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Question 1
Correct
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In which of the following conditions will the oxygen-haemoglobin dissociation curve shift to the right?
Your 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 2
Correct
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Chronic obstructive pulmonary disease (COPD) is likely to result in:
Your Answer: Respiratory acidosis
Explanation:COPD leads to respiratory acidosis (chronic). This occurs due to hypoventilation which involves multiple causes, such as poor responsiveness to hypoxia and hypercapnia, increased ventilation/perfusion mismatch leading to increased dead space ventilation and decreased diaphragm function.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 3
Incorrect
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Which of the following key features will be seen in an organ undergoing atrophy?
Your Answer: A greater number of myofilaments
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 4
Correct
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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: 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 5
Incorrect
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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: 10 mmHg
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 6
Correct
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What is the normal duration of the ST segment?
Your 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 7
Correct
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What is the normal duration of PR interval on an electrocardiogram of a healthy individual?
Your 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 8
Correct
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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: 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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Question 9
Correct
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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: 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 10
Incorrect
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A child presents with hypertension. Serum potassium analysis shows hypokalaemia. What is the most likely diagnosis?
Your Answer: Bartter Syndrome
Correct Answer: Liddle syndrome
Explanation:Liddle’s syndrome, is an autosomal dominant disorder, that is characterized by early, and frequently severe, high blood pressure associated with low plasma renin activity, metabolic alkalosis, low blood potassium, and normal to low levels of aldosterone. Liddle syndrome involves abnormal kidney function, with excess reabsorption of sodium and loss of potassium from the renal tubule. Bartter Syndrome also presents with hypokalaemia, however blood pressure of these patients is usually low or normal.
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This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
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Question 11
Correct
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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: 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 12
Correct
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Which organ is most vulnerable to haemorrhagic shock?
Your Answer: Kidneys
Explanation:At rest, the brain receives 15% cardiac output, muscles 15%, gastrointestinal tract 30% and kidneys receive 20%. However, if normalised by weight, the largest specific blood flow is received by the kidneys at rest (400 ml/min x 100g), making them highly vulnerable in the case of a haemorrhagic shock.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 13
Correct
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What is the normal glomerular filtration rate?
Your 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 14
Correct
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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: 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 15
Correct
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Cyclophosphamide is used as a chemotherapy and immunosuppressant agent and is indicated in various diseases. One of the most severe complications of its use is cancer of the:
Your Answer: Urinary bladder
Explanation:Cyclophosphamide is used to treat various types of cancer and autoimmune disorders. The main use of cyclophosphamide is in combination with other chemotherapy agents in the treatment of lymphomas, some forms of leukaemia and some solid tumours. Side-effects include nausea and vomiting, bone marrow suppression, stomach ache, diarrhoea, darkening of the skin/nails, alopecia, lethargy, and haemorrhagic cystitis. Cyclophosphamide is itself carcinogenic, potentially causing transitional cell carcinoma of the bladder as a long-term complication.
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This question is part of the following fields:
- Pathology
- Pharmacology
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Question 16
Correct
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Low molecular weight heparin (LMWH) has less side effects than heparin and is used in the prophylaxis and treatment of venous and arterial thrombotic disorders. Which of the following is LMWHs mechanism of action?
Your Answer: Inhibition of factor Xa
Explanation:Low molecular weight heparin (LMWH) is a anticoagulant that differs from normal heparin in that it has only short chains of polysaccharide. LMWH inhibits thrombin formation by converting antithrombin from a slow to a rapid inactivator of coagulation factor Xa.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 17
Correct
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What is the most likely cause of bitemporal hemianopia?
Your 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 18
Correct
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A 31-year-old woman is diagnosed with adrenal hyperplasia, and laboratory samples are taken to measure serum aldosterone and another substance. Which is most likely to be the other test that was prescribed to this patient?
Your Answer: Plasma renin
Explanation:The evaluation of a patient in whom hyperaldosteronism is first to determine that hyperaldosteronism is present (serum aldosterone) and, if it is present, to differentiate primary from secondary causes of hyperaldosteronism. The aldosterone-to-renin ratio (ARR) is the most sensitive means of differentiating primary from secondary causes of hyperaldosteronism as it is abnormally increased in primary hyperaldosteronism, and decreased or normal but with high renin levels in secondary hyperaldosteronism.
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This question is part of the following fields:
- Physiology
- Renal
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Question 19
Correct
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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: 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 20
Correct
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A 15 month old boy has a history of repeated bacterial pneumonia, failure to thrive and a sputum culture positive for H.influenzea and S.pneumoniae. There is no history of congenital anomalies. He is most likely suffering from?
Your Answer: X-linked agammaglobulinemia
Explanation:Recurrent bacterial infections may be due to lack of B-cell function, consequently resulting in a lack of gamma globulins production. Once the maternal antibodies have depleted, the disease manifests with greater severity and is called x-linked agammaglobulinemia also known as ‘X-linked hypogammaglobulinemia’, ‘XLA’ or ‘Bruton-type agammaglobulinemia. it is a rare x linked genetic disorder that compromises the bodies ability to fight infections.
Acute leukaemia causes immunodeficiency but not so specific.
DiGeorge syndrome is due to lack of T cell function.
Aplastic anaemia and EBV infection does not cause immunodeficiency.
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This question is part of the following fields:
- Inflammation & Immunology; Respiratory
- Pathology
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Question 21
Correct
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From which of the following cells is heparin produced?
Your Answer: Mast cells
Explanation:Heparin is a natural highly-sulphated glycosaminoglycan that has anticoagulant functions. It is produced by the body basophils and mast cells.
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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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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: 30 mmHg
Correct 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 23
Incorrect
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What's the nodal stage of a testicular seminoma if several lymph nodes between 2cm and 5cm are found?
Your Answer: N3
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 24
Correct
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Causes of metabolic acidosis with a normal anion gap include:
Your 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 25
Correct
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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: 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 26
Correct
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A 41 year old women presents with a history of carcinoma involving the right breast with enlarged axillary nodes on the same side. She underwent mastectomy and axillary node clearance. These were sent for histopathological examination. They showed no signs of metastasis. What could be cause of this enlargement in the lymph nodes?
Your Answer: Sinus histiocytosis
Explanation:Sinus histiocytosis also referred to as reticular hyperplasia, refers to the enlargement, distention and prominence of the sinusoids of the lymph nodes. This is a non-specific form of hyperplasia characteristically seen in lymph nodes that drain tumours. The endothelial lining of the lymph node becomes markedly hypertrophied, along with an increase in the number of macrophages which results in the distortion, distention and enlargement of the sinus. In this scenario there is no evidence that an infection or another malignancy could account for the enlargement. Paracortical lymphoid hyperplasia is caused by an immune response.
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This question is part of the following fields:
- Inflammation & Immunology; Female Health
- Pathology
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Question 27
Correct
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An experiment is carried out to observe engulfment and phagocytosis of microbes. Strep pneumoniae are added to a solution of leukocytes with a substance added to enhance the process of phagocytosis. What is this substance?
Your Answer: Complement C3b
Explanation:C3b is cleaved from C3 complement with the help of the enzyme C3- convertase. It binds to the cell surface of the offending substance and opsonizes it. This makes it easy for the phagocytes to detect and eliminate them.
IgM does not act as an opsonin but igG does.
Selectins aid leukocytes to bind to the endothelial surfaces.
C5a is a chemo-attractant and histamine a vasodilator.
NADPH oxidises offending substance after phagocytosis within the macrophage.
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This question is part of the following fields:
- Inflammation & Immunology
- Pathology
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Question 28
Correct
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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: 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 29
Incorrect
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Cancer of the testis most likely metastases to which set of lymph nodes?
Your Answer: Superficial inguinal
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 30
Correct
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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: 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 31
Correct
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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: 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 32
Correct
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As per the Poiseuille-Hagen formula, doubling the diameter of a vessel will change the resistance of the vessel from 16 peripheral resistance units (PRU) to:
Your Answer: 1 PRU
Explanation:Poiseuille-Hagen formula for flow in along narrow tube states that F = (PA– PB) × (Π/8) × (1/η) × (r4/l) where F = flow, PA– PB = pressure difference between the two ends of the tube, η = viscosity, r = radius of tube and L = length of tube. Also, flow is given by pressure difference divided by resistance. Hence, R = 8ηL ÷ Πr4. Hence, the resistance of the vessel changes in inverse proportion to the fourth power of the diameter. So, if the diameter of the vessel is increased to twice the original, it will lead to decrease in resistance to one-sixteenth its initial value.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 33
Correct
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The extent of cancer growth can be described through staging. What is taken into consideration when staging a cancer?
Your Answer: Local invasion
Explanation:Cancer stage is based on four characteristics: the size of cancer, whether the cancer is invasive or non-invasive, whether the cancer has spread to the lymph nodes, and whether the cancer has spread to other parts of the body, in this case beyond the breast. Staging is important as it is often a good predictor of outcomes and treatment is adjusted accordingly.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 34
Correct
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Diuretics that act on the ascending limb of the loop of Henle produce:
Your 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 35
Incorrect
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A week after a renal transplant the patient received antilymphocyte globulins. Shortly after she developed fever and hypotension. Which of the following mechanisms is involved in this response?
Your Answer: Type IV hypersensitivity
Correct Answer: Type III hypersensitivity
Explanation:Type III hypersensitivity is characterized by soluble immune complexes which are aggregations of IgG and IgM antibodies with antigens that deposit in different tissues e.g. the skin, joints, kidneys. They can then trigger an immune response by activating the complement cascade. This reaction can take hours to develop and examples include: immuno-complex glomerulonephritis, rheumatoid arthritis, SLE, subacute bacterial endocarditis, arthus reaction and serum sickness.
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This question is part of the following fields:
- Inflammation & Immunology
- Pathology
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Question 36
Incorrect
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A biopsy taken from the respiratory passage of a 37 year old male, chronic smoker will mostly likely show which cellular adaptation?
Your Answer: Mucous hyperplasia
Correct Answer: Stratified squamous metaplasia
Explanation:Metaplasia is a change in the cell type caused in part due to an extrinsic stress on the organ. It involves a change in the surface epithelium from one cell type to the another, most commonly squamous to columnar. This is a reversible process, and removal of the stress should theoretically reverse the surface epithelium back to normal morphology. Respiratory tract metaplasia is a classic example, in which the normal pseudostratified columnar epithelium is replaced by stratified squamous epithelium to better cope with the stress. Under continuous stress metaplasia can progress to dysplasia which is a disordered growth of cells eventually leading to the development of carcinoma.
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This question is part of the following fields:
- Cell Injury & Wound Healing; Respiratory
- Pathology
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Question 37
Incorrect
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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: 120 ml
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 38
Incorrect
-
Which of the following is the most abundant WBC seen in a smear from a healthy person.
Your Answer: Lymphocytes
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 39
Correct
-
During strenuous exercise, what else occurs besides tachycardia?
Your 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 40
Incorrect
-
Out of the following options, which malignancy has the highest potential for multicentricity?
Your Answer: Renal cell carcinoma
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 41
Correct
-
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: 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 42
Correct
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Question 43
Correct
-
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: 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 44
Correct
-
Loperamide is a drug used to treat diarrhoea. What is the mechanism of action of loperamide?
Your Answer: Opiate agonist
Explanation:Loperamide is an opioid-receptor agonist and acts on the mu opioid receptors in the myenteric plexus of large intestine. It works by decreasing the motility of the circular and longitudinal smooth muscles of the intestinal wall. It is often used for this purpose in gastroenteritis, inflammatory bowel disease, and short bowel syndrome.
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This question is part of the following fields:
- Pathology
- Pharmacology
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Question 45
Correct
-
A 54-year-old woman with amyotrophic lateral sclerosis is diagnosed with respiratory acidosis. The patient’s renal excretion of potassium would be expected to:
Your Answer: Fall, since tubular secretion of potassium is inversely coupled to acid secretion
Explanation:Respiratory acidosis is a medical emergency in which decreased ventilation (hypoventilation) increases the concentration of carbon dioxide in the blood and decreases the blood’s pH (a condition generally called acidosis). Secretion of acid and potassium by the renal tubule are inversely related. So, increased excretion of H+ during renal compensation for respiratory acidosis will result in decreased secretion (or increased retention) of potassium ions, with the result that the body’s potassium store rises. An increase in K+ excretion would be associated with renal compensation for respiratory alkalosis. The filtered load of K+depends only on K+ plasma concentration and glomerular filtration rate, not on plasma pH.
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This question is part of the following fields:
- Physiology
- Renal
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Question 46
Correct
-
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: 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 47
Correct
-
A 30-year-old female was alarmed when she started to experience hair loss and balding, however, she also noted increased hair on her face and body and developed an acne breakout. Deepening of her voice also became prominent. She was referred to an oncologist and was diagnosed with a hormone-producing tumour. What is the most likely diagnosis ?
Your Answer: Arrhenoblastoma
Explanation:Arrhenoblastoma, known as ‘Sertoli–Leydig tumour’ is a rare ovarian stromal neoplasm that secretes testosterone. It is mostly seen in women in the reproductive years. The key clinical features of this tumour is due to excessive production of testosterone which leads to progressive masculinisation in a woman who was typical normal beforehand. The lesion tends to grow slowly and rarely metastasises. Treatment is surgical removal of the tumour and the prognosis is generally good.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 48
Correct
-
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: 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
Correct
-
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: 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 50
Incorrect
-
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: Minimal-change disease
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 51
Incorrect
-
A 17-year-old boy, who had developed shortness of breath and a loss of appetite over the last month, was referred to a haematologist because he presented with easy bruising and petechiae. His prothrombin time, platelet count, partial thromboplastin and bleeding time were all normal. Which of the following would explain the presence of the petechiae and easy bruising tendency?
Your Answer:
Correct Answer: Scurvy
Explanation:Scurvy is a condition caused by a dietary deficiency of vitamin C, also known as ascorbic acid. Humans are unable to synthesize vitamin C, therefore the quantity of it that the body needs has to come from the diet. The presence of an adequate quantity of vitamin C is required for normal collagen synthesis. In scurvy bleeding tendency is due to capillary fragility and not coagulation defects, therefore blood tests are normal.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 52
Incorrect
-
A 40 year old man suffered severe trauma following an MVA. His BP is 70/33 mmhg, heart rate of 140 beats/mins and very feeble pulse. He was transfused 3 units of blood resulting in his BP returning to 100/70 and his heart rate to 90 beats/min. What decreased following transfusion?
Your Answer:
Correct Answer: Total peripheral resistance
Explanation:The patient is in hypovolemic shock, he is transfused with blood to replace the volume lost. It is important not only to replace fluids but stop active bleeding in resuscitation. Fluid replacement will result in a decreased sympathetic discharge and adequate ventricular filling thus reducing total peripheral resistance and increasing cardiac output and cardiac filling pressures.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 53
Incorrect
-
When a penile tumour invades the subepithelial connective tissue of the penis, what is its stage?
Your Answer:
Correct Answer: T1
Explanation:The TNM staging used for penile cancer is as follows:
TX: primary tumour cannot be assessed
T0: primary tumour is not evident
Tis: carcinoma in situ is present
Ta: non-invasive verrucous carcinoma is present
T1: tumour is invading subepithelial connective tissue
T2: tumour is invading the corpora spongiosum or cavernosum
T3: tumour invading the urethra or prostate
T4: tumour invading other adjacent structures.
In this case, the patient has a T1 tumour.
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This question is part of the following fields:
- Pathology
- Urology
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Question 54
Incorrect
-
The most important difference between interstitial fluid and plasma is the:
Your Answer:
Correct Answer: Protein concentration
Explanation:Interstitial fluid (or tissue fluid or intercellular fluid) is a solution that surrounds the cells of multicellular animals. It is the main component of the extracellular fluid, which also includes plasma, lymph and transcellular fluid. Plasma, the major component in blood, communicates freely with interstitial fluid through pores and intercellular clefts in capillary endothelium. Interstitial fluid consists of a water solvent containing amino acids, sugars, fatty acids, coenzymes, hormones, neurotransmitters, salts, as well as waste products from the cells. Red blood cells, platelets and plasma proteins cannot pass through the walls of the capillaries. The resulting mixture that does pass through is essentially blood plasma without the plasma proteins. Tissue fluid also contains certain types of white blood cells. Once the extracellular fluid collects into small vessels it is considered to be lymph, and the vessels that carry it back to the blood are called the lymphatic vessels. The lymphatic system returns protein and excess interstitial fluid to the circulation.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 55
Incorrect
-
A 35-year old lady presents to her GP with vague abdominal symptoms. Examination reveals a normal size spleen. Which of the following is the likely diagnosis?
Your Answer:
Correct Answer: Idiopathic thrombocytopenic purpura
Explanation:Idiopathic thrombocytopenic purpura (ITP) is a disease caused due to development of an antibody against a platelet antigen (autoantibody). In childhood disease, the autoantibody gets triggered by binding of viral antigen to the megakaryocytes. Presentation includes unexplained thrombocytopenia, petechiae and bleeding from mucosal surfaces. The spleen usually does not enlarge in size. However, splenomegaly can occur due to coexisting viral infection. Marrow examination reveals normal or increased number of megakaryocytes. Diagnosis is by exclusion of other thrombocytopenic disorders.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 56
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 57
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 58
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 59
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 60
Incorrect
-
A 65-year old man, known with Type 2 diabetes and chronic renal failure, is likely to eventually present with which of the following conditions?
Your Answer:
Correct Answer: Secondary hyperparathyroidism
Explanation:When the parathyroid glands secrete excess parathyroid hormone (PTH) in response to hypocalcaemia, it is known as Secondary hyperparathyroidism and is often seen in patients with renal failure. In chronic renal failure, the kidneys fail to excrete adequate phosphorus and also fail to convert enough vitamin D to its active form. This leads to formation of insoluble calcium phosphate in the body which ultimately causes hypocalcaemia. The glands then undergo hyperplasia and hypertrophy leading to secondary hyperparathyroidism. Symptoms include bone and joint pains, along with limb deformities. The raised PTH also results in pleiotropic effects on blood, the immune system and nervous system.
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This question is part of the following fields:
- Endocrine
- Pathology
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Question 61
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 62
Incorrect
-
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 63
Incorrect
-
Decreased velocity of impulse conduction through the atrioventricular node (AV node) in the heart will lead to:
Your Answer:
Correct Answer: Increased PR interval
Explanation:AV node damage may lead to an increase in the PR interval to as high as 0.25 – 0.40 s (normal = 0.12 – 0.20 s). In the case of severe impairment, there might be a complete failure of passage of impulses leading to complete block. In this case, the atria and ventricles will beat independently of each other.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 64
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 65
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 66
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 67
Incorrect
-
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 68
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 69
Incorrect
-
Abnormal breathing is noticed in a of victim of a road traffic accident, who sustained a head injury. The breathing pattern is characterised by alternate periods of waxing and waning tidal volumes with interspersed periods of apnoea. This breathing pattern is known as:
Your Answer:
Correct Answer: Cheyne–Stokes breathing
Explanation:Cheyne-Stokes breathing is an abnormal breathing pattern with breathing periods of gradually waxing and waning tidal volumes, with apnoeic periods interspersed. It is usually the first breathing pattern to be seen with a rise in intracranial pressure and is caused by failure of the respiratory centre in the brain to compensate quickly enough to changes in serum partial pressure of oxygen and carbon dioxide. The aetiology includes strokes, head injuries, brain tumours and congestive heart failure. It is also a sign of altitude sickness in normal people, a symptom of carbon monoxide poisoning or post-morphine administration. Biot’s respiration (cluster breathing) is characterized by cluster of quick, shallow inspirations followed by regular or irregular periods of apnoea. It is different from ataxic respiration, which has completely irregular breaths and pauses. It results due to damage to the medulla oblongata by any reason (stroke, uncal herniation, trauma) and is a poor prognostic indicator. Kussmaul breathing, also known as ‘air hunger’, is basically respiratory compensation for metabolic acidosis and is characterized by quick, deep and laboured breathing. It is most often seen in in diabetic ketoacidosis. Due to forced inspiratory rate, the patients will show a low p(CO2). Ondine’s curse is congenital central hypoventilation syndrome or primary alveolar hypoventilation, which can be fatal and leads to sleep apnoea. It involves an inborn failure to control breathing autonomically during sleep and in severe cases, can affect patients even while awake. It is known to occur in 1 in 200000 liveborn children. Treatment includes tracheostomies and life long mechanical ventilator support.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 70
Incorrect
-
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 71
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 72
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 73
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 74
Incorrect
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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 75
Incorrect
-
Hepatomegaly with greatly increased serum alpha-fetoprotein is seen in which of the following conditions?
Your Answer:
Correct Answer: Hepatocellular carcinoma
Explanation:Hepatocellular carcinoma or hepatoma affects people with pre-existing cirrhosis and is more common in areas with higher prevalence of hepatitis B and C. Diagnosis include raise alpha-fetoprotein levels, imaging and liver biopsy if needed. Patients at high-risk for developing this disease can undergo screening by periodic AFP measurement and abdominal ultrasonography. The malignancy carries poor prognosis (see also Answer to 10.4).
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This question is part of the following fields:
- Gastrointestinal; Hepatobiliary
- Pathology
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Question 76
Incorrect
-
A 14-year old girl presented with a 2cm, mobile, cystic mass in the midline of her neck. Fine needle aspiration of the mass revealed clear fluid. This is most likely a case of:
Your Answer:
Correct Answer: Thyroglossal duct cyst
Explanation:Thyroglossal cyst is the most common congenital thyroid anomaly which is clinically significant and affects women more than men. It is a vestigial remnant of developing thyroid. Although the thyroglossal cyst can develop anywhere along the thyroglossal duct, the most common site is in the midline between the isthmus of thyroid and hyoid bone, or just above the hyoid. Thyroglossal cysts are also associated with ectopic thyroid tissue. Clinically, the cyst moves upward with protrusion of the tongue. Rarely, the persistent duct can become malignant (thyroglossal duct carcinoma) where the cancerous cells arise in the ectopic thyroid tissue that are deposited along the duct. Exposure to radiation is a predisposing factor.
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This question is part of the following fields:
- Endocrine
- Pathology
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Question 77
Incorrect
-
A 25 year old women is pregnant with her second child. She is A- blood group. Her first child was Rh+ and the father is also Rh+. The second child is at a risk of developing which condition?
Your Answer:
Correct Answer: Haemolytic disease of the new-born
Explanation:This infant is at risk for haemolytic disease of the new born also known as erythroblastosis fetalis. In the pregnancy, Rh-positive RBC’s cross the placenta and enter the mothers blood system. She then becomes sensitised and forms IgG antibodies/anti-Rh antibodies against them. The second child is at a greater risk for this disease than the first child with Rh-positive blood group as during the second pregnancy, a more powerful response is produced. IgG has the ability to cross the placenta and bind to the fetal RBCs (type II hypersensitivity reaction) which are phagocytosed by the macrophages.
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This question is part of the following fields:
- Inflammation & Immunology; Haematology
- Pathology
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Question 78
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 79
Incorrect
-
Which of the following conditions may cause hypervolaemic hyponatraemia?
Your Answer:
Correct Answer: Cirrhosis
Explanation:Hypovolaemic hyponatraemia: reduced extracellular fluid
Renal loss of sodium and water; urine Na >20 mmol/day
Causes:
Diuretic use
Salt wasting nephropathy
Cerebral salt wasting
Mineralocorticoid deficiency/adrenal insufficiency
Renal tubular acidosis
Extrarenal loss of sodium and water with renal conservation; urine Na <20 mmol/day
Causes:
Burns
Gastrointestinal loss
Pancreatitis
Blood loss
3rd space loss (bowel obstruction, peritonitis)Hypervolaemic hyponatraemia: expanded intracellular fluid and extracellular fluid but reduced effective arterial blood volume
Causes:
Congestive cardiac failure
Cirrhosis
Nephrotic syndromeEuvolaemic hyponatraemia: expanded intracellular and extracellular fluid but oedema absent
Causes:
Thiazide diuretics (can be euvolaemic or hypovolaemic)
Hypothyroidism
Adrenal insufficiency (can be euvolaemic or hypovolaemic)
SIADH (cancer, central nervous system disorders, drugs, pulmonary disease, nausea, postoperative pain, HIV, infection, Guillain‐Barre syndrome, acute intermittent porphyria)
Decreased solute ingestion (beer potomania/tea and toast diet) -
This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
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Question 80
Incorrect
-
A 22-year old man presented with a mass in his left scrotum which was more prominent when standing and felt like a 'bag of worms'. Examination revealed a non-tender mass along the spermatic cord. Also, the right testis was larger than the left testis. What is the likely diagnosis?
Your Answer:
Correct Answer: Varicocele
Explanation:Varicocele refers to dilatation and increased tortuosity of the pampiniform plexus – which is a network of veins found in spermatic cord that drain the testicle. Defective valves or extrinsic compression can result in outflow obstruction and cause dilatation near the testis. Normal diameter of the small vessels ranges from 0.5 – 1.5mm. A varicocele is a dilatation more than 2mm.
The plexus travels from the posterior aspect of testis into the inguinal canal with other structures forming the spermatic cord. They then form the testicular veins out of which the right testicular vein drains into the inferior vena cava and the left into the left renal vein.
It affects 15-20% men, and 40% of infertile males. Usually diagnosed in 15-25 years of age, they are rarely seen after 40 years of age. Because of the vertical path taken by the left testicular vein to drain into left renal vein, 98% idiopathic varicoceles occur on the left side. It is bilateral in 70% cases. Right-sided varicoceles are rare.
Symptoms include pain or heaviness in the testis, infertility, testicular atrophy, a palpable mass, which is non-tender and along the spermatic cord (resembling a ‘bag of worms’). The testis on the affected side might be smaller.
Diagnosis can be made by ultrasound. Provocative measures such as Valsalva manoeuvre or making the patient stand up to increase the dilatation by increasing the intra-abdominal venous pressure.
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This question is part of the following fields:
- Pathology
- Urology
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Question 81
Incorrect
-
A 4-year-old child was brought to a paediatrician for consult due to a palpable mass in his abdomen. The child has poor appetite and regularly complains of abdominal pain. The child was worked up and diagnosed with a tumour. What is the most likely diagnosis ?
Your Answer:
Correct Answer: Nephroblastoma
Explanation:Nephroblastoma is also known as Wilms’ tumour. It is a cancer of the kidneys that typically occurs in children. The median age of diagnose is approximately 3.5 years. With the current treatment, approximately 80-90% of children with Wilms’ tumour survive.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 82
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 83
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 84
Incorrect
-
A 77-year-old woman's renal function is tested. The following results were obtained during a 24-h period:
Urine flow rate: 2. 0 ml/min
Urine inulin: 0.5 mg/ml
Plasma inulin: 0.02 mg/ml
Urine urea: 220 mmol/l
Plasma urea: 5 mmol/l.
What is the urea clearance?Your Answer:
Correct Answer: 88 ml/min
Explanation:Urea is reabsorbed in the inner medullary collecting ducts of the nephrons. 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. So, glomerular filtration rate = (0.220 × 2. 0)/0.005 = 88 ml/min.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 85
Incorrect
-
The renal cortex and medulla, if seen under the microscope, is lacking one of the following:
Your Answer:
Correct Answer: Squamous epithelium
Explanation:Capillaries, Henle’s loop, collecting ducts, Bertin columns and type IV collagen in glomerular basement membrane are all structures present in the renal cortex or medulla. The squamous epithelium is the only one that is lacking in both the renal cortex and medulla, because normally it is not found above the outer urethra.
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This question is part of the following fields:
- Pathology
- Renal
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Question 86
Incorrect
-
A 30 year old man suffered severe blood loss, approx. 20-30% of his blood volume. What changes are most likely seen in the pulmonary vascular resistance (PVR) and pulmonary artery pressure (PAP) respectively following this decrease in cardiac output?
Your Answer:
Correct Answer: Increase Decrease
Explanation:Hypovolemia will result in the activation of the sympathetic adrenal discharge resulting is a decrease pulmonary artery pressure and an elevated pulmonary vascular resistance.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 87
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 88
Incorrect
-
The Henderson–Hasselbalch equation describes the derivation of pH as a measure of acidity. According to this equation, the buffering capacity of the system is at maximum when the number of free anions compared with undissociated acid is:
Your Answer:
Correct Answer: Equal
Explanation:In 1908, Lawrence Joseph Henderson wrote an equation describing the use of carbonic acid as a buffer solution. Later, Karl Albert Hasselbalch re-expressed that formula in logarithmic terms, resulting in the Henderson–Hasselbalch equation. The equation is also useful for estimating the pH of a buffer solution and finding the equilibrium pH in acid–base reactions. Two equivalent forms of the equation are: pH = pKa + log10 [A–]/[HA] or pH = pKa + log10 [base]/[acid]. Here, pKa is − log10(Ka) where Ka is the acid dissociation constant, that is: pKa = –log10(Ka) = –log10 ([H3 O+][A–]/[HA]) for the reaction: HA + H2 O ≈ A– + H3 O+ In these equations, A– denotes the ionic form of the relevant acid. Bracketed quantities such as [base] and [acid] denote the molar concentration of the quantity enclosed. Maximum buffering capacity is found when pH = pKa or when the number of free anions to undissociated acid is equal and buffer range is considered to be at a pH = pKa ± 1.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 89
Incorrect
-
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:
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 90
Incorrect
-
A 27-year old lady presented with dull, abdominal pain and some pain in her lower limbs. On enquiry, it was revealed that she has been suffering from depression for a few months. Physical examination and chest X-ray were normal. Further investigations revealed serum calcium 3.5 mmol/l, albumin 3.8 g/dl and phosphate 0.65 mmol/l. What is the diagnosis?
Your Answer:
Correct Answer: Parathyroid adenoma
Explanation:Hypercalcaemia with hypophosphatemia indicates parathyroid disorder and adenomas are more common than hyperplasia. In this young age group, metastatic disease is unlikely. Solitary adenomas are responsible for 80-85% cases of primary hyperparathyroidism. 10-15% cases are due to parathyroid hyperplasia and carcinomas account for 2-3% cases. Symptoms include bone pain (bones), nephrolithiasis (stones), muscular aches, peptic ulcer disease, pancreatitis (groans), depression (moans), anxiety and other mental disturbances.
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This question is part of the following fields:
- Endocrine
- Pathology
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Question 91
Incorrect
-
Arterial blood gas analysis of a man admitted with acute exacerbation of chronic obstructive pulmonary disease (COPD) showed the following: pH = 7.28, p(CO2) = 65.5 mmHg, p(O2)= 60 mmHg and standard bicarbonate = 30.5 mmol/l. This patient had:
Your Answer:
Correct Answer: Respiratory acidosis
Explanation:Acidosis with high p(CO2) and normal standard bicarbonate indicates respiratory acidosis, commonly seen in acute worsening of COPD patients. 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. In acute respiratory acidosis, the p(CO2) is raised above the upper limit of normal (over 45 mm Hg) with a 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).
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 92
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 93
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 94
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 95
Incorrect
-
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 96
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 97
Incorrect
-
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 98
Incorrect
-
A 45 -year-old female is recently diagnosed with breast cancer. She has a 8-cm-diameter mass in her left breast, with enlarged left axillary node. What is the most likely stage of her disease?
Your Answer:
Correct Answer: IIIA
Explanation:Stage IIIA breast cancer is T0–2 N2 M0 or T3 N1-2 M0 disease. It describes invasive breast cancer in which either: the tumour is smaller than 5 cm in diameter and has spread to 4 to 9 axillary lymph nodes; or it is found through imaging studies or clinical exam to have spread to internal mammary nodes (near the breastbone found during imaging tests or a physical exam); or the tumour is larger than 5 cm and has spread to 1 to 9 axillary nodes, or to internal mammary nodes. In this stage, the cancer has not metastasized (spread to distant sites).
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 99
Incorrect
-
Which of these is secreted by both macrophages and muscle cells?
Your Answer:
Correct Answer: Interleukin-6
Explanation:IL-6 is secreted by the T cells and macrophages and is a pro inflammatory cytokine. It is secreted in response to trauma e.g. burns and tissue damage that leads to inflammation. Apart from this its is also a myokine and is elevated due to muscle contraction. Other functions include: stimulate osteoclast formation when secreted by osteoblasts, mediate fever in acute phase response and are responsible for energy metabolism in muscle and fatty tissues. Inhibitors of IL-6 e.g. oestrogen are used as a treatment for postmenopausal osteoporosis.
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This question is part of the following fields:
- Inflammation & Immunology
- Pathology
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Question 100
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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SESSION STATS - PERFORMANCE PER SPECIALTY
