from the catheter shows an oxygen saturation of 70%, and the pressure ranging from 12 to 24 mm Hg, it implies that the catheter tip is located in the:
Your Answer:
Correct Answer: Pulmonary artery
Explanation:Normal values for various parameters are as follows:
Systolic arterial blood pressure (SBP): 90–140 mmHg.
Diastolic arterial blood pressure: 60–90 mmHg.
Mean arterial blood pressure (MAP): SBP + (2 × DBP)/3 (normal range: 70-105 mmHg).
Right atrial pressure (RAP): 2–6 mmHg.
Systolic right ventricular pressure (RVSP): 15–25 mmHg.
Diastolic right ventricular pressure (RVDP): 0–8 mmHg.
Pulmonary artery pressure (PAP): Systolic (PASP) is 15-25 mmHg and Diastolic (PADP) is 8–15 mmHg.
Pulmonary artery wedge pressure (PAWP): 6–12 mmHg.
Left atrial pressure (LAP): 6–12 mmHg.
Thus, the given value indicates that the position of catheter tip is likely to be in the pulmonary artery.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 116
Incorrect
-
Which of the following changes in the histology of the cell is most likely to be accompanied by disruption of the cell membrane following an injury?
Your Answer:
Correct Answer: Coagulative necrosis
Explanation:The process of necrosis ends with the rupture of the cell membrane and the consequent release of the cellular components into the surrounding tissue. Apoptosis, pyknosis and karyorrhexis are not reversible events but the cell membrane remains intact. Cloudy swelling and hydropic changes are also reversible but again the cell membrane remains intact and they are therefore different and distinct from necrosis.
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This question is part of the following fields:
- Cell Injury & Wound Healing
- Pathology
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Question 117
Incorrect
-
For calculation of cardiac output by Fick's principle, which of the following vessels is the best source of venous blood to determine the arterial-to-venous oxygen tension difference?
Your Answer:
Correct Answer: Pulmonary artery
Explanation:Fick’s principle states that the total uptake (or release) of a substance by peripheral tissues is equal to the product of the blood flow to the peripheral tissues and the arterial– venous concentration difference (gradient) of the substance. It is used to measure the cardiac output, and the formula is Cardiac output = oxygen consumption divided by arteriovenous oxygen difference. Assuming there are no shunts across the pulmonary system, the pulmonary blood flow equals the systemic blood flow. The arterial and venous blood oxygen content is measured by sampling from the pulmonary artery (low oxygen content) and pulmonary vein (high oxygen content). Peripheral arterial blood is used as a surrogate for the pulmonary vein.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 118
Incorrect
-
A 55-year-old man underwent CT scan of the whole abdomen. The result showed renal cell carcinoma with a tumour size of 7cm and extension into the regional lymph. What is the clinical stage of his renal cell cancer?
Your Answer:
Correct Answer: Stage III
Explanation:Renal cell carcinoma is a kidney cancer that originates in the lining of the proximal convoluted tubule. It is the most common type of kidney cancer in adults, responsible for approximately 90–95% of cases. Renal cell carcinomas can be staged by using the American Joint Committee on Cancer (AJCC) TNM (tumour-node-metastasis) classification, as follows: Stage I: tumours that are 7 cm or smaller and confined to the kidney, Stage II: tumours that are larger than 7 cm but still confined to the kidney, Stage III: tumours extending into the renal vein or vena cava, involving the ipsilateral adrenal gland and/or perinephric fat, or which have spread to one local lymph node and Stage IV: tumours extending beyond Gerota’s fascia, to more than one local node, or with distant metastases Recent literature has questioned whether the cut-off in size between stage I and stage II tumours should be 5 cm instead of 7 cm. The patient’s cancer in this case is stage III.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 119
Incorrect
-
A 30-year-old woman known with Von Willebrand disease (vWD) has to undergo surgery. Which of these complications is most unlikely in this patient?
Your Answer:
Correct Answer: Hemarthrosis
Explanation:Von Willebrand disease (vWD) is an inherited haemorrhagic disorder characterised by the impairment of primary haemostasis. It is caused by the deficiency or dysfunction of a protein named von Willebrand factor. The most common manifestation due to the condition is abnormal bleeding. Complications include easy bruising, hematomas, epistaxis, menorrhagia, prolonged bleeding and severe haemorrhage. Hemarthrosis is a complication that is more commonly found in haemophilia.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 120
Incorrect
-
Question 121
Incorrect
-
If a catheter is placed in the main pulmonary artery of a healthy 30-year-old woman, which of the following will be its mean pulmonary arterial pressure?
Your Answer:
Correct Answer: 15 mmHg
Explanation:The pulmonary artery pressure (PA pressure) is a measure of the blood pressure found in the main pulmonary artery. The hydrostatic pressure of the pulmonary circulation refers to the actual pressure inside pulmonary vessels relative to atmospheric pressure. Hydrostatic (blood pressure) in the pulmonary vascular bed is low compared with that of similar systemic vessels. The mean pulmonary arterial pressure is about 15 mmHg (ranging from about 13 to 19 mmHg) and is much lower than the average systemic arterial pressure of 90 mmHg.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 122
Incorrect
-
A 38-year old lady presented to the hospital with abnormal passing of blood per vagina. On examination, she was found to have an endocervical polypoidal mass. On enquiry, she gave history of oral contraceptive usage for 3 years. What finding is expected on the histopathology report of biopsy of the mass?
Your Answer:
Correct Answer: Microglandular hyperplasia
Explanation:Endocervical polyps or microglandular hyperplasia are benign growths occurring in the endocervical canal, in about 2-5% women and occur secondary to use of oral contraceptives. They are usually < 1cm in size, friable and reddish-pink. Usually asymptomatic, they can cause bleeding or become infected, leading to leucorrhoea (purulent vaginal discharge). They are usually benign but need to be differentiated from adenocarcinomas by histology.
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This question is part of the following fields:
- Pathology
- Women's Health
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Question 123
Incorrect
-
A 30-year-old man is brought to the emergency department suffering from extreme dehydration, and subsequent hypotension and tachycardia. Which part of the kidney will compensate for this loss?
Your Answer:
Correct Answer: Collecting ducts
Explanation:The collecting duct system of the kidney consists of a series of tubules and ducts that physically connect nephrons to a minor calyx or directly to the renal pelvis. The collecting duct system is the last component of the kidney to influence the body’s electrolyte and fluid balance. In humans, the system accounts for 4–5% of the kidney’s reabsorption of sodium and 5% of the kidney’s reabsorption of water. At times of extreme dehydration, over 24% of the filtered water may be reabsorbed in the collecting duct system.
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This question is part of the following fields:
- Physiology
- Renal
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Question 124
Incorrect
-
Gastric acid secretion is stimulated by which of the following?
Your Answer:
Correct Answer: Gastrin
Explanation:Gastric acid secretion is stimulated by three factors:
– Acetylcholine, from parasympathetic neurones of the vagus nerve that innervate parietal cells directly
– Gastrin, produced by pyloric G-cells
– Histamine, produced by mast cells.
Gastric acid is inhibited by three factors:
– Somatostatin
– Secretin
– Cholecystokinin
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This question is part of the following fields:
- Endocrine
- Physiology
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Question 125
Incorrect
-
A 72 year old man suffered a MI. What is the approximate time needed by the scar tissue of the MI to recover and attain full strength?
Your Answer:
Correct Answer: Several months
Explanation:A week following a MI attack, a little collagen starts to form and deposit. By the end of the 2nd week, neovascularisation of the scar occurs, with some collagen being laid down in a haphazard fashion. By this time the scar attains some strength. During the next 6 months, collagen is constantly being laid down and is rearranged in order to shrink the scar. Most of the blood vessels by this time have regenerated, decreasing vascularity of the scar reaching full maturity.
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This question is part of the following fields:
- Cell Injury & Wound Healing; Cardiovascular
- Pathology
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Question 126
Incorrect
-
A young male was diagnosed with hepatitis A, which clinically resolved in 2 weeks. What will his liver biopsy done after 6 months show?
Your Answer:
Correct Answer: Normal architecture
Explanation:Hepatitis A is the most common acute viral hepatitis, more common in children and young adults. It is caused by Hepatitis A virus, which is a single-stranded RNA picornavirus. The primary route of spread of Hepatitis A is the faecal-oral route. Consumption of contaminated raw shellfish is also a likely causative factor. The shedding of the virus in faecal matter occurs before the onset of symptoms and continues a few days after. Hepatitis A does not lead to chronic hepatitis or cirrhosis, and there is no known chronic carrier state. Hence, a biopsy performed after recovery will show normal hepatocellular architecture.
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This question is part of the following fields:
- Gastrointestinal; Hepatobiliary
- Pathology
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Question 127
Incorrect
-
A 55-year-old male has a mass on the right lung with involvement of the right mediastinal lymph nodes. What is the nodal staging according to the TNM staging?
Your Answer:
Correct Answer: N2
Explanation:The N stages for lung cancer are from NO to N3 : NO, there is no lung cancer in any lymph nodes: N1, there is lung cancer in the proximal lymph nodes: N2, there is lung cancer in the mediastinal hilar lymph nodes, but on the same side as the affected lung or there is lung cancer in the carinal lymph nodes: N3, there is metastatic lung cancer in lymph nodes on the opposite side of the chest, in the cervical or apical lymph nodes. In this patient the ipsilateral mediastinal node is involved, thus it is classified as N2.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 128
Incorrect
-
A 70-year-old male who has smoked since his teens complains of progressive shortness of breath and a persistent cough. He is diagnosed with COPD. Which of the following abnormalities is most likely to be present in his pulmonary function tests?
Your Answer:
Correct Answer: Increased residual volume
Explanation:Chronic obstructive pulmonary disease (COPD) is a type of obstructive lung disease characterized by long-term poor airflow. The main symptoms include shortness of breath and cough with sputum production. The best diagnostic test for evaluating patients with suspected chronic obstructive pulmonary disease (COPD) is lung function measured with spirometry. Key spirometrical measures may be obtained with a portable office spirometer and should include forced vital capacity (FVC) and the normal forced expiratory volume in the first second of expiration (FEV1). The ratio of FEV1 to forced vital capacity (FEV1/FVC) normally exceeds 0.75. Patients with COPD typically present with obstructive airflow. Complete pulmonary function testing may show increased total lung capacity, functional residual capacity and residual volume. A substantial loss of lung surface area available for effective oxygen exchange causes diminished carbon monoxide diffusion in the lung (DLco) in patients with emphysema. Tobacco smoking is the most common cause of COPD, with factors such as air pollution and genetics playing a smaller role.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 129
Incorrect
-
Multiple, non-tender lymphadenopathy with biopsy showing several crowded follicles of small, monomorphic lymphocytes and the absence of Reed-Sternberg cells is seen in which of the following?
Your Answer:
Correct Answer: Poorly differentiated lymphocytic lymphoma
Explanation:Malignant lymphoma usually causes non-tender lymphadenopathy, unlike the tender lymphadenopathy caused by infections (including infectious mononucleosis caused by Epstein-Barr virus). Also, the lymphoid hyperplasia seen in infectious mononucleosis is benign and polyclonal.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 130
Incorrect
-
A 5-year-old child is rushed to the emergency department because of fever, erythema and neck stiffness. The child is also hypotensive upon physical examination. Which is the most likely toxin responsible for this child's condition?
Your Answer:
Correct Answer: Endotoxin
Explanation:An endotoxin is part of the bacterial outer membrane, and it is not released until the bacterium is killed by the immune system. The body’s response to an endotoxin can involve severe inflammation. In general, the inflammation process is usually considered beneficial to the infected host, but if the reaction is severe enough, it can lead to sepsis.
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This question is part of the following fields:
- Microbiology
- Pathology
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Question 131
Incorrect
-
A 7-year-old boy is diagnosed with metabolic acidosis as a result of severe dehydration. Which of the following conditions is most likely to cause severe dehydration and metabolic acidosis?
Your Answer:
Correct Answer: Severe diarrhoea
Explanation:Diarrhoea is defined as having three or more loose or liquid stools per day, or as having more stools than is normal for that person. Severe diarrhoea, causing fluid loss and loss of bicarbonate, will result in marked dehydration and metabolic acidosis.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 132
Incorrect
-
An ECG of a 30 year old woman revealed low voltage QRS complexes. This patient is most probably suffering from?
Your Answer:
Correct Answer: Pericardial effusion
Explanation:The QRS complex is associated with current that results in the contraction of both the ventricles. As ventricles have more muscle mass than the atria, they result in a greater deflection on the ECG. The normal duration of a QRS complex is 10s. A wide and deep Q wave depicts myocardial infarction. Abnormalities in the QRS complex maybe indicative of a bundle block, ventricular tachycardia or hypertrophy of the ventricles. Low voltage QRS complexes are characteristic of pericarditis or a pericardial effusion.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 133
Incorrect
-
A 76-year-old man with a urinary tract obstruction due to prostatic hyperplasia develops acute renal failure. Which of the following physiological abnormalities of acute renal failure will be most life threatening for this patient?
Your Answer:
Correct Answer: Acidosis
Explanation:Acute renal failure (ARF) is a rapid loss of renal function due to damage to the kidneys, resulting in retention of nitrogenous (urea and creatinine) and non-nitrogenous waste products that are normally excreted by the kidney. This accumulation may be accompanied by metabolic disturbances, such as metabolic acidosis and hyperkalaemia, changes in body fluid balance and effects on many other organ systems. Metabolic acidosis and hyperkalaemia are the two most serious biochemical manifestations of acute renal failure and may require medical treatment with sodium bicarbonate administration and antihyperkalaemic measures. If not appropriately treated these can be life-threatening. ARF is diagnosed on the basis of characteristic laboratory findings, such as elevated blood urea nitrogen and creatinine, or inability of the kidneys to produce sufficient amounts of urine.
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This question is part of the following fields:
- Physiology
- Renal
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Question 134
Incorrect
-
Glucose is the most important source of energy for cellular respiration. The transport of glucose in the renal tubular cells occurs via:
Your Answer:
Correct Answer: Secondary active transport with sodium
Explanation:In 1960, Robert K. Crane presented for the first time his discovery of the sodium-glucose cotransport as the mechanism for glucose absorption. Glucose transport through biological membranes requires specific transport proteins. Transport of glucose through the apical membrane of renal tubular as well as intestinal epithelial cells depends on the presence of secondary active Na+–glucose symporters, SGLT-1 and SGLT-2, which concentrate glucose inside the cells, using the energy provided by co-transport of Na+ ions down their electrochemical gradient.
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This question is part of the following fields:
- Physiology
- Renal
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Question 135
Incorrect
-
Different portions of the renal tubule have varying degrees of water permeability. Which of the following renal sites is characterised by low water permeability under normal circumstances?
Your Answer:
Correct Answer: Thick ascending limb of the loop of Henlé
Explanation:Within the nephron of the kidney, the ascending limb of the loop of Henle is a segment of the loop of Henle downstream of the descending limb, after the sharp bend of the loop. Both the thin and the thick ascending limbs of the loop of Henlé have very low permeability to water. Since there are no regulatory mechanisms to alter its permeability, it remains poorly permeable to water under all circumstances. Sodium and chloride are transported out of the luminal fluid into the surrounding interstitial spaces, where they are reabsorbed. Water must remain behind because it is not reabsorbed, so the solute concentration becomes less and less (the luminal fluid becomes more dilute). This is one of the principal mechanisms (along with diminution of ADH secretion) for the production of a dilute, hypo-osmotic urine (water diuresis).
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This question is part of the following fields:
- Physiology
- Renal
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Question 136
Incorrect
-
A 59-year-old woman with hyperaldosteronism is prescribed a diuretic. Which of the following diuretics promotes diuresis by opposing the action of aldosterone?
Your Answer:
Correct Answer: Potassium-sparing diuretic
Explanation:The term potassium-sparing refers to an effect rather than a mechanism or location. Potassium-sparing diuretics act by either antagonising the action of aldosterone (spironolactone) or inhibiting Na+ reabsorption in the distal tubules (amiloride). This group of drugs is often used as adjunctive therapy, in combination with other drugs, for the management of chronic heart failure. Spironolactone, the first member of the class, is also used in the management of hyperaldosteronism (including Conn’s syndrome) and female hirsutism (due to additional antiandrogen actions).
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This question is part of the following fields:
- Physiology
- Renal
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Question 137
Incorrect
-
Which of the following is most likely to cause hypovolaemic hypernatremia:
Your Answer:
Correct Answer: Hyperalimentation
Explanation:Hypernatremia, characterised by a high serum sodium concentration, is rarely associated with volume overload (hypervolemia). A hypovolaemic hypernatremia may be seen during excessive administration of hypertonic sodium bicarbonate, hypertonic saline or hyperalimentation.
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This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
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Question 138
Incorrect
-
A 78-year-old diabetic man undergoes renal function tests. Which of the following substances will be the most accurate for measuring glomerular filtration rate (GFR)?
Your Answer:
Correct Answer: Inulin
Explanation:Glomerular filtration rate (GFR) is the volume of fluid filtered from the renal glomerular capillaries into the Bowman’s capsule per unit time. Clinically, this is often measured to determine renal function. Inulin was originally used as it is not reabsorbed by the kidney after glomerular filtration, therefore its rate of excretion is directly proportional to the rate of filtration of water and solutes across the glomerular filter. However, in clinical practice, creatinine clearance is used to measure GFR. Creatinine is an endogenous molecule, synthesised in the body, that is freely filtered by the glomerulus (but also secreted by the renal tubules in very small amounts). Creatinine clearance exceeds GFR due to creatinine secretion, and is therefore a close approximation of the GFR.
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This question is part of the following fields:
- Physiology
- Renal
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Question 139
Incorrect
-
Risk factors for hepatocellular carcinoma do NOT include:
Your Answer:
Correct Answer: Heptitis E
Explanation:Risk factors for hepatocellular carcinoma include hepatitis B and C, aflatoxin, anabolic steroids, alcohol cirrhosis and primary liver disease.
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This question is part of the following fields:
- Gastrointestinal; Hepatobiliary
- Pathology
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Question 140
Incorrect
-
The anatomical dead space in a patient with low oxygen saturation, is 125 ml, with a tidal volume of 500 ml and pa(CO2) of 40 mm Hg. The dead space was determined by Fowler's method. If we assume that the patient's lungs are healthy, what will his mixed expired CO2 tension [pE(CO2)] be?
Your Answer:
Correct Answer: 30 mmHg
Explanation:According to Bohr’s equation, VD/VT = (pA(CO2) − pE(CO2))/pA(CO2), where pE(CO2) is mixed expired CO2 and pA(CO2) is alveolar CO2pressure. Normally, the pa(CO2) is virtually identical to pA(CO2). Thus, VD/VT = (pa(CO2)) − pE(CO2)/pa(CO2). By Fowler’s method, VD/VT= 0.25. In the given problem, (pa(CO2) − pE(CO2)/pa(CO2) = (40 − pE(CO2)/40 = 0.25. Thus, pE(CO2) = 30 mmHg. If there is a great perfusion/ventilation inequality, pE(CO2) could be significantly lower than 30 mm Hg, and the patient’s physiological dead space would exceed the anatomical dead space.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 141
Incorrect
-
A 15-day old male baby was brought to the emergency department with sweating and his lips turning blue while feeding. He was born full term. On examination, his temperature was 37.9°C, blood pressure 75/45 mmHg, pulse was 175/min, and respiratory rate was 42/min. A harsh systolic ejection murmur could be heard at the left upper sternal border. X-ray chest showed small, boot-shaped heart with decreased pulmonary vascular markings. He most likely has:
Your Answer:
Correct Answer: Tetralogy of Fallot
Explanation:The most common congenital cyanotic heart disease and the most common cause of blue baby syndrome, Tetralogy of Fallot shows four cardiac malformations occurring together. These are ventricular septal defect (VSD), pulmonary stenosis (right ventricular outflow obstruction), overriding aorta (degree of which is variable), and right ventricular hypertrophy. The primary determinant of severity of disease is the degree of pulmonary stenosis. Tetralogy of Fallot is seen in 3-6 per 10,000 births and is responsible for 5-7% congenital heart defects, with slightly higher incidence in males. It has also been associated with chromosome 22 deletions and DiGeorge syndrome. It gives rise to right-to-left shunt leading to poor oxygenation of blood. Primary symptom is low oxygen saturation in the blood with or without cyanosis at birth of within first year of life. Affected children ay develop acute severe cyanosis or ‘tet spells’ (sudden, marked increase in cyanosis, with syncope, and may result in hypoxic brain injury and death). Other symptoms include heart murmur, failure to gain weight, poor development, clubbing, dyspnoea on exertion and polycythaemia. Chest X-ray reveals characteristic coeur-en-sabot (boot-shaped) appearance of the heart. Treatment consists of immediate care for cyanotic spells and Blalock–Taussig shunt (BT shunt) followed by corrective surgery.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 142
Incorrect
-
Increased resistance to flow of blood in cerebral vessels is most likely seen in:
Your Answer:
Correct Answer: Elevation in systemic arterial pressure from 100 to 130 mmHg
Explanation:Constant cerebral blood flow is maintained by autoregulation in the brain, which causes an increase in local vascular resistance to offset an increase in blood pressure. There will be an increase in cerebral blood flow (and decrease in resistance to cerebral blood flow) with a decrease in arterial oxygen or an increase in arterial CO2. Similarly, a decrease in viscosity will also increase the blood flow. Due to increased brain metabolism and activity during a seizure, there will also be an increase in the cerebral blood flow.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 143
Incorrect
-
A 5-year-old child presents with fever and otalgia. Greenish pustular discharge was seen in his left ear during physical examination. The patient is diagnosed with otitis externa. Which of the following organisms most likely caused the infection?
Your Answer:
Correct Answer: Pseudomonas aeruginosa
Explanation:P. aeruginosa is a multidrug resistant pathogen recognised for its ubiquity, its advanced antibiotic resistance mechanisms and its association with serious illnesses – especially hospital-acquired infections such as ventilator-associated pneumonia and various septic syndromes. The species name aeruginosa is a Latin word meaning verdigris (copper rust), referring to the blue-green colour of laboratory cultures of the species. This blue-green pigment is a combination of two metabolites of P. aeruginosa, pyocyanin (blue) and pyoverdine (green), which impart the blue-green characteristic colour of cultures.
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This question is part of the following fields:
- Microbiology
- Pathology
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Question 144
Incorrect
-
When does the heart rate decrease?
Your Answer:
Correct Answer: Pressure on the eyeball
Explanation:Various vagotonic manoeuvres (e.g. Valsalva manoeuvre, carotid sinus massage, pressure on eyeballs, ice-water facial immersion, swallowing of ice-cold water) result in increased parasympathetic tone through the vagus nerve which results in a decrease in heart rate. These manoeuvres may be clinically useful in terminating supraventricular arrhythmias.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 145
Incorrect
-
A 50-year-old man is diagnosed with emphysema and cirrhosis of the liver. Which of the following condition may be the cause of both cirrhosis and emphysema in this patient?
Your Answer:
Correct Answer: Alpha1-antitrypsin deficiency
Explanation:Alpha-1 antitrypsin (A1AT) deficiency is a condition characterised by the lack of a protein that protects the lungs and liver from damage, called alpha1-antytripsin. The main complications of this condition are liver diseases such as cirrhosis and chronic hepatitis, due to accumulation of abnormal alpha 1-antytripsin and emphysema due to loss of the proteolytic protection of the lungs.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 146
Incorrect
-
Under normal conditions, what is the major source of energy of cardiac muscles?
Your Answer:
Correct Answer: Fatty acids
Explanation:Under basal conditions, most of the energy needed by cardiac muscle for metabolism is derived from fats (60%), 35% by carbohydrates, and 5% by ketones and amino acids. However, after intake of large amounts of glucose, lactate and pyruvate are mainly used. During prolonged starvation, fat acts as the primary source. 50% of the used lipids are sourced from circulating fatty acids.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 147
Incorrect
-
During routine laboratory tests, a 66-year-old man is found to be suffering from hypercholesterolaemia and is prescribed atorvastatin. What is the mechanism of action of atorvastatin?
Your Answer:
Correct Answer: Inhibits cholesterol synthesis
Explanation:Atorvastatin is a member of the drug class of statins, used for lowering cholesterol. The mode of action of statins is inhibition of 3-hydroxy-3-methylglutaryl-coenzyme A (HMG-CoA) reductase. This enzyme is needed by the body to make cholesterol. The primary uses of atorvastatin is for the treatment of dyslipidaemia and the prevention of cardiovascular disease.
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This question is part of the following fields:
- Pathology
- Pharmacology
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Question 148
Incorrect
-
What causes a reduction in pulmonary functional residual capacity?
Your Answer:
Correct Answer: Pulmonary fibrosis
Explanation:Pulmonary functional residual capacity (FRC) is = volume of air present in the lungs at the end of passive expiration.
Obstructive diseases (e.g. emphysema, chronic bronchitis, asthma) = an increase in FRC due to an increase in lung compliance and air trapping.
Restrictive diseases (e.g. pulmonary fibrosis) result in stiffer, less compliant lungs and a reduction in FRC.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 149
Incorrect
-
Cancer of the testis most likely metastases to which set of lymph nodes?
Your Answer:
Correct Answer: Aortic
Explanation:The lymphatic drainage of an organ is related to its blood supply. The lymphatic drainage of the testis drains along the testicular artery to reach the lymph nodes along the aorta.
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This question is part of the following fields:
- Endocrine
- Pathology
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Question 150
Incorrect
-
A 27 year old women had developed a darker complexion following a vacation to India. She had no erythema or tenderness. Her skin colour returned to normal over a period of 1 month. Which of the these substances is related to the biochemical change mentioned above?
Your Answer:
Correct Answer: Tyrosine
Explanation:The tanning process can occur due to UV light exposure as a result of oxidation of tyrosine to dihydrophenylalanine with the help of the tyrosinase enzyme within the melanocytes. Hemosiderin can impart a brown colour due to breakdown of RBC but its usually due to a trauma and is known as haemochromatosis.
Lipofuscin gives a golden brown colour to the cell granules not the skin.
Homogentisic acid is part of a rare disease alkaptonuria, with characteristic black pigment deposition within the connective tissue.
Copper can impart a brown golden colour, but is not related to UV light exposure.
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This question is part of the following fields:
- Cell Injury & Wound Healing
- Pathology
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Question 151
Incorrect
-
Normally, the O2 transfer in the lungs from alveolar to capillary is perfusion-limited. In which of the following situations does it become a diffusion-limited process?
Your Answer:
Correct Answer: Pulmonary oedema
Explanation:Normally, the transfer of oxygen from air spaces to blood takes place across the alveolar-capillary membrane by simple diffusion and depends entirely on the amount of blood flow (perfusion-limited process). Diseases that affect this diffusion will transform the normal process to a diffusion limited process. Thus, the diseases which cause a thickened barrier (such as pulmonary oedema due to increased extravascular lung water or asbestosis) will limit the diffusion of oxygen. Chronic obstructive lung diseases will have little effect on diffusion. Inhaling hyperbaric gas mixtures might overcome the diffusion limitation in patients with mild asbestosis or interstitial oedema, by increasing the driving force. Strenuous (not mild) exercise might also favour diffusion limitation and decrease passage time. Increasing the rate of ventilation will not have this affect but will only maintain a high oxygen gradient from air to blood.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 152
Incorrect
-
Work of breathing (WOB) is the energy expended to inhale and exhale a breathing gas. Normally, maximal amount of work of breathing is required to overcome:
Your Answer:
Correct Answer: Elastic lung compliance
Explanation:The forces of elastance (compliance), frictional resistance and inertia have been identified as the forces that oppose lung inflation and deflation. The normal relaxed state of the lung and chest is partially empty. Further exhalation requires muscular work. Inhalation is an active process requiring work. About 60–66% of the total work performed by the respiratory muscles is used to overcome the elastic or compliance characteristics of the lung–chest cage, 30–35% is used to overcome frictional resistance and only 2–5% of the work is used for inertia.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 153
Incorrect
-
A 32-year-old man presents to the doctor complaining of pain in his left calf whilst walking. He says that the pain goes away after a short period of rest but starts again during exercise or walking. The man reveals he has been a smoker for the last 15 years. His blood pressure, blood sugar and cholesterol level are normal. Artery biopsy shows intraluminal thrombosis and vasculitis. What's is the most likely cause of these findings?
Your Answer:
Correct Answer: Buerger's disease
Explanation:Thromboangiitis obliterans, also known as Buerger’s disease, is a rare type of occlusive peripheral arterial disease, usually seen in smokers, most commonly in men aged 20 to 40. Symptoms most often include intermittent claudication, skin changes, painful ulcers on extremities, pain in the extremities during rest and gangrene. Diagnosis is based on clinical findings, arteriography, echocardiography, and computed tomography angiography. A difference in blood pressure between arms, or between the arms and legs is a common finding. Electrocardiographic findings include nonspecific abnormality or normal results.
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This question is part of the following fields:
- Cardiovascular
- Pathology
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Question 154
Incorrect
-
A 65-year-old man complains of headaches, weakness, cramps, and confusion; blood tests reveal he has severe hyponatremia. The most likely cause is:
Your Answer:
Correct Answer: Severe diarrhoea or vomiting
Explanation:Hyponatraemia occurs when the sodium level in the plasma falls below 135 mmol/l. Hyponatraemia is an abnormality that can occur in isolation or, more commonly as a complication of other medical illnesses. Severe hyponatraemia may cause osmotic shift of water from the plasma into the brain cells. Typical symptoms include nausea, vomiting, headache and malaise. As the hyponatraemia worsens, confusion, diminished reflexes, convulsions, stupor or coma may occur. The cause of hyponatremia is typically classified by a person’s fluid status into low volume, normal volume, and high volume. Low volume hyponatremia can occur from diarrhoea, vomiting, diuretics, and sweating.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 155
Incorrect
-
Calculate the cardiac output of a patient with the following measurements: oxygen uptake 200 ml/min, oxygen concentration in the peripheral vein 7 vol%, oxygen concentration in the pulmonary artery 10 vol% and oxygen concentration in the aorta 15 vol%.
Your Answer:
Correct Answer: 4000 ml/min
Explanation:The Fick’s principle states that the uptake of a substance by an organ equals the arteriovenous difference of the substance multiplied by the blood flowing through the organ. We can thus calculate the pulmonary blood flow with pulmonary arterial (i.e., mixed venous) oxygen content, aortic oxygen content and oxygen uptake. The pulmonary blood flow, systemic blood flow and cardiac output can be considered the same assuming there are no intracardiac shunts. Thus, we can calculate the cardiac output. Cardiac output = oxygen uptake/(aortic − mixed venous oxygen content) = 200 ml/min/(15 ml O2/100 ml − 10 ml O2/100 ml) = 200 ml/min/(5 ml O2/100 ml) = 200 ml/min/0.05 = 4000 ml/min.
It is crucial to remember to use pulmonary arterial oxygen content and not peripheral vein oxygen content, when calculating the cardiac output by Fick’s method.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 156
Incorrect
-
An elderly, diabetic man has firm, tender nodules at the base of his left middle and ring fingers, which he can't extend fully. What's the most likely diagnosis?
Your Answer:
Correct Answer: Fibromatosis
Explanation:This case is suggestive of Dupuytren’s contracture due to palmar fibromatosis. Its incidence is higher in men over the age of 45 years, and it increases in patients with diabetes, alcoholism, or epilepsy. These nodules are benign, usually appearing as a tender nodule in the palm which becomes painless. The disease has an aggressive clinical behaviour and recurs frequently.
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This question is part of the following fields:
- Orthopaedics
- Pathology
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Question 157
Incorrect
-
When a patient that is less than 21 years of age develops a bone tumour. What is the most common benign bone tumour that would be considered in individuals below 21 years?
Your Answer:
Correct Answer: Osteochondroma
Explanation:Osteochondroma is a benign new bone growth that protrudes from the outer contour of bones and is capped by growing cartilage. Nearly 80% of these lesions are noted before the age of 21 years.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 158
Incorrect
-
Calculate the stroke volume in an adult male with the following parameters:
Heart rate 70 beats/min
Arterial [O2] 0.24 ml O2/min
Venous [O2] 0.16 ml O2/mi
Whole body O2 consumption 500 ml/min
Pulmonary diastolic pressure 15 mmHg
Pulmonary systolic pressure 25 mmHg
Wedge pressure 5 mmHg.Your Answer:
Correct Answer: 90 ml
Explanation:Fick’s principle states that, VO2 = (CO × CAO2) – (CO × CVO2) where VO2 = oxygen consumption, CO = cardiac output, CAO2 = oxygen concentration of arterial blood and CVO2 = oxygen concentration of venous blood. Thus, CO = VO2/CAO2– CVO2, CO = 500/0.24–0.16, CO = 500/0.8, CO = 6.25 l/min. Cardiac output is also given by product of stroke volume and heart rate. Thus, stroke volume = cardiac output / heart rate = 6.25/70 × 1000 stroke volume = 90 ml approximately.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 159
Incorrect
-
A 40-year old woman presents with tightening of the skin over her fingers which makes movement of her fingers difficult.. She also gives a history of her fingers turning blue on exposure to low temperatures. She admits to gradual weight loss. Investigations reveal negative rheumatoid factor, negative antinuclear antibody and a positive anticentromere body. Which of the following conditions is she likely to have?
Your Answer:
Correct Answer: Oesophageal stricture
Explanation:Scleroderma is a connective tissue disorder that ranges in severity and progression. The disease could show generalised skin thickening with rapid, fatal, visceral involvement; or only cutaneous involvement (typically fingers and face). The slow progressive form is also known as ‘limited cutaneous scleroderma’ or CREST syndrome (calcinosis cutis, Raynaud’s phenomenon, (o)oesophageal dysmotility, sclerodactyly, and telangiectasia).
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This question is part of the following fields:
- Gastrointestinal; Hepatobiliary
- Pathology
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Question 160
Incorrect
-
Which of the following malignancies is likely to have the best prognosis?
Your Answer:
Correct Answer: Papillary carcinoma of the thyroid
Explanation:Papillary carcinoma accounts for 70-80% of all thyroid cancers and is seen commonly in people aged 30-60 years. It is more aggressive in elderly patients. 10-20% cases may have recurrence or persistent disease. More common in females with a female to male ratio of 3:1. Papillary carcinomas can also contain follicular carcinomas. The common route of spread is through lymphatics to regional nodes in one-third cases and pulmonary metastasis can also occur. Papillary carcinomas of the thyroid have the best prognosis, especially in patients less than 45 years of age with small tumours confined to the thyroid gland.
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This question is part of the following fields:
- Endocrine
- Pathology
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Question 161
Incorrect
-
A 49-year-old woman with acute renal failure has a total plasma [Ca2+] = 2. 5 mmol/l and a glomerular filtration rate of 160 l/day. What is the estimated daily filtered load of calcium?
Your Answer:
Correct Answer: 240 mmol/day
Explanation:Calcium is the most abundant mineral in the human body. The average adult body contains in total approximately 1 kg of calcium of which 99% is in the skeleton in the form of calcium phosphate salts. The extracellular fluid (ECF) contains approximately 22 mmol, of which about 9 mmol is in the plasma. About 40% of total plasma Ca2+ is bound to proteins and not filtered at the glomerular basement membrane. Therefore, the estimated daily filtered load is 1.5 mmol/l × 160 l/day = 240 mmol/day. The exact amount of free versus total Ca2+ depends on the blood pH: free Ca2+ increases during acidosis and decreases during alkalosis.
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This question is part of the following fields:
- Physiology
- Renal
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Question 162
Incorrect
-
A 68-year-old man underwent sigmoid resection with diverting colostomy for a ruptured sigmoid diverticulum 10 days ago. He received gentamicin and ampicillin post-op. 2 days after he was discharged from the hospital, he was readmitted because of high grade fever and chills. His blood culture grew Gram-negative bacilli. Which organism is most likely responsible for the patient's infection?
Your Answer:
Correct Answer: Bacteroides fragilis
Explanation:Bacteroides fragilis is an anaerobic, Gram-negative, rod-shaped bacterium. It is part of the normal flora of the human colon and is generally a commensal, but can cause infection if displaced into the bloodstream or surrounding tissue following surgery, disease, or trauma.
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This question is part of the following fields:
- Microbiology
- Pathology
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Question 163
Incorrect
-
A 63-year old lady presents with pain in multiple bones and renal failure. On enquiry, there is history of recurrent pneumonia in the past. What will be the likely finding on her bone marrow biopsy?
Your Answer:
Correct Answer: Plasma cells
Explanation:Multiple myeloma is a plasma cell malignancy that produce excessive monoclonal immunoglobulins. The disease presents with bone pains, renal dysfunction, increased calcium, anaemia and recurrent infections. Diagnosis is by demonstrating the presence of M-protein in urine or serum, lytic bone lesions, light chain proteinuria or excessive plasma cells on marrow biopsy.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 164
Incorrect
-
Following a bee sting, a women develops a 2cm red, raised, swollen lesion at the site of the sting . Which of the following findings is likely to be seen in this lesion?
Your Answer:
Correct Answer: Vasodilation
Explanation:Inflammation is the immediate response of the body towards infections or irritations. The cardinal signs of inflammation are 1. redness/rubor, 2. tumour/swelling, 3.dolar/pain, 4.calor/heat and organ dysfunction (function laesa). Inflammation has 2 components; vascular and cellular. Blood vessels dilate upstream of the inflamed area leading to the rubor and calor and constrict downstream, increasing pressure and causing fluid to leak out of the capillary, resulting in swelling. The cellular component includes infiltration by neutrophils. Leukocyte arrival and functions include; 1. margination: cells marginated from the centre to the periphery of the vessel, 2. rolling: selectins are upregulated on the vessel walls, 3. adhesion: upregulation of the adhesion molecules ICAM and VCAM on the endothelium interact with integrins on the leukocytes resulting in adhesion, 4. diapedesis and chemotaxis: diapedesis is the transmigration of the leukocyte across the endothelium of the capillary and towards a chemotactic product and 5. phagocytosis: engulfing the offending substance/cell.
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This question is part of the following fields:
- Inflammation & Immunology
- Pathology
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Question 165
Incorrect
-
Which of the following is the most accurate test for the diagnosis of primary syphilis?
Your Answer:
Correct Answer: Dark-field microscopy
Explanation:Primary syphilis is transmitted via sexual contact. Lesions on genitalia, called a chancre occur after an asymptomatic incubation period of 10-90 days (average 21 days) after exposure. This chancre is a typically solitary (can be multiple), firm, painless, ulceration over the skin at the point of exposure to spirochete, seen on penis, vagina or rectum. It heals spontaneously after 4-6 weeks. Local lymphadenopathy can be seen.
Diagnosis is made by microscopy of fluid from lesion using dark-field illumination, taking care to not confuse with other treponemal disease. Screening tests include rapid plasma regain (RPR) and Venereal Diseases Research Laboratory (VDRL) tests. False positives are known to occur with these tests and can be seen in viral infections like hepatitis, varicella, Epstein-Barr virus, tuberculosis, lymphoma, pregnancy and IV drug use. More specific tests should therefore be carried out in case these screening tests are positive.
The Treponema pallidum hemagglutination assay (TPHA) and the fluorescent treponemal antibody absorption (FTAABS) test are based on monoclonal antibodies and immunofluorescence and are more specific. However, they can too show false positives with other treponemal diseases like yaws or pinta. Other confirmatory tests include those based on enzyme-linked immunoassays.
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This question is part of the following fields:
- Pathology
- Urology
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Question 166
Incorrect
-
What occurs during cellular atrophy?
Your Answer:
Correct Answer: Cell size decreases
Explanation:Atrophy is the decrease in the size of cells, tissues, or organs. There are several causes including inadequate nutrition, poor circulation, loss of hormonal support or nerve supply, disuse, lack of exercise, or disease. An increase in cell size is termed hypertrophy which is distinguished from hyperplasia, in which the cells remain approximately the same size but increase in number.
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This question is part of the following fields:
- Cell Injury & Wound Healing; Urology
- Pathology
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Question 167
Incorrect
-
A 49-year-old man, smoker, complains of a persisting and worsening cough over the past few months. He also has noted blood in his sputum. The patient has no other major health conditions. Which of the following investigative procedures should be done first?
Your Answer:
Correct Answer: Sputum cytology
Explanation:Sputum cytology is a diagnostic test used for the examination of sputum under a microscope to determine if abnormal cells are present. It may be used as the first diagnostic procedure to help detect a suspected lung cancer or certain non-cancerous lung conditions.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 168
Incorrect
-
A 29-year-old pregnant woman suffering from hyperemesis gravidarum is prescribed metoclopramide. What is the mechanism of action of metoclopramide?
Your Answer:
Correct Answer: Dopamine antagonist
Explanation:Metoclopramide is a potent dopamine-receptor antagonist with anti-emetic and prokinetic properties. It is therefore commonly used to treat nausea and vomiting, and to facilitate gastric emptying in patients with gastric stasis. The anti-emetic action of metoclopramide is due to its antagonist activity at D2 receptors in the chemoreceptor trigger zone (CTZ) in the central nervous system. Common adverse drug reactions associated with metoclopramide include restlessness (akathisia), and focal dystonia.
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This question is part of the following fields:
- Pathology
- Pharmacology
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Question 169
Incorrect
-
A 60-year-old male who was admitted due to cerebrovascular disease on his 5th hospital stay developed pneumonia. The most likely organism that causes hospital acquired pneumonia is pseudomonas aeruginosa. What is the most likely mechanism for the pathogenesis on pseudomonas infection?
Your Answer:
Correct Answer: Exotoxin
Explanation:Pseudomonas aeruginosa is a common Gram-negative, rod-shaped bacterium that can cause disease in plants and animals, including humans. It is citrate, catalase, and oxidase positive. P. aeruginosa uses the virulence factor exotoxin A to inactivate eukaryotic elongation factor 2 via ADP-ribosylation in the host cell, much the same as the diphtheria toxin does. Without elongation factor 2, eukaryotic cells cannot synthesize proteins and necrotise. The release of intracellular contents induces an immunologic response in immunocompetent patients.
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This question is part of the following fields:
- Microbiology
- Pathology
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Question 170
Incorrect
-
There are several mechanisms involved in the transport of sodium ions from blood to interstitial fluid of the muscle cells. Which of the following mechanisms best describes this phenomenon?
Your Answer:
Correct Answer: Diffusion through channels between endothelial cells
Explanation:Capillaries are the smallest of the body’s blood vessels, measuring 5–10 μm and they help to enable the exchange of water, oxygen, carbon dioxide, and many other nutrients and waste substances between the blood and the tissues surrounding them. The walls of capillaries are composed of only a single layer of cells, the endothelium. Ion channels are pore-forming proteins that help to establish and control the small voltage gradient that exists across the plasma membrane of all living cells by allowing the flow of ions down their electrochemical gradient. An ion channel is an integral membrane protein or more typically an assembly of several proteins. The archetypal channel pore is just one or two atoms wide at its narrowest point. It conducts a specific ion such as sodium or potassium and conveys them through the membrane in single file.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 171
Incorrect
-
A 56-year-old woman weighs 75 kg. In this patient, total body water, intracellular fluid and extracellular fluid are respectively:
Your Answer:
Correct Answer: 45 l, 30 l, 15 l
Explanation:The percentages of body water contained in various fluid compartments add up to total body water (TBW). This water makes up a significant fraction of the human body, both by weight and by volume. The total body water (TBW) content of humans is approximately 60% of body weight. Two-thirds is located in the intracellular and one-third in the extracellular compartment. So, in a 75-kg individual, TBW = 60 × 75/100 = 45 l. Intracellular content = 2/3 × 45 = 30 l and extracellular content = 1/3 × 45 = 15 l.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 172
Incorrect
-
A 5 year-old-child with fever complains of sore throat . She was brought to her paediatrician for consult because she has also developed a rash and has swollen lymph nodes. Upon physical examination she cried when her liver was palpated and the tip of her spleen is slightly palpable. Full blood count shows haemoglobin 13 g/dL, Haematocrit 40%, white blood cell count 13x109/L with a WBC differential count of 45 neutrophils, 4 bands, 26 lymphocytes, 15 atypical lymphocytesm, 10 monocytes and 1 eosinophil. Whick is the most likely infectious agent that is responsible for the patient's condition?
Your Answer:
Correct Answer: Epstein–Barr virus
Explanation:Epstein-Barr virus (EBV), also known as human herpes virus 4,is a member of the herpes virus family. EBV spreads most commonly through bodily fluids, primarily saliva. EBV can cause infectious mononucleosis. Symptoms of EBV can include fatigue, fever, inflamed throat, swollen lymph nodes in the neck, enlarged spleen, swollen liver and rash.
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This question is part of the following fields:
- Microbiology
- Pathology
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Question 173
Incorrect
-
A 62-year-old man presented with a persistent cough and weight loss. Chest x-ray demonstrated widespread nodular opacities. After a bronchoalveolar lavage, atypical cells were detected. Which is the most probable diagnosis?
Your Answer:
Correct Answer: Bronchioalveolar carcinoma
Explanation:Bronchioloalveolar carcinoma (BAC) is a term used to define a particular subtype of adenocarcinoma which develops in cells near the alveoli, in the outer regions of the lungs. On a chest X-ray it can appear as a single peripheral spot or as scattered spots throughout the lungs. Symptoms include cough, haemoptysis, chest pain, dyspnoea and loss of weight.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 174
Incorrect
-
A teenager presents with pain and swelling in a limb which increases after activity. X-rays reveal an expansible, eccentric, lytic lesion in the metaphysis distally in the affected bone surrounded by new bone. What is the most likely diagnosis?
Your Answer:
Correct Answer: Aneurysmal bone cyst
Explanation:Aneurysmal bone cysts tend to develop in patients younger than 20 years old. They usually occur in the metaphyseal region of long bones, and are cystic lesions composed of numerous blood filled channels that grow slowly. In X-rays, they show up as circumscribed lesions, sometimes surrounded by new bone.
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This question is part of the following fields:
- Orthopaedics
- Pathology
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Question 175
Incorrect
-
Intravenous diazepam was administered to a man who was brought to the emergency department with status epilepticus. He was administered 15 l/min oxygen via a reservoir bag mask. Blood investigations showed sodium = 140 mmol/l, potassium = 4 mmol/l and chloride = 98 mmol/l. His arterial blood gas analysis revealed pH 7.08, p(CO2)= 61.5 mmHg, p(O2) = 111 mmHg and standard bicarbonate = 17 mmol/l. This patient had:
Your Answer:
Correct Answer: Mixed acidosis
Explanation:Acidosis with high p(CO2) and low standard bicarbonate indicates mixed acidosis. Lower p(O2) is due to breathing of 70% oxygen. The prolonged seizures lead to lactic acidosis and the intravenous diazepam is responsible for the respiratory acidosis. Treatment includes airway manoeuvres and oxygen, assisted ventilation if needed, and treatment with fluids.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 176
Incorrect
-
Which is a feature of the action of insulin?
Your Answer:
Correct Answer: Promotes protein synthesis
Explanation:Insulin is produced by the beta-cells of the islets of Langerhans in the pancreas. Its actions include:
– promoting uptake of glucose into cells
– glycogen synthesis (glycogenesis)
– protein synthesis
– stimulation of lipogenesis (fat formation).
– driving potassium into cells – used to treat hyperkaelamia.
Parathyroid hormone and activated vitamin D are the principal hormones involved in calcium/phosphate metabolism, rather than insulin.
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This question is part of the following fields:
- Endocrine
- Physiology
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Question 177
Incorrect
-
Question 178
Incorrect
-
Which of the following diseases is caused by intra-articular and/or extra-articular deposition of calcium pyrophosphate dihydrate (CPPD) crystals, due to unknown causes?
Your Answer:
Correct Answer: Pseudogout
Explanation:Pseudogout or chondrocalcinosis is a rheumatological disease caused by the accumulation of crystals of calcium pyrophosphate dihydrate (CPPD) in the connective tissues. It is frequently associated with other conditions, such as trauma, amyloidosis, gout, hyperparathyroidism and old age, which suggests that it is secondary to degenerative or metabolic changes in the tissues. The knee is the most commonly affected joint. It causes symptoms similar to those of rheumatoid arthritis or osteoarthritis.
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This question is part of the following fields:
- Orthopaedics
- Pathology
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Question 179
Incorrect
-
A 37-year-old woman with a history of rheumatic heart disease presents with 10 days recurrent low fever. Patient underwent laboratory work up and was diagnosed with infective endocarditis. What is the most likely organism that caused the infective endocarditis in this patient?
Your Answer:
Correct Answer: Streptococcus viridans
Explanation:Subacute bacterial endocarditis is often due to streptococci of low virulence, mainly streptococcus viridans. It is a mild to moderate illness which progresses slowly over weeks and months (>2weeks) and has low propensity to hematogenously seed to extracardiac sites.
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This question is part of the following fields:
- Microbiology
- Pathology
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Question 180
Incorrect
-
Calculate the cardiac output in an adult male with the following parameters:
Heart rate 70 beats/min
Arterial [O2] 0.24 ml O2/min
Venous [O2] 0.16 ml O2/mi
Whole body O2 consumption 500 ml/min
Pulmonary diastolic pressure 15 mmHg
Pulmonary systolic pressure 25 mmHg
Wedge pressure 5 mmHg.Your Answer:
Correct Answer: 6.25 l/min
Explanation:As per Fick’s principle, VO2 = (CO × CAO2) – (CO × CVO2) where VO2 = oxygen consumption, CO = cardiac output, CAO2 = oxygen concentration of arterial blood and CVO2 = oxygen concentration of venous blood. Thus, CO = VO2/CAO2– CVO2, CO = 500/0.24 – 0.16, CO = 500/0.8, CO = 6.25 l/min.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 181
Incorrect
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In which of the following conditions will the oxygen-haemoglobin dissociation curve shift to the right?
Your Answer:
Correct Answer: Exercise
Explanation:The oxygen-haemoglobin dissociation curve plots saturated haemoglobin against the oxygen tension and is usually a sigmoid plot. Each molecule of haemoglobin can bind to four molecules of oxygen reversibly. Factors that can influence the binding include: pH, concentration of 2,3-diphosphoglycerate (2,3-DPG), temperature, type of haemoglobin molecules, and presence of toxins, especially carbon monoxide. Shape of the curve is due to interaction of bound oxygen molecules with the incoming molecules. The binding of first molecule is difficult, with easier binding of the second and third molecule and increase in difficulty with the fourth molecule – partly as a result of crowding and partly as a natural tendency of oxygen to dissociate.
Left shift of curve indicates haemoglobin’s increased affinity for oxygen (seen at lungs). Right shift indicates decreased affinity and is seen with an increase in body temperature, hydrogen ions, 2,3-diphosphoglycerate (DPG), carbon dioxide concentration and exercise. Under normal resting conditions in a healthy individual, the normal position of the curve is at a pH of 7.4. A shift in the position of the curve with a change in pH is called the Bohr effect. Left shift occurs in acute alkalosis, decrease in p(CO2), decrease in temperature and decrease in 2,3-DPG. The fetal haemoglobin curve is to the left of the adult haemoglobin to allow for oxygen diffusion across the placenta. The curve for myoglobin is even further to the left. Carbon monoxide has a much higher affinity for haemoglobin than oxygen does. Thus, carbon monoxide poisoning leads to hypoxia.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 182
Incorrect
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What is the normal amount of oxygen that is carried in the blood?
Your Answer:
Correct Answer: 20 ml oxygen/100 ml blood
Explanation:Normally, 100 ml of blood contains 15g haemoglobin and a single gram of haemoglobin can bind to 1.34 ml oxygen when 100% saturated. Thus, 15 × 1.34 = 20 ml O2/100 ml blood. The haemoglobin in venous blood that is leaving the tissues is about 75% saturated with oxygen, and hence it carries about 15 ml O2/100 ml venous blood. This implies that for each 10 ml of blood, 5 ml oxygen is transported to the tissues. With a p(O2) > 100 mm Hg, only 3 ml of oxygen is dissolved in every one litre of plasma. By increasing the pA(O2) by breathing 100% oxygen, one can add an extra amount of oxygen in the plasma, but the amount of oxygen carried by haemoglobin will not increase significantly as it is already > 95% saturated.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 183
Incorrect
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A 58-year-old woman has had a headache, fever, lethargy and nausea for the last 10 days. He undergoes a CT scan which reveals a lesion in his frontal lobe, which, after a biopsy, is found to be formed by granulation tissue with collagenisation, gliosis and oedema. What's the most likely diagnosis?
Your Answer:
Correct Answer: Chronic brain abscess
Explanation:A cerebral abscess can result from direct extension of cranial infections, penetrating head trauma, haematogenous spread, or for unknown causes. An abscess forms when an area of cerebral inflammation becomes necrotic and encapsulated by glial cells and fibroblasts. Oedema around the abscess can increase the intracranial pressure. Symptoms result from increased intracranial pressure and mass effects. In a CT scan, an abscess appears as an oedematous mass with ring enhancement.
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This question is part of the following fields:
- Neurology
- Pathology
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Question 184
Incorrect
-
Which of these substances is secreted by pericytes in the juxtaglomerular cells?
Your Answer:
Correct Answer: Renin
Explanation:The juxtaglomerular cells synthesise, store and secrete the enzyme renin in the kidney. They are specialised smooth muscle cells in the wall of the afferent arteriole that delivers blood to the glomerulus and thus play a critical role in the renin– angiotensin system and so in renal autoregulation.
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This question is part of the following fields:
- Physiology
- Renal
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Question 185
Incorrect
-
Which of the following accumulates within a cell due to the aging process?
Your Answer:
Correct Answer: Lipofuscin
Explanation:Lipofuscin , also known as lipochrome, is a wear and tear pigment or an aging pigment. It represents free radical injury or lipid peroxidation. On microscopic examination is appears as a yellowish brown pigment around the nucleus (perinuclear pigment). It is often seen in cells which are undergoing regressive changes, commonly in the liver and heart of old patients or patients with cancer, cachexia or severe malnutrition.
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This question is part of the following fields:
- Cell Injury & Wound Healing
- Pathology
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Question 186
Incorrect
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Which of the following abnormalities can be seen in patients with hypermagnesemia?
Your Answer:
Correct Answer: Respiratory depression
Explanation:Hypermagnesemia is an electrolyte disturbance in which there is a high level of magnesium in the blood. It is defined as a level greater than 1.1 mmol/L. Symptoms include weakness, confusion, decreased breathing rate, and cardiac arrest.
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This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
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Question 187
Incorrect
-
Langhans giant cells are characteristically seen in which type of inflammation?
Your Answer:
Correct Answer: Granulomatous inflammation
Explanation:Langhans giant cells are characteristically seen in granulomatous inflammation. They form when epithelioid cells fuse together. They usually contain a nuclei with a horseshoe-shaped pattern in the periphery of the cell.
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This question is part of the following fields:
- Inflammation & Immunology
- Pathology
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Question 188
Incorrect
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Which of the following terms best describes the movement of leukocytes towards a specific target?
Your Answer:
Correct Answer: Chemotaxis
Explanation:The movement of leukocytes towards a chemical mediator is termed chemotaxis and the mediators likewise called chemoattractants.
Diapedesis is the squeezing of the leukocytes from the capillary wall into the intercellular space.
Endocytosis is engulfing of a small substance by the cells e.g. glucose, protein, fats.
Margination is lining of the WBC along the periphery of the blood vessel.
Adhesion is attachment with the vessel wall.
Phagocytosis is described as engulfing the bacteria or the offending substance.
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This question is part of the following fields:
- Inflammation & Immunology
- Pathology
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Question 189
Incorrect
-
A 45-year old farmer was referred to the surgical clinic with complaints of pain in his right hypochondrium. Investigations confirmed the diagnosis of hepatocellular carcinoma with malignant ascites. According to you, what is the most likely cause of HCC in this patient?
Your Answer:
Correct Answer: Aflatoxin
Explanation:Aflatoxins are naturally occurring toxins produced by the Aspergillus fungus (most often, A. flavus and A. parasiticus). These organisms are common and their native habitat is soil, decaying vegetation and grains. They can contaminate the grain before harvest or after storage, more likely in high-humidity (at least 7%) or high temperature environment of after stressful conditions like drought. Aflatoxins are mycotoxins and also carcinogenic. They get metabolized in the liver to an epoxide, aflatoxin M1. High exposure can lead to acute necrosis, cirrhosis or liver carcinoma. These substances can cause haemorrhage, acute liver damage, oedema, and alteration in digestion, absorption and/or metabolism of nutrients. Although humans are susceptible to these toxins like all other animals, they have a high tolerance level and hence, rarely develop acute aflatoxicosis. However, children are particularly susceptible to exposure leading to growth impairment and delayed development. Chronic exposure carries a high risk of hepatic cancer, due to intercalation of its metabolite aflatoxin M1 into the DNA and alkylation of the bases because of its epoxide moiety.
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This question is part of the following fields:
- Gastrointestinal; Hepatobiliary
- Pathology
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Question 190
Incorrect
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Which of the following is the cause of flattened (notched) T waves on electrocardiogram (ECG)?
Your Answer:
Correct Answer: Hypokalaemia
Explanation:The T-wave is formed due to ventricular repolarisation. Normally, it is seen as a positive wave. It can be normally inverted (negative) in V1 (occasionally in V2-3 in African-Americans/Afro-Caribbeans). Hyperacute T-waves are the earliest ECG change of acute myocardial infarction. ECG findings of hyperkalaemia include high, tent-shaped T-waves, a small P-wave and a wide QRS complex. Hypokalaemia results in flattened (notched) T-waves, U-waves, ST-segment depression and prolonged QT interval.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 191
Incorrect
-
Skin infiltration by neoplastic T lymphocytes is seen in:
Your Answer:
Correct Answer: Mycosis fungoides
Explanation:Mycosis fungoides is a chronic T-cell lymphoma that involves the skin and less commonly, the internal organs such as nodes, liver, spleen and lungs. It is usually diagnosed in patients above 50 years and the average life expectancy is 7-10 years. It is insidious in onset and presents as a chronic, itchy rash, eventually spreading to involve most of the skin. Lesions are commonly plaque-like, but can be nodular or ulcerated. Symptoms include fever, night sweats and weight loss. Skin biopsy is diagnostic. However, early cases may pose a challenge due to fewer lymphoma cells. The malignant cells are mature T cells (T4+, T11+, T12+). The epidermis shows presence of characteristic Pautrier’s micro abscesses are present in the epidermis.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 192
Incorrect
-
A 35-year-old ultra marathon runner becomes severely dehydrated and collapses. This patient most likely has:
Your Answer:
Correct Answer: Decreased baroreceptor firing rate
Explanation:Baroreceptors are sensors located in the blood vessels of all vertebrate animals. They sense the blood pressure and relay the information to the brain, so that a proper blood pressure can be maintained. Acute dehydration results in decreased plasma volume and increased plasma osmolarity, since more water than salt is lost in sweat. The decrease in plasma volume leads to an inhibition of the baroreceptors and a lower firing rate. The increase in plasma osmolarity leads to increased ADH secretion and high plasma ADH levels, which increases water permeability of collecting duct cells. Therefore more water is reabsorbed by the kidneys and renal water excretion is low.
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This question is part of the following fields:
- Physiology
- Renal
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Question 193
Incorrect
-
A 35-year-old woman in her 37th week of pregnancy complains of urinary incontinence. She is most likely to have:
Your Answer:
Correct Answer: Stress incontinence
Explanation:Urinary incontinence is the involuntary excretion of urine from one’s body. It is often temporary and it almost always results from an underlying medical condition. Several types include:
– Stress incontinence is the voiding of urine following increased abdominal pressure e.g. laughing, coughing, pregnancy etc. It is the most common form of incontinence in women, most commonly due to pelvic floor muscle weakness, physical changes from pregnancy, childbirth and menopause. In men it is a common problem following a prostatectomy. Most lab results such as urine analysis, cystometry and postvoid residual volume are normal.
– Urge incontinence is involuntary loss of urine occurring for no apparent reason while suddenly feeling the need or urge to urinate. The most common cause of urge incontinence are involuntary and inappropriate detrusor muscle contractions.
– Functional incontinence – occurs when a person does not recognise the need to go to the toilet, recognise where the toilet is or get to the toilet in time. The urine loss may be large. Causes of functional incontinence include confusion, dementia, poor eyesight, poor mobility, poor dexterity or unwillingness. t
– Overflow incontinence – sometimes people find that they cannot stop their bladders from constantly dribbling or continuing to dribble for some time after they have passed urine.
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This question is part of the following fields:
- Physiology
- Renal
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Question 194
Incorrect
-
Calculate the pulmonary vascular resistance in an adult male with the following parameters:
Heart rate 70 beats/min
Arterial [O2] 0.24 ml O2/min
Venous [O2] 0.16 ml O2/mi
Whole body O2 consumption 500 ml/min
Pulmonary diastolic pressure 15 mmHg
Pulmonary systolic pressure 25 mmHg
Wedge pressure 5 mmHg.Your Answer:
Correct Answer: 2.0 resistance units (mmHg/l per min)
Explanation:Pulmonary vascular resistance (PVR) = (Mean pulmonary artery pressure – Pulmonary capillary wedge pressure) divided by Cardiac output. To get cardiac output, Fick’s principle needs to be applied which states that VO2 = (CO × CAO2) – (CO × CVO2) where VO2 = oxygen consumption, CO = cardiac output, CA = oxygen concentration of arterial blood and CVO2 = oxygen concentration of venous blood. Thus, CO = VO2/CAO2– CVO2, CO = 500/0.24–0.16, CO = 500/0.8, CO = 6.25 l/min. To calculate mean pulmonary artery pressure, we use the formula: Mean pulmonary artery pressure = Diastolic pressure + 1/3(Systolic pressure – Diastolic pressure). Thus, Mean pulmonary artery pressure = 15 + 1/3(25 – 15) = 15 + 3. 33 = 18.33. Substituting these values in the first formula, PVR = 18.3–5/6.25 = 13.5/5.25 = 2.0 resistance units (mmHg/l per min) approximately.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 195
Incorrect
-
A 50-year-old woman goes to the doctor complaining of myalgia, muscle cramps, and weakness; she is diagnosed with severe hypokalaemia. Which of the following is the most common cause of hypokalaemia?
Your Answer:
Correct Answer: Prolonged vomiting
Explanation:Potassium is one of the body’s major ions. Nearly 98% of the body’s potassium is intracellular. The ratio of intracellular to extracellular potassium is important in determining the cellular membrane potential. Small changes in the extracellular potassium level can have profound effects on the function of the cardiovascular and neuromuscular systems. Hypokalaemia may result from conditions as varied as renal or gastrointestinal (GI) losses, inadequate diet, transcellular shift (movement of potassium from serum into cells) and medications. The important causes of hypokalaemia are:
Renal losses: renal tubular acidosis, hyperaldosteronism, magnesium depletion, leukaemia (mechanism uncertain).
GI losses: vomiting or nasogastric suctioning, diarrhoea, enemas or laxative use, ileal loop.
Medication effects: diuretics (most common cause), β-adrenergic agonists, steroids, theophylline, aminoglycosides.
Transcellular shift: insulin, alkalosis.
Severe hypokalaemia, with serum potassium concentrations of 2.5–3 meq/l, may cause muscle weakness, myalgia, tremor, muscle cramps and constipation.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 196
Incorrect
-
A 30-year-old woman feels thirsty. This thirst is probably due to:
Your Answer:
Correct Answer: Increased level of angiotensin II
Explanation:Thirst is the basic need or instinct to drink. It arises from a lack of fluids and/or an increase in the concentration of certain osmolites such as salt. If the water volume of the body falls below a certain threshold or the osmolite concentration becomes too high, the brain signals thirst. Excessive thirst, known as polydipsia, along with excessive urination, known as polyuria, may be an indication of diabetes. Angiotensin II is a hormone that is a powerful dipsogen (i.e. it stimulates thirst) that acts via the subfornical organ. It increases secretion of ADH in the posterior pituitary and secretion of ACTH in the anterior pituitary.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 197
Incorrect
-
The renal tubule is the portion of the nephron that contains the fluid that has been filtered by the glomerulus. Which of the following substances is actively secreted into the renal tubules?
Your Answer:
Correct Answer: Potassium
Explanation:The renal corpuscle filters out solutes from the blood, delivering water and small solutes to the renal tubule for modification. In normal circumstances more than 90% of the filtered load of K is reabsorbed by the proximal tubules and loops of Henlé and almost all K appearing in the urine has been secreted by the late distal tubules and collecting tubules. So the rate of excretion is usually independent of the rate of filtration, but is closely tied to the rate of secretion and control of K excretion, largely accomplished by control of the secretion rate. Around 65–70% of the filtered potassium is reabsorbed along with water in the proximal tubule and the concentration of potassium in the tubular fluid varies little from that of the plasma.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 198
Incorrect
-
Which of the following key features will be seen in an organ undergoing atrophy?
Your Answer:
Correct Answer: A greater number of autophagic vacuoles
Explanation:Atrophy is characterised by the breakdown of intracellular components along with organelles and packing them into vacuoles known as autophagic vacuoles. This is an adaptive response that separates the damaged cellular structures from the rest of the cells.
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This question is part of the following fields:
- Cell Injury & Wound Healing
- Pathology
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Question 199
Incorrect
-
If the blood flow is constant, oxygen extraction by tissues will show the greatest decrease due to which of the following interventions?
Your Answer:
Correct Answer: Tissue cooling
Explanation:With a constant blood flow to a given tissue bed, there will be an increase in oxygen extraction by the tissue with the following; an increase in tissue metabolism and oxygen requirements: warming (or fever), exercise, catecholamines and thyroxine. With cooling, the demand for oxygen decreases, leading to decreased oxygen extraction.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 200
Incorrect
-
A 36-year-old female was advised to undergo genetic testing of BRCA1 and BRCA2. BRCA1 and BRCA2 are tumour markers specifically for which organ?
Your Answer:
Correct Answer: Breast
Explanation:Women with harmful mutations in either BRCA1 or BRCA2 have a risk of breast cancer that is about five times the normal risk.
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This question is part of the following fields:
- Neoplasia
- Pathology
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SESSION STATS - PERFORMANCE PER SPECIALTY
