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Question 1
Correct
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A 34-year-old woman with pelvic inflammatory disease is administered ceftriaxone. The subclass of antibiotics that ceftriaxone belongs to is:
Your Answer: Third-generation cephalosporins
Explanation:Ceftriaxone is a third-generation cephalosporin antibiotic. It has a broad spectrum of activity against Gram-positive and Gram-negative bacteria. Its indications include middle ear infections, endocarditis, meningitis, pneumonia, bone and joint infections, intra-abdominal infections, skin infections, urinary tract infections, gonorrhoea, and pelvic inflammatory disease.
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This question is part of the following fields:
- Pathology
- Pharmacology
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Question 2
Correct
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Ventricular filling follows a delay caused by?
Your Answer: AV node
Explanation:The AV node is a conducting tissue found between the atria and the ventricles of the heart. It conducts electrical signal from the atria to the ventricles and acts a delaying mechanism preventing the atria and the ventricles from contracting at the same time. This decremental conduction prevents premature ventricular contraction in cases such as atrial fibrillation. A delay in the AV node is the reason for the PR segment seen on the ECG. In certain types of supraventricular tachycardia, a person could have two AV nodes; this will cause a loop in electrical current and uncontrollably rapid heart beat. When this electricity catches up with itself, it will dissipate and return to a normal heart rate.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 3
Correct
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Normally, the O2 transfer in the lungs from alveolar to capillary is perfusion-limited. In which of the following situations does it become a diffusion-limited process?
Your Answer: Pulmonary oedema
Explanation:Normally, the transfer of oxygen from air spaces to blood takes place across the alveolar-capillary membrane by simple diffusion and depends entirely on the amount of blood flow (perfusion-limited process). Diseases that affect this diffusion will transform the normal process to a diffusion limited process. Thus, the diseases which cause a thickened barrier (such as pulmonary oedema due to increased extravascular lung water or asbestosis) will limit the diffusion of oxygen. Chronic obstructive lung diseases will have little effect on diffusion. Inhaling hyperbaric gas mixtures might overcome the diffusion limitation in patients with mild asbestosis or interstitial oedema, by increasing the driving force. Strenuous (not mild) exercise might also favour diffusion limitation and decrease passage time. Increasing the rate of ventilation will not have this affect but will only maintain a high oxygen gradient from air to blood.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 4
Incorrect
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A 49-year-old man, smoker, complains of a persisting and worsening cough over the past few months. He also has noted blood in his sputum. The patient has no other major health conditions. Which of the following investigative procedures should be done first?
Your Answer: Bronchoalveolar lavage
Correct Answer: Sputum cytology
Explanation:Sputum cytology is a diagnostic test used for the examination of sputum under a microscope to determine if abnormal cells are present. It may be used as the first diagnostic procedure to help detect a suspected lung cancer or certain non-cancerous lung conditions.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 5
Incorrect
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A 31 -year-old female patient had a blood gas done on presentation to the emergency department. She was found to have a metabolic acidosis and decreased anion gap. The most likely cause of these findings in this patient would be?
Your Answer: Hypomagnesaemia
Correct Answer: Hypoalbuminemia
Explanation:A low anion gap might be caused by alterations in serum protein levels, primarily albumin (hypoalbuminemia), increased levels of calcium (hypercalcaemia) and magnesium (hypermagnesemia) or bromide and lithium intoxication. However, the commonest cause is hypoalbuminemia, thus if the albumin concentration falls, the anion gap will also be lower. The anion gap should be corrected upwards by 2.5 mmol/l for every 10g/l fall in the serum albumin.
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This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
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Question 6
Incorrect
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A medical student is told a substance is freely filtered but is not metabolised, secreted, or stored in the kidney. It has a plasma concentration of 1000 mg/l and its urine excretion rate is 25 mg/min, and the inulin clearance is 100 ml/min. What is the rate of tubular reabsorption of the substance?
Your Answer: 65 mg/min
Correct Answer: 75 mg/min
Explanation:Reabsorption or tubular reabsorption is the process by which the nephron removes water and solutes from the tubular fluid (pre-urine) and returns them to the circulating blood. To calculate the reabsorption rate of substance Z we use the following equation: excretion = (filtration + secretion) – reabsorption. As this substance is freely filtered, its filtration rate is equal to that of inulin. So 25 = (100 + 0) – reabsorption. Reabsorption = 100 – 25 therefore reabsorption = 75 mg/min.
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This question is part of the following fields:
- Physiology
- Renal
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Question 7
Incorrect
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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: Metabolic acidosis
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 8
Incorrect
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A 30-year-old man is brought to the emergency department suffering from extreme dehydration, and subsequent hypotension and tachycardia. Which part of the kidney will compensate for this loss?
Your Answer: Proximal convoluted tubule
Correct Answer: Collecting ducts
Explanation:The collecting duct system of the kidney consists of a series of tubules and ducts that physically connect nephrons to a minor calyx or directly to the renal pelvis. The collecting duct system is the last component of the kidney to influence the body’s electrolyte and fluid balance. In humans, the system accounts for 4–5% of the kidney’s reabsorption of sodium and 5% of the kidney’s reabsorption of water. At times of extreme dehydration, over 24% of the filtered water may be reabsorbed in the collecting duct system.
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This question is part of the following fields:
- Physiology
- Renal
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Question 9
Incorrect
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A 48-year-old woman has a mass in her right breast and has right axillary node involvement. She underwent radical mastectomy of her right breast. The histopathology report described the tumour to be 4 cm in its maximum diameter with 3 axillary lymph nodes with evidence of tumour. The most likely stage of cancer in this patient is:
Your Answer: IIA
Correct Answer: IIB
Explanation:Stage IIB describes invasive breast cancer in which: the tumour is larger than 2 centimetres but no larger than 5 centimetres; small groups of breast cancer cells — larger than 0.2 millimetre but not larger than 2 millimetres — are found in the lymph nodes OR the tumour is larger than 2 centimetres but no larger than 5 centimetres; cancer has spread to 1 to 3 axillary lymph nodes or to lymph nodes near the breastbone (found during a sentinel node biopsy) OR the tumour is larger than 5 centimetres but has not spread to the axillary lymph nodes.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 10
Incorrect
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Leukotrienes normally function during an asthma attack and work to sustain inflammation. Which of the following enzymes would inhibit their synthesis?
Your Answer: Cyclooxygenase-2
Correct Answer: 5-lipoxygenase
Explanation:Leukotrienes are produced from arachidonic acid with the help of the enzyme 5-lipoxygenase. This takes place in the eosinophils, mast cells, neutrophils, monocytes and basophils. They are eicosanoid lipid mediators and take part in allergic and asthmatic attacks. They are both autocrine as well as paracrine signalling molecules to regulate the body’s response and include: LTA4, LTB4, LTC4, LTD4, LTE4 and LTF4.
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This question is part of the following fields:
- Inflammation & Immunology
- Pathology
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Question 11
Correct
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Which of the following muscles aid in inspiration?
Your Answer: Diaphragm and external intercostals
Explanation:The diaphragm and external intercostals are muscles of inspiration as they increase the volume of thoracic cavity and reduce the intrathoracic pressure. Muscles of expiration include abdominal muscles and internal intercostals.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 12
Correct
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A 55-year old lady underwent a major surgery for repair of an aortic aneurysm. Her blood pressure was low throughout the intra-operative and the post-operative period, along with increasing serum creatinine and urea. Microscopic examination of her urine showed multiple granular and hyaline casts. What is the likely condition the patient is suffering from?
Your Answer: Acute tubular necrosis
Explanation:The most common predisposing factor leading to acute tubular necrosis is ischemia, typically seen in hospitalized patients with low blood pressure.
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This question is part of the following fields:
- Pathology
- Renal
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Question 13
Correct
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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: 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 14
Incorrect
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Hyperplastic arteriosclerosis with fibrinoid necrosis, petechial haemorrhages, microinfarcts in the kidneys and elevated plasma renin are common findings in which of the following patients?
Your Answer: A 42-year-old man with chronic renal disease
Correct Answer: A 45-year-old woman with scleroderma
Explanation:Scleroderma, also known as systemic sclerosis, is a chronic disease of the connective tissue. Involvement of the kidneys occurs in patients with diffuse scleroderma, causing rapid onset of high blood pressure with hyperreninemia, thrombotic microangiopathy, and progressive renal failure.
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This question is part of the following fields:
- Pathology
- Renal
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Question 15
Incorrect
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Which part of the nephron would have to be damaged to stop the reabsorption of the majority of salt and water?
Your Answer: Thin ascending limb of loop of Henle
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 16
Correct
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Which of these substances is secreted by pericytes in the juxtaglomerular cells?
Your Answer: Renin
Explanation:The juxtaglomerular cells synthesise, store and secrete the enzyme renin in the kidney. They are specialised smooth muscle cells in the wall of the afferent arteriole that delivers blood to the glomerulus and thus play a critical role in the renin– angiotensin system and so in renal autoregulation.
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This question is part of the following fields:
- Physiology
- Renal
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Question 17
Correct
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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: 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 18
Correct
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A 59-year old gentleman admitted for elective cholecystectomy was found to have a haemoglobin 12.5 g/dl, haematocrit 37%, mean corpuscular volume 90 fl, platelet count 185 × 109/l, and white blood cell count 32 × 109/l; along with multiple, small mature lymphocytes on peripheral smear. The likely diagnosis is:
Your Answer: Chronic lymphocytic leukaemia
Explanation:CLL or chronic lymphocytic leukaemia is the most common leukaemia seen in the Western world. Twice more common in men than women, the incidence of CLL increases with age. About 75% cases are seen in patients aged more than 60 years. The blood, marrow, spleen and lymph nodes all undergo infiltration, eventually leading to haematopoiesis (anaemia, neutropenia, thrombocytopenia), hepatomegaly, splenomegaly and decreased production of immunoglobulin. In 98% cases, CD+5 B cells undergo malignant transformation.
Often diagnosed on blood tests while being evaluated for lymphadenopathy, CLL causes symptoms like fatigue, anorexia, weight loss, pallor, dyspnoea on exertion, abdominal fullness or distension. Findings include multiple lymphadenopathy with minimal-to- moderate hepatomegaly and splenomegaly. Increased susceptibility to infections is seen. Herpes Zoster is common. Diffuse or maculopapular skin infiltration can also be seen in T-cell CLL.
Diagnosis is by examination of peripheral blood smear and marrow: hallmark being a sustained, absolute leucocytosis (>5 ×109/l) and increased lymphocytes in the marrow (>30%). Other findings can include hypogammaglobulinemia (<15% of cases) and, rarely, raised lactate dehydrogenase (LDH). Only 10% cases demonstrate moderate anaemia and/or thrombocytopenia.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 19
Correct
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QT interval in the electrocardiogram of a healthy individual is normally:
Your Answer: 0.40 s
Explanation:QT interval extends from beginning of the QRS complex till the end of he T-wave and normally lasts for 0.40 s. It is important in the diagnosis of long-QT and short-QT syndrome. The QT interval varies on the basis of heart rate and may need to be corrected.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 20
Incorrect
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A 50-year old lady presented to the clinic with chronic pain in the abdomen. On physical examination, she was found to be pale. Further investigations revealed a decrease in both serum iron and total iron-binding capacity, along with an increase in serum ferritin. These findings are seen in:
Your Answer: Autoimmune haemolytic anaemia
Correct Answer: Anaemia of chronic disease
Explanation:Anaemia of chronic disease is characterized by low serum iron, iron-binding capacity and saturation with increased ferritin (storage iron). Haemolytic anaemia is characterized by normal iron levels as the haemoglobin released from the haemolysed red blood cells is recycled. Anaemia due to chronic blood loss leads to low serum iron, low ferritin and high total iron-binding capacity (TIBC). Malabsorption, especially with duodenal involvement can also lead to iron deficiency anaemia with low ferritin and high TIBC. Megaloblastic anaemia due to vitamin B12 and folate deficiency is not associated with abnormalities in metabolism of iron.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 21
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 Henl̩ Рlow permeability to ions and urea, while being highly permeable to water
2. Thin ascending limb of loop of Henl̩ Рnot 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 22
Incorrect
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A 47-year-old male smoker, who had been self-medicating with oral steroids for the last two years due to persistent breathlessness presented to the doctor complaining of a productive cough, fever and chest pain. A chest X-ray revealed bilateral patchy opacities. He was diagnosed with bilateral bronchopneumonia. Which of these organisms is most probably causing these findings?
Your Answer: Streptococcus pneumoniae
Correct Answer: Nocardia asteroides
Explanation:Nocardia is a Gram-positive aerobic actinomycete. Several species have been identified but the most common human pathogen is Nocardia asteroides. The predominant clinical finding in the majority of patients affected by nocardiosis is pulmonary disease. Predisposing factors for pulmonary nocardiosis include leukaemia, human immunodeficiency virus (HIV) infection, organ transplantation, diabetes and receiving prolonged corticosteroids.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 23
Correct
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A biopsy is performed on a 67-year-old-man with testicular seminoma; it reveals that the tumour affects the tunica vaginalis. The tumour stage in this case is:
Your Answer: T2
Explanation:The primary tumour staging for testicular seminoma is as follows, according to AJCC guidelines:
Tis: intratubular germ cell neoplasia (carcinoma in situ)
T1: tumour limited to testis/epididymis without vascular or lymphatic invasion; the tumour can invade into the tunica albuginea but not the tunica vaginalis
T2: tumour limited to testis/epididymis with vascular or lymphatic invasion or tumour extending through the tunica albuginea with involvement of the tunica vaginalis
T3: tumour invading the spermatic cord, with or without vascular/lymphatic invasion
T4: tumour invading the scrotum, with or without vascular/lymphatic invasion.
According to these guidelines, the tumour in this case has a T2 stage.
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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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Which of the following malignancies is associated with the development of Lambert-Eaton myasthenic syndrome?
Your Answer: Lung cancer
Explanation:Lambert–Eaton myasthenic syndrome is a rare disorder of the neuromuscular junction. It can occur as a solitary diagnosis but it can also occur as a paraneoplastic syndrome associated with lung cancer, particularly small-cell histology. It can also be associated with other cancers such as lymphoma, non-Hodgkin’s lymphoma, T-cell leukaemia, non-small-cell lung cancer, prostate cancer and thymoma.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 25
Correct
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Which is a feature of the action of insulin?
Your Answer: Promotes protein synthesis
Explanation:Insulin is produced by the beta-cells of the islets of Langerhans in the pancreas. Its actions include:
– promoting uptake of glucose into cells
– glycogen synthesis (glycogenesis)
– protein synthesis
– stimulation of lipogenesis (fat formation).
– driving potassium into cells – used to treat hyperkaelamia.
Parathyroid hormone and activated vitamin D are the principal hormones involved in calcium/phosphate metabolism, rather than insulin.
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This question is part of the following fields:
- Endocrine
- Physiology
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Question 26
Correct
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A 5-year-old child diagnosed with nephrotic syndrome developed generalised oedema. What is the mechanism for the development of oedema in patients with nephrotic syndrome?
Your Answer: Decreased colloid osmotic pressure
Explanation:The development of oedema in nephrotic syndrome has traditionally been viewed as an underfill mechanism. According to this view, urinary loss of protein results in hypoalbuminemia and decreased plasma oncotic pressure. As a result, plasma water translocates out of the intravascular space and results in a decrease in intravascular volume. In response to the underfilled circulation, effector mechanisms are then activated that signal the kidney to secondarily retain salt and water. While an underfill mechanism may be responsible for oedema formation in a minority of patients, recent clinical and experimental findings would suggest that oedema formation in most nephrotic patients is the result of primary salt retention. Direct measurements of blood and plasma volume or measurement of neurohumoral markers that indirectly reflect effective circulatory volume are mostly consistent with either euvolemia or a volume expanded state. The ability to maintain plasma volume in the setting of a decreased plasma oncotic pressure is achieved by alterations in transcapillary exchange mechanisms known to occur in the setting of hypoalbuminemia that limit excessive capillary fluid filtration.
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This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
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Question 27
Incorrect
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The proximal tubule is the portion of the ductal system of the nephron of the kidney which leads from Bowman's capsule to the loop of Henle. Which of the following is most likely to be seen in a sample of fluid leaving the proximal tubule?
Your Answer: It will have more HCO3 – ions than plasma
Correct Answer: It will have no amino acids
Explanation:The proximal tubule is the portion of the duct system of the nephron leading from Bowman’s capsule to the loop of Henlé. The most distinctive characteristic of the proximal tubule is its brush border (or ‘striated border’). The luminal surface of the epithelial cells of this segment of the nephron is covered with densely packed microvilli forming a border which greatly increases the luminal surface area of the cells, presumably facilitating their reabsorptive function. Glucose, amino acids, inorganic phosphate, and some other solutes are100% reabsorbed via secondary active transport through co-transporters driven by the sodium gradient out of the nephron.
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This question is part of the following fields:
- Physiology
- Renal
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Question 28
Incorrect
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A 45-year old man presents with sclerosing cholangitis, blood in his stools and apparent iron deficiency anaemia. What will be the most likely finding on his colonic biopsy?
Your Answer: Villous adenoma
Correct Answer: Pseudopolyps
Explanation:Sclerosing cholangitis along with the passage of blood in stools suggests ulcerative colitis affecting the mucosa and submucosa of rectum and colon, with a sharp demarcation with the normal tissue. The musclaris layer is involved in severe cases. Initially, the mucosa is erythematous, friable with scattered haemorrhagic areas and loss of normal vascular pattern. Severe disease is indicated by presence of large mucosal ulcers with purulent exudate. There can be islands of normal mucosa between the ulcerated mucosa, along with few hyperplastic inflammatory mucosal lesions (pseudopolyps). Ulcerative colitis does not lead to development of fistulas or abscesses.
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This question is part of the following fields:
- Gastrointestinal; Hepatobiliary
- Pathology
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Question 29
Correct
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Purkinje fibres in the heart conduct action potentials at the rate of:
Your Answer: 1.5–4.0 m/s
Explanation:Purkinje fibres control the heart rate along with the sinoatrial node (SA node) and the atrioventricular node (AV node). The QRS complex is associated with the impulse passing through the Purkinje fibres. These fibres conduct action potential about six times faster than the velocity in normal cardiac muscle.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 30
Incorrect
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Renal function is an indication of the state of the kidney, measured by glomerular filtration rate (GFR). In a healthy person, GFR would be greatly increased by:Â
Your Answer: Vasoconstriction of glomerular afferent arterioles
Correct Answer: Substantial increases in renal blood flow
Explanation:An increase in the rate of renal blood flow (RBF) greatly increases the glomerular filtration rate (GFR). The more plasma available (from increased RBF), the more filtrate is formed. Glomerular filtration rate (GFR) is the volume of fluid filtered from the renal (kidney) glomerular capillaries into the Bowman’s capsule per unit time. Central to the physiologic maintenance of GFR is the differential basal tone of the afferent and efferent arterioles.
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This question is part of the following fields:
- Physiology
- Renal
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Question 31
Incorrect
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A 58-year-old woman presents with signs of inflammation in the first metatarsophalangeal joint: redness, swelling and pain. The analysis of synovial fluid reveals needle-shaped, strongly negatively birefringent crystals. What's the most likely diagnosis in this case?
Your Answer: Pseudogout
Correct Answer: Gout
Explanation:Gout is a rheumatic disease caused by the precipitation of monosodium urate crystals into tissues, usually joints. This causes acute or chronic pain; the acute illness initially affects only one joint, often the first metatarsophalangeal joint. The diagnosis of the disease requires the identification of crystal in the synovial fluid. These crystals are needle-shaped and strongly negatively birefringent.
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This question is part of the following fields:
- Orthopaedics
- Pathology
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Question 32
Correct
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A patient is diagnosed with lung cancer. His physician told him that his lung cancer type is aggressive. It can grow rapidly and may undergo early metastasis, however it is very sensitive to chemotherapy and radiotherapy. Which lung cancer type is most likely present
Your Answer: Small-cell carcinoma
Explanation:Small-cell lung carcinoma (SCLC) is a type of highly malignant cancer that most commonly arises within the lung. SCLC usually metastasizes widely very early on in the natural history of the tumour, and in nearly all cases responds dramatically to chemotherapy and/or radiotherapy. Surgery has no role in the treatment of this disease.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 33
Correct
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A 68-year-old man underwent sigmoid resection with diverting colostomy for a ruptured sigmoid diverticulum 10 days ago. He received gentamicin and ampicillin post-op. 2 days after he was discharged from the hospital, he was readmitted because of high grade fever and chills. His blood culture grew Gram-negative bacilli. Which organism is most likely responsible for the patient's infection?
Your Answer: Bacteroides fragilis
Explanation:Bacteroides fragilis is an anaerobic, Gram-negative, rod-shaped bacterium. It is part of the normal flora of the human colon and is generally a commensal, but can cause infection if displaced into the bloodstream or surrounding tissue following surgery, disease, or trauma.
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This question is part of the following fields:
- Microbiology
- Pathology
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Question 34
Correct
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A 50-year-old man is diagnosed with emphysema and cirrhosis of the liver. Which of the following condition may be the cause of both cirrhosis and emphysema in this patient?
Your Answer: Alpha1-antitrypsin deficiency
Explanation:Alpha-1 antitrypsin (A1AT) deficiency is a condition characterised by the lack of a protein that protects the lungs and liver from damage, called alpha1-antytripsin. The main complications of this condition are liver diseases such as cirrhosis and chronic hepatitis, due to accumulation of abnormal alpha 1-antytripsin and emphysema due to loss of the proteolytic protection of the lungs.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 35
Correct
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The renal tubule is the portion of the nephron that contains the fluid that has been filtered by the glomerulus. Which of the following substances is actively secreted into the renal tubules?
Your Answer: Potassium
Explanation:The renal corpuscle filters out solutes from the blood, delivering water and small solutes to the renal tubule for modification. In normal circumstances more than 90% of the filtered load of K is reabsorbed by the proximal tubules and loops of Henlé and almost all K appearing in the urine has been secreted by the late distal tubules and collecting tubules. So the rate of excretion is usually independent of the rate of filtration, but is closely tied to the rate of secretion and control of K excretion, largely accomplished by control of the secretion rate. Around 65–70% of the filtered potassium is reabsorbed along with water in the proximal tubule and the concentration of potassium in the tubular fluid varies little from that of the plasma.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 36
Incorrect
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Which of the following is found to be elevated in a case of hepatocellular carcinoma?
Your Answer: hCG
Correct Answer: AFP
Explanation:Alpha-fetoprotein (AFP) is a glycoprotein that is normally produced by the yolk sac of the embryo, and then the fetal liver. It is elevated in the new-born and thus, also in the pregnant women. Eventually, it decreases in the first year of life to reach the adult normal value of < 20 ng/ml by 1 year of age. Markedly elevated levels (>500 ng/ml) in a high-risk patient is considered diagnostic for primary hepatocellular carcinoma (HCC). Moreover, due to smaller tumours secreting less quantities of AFP, rising levels can be a better indication. However, not all hepatocellular carcinomas produce AFP. Also, the level of AFP is not a prognostic factor. Populations where hepatitis B and HCC are common (e.g.: sub-Saharan Africans, ethnic Chinese) can see AFP levels as high as 100,000 ng/ml, whereas levels are low (about 3000 ng/ml) in regions with lesser incidences of HCC.
AFP can also be elevated up to 500 ng/ml in conditions like embryonic teratocarcinomas, hepatoblastomas, fulminant hepatitis, hepatic metastases from gastrointestinal tract cancers, some cholangiocarcinomas). Lesser values are seen in acute and chronic hepatitis.
Overall, the sensitivity of AFP value ≥20 ng/ml is 39-64% and the specificity is 76%–91%. Value of 500 ng/ml is considered as the diagnostic cut-off level for HCC.
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This question is part of the following fields:
- Gastrointestinal; Hepatobiliary
- Pathology
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Question 37
Correct
-
Which of the following is the cause of flattened (notched) T waves on electrocardiogram (ECG)?
Your Answer: Hypokalaemia
Explanation:The T-wave is formed due to ventricular repolarisation. Normally, it is seen as a positive wave. It can be normally inverted (negative) in V1 (occasionally in V2-3 in African-Americans/Afro-Caribbeans). Hyperacute T-waves are the earliest ECG change of acute myocardial infarction. ECG findings of hyperkalaemia include high, tent-shaped T-waves, a small P-wave and a wide QRS complex. Hypokalaemia results in flattened (notched) T-waves, U-waves, ST-segment depression and prolonged QT interval.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 38
Incorrect
-
A 45 year-old male, with behavioural changes developed euvolemic hyponatraemia. Which of the following conditions most likely predisposed the patient to develop euvolemic hyponatraemia?
Your Answer: Protracted vomiting
Correct Answer: Psychosis
Explanation:In euvolemic hyponatraemia, there is volume expansion in the body, there is no oedema, but hyponatremia occurs. Causes include: state of severe pain or nausea, psychosis, brain trauma, SIADH, hypothyroidism and glucocorticoid deficiency.
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This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
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Question 39
Incorrect
-
Some substances, such as Chromium-51 and Technetium-99, are freely filtered but not secreted or absorbed by the kidney. In these cases, their clearance rate is equal to:
Your Answer: Filtration fraction
Correct Answer: Glomerular filtration rate
Explanation:If a substance passes through the glomerular membrane with perfect ease, the glomerular filtrate contains virtually the same concentration of the substance as does the plasma and if the substance is neither secreted nor reabsorbed by the tubules, all of the filtered substance continues on into the urine. Glomerular filtration rate (GFR) describes the flow rate of filtered fluid through the kidney.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 40
Incorrect
-
A 27-year-old female was admitted due to severe dehydration. The patient also complained of chest tightness, thus an ECG was requested. The ECG strip showed an isoelectric ST segment, upright T wave, with prominent U waves. What is the most likely electrolyte abnormality responsible for these ECG tracing?
Your Answer: Hyperkalaemia
Correct Answer: Hypokalaemia
Explanation:U waves are prominent if it is >1-2mm or 25% of the height of the T wave. Abnormally prominent U waves are characteristically seen in severe hypokalaemia.
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This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
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Question 41
Correct
-
Increased resistance to flow of blood in cerebral vessels is most likely seen in:
Your Answer: Elevation in systemic arterial pressure from 100 to 130 mmHg
Explanation:Constant cerebral blood flow is maintained by autoregulation in the brain, which causes an increase in local vascular resistance to offset an increase in blood pressure. There will be an increase in cerebral blood flow (and decrease in resistance to cerebral blood flow) with a decrease in arterial oxygen or an increase in arterial CO2. Similarly, a decrease in viscosity will also increase the blood flow. Due to increased brain metabolism and activity during a seizure, there will also be an increase in the cerebral blood flow.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 42
Correct
-
Which of the following compensatory parameters is responsible for causing an increase in the blood pressure in a 30 year old patient with a BP of 40 mmHg?
Your Answer: Baroreceptor reflex
Explanation:The baroreflex or baroreceptor reflex is one of the body’s homeostatic mechanisms for regulating blood pressure. It provides a negative feedback response in which an elevated blood pressure will causes blood pressure to decrease; similarly, decreased blood pressure depresses the baroreflex, causing blood pressure to rise. The system relies on specialised neurones (baroreceptors) in the aortic arch, carotid sinuses and elsewhere to monitor changes in blood pressure and relay them to the brainstem. Subsequent changes in blood pressure are mediated by the autonomic nervous system. Baroreceptors include those in the auricles of the heart and vena cava, but the most sensitive baroreceptors are in the carotid sinuses and aortic arch. The carotid sinus baroreceptors are innervated by the glossopharyngeal nerve (CN IX); the aortic arch baroreceptors are innervated by the vagus nerve (CN X).
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 43
Incorrect
-
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: 1 ml
Correct Answer: 60 ml
Explanation:By Fick’s principle, cardiac output can be calculated as follows: VO2 = CO × (CAO2– CVO2) where VO2= oxygen consumption, CO = cardiac output, CAO2 = arterial oxygen content and CvO2 = mixed venous oxygen content. Thus, in the given problem, 300 ml/min = CO × (20 – 15) ml/100 ml CO = 300 × 100/5 ml/min CO = 6000 ml/min. Also, cardiac output = stroke volume × heart rate. Thus, 6000 ml/min = stroke volume × 100 beats/min. Hence, stroke volume = 6000/100 ml/min which is 60 ml/min.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 44
Correct
-
Extracellular body fluid as compared with intracellular body fluid:
Your Answer: Is relatively rich in glucose
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. Ensuring the right amount of body water is part of fluid balance, an aspect of homeostasis. The extracellular fluid (ECF) includes all fluids outside the cells. This fluid can be divided into three fluid departments: interstitial (in the tissue spaces) fluid, blood plasma and lymph, and specialised compartments called transcellular fluid. The extracellular fluid surrounds all the cells in the body and is in equilibrium with the intracellular fluid. So, its composition must remain fairly constant even though substances are passing into and out of the cells. The interstitial fluid, though called a fluid, is in a reality a gel-like composition made up of: water, proteoglycan molecules and collagen. The extracellular fluid constitutes 40% of total body water, with intracellular fluid making up the remaining 60%. It is relatively rich in glucose.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 45
Correct
-
Lung compliance is increased by:
Your 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 46
Correct
-
A 14 year-old girl is found to have haemophilia B. What pathological problem does she have?
Your Answer: Deficiency of factor IX
Explanation:Haemophilia B (also known as Christmas disease) is due to a deficiency in factor IX. Haemophilia A is due to a deficiency in factor VIII.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 47
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 48
Correct
-
A 78-year-old diabetic man undergoes renal function tests. Which of the following substances will be the most accurate for measuring glomerular filtration rate (GFR)?
Your Answer: Inulin
Explanation:Glomerular filtration rate (GFR) is the volume of fluid filtered from the renal glomerular capillaries into the Bowman’s capsule per unit time. Clinically, this is often measured to determine renal function. Inulin was originally used as it is not reabsorbed by the kidney after glomerular filtration, therefore its rate of excretion is directly proportional to the rate of filtration of water and solutes across the glomerular filter. However, in clinical practice, creatinine clearance is used to measure GFR. Creatinine is an endogenous molecule, synthesised in the body, that is freely filtered by the glomerulus (but also secreted by the renal tubules in very small amounts). Creatinine clearance exceeds GFR due to creatinine secretion, and is therefore a close approximation of the GFR.
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This question is part of the following fields:
- Physiology
- Renal
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Question 49
Incorrect
-
A 69 Year old lady presented to the emergency department following a massive myocardial infarction. She was found to be in hypotensive shock with focal neurological signs. Unfortunately the patient demised. What would be the expected findings on the brain biopsy?
Your Answer: Caseous necrosis
Correct Answer: Liquefactive necrosis
Explanation:Liquefactive necrosis is often associated with bacterial or fungal infections. However, hypoxic death of cells within the central nervous system can also result in liquefactive necrosis. The focal area is soft with a liquefied centre containing necrotic debris and dead white cells. This may later be enclosed by a cystic wall
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This question is part of the following fields:
- Cell Injury & Wound Healing; Neurology
- Pathology
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Question 50
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: 22 mmHg × min/l
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 51
Correct
-
A 25-year-old woman complains of generalised swelling and particularly puffiness around the eyes which is worst in the morning. Laboratory studies showed:
Blood urea nitrogen (BUN) = 30 mg/dl
Creatinine = 2. 8 mg/dl
Albumin = 2. 0 mg/dl
Alanine transaminase (ALT) = 25 U/l
Bilirubin = 1 mg/dl
Urine analysis shows 3+ albumin and no cells.
Which of the following is the most likely diagnosis?Your Answer: Nephrotic syndrome
Explanation:Nephrotic syndrome is a disorder in which the glomeruli have been damaged, characterized by:
– Proteinuria (>3.5 g per 1.73 m2 body surface area per day, or > 40 mg per square meter body surface area per hour in children)
– Hypoalbuminemia (< 2,5 g/dl) – Hyperlipidaemia, and oedema (generalized anasarca).
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This question is part of the following fields:
- Physiology
- Renal
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Question 52
Correct
-
Out of the given options, which of the following is the most likely diagnosis in a 25-year old gentleman presenting with a testicular germ cell tumour?
Your Answer: Embryonal carcinoma
Explanation:Embryonal carcinoma is a non-seminomatous germ cell tumour of the testis, accounting for 25% testicular tumours. Other germ cell tumours include seminoma, teratoma and choriocarcinoma. Embryonal carcinomas are known to occur in men aged 25-35 years, and occasionally in teens. They are rarely seen in ovaries of females. It can spread to the vas deferens and also to the aortic lymph nodes. Embryonal carcinomas are known to have elements of fetal origin such as cartilage. Usually, the main tumour is about 2.5cm long, with an extension of 8-9cm along the testicular cord. Contiguous spread to the testicle is seen in less than 1% cases.
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This question is part of the following fields:
- Pathology
- Urology
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Question 53
Correct
-
During cardiac catheterisation in a 20-year old man, the following data is obtained: Pressure (mmHg), O2 saturation (%) Right atrium 7 (N = 5) 90 (N = 75), Right ventricle 35/7 (N = 25/5) 90 (N = 75), Pulmonary artery 35/8 (N = 25/15), 90 (N = 75), Left atrium 7 (N = 9) 95 (N = 95), Left ventricle 110/7 (N = 110/9) 95 (N = 95), Aorta 110/75 (N = 110/75) 95 (N = 95) where N = Normal value. What is the likely diagnosis?
Your Answer: Atrial septal defect
Explanation:A congenital heart disease, ASD or atrial septal defect leads to a communication between the right and left atria due to a defect in the interatrial septum. This leads to mixing of arterial and venous blood from the right and left side of the heart. The hemodynamic significance of this defect depends on the presence of shunting of blood. Normally, the left side of the heart has higher pressure than the right as the left side has to pump blood throughout the body. A large ASD (> 9 mm) will result in a clinically significant left-to-right shunt, causing volume overload of the right atrium and ventricle, eventually leading to heart failure. Cardiac catheterization would reveal very high oxygen saturation in the right atrium, right ventricle and pulmonary artery. Eventually, the left-to-right shunt will lead to pulmonary hypertension and increased afterload in the right ventricle, along with the increased preload due to the shunted blood. This will either cause right ventricular failure, or raise the pressure in the right side of the heart to equal or more than that in the left. Elevation of right atrial pressure to that of left atrial pressure would thus lead to diminishing or complete cessation of the shunt. If left uncorrected, there will be reversal of the shunt, known as Eisenmenger syndrome, resulting in clinical signs of cyanosis as the oxygen-poor blood form right side of the heart will mix with the blood in left side and reach the peripheral vascular system.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 54
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: Foramen ovale
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 55
Correct
-
Glucose is the most important source of energy for cellular respiration. The transport of glucose in the renal tubular cells occurs via:
Your Answer: Secondary active transport with sodium
Explanation:In 1960, Robert K. Crane presented for the first time his discovery of the sodium-glucose cotransport as the mechanism for glucose absorption. Glucose transport through biological membranes requires specific transport proteins. Transport of glucose through the apical membrane of renal tubular as well as intestinal epithelial cells depends on the presence of secondary active Na+–glucose symporters, SGLT-1 and SGLT-2, which concentrate glucose inside the cells, using the energy provided by co-transport of Na+ ions down their electrochemical gradient.
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This question is part of the following fields:
- Physiology
- Renal
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Question 56
Incorrect
-
A 18-year old girl presents to her doctor with an excessively enlarged left breast as compared to the right breast since puberty. The most likely cause for this is:
Your Answer: Fibrocystic disease
Correct Answer: Virginal breast hypertrophy
Explanation:Virginal breast hypertrophy’ is the term assigned to excessive growth of breasts during puberty and is a common phenomenon. It is also known as ‘juvenile macromastia’ or ‘ juvenile gigantomastia’. The breast hypertrophy often starts with menarche and occasionally occurs in growth spurts. These spurts can cause physical discomfort, red and itchy skin or pain in the breasts. The breasts can also grow continuously over several years and lead to overdevelopment of a normal breast. Nipples also undergo enlargement along with the breasts.
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This question is part of the following fields:
- Pathology
- Women's Health
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Question 57
Correct
-
Which of the following substances is most likely to cause pulmonary vasodilatation?
Your Answer: Nitric oxide
Explanation:In the body, nitric oxide is synthesised from arginine and oxygen by various nitric oxide synthase (NOS) enzymes and by sequential reduction of inorganic nitrate. The endothelium of blood vessels uses nitric oxide to signal the surrounding smooth muscle to relax, so dilating the artery and increasing blood flow. Nitric oxide/oxygen blends are used in critical care to promote capillary and pulmonary dilation to treat primary pulmonary hypertension in neonatal patients post-meconium aspiration and related to birth defects.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 58
Correct
-
A young 16 year old boy presented to the ENT clinic with a history of sore throat for the past 1 day. On examination there was a pharyngeal purulent discharge. Which of the following types of inflammation is seen in this boy?
Your Answer: Acute inflammation
Explanation:A 1 day history suggests the purulent discharge is due to acute inflammation. Acute inflammation has 3 features:
1) the affected area is occupied by a purulent discharge composed of proteins, fluids and cells from local blood vessels
2) the infective agent i.e. bacteria is present in the affected area
3) the damaged tissue can be liquified and the debris removed from the site.
If the inflammation lasts over weeks or months, then it is termed as chronic inflammation.
Granulomatous inflammation is characterised by the presence and formation of granulomas.
Exudate is not a feature of resolution or a complication of inflammation.
Abscess formation takes more than 1 day to form and is usually within a capsule/cavity.
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This question is part of the following fields:
- Inflammation & Immunology; Respiratory
- Pathology
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Question 59
Correct
-
An excised lesion is found to be a premalignant during examination by the pathologist. What is the most likely histopathology report of this lesion?
Your Answer: Solar keratosis
Explanation:Premalignant condition is a state of disordered morphology of cells that is associated with an increased risk of cancer. If this condition is left untreated, it may lead to the development of cancer. The following are examples of pre-malignant lesions: actinic keratosis, Barret’s oesophagitis, atrophic gastritis, ductal carcinoma in situ, dyskeratosis congenita, sideropenic dysphagia, lichen planus, oral submucous fibrosis, solar elastosis, cervical dysplasia, leucoplakia and erythroplakia.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 60
Correct
-
A 55-years-old man presented to the emergency department complaining of a squeezing sensation in his chest that has spread to his neck with associated worsening shortness of breath. Which of these laboratory tests would you ask for in this patient:
Your Answer: Creatine kinase-MB
Explanation:Creatine kinase-MB is a test that usually is ordered when the patient has chest pain as a cardiac marker. When a heart attack is suspected and a troponin test (which is more specific for heart damage), is not available CK-MB is ordered. There are 3 forms of CK: CK-MM, CK-BB and CK-MB. CK-MB is commonly found in heart tissue, therefore injured heart muscle cells release CK-MB into the blood. Elevated CK-MB levels indicate that it is probable that a person has recently had a heart attack.
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This question is part of the following fields:
- Cardiovascular
- Pathology
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Question 61
Correct
-
The bronchial circulation is a part of the circulatory system that supplies nutrients and oxygen to the pulmonary parenchyma. What percentage of cardiac output is received by bronchial circulation?
Your Answer: 2%
Explanation:The bronchial circulation is part of the systemic circulation and receives about 2% of the cardiac output from the left heart. Bronchial arteries arise from branches of the aorta, intercostal, subclavian or internal mammary arteries. The bronchial arteries supply the tracheobronchial tree with both nutrients and O2. It is complementary to the pulmonary circulation that brings deoxygenated blood to the lungs and carries oxygenated blood away from them in order to oxygenate the rest of the body.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 62
Correct
-
What causes a reduction in pulmonary functional residual capacity?
Your Answer: Pulmonary fibrosis
Explanation:Pulmonary functional residual capacity (FRC) is = volume of air present in the lungs at the end of passive expiration.
Obstructive diseases (e.g. emphysema, chronic bronchitis, asthma) = an increase in FRC due to an increase in lung compliance and air trapping.
Restrictive diseases (e.g. pulmonary fibrosis) result in stiffer, less compliant lungs and a reduction in FRC.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 63
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: Plasma creatinine concentration
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 64
Incorrect
-
A 55-year-old male chronic smoker is diagnosed with non-small-cell cancer. His right lung underwent complete atelectasis and he has a 7cm tumour involving the chest wall. What is the stage of the lung cancer of this patient?
Your Answer: T4
Correct Answer: T3
Explanation:Non-small-cell lung cancer is staged through TNM classification. The stage of this patient is T3 because based on the TNM classification the tumour is staged T3 if > 7 cm or one that directly invades any of the following: Chest wall (including superior sulcus tumours), diaphragm, phrenic nerve, mediastinal pleura, or parietal pericardium; or the tumour is in the main bronchus < 2 cm distal to the carina but without involvement of the carina, Or it is associated with atelectasis/obstructive pneumonitis of the entire lung or separate tumour nodule(s) in the same lobe.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 65
Correct
-
In a cardiac cycle, what event does the closing of atrioventricular (AV) valves coincide with?
Your Answer: First heart sound
Explanation:In the cardiac cycle, the closing of the atrioventricular (AV) valves coincides with the onset of ventricular systole. This event marks the beginning of the isovolumetric contraction phase, where the ventricles begin to contract, but the volume of blood in the ventricles remains the same because both the AV valves and the semilunar valves (aortic and pulmonary valves) are closed. The closing of the AV valves produces the first heart sound, known as “S1” or “lub.”
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 66
Correct
-
A diabetic 58-year-old man, after the injection of radiographic contrast, has a decreased urine output and decreased level of consciousness. Which of the following conditions has he most likely developed
Your Answer: Acute tubular necrosis
Explanation:Acute tubular necrosis (ATN) involves damage to the tubule cells of the kidneys and is the most common cause of acute kidney injury. ATN in the majority of the cases is caused by ischaemia of the kidneys due to lack of perfusion and oxygenation but it may also occur due to poison or harmful substance. Contrast used for radiology may cause ATN in patients with several risk factors e.g. diabetic nephropathy. Symptoms may include oliguria, nausea, fluid retention, fatigue and decreased consciousness.
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This question is part of the following fields:
- Pathology
- Renal
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Question 67
Incorrect
-
A 14 year old girl suffers from haemophilia A and chronic knee pain with progressive swelling and deformity over the last 4 years. Test results reveal a significantly reduced factor VIII activity. Which of the following is seen in the knee joint space after an acute painful episode?
Your Answer: Curschmann's spirals
Correct Answer: Cholesterol crystals
Explanation:Due to breakdown of the red blood cell membrane in haemophilic patients, cholesterol crystals are formed by the lipids. On the other hand lipofuscin deposition does not occur in haemolysis or haemorrhage. Neutrophil accumulation suggests acute inflammation. Anthracotic pigment is an exogenous carbon pigment that deposits in the lung from dust. Russell bodies are intracellular accumulations of immunoglobins in plasma cells. Curschmann’s spirals and Charcot Leyden crystals are pathognomonic of asthma.
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This question is part of the following fields:
- Cell Injury & Wound Healing; Haematology
- Pathology
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Question 68
Incorrect
-
An 8 year old boy presents with a history of repeated bacterial/pyogenic infections. He had a normal recovery from chickenpox and measles and shows normal antibody response. A decrease in which of the cell types can best explain this history of repeated pyogenic infections?
Your Answer: B lymphocytes
Correct Answer: Neutrophils
Explanation:A decrease in the number of granulocytes, particularly neutrophils is known as agranulocytosis and it increases the susceptibility of an individual towards recurrent infections. Neutropenia can be either due to decreased production or increased elimination of neutrophils.
Ineffective agranulopoiesis is seen in: 1. myeloid stem cell suppression, 2. disease conditions associated with granulopoiesis such as megaloblastic anaemia and myelodysplastic syndromes, 3. rare genetic diseases, 4. splenic sequestration and 5. increased peripheral utilization.
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This question is part of the following fields:
- Inflammation & Immunology
- Pathology
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Question 69
Correct
-
A 7-year-old girl is given cephalexin to treat an infection and develops hives, with localised facial oedema. Which of the following conditions will cause localised oedema?
Your Answer: Angio-oedema
Explanation:Angio-oedema, is the rapid swelling of the skin, mucosa and submucosal tissues. The underlying mechanism typically involves histamine or bradykinin. The version related to histamine is to due an allergic reaction to agents such as insect bites, food, or medications. The version related to bradykinin may occur due to an inherited C1 esterase inhibitor deficiency, medications e.g. angiotensin converting enzyme inhibitors, or a lymphoproliferative disorder.
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This question is part of the following fields:
- Physiology
- Renal
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Question 70
Correct
-
Glucose is not secreted by the kidneys, and is filtered without a limit. What is the transport maximum for glucose?
Your Answer: 300 mg/dl
Explanation:Transport maximum (or Tm) refers to the point at which increases in concentration do not result in an increase in movement of a substance across a membrane. Glucose is not secreted, thus excretion = filtration – reabsorption. Both filtration and reabsorption are directly proportional to the concentration of glucose in the plasma. However, reabsorption has a transport maximum of about 300 mg/dl in healthy nephrons, while filtration has effectively no limit (within reasonable physiological ranges). So, if the concentration rises above 300 mg/dl, the body cannot retain all the glucose, leading to glucosuria. Glucosuria 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:
- Physiology
- Renal
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Question 71
Correct
-
A 65-year-old smoker presents with painless haematuria, urinary frequency and urgency. He is diagnosed with bladder cancer. Which is the most likely type?
Your Answer: Transitional cell carcinoma
Explanation:90% of bladder cancers are transitional cell carcinomas derived from the bladder urothelium. Risk factors include industrial chemicals, smoking and infection. Schistosomiasis and bladder stones predispose to the squamous cell variety.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 72
Correct
-
The physician suggested lifestyle modification for his patient because his present condition could increase his risk for the development of adenocarcinoma of the oesophagus. What is the most common predisposing factor for the development of adenocarcinoma of the oesophagus?
Your Answer: Gastro-oesophageal reflux disease
Explanation:Barret’s oesophagus is attributed primarily to gastro-oesophageal reflux disease. The chronic acidic environment damages the squamous epithelial lining of the oesophagus, and subsequently undifferentiated pluripotent stem cells develop into columnar epithelium, this is then known as Barret’s oesophagitis.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 73
Correct
-
Which of the following chemotherapy regimens is most likely to be used in colorectal cancer?
Your Answer: FOLFOX
Explanation:Chemotherapy regimens are often identified by acronyms, identifying the agents used in the drug combination. However, the letters used are not consistent across regimens. FOLFOX is a chemotherapy regimen used for the treatment of colorectal cancer, made up of the following drugs: • FOL: fluorouracil (5-fluorouracil or 5-FU) • F: folinic acid (leucovorin) • OX: oxaliplatin (Eloxatin®).
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This question is part of the following fields:
- Pathology
- Pharmacology
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Question 74
Correct
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The most important difference between interstitial fluid and plasma is the:
Your 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 75
Incorrect
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Renin is secreted by pericytes in the vicinity of the afferent arterioles of the kidney from the juxtaglomerular cells. Plasma renin levels are decreased in patients with:
Your Answer: Heart failure
Correct Answer: Primary aldosteronism
Explanation:Primary aldosteronism, also known as primary hyperaldosteronism or Conn’s syndrome, is excess production of the hormone aldosterone by the adrenal glands resulting in low renin levels. Most patients with primary aldosteronism (Conn’s syndrome) have an adrenal adenoma. The increased plasma aldosterone concentration leads to increased renal Na+ reabsorption, which results in plasma volume expansion. The increase in plasma volume suppresses renin release from the juxtaglomerular apparatus and these patients usually have low plasma renin levels. Salt restriction and upright posture decrease renal perfusion pressure and therefore increases renin release from the juxtaglomerular apparatus. Secondary aldosteronism is due to elevated renin levels and may be caused by heart failure or renal artery stenosis.
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This question is part of the following fields:
- Physiology
- Renal
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Question 76
Incorrect
-
Which of the following physiological changes will you see in a young man who has been trekking in the Himalayas for 3 years?
Your Answer: Pulmonary vasodilatation
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 77
Incorrect
-
Which of the following abnormalities can be seen in patients with hypermagnesemia?
Your Answer: Generalised seizures
Correct Answer: Respiratory depression
Explanation:Hypermagnesemia is an electrolyte disturbance in which there is a high level of magnesium in the blood. It is defined as a level greater than 1.1 mmol/L. Symptoms include weakness, confusion, decreased breathing rate, and cardiac arrest.
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This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
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Question 78
Incorrect
-
A 75-year old patient is in atrial fibrallation but has never been on anticoagulation therapy. To reduce the risk of future emboli, she would benefit from starting on long-term warfarin. Arterial emboli leading to acute limb ischaemia most commonly lodge at which one of the following sites?
Your Answer:
Correct Answer: Common femoral artery
Explanation:The common femoral artery is the commonest site of arterial emboli causing acute limb ischemia. The treatment of choice is urgent femoral embolectomy.
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This question is part of the following fields:
- Cardiovascular
- Pathology
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Question 79
Incorrect
-
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:
Correct 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 80
Incorrect
-
A 10-year-old boy was sent for an x-ray of the leg because he was complaining of pain and swelling. The x-ray showed the classic sign of Codman's triangle. What is the most likely diagnosis of this patient?
Your Answer:
Correct Answer: Osteosarcoma
Explanation:Codman’s triangle is the triangular area of new subperiosteal bone that is created when a lesion, often a tumour, raises the periosteum away from the bone. The main causes for this sign are osteosarcoma, Ewing’s sarcoma, eumycetoma, and a subperiosteal abscess.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 81
Incorrect
-
What is the primary function of the Kupffer cells found in the liver?
Your Answer:
Correct Answer: Recycling of old red blood cells
Explanation:Kupffer cells found in the liver are part of the monocyte-reticular system. They are specialised macrophages and primarily function to recycle old and damaged RBCs. The RBCs are phagocytosed and the haemoglobin is broken down into haem and globin. The haem is further broken down into iron that is recycled and bilirubin that is conjugated with glucuronic acid and excreted in the bile.
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This question is part of the following fields:
- Inflammation & Immunology; Hepatobiliary
- Pathology
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Question 82
Incorrect
-
A 24-year-old patient with recurrent episodes of deep vein thrombosis presents again to the clinic. Deficiency of which of the following blood proteins is the most probable cause of this episode?
Your Answer:
Correct Answer: Antithrombin III
Explanation:Antithrombin III (ATIII) is a blood protein that acts by inhibiting blood coagulation by neutralizing the enzymatic activity of thrombin.
Antithrombin III deficiency is an autosomal dominant disorder that leads to an increased risk of venous and arterial thrombosis. Clinical manifestations typically appear in young adulthood.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 83
Incorrect
-
A 55 year-old construction worker is diagnosed with malignant mesothelioma. Exposure to which substance increased his risk in developing mesothelioma?
Your Answer:
Correct Answer: Asbestos
Explanation:Mesothelioma is a rare, aggressive form of cancer that develops in the lining of the lungs, abdomen or heart. It is linked to inhalation of asbestos commonly used in ship building and the insulation industry. It has no known cure and has a very poor prognosis.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 84
Incorrect
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Question 85
Incorrect
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A 45-year-old pregnant woman develops high blood pressure at 20 weeks. She complains of headaches and swollen feet, and a test reveals proteinuria (350 mg/day). Which of the following is the most likely diagnosis?
Your Answer:
Correct Answer: Pre-eclampsia
Explanation:Pre-eclampsia (PE) is a disorder of pregnancy characterized by the onset of high blood pressure (two separate readings taken at least 6 h apart of 140/90 or more) and often a significant amount of protein in the urine (>300 mg of protein in a 24-h urine sample). While blood pressure elevation is the most visible sign of the disease, it involves generalised damage to the maternal endothelium of the kidneys and liver, with the release of vasopressive factors only secondary to the original damage. Pre-eclampsia may develop at varying times within pregnancy and its progress differs among patients; most cases present pre-term. It has no known cure apart from ending the pregnancy (induction of labour or abortion). It may also present up to 6 weeks post partum. Risk factors for pre-eclampsia include obesity, prior hypertension, older age, and diabetes mellitus.
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This question is part of the following fields:
- Physiology
- Renal
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Question 86
Incorrect
-
Loperamide is a drug used to treat diarrhoea. What is the mechanism of action of loperamide?
Your Answer:
Correct 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 87
Incorrect
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A 30 year old female presented in the emergency with an irregular pulse. Her ECG showed absent P-waves with irregular RR interval. What is the most likely diagnosis?
Your Answer:
Correct Answer: Atrial fibrillation
Explanation:Atrial fibrillation is one of the most common cardiac arrhythmias. It is often asymptomatic but may present with symptoms of palpitations, fainting, chest pain and heart failure. Characteristic findings are: absence of P-waves, unorganised electrical activity in their place, irregularity of RR interval due to irregular conduction of impulses to the ventricles and if paroxysmal AF is suspected, episodes may be documented with the use of Holter monitoring
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 88
Incorrect
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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:
Correct Answer: Compliance of the lung
Explanation:Acute or adult respiratory distress syndrome (ARDS) is a reaction to several forms of lung injuries and is commonly associated with sepsis and SIRS (systemic inflammatory response syndrome), severe traumatic injury, severe head injury, narcotics overdose, drowning, pulmonary contusion, and multiple blood transfusions. There is an increase in risk due to pre-existing liver disease or coagulation abnormalities. It results due to indirect toxic effects of neutrophil-derived inflammatory mediators in the lungs. ARDS is defined by the 1994 American–European Consensus Committee as the acute onset of bilateral infiltrates on chest X-ray, a partial pressure of arterial oxygen (pa(O2)) to fraction of inspired oxygen Fi(O2) ratio of less than 200 mmHg and a pulmonary artery occlusion pressure of less than 18 or the absence of clinical evidence of left arterial hypertension. ARDS is basically pulmonary oedema in the absence of volume overload or poor left ventricular function. This is different from acute lung injury, which shows a pa(O2)/Fi(O2) ratio of less than 300 mmHg. Pathogenesis of ARDS starts from damage to alveolar epithelium and vascular endothelium, causing increased permeability. Damage to surfactant-producing type II cells disrupts the production and function of pulmonary surfactant, causing micro atelectasis and poor gas exchange. There is a decrease in lung compliance and increase in work of breathing. Eventually, there is resorption of alveolar oedema, regeneration of epithelial cells, proliferation and differentiation of type II alveolar cells and alveolar remodelling. Some show resolution and some progress to fibrosing alveolitis, which involves the deposition of collagen in alveolar, vascular and interstitial spaces.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 89
Incorrect
-
Leakage from a silicone breast implant can lead to:
Your Answer:
Correct Answer: Pain and contracture
Explanation:Breast implants are mainly: saline-filled and silicone gel-filled. Complications include haematoma, fluid collections, infection at the surgical site, pain, wrinkling, asymmetric appearance, wound dehiscence and thinning of the breast tissue.
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This question is part of the following fields:
- Pathology
- Women's Health
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Question 90
Incorrect
-
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:
Correct 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 91
Incorrect
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After finding elevated PSA levels, a 69-year-old man undergoes a needle biopsy and is diagnosed with prostatic cancer. What is the stage of this primary tumour?
Your Answer:
Correct Answer: T1c
Explanation:The AJCC uses a TNM system to stage prostatic cancer, with categories for the primary tumour, regional lymph nodes and distant metastases:
TX: cannot evaluate the primary tumour T0: no evidence of tumour
T1: tumour present, but not detectable clinically or with imaging T1a: tumour was incidentally found in less than 5% of prostate tissue resected (for other reasons)
T1b: tumour was incidentally found in more than 5% of prostate tissue resected
T1c: tumour was found in a needle biopsy performed due to an elevated serum prostate-specific antigen
T2: the tumour can be felt (palpated) on examination, but has not spread outside the prostate
T2a: the tumour is in half or less than half of one of the prostate gland’s two lobes
T2b: the tumour is in more than half of one lobe, but not both
T2c: the tumour is in both lobes
T3: the tumour has spread through the prostatic capsule (if it is only part-way through, it is still T2)
T3a: the tumour has spread through the capsule on one or both sides
T3b: the tumour has invaded one or both seminal vesicles
T4: the tumour has invaded other nearby structures.
In this case, the tumour has a T1c stage.
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This question is part of the following fields:
- Pathology
- Urology
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Question 92
Incorrect
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Question 93
Incorrect
-
A 65 year old man with a history of diabetes and hypertension presented with a stroke a few months ago severely affecting his speech and movement in the right arm and leg. A cerebral angiogram revealed a middle cerebral artery occlusion. A recent CT scan was done which revealed a 5 cm cystic space in his left parietal lobe. This lesion is a result of which of the following forms of resolution?
Your Answer:
Correct Answer: Liquefactive necrosis
Explanation:Characteristically, the brain will undergo liquefactive necrosis following ischaemic injury. This leaves a cystic space in that region which would show up on a CT scan. Atrophy would result in a generalized decrease in the brain size. Coagulative necrosis typically occurs in parenchymal organs e.g. the spleen or kidney which have a lower lipid content. Caseous necrosis is typical in granulomatous tuberculosis infection. Apoptosis will not form a cystic area as it is programmed cell death involving a individual cells. Gangrenous necrosis is characteristic of ischaemic injury of the lower limb and GI tract. Fibrinous necrosis results from necrotic damage to the blood vessels with the leaking of proteins into the vessel, appearing bright pink on H & E staining.
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This question is part of the following fields:
- Cell Injury & Wound Healing; Neurology
- Pathology
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Question 94
Incorrect
-
What's the nodal stage of a testicular seminoma if several lymph nodes between 2cm and 5cm are found?
Your Answer:
Correct Answer: N2
Explanation:According to the American Joint Committee on Cancer (AJCC) 2002 guidelines, the nodal staging of testicular seminoma is the following:
N0: no regional lymph node metastases
N1: metastasis with lymph nodes 2 cm or less in their greatest dimension or multiple lymph nodes, none more than 2 cm
N2: metastasis with lymph nodes greater than 2 cm but not greater than 5 cm in their greatest dimension, or multiple lymph nodes, any one mass greater than 2 cm, but not more than 5 cm
N3: metastasis with lymph nodes greater than 5 cm in their greatest dimension.
The patient in this case has N2 testicular seminoma. This TNM staging is extremely important because treatment options are decided depending on this classification.
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This question is part of the following fields:
- Pathology
- Urology
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Question 95
Incorrect
-
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:
Correct 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 96
Incorrect
-
A 38-year old woman presents to the clinic with a 2 cm eczema-like lesion on the areolar region of her left breast, for 5 months. Biopsy of the lesion showed large cells at the dermal-epidermal junction with positive staining for mucin. What is the likely diagnosis?
Your Answer:
Correct Answer: Paget’s disease of the breast
Explanation:Paget’s disease of the breast or nipple resembles eczema in appearance with an underlying carcinoma typically. The disease is usually unilateral and presents with inflammation, oozing and crusting along with a non-healing ulcer. Treatment is often delayed due to the innocuous appearance but can be fatal. It results due to spread of neoplastic cells from the ducts of the mammary gland to the epithelium.
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This question is part of the following fields:
- Pathology
- Women's Health
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Question 97
Incorrect
-
A 60-year-old woman has had persistent diarrhoea for a week. A stool test reveals an infection by Clostridium difficile. Which of the following antibiotics could be used to treat the infection?
Your Answer:
Correct Answer: Oral vancomycin
Explanation:Three antibiotics are effective against Clostridium difficile:
Metronidazole 500 mg orally three times daily is the drug of choice, because of superior tolerability, lower price and comparable efficacy.
Oral vancomycin 125 mg four times daily is second-line therapy in particular cases of relapse or where the infection is unresponsive to metronidazole treatment.
Thirdly, the use of linezolid might also be considered.
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This question is part of the following fields:
- Pathology
- Pharmacology
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Question 98
Incorrect
-
A 24-year old, lactating mother presents to the clinic with a tender, 1.5cm mass just below the right nipple, which shows multiple fissures. What finding is likely associated with her condition?
Your Answer:
Correct Answer: Staphylococcus aureus infection
Explanation:Breast abscess occur commonly in lactating mothers in the postpartum period due to cracking of the nipple. It is commonly caused due to Staphylococcus aureus infection. Fat necrosis usually results from trauma wherein an ill-defined mass is formed. Ductal carcinomas are malignant masses which are not tender usually, and rare in the young age group. Plasma cell mastitis affect women in an older age group. Sclerosing adenosis is a type of fibrocystic disease which can lead to a tender, cystic mass but no fissuring or cracks are seen in the nipple. Fibroadenoma and lipomas are non-tender, well-defined masses.
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This question is part of the following fields:
- Pathology
- Women's Health
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Question 99
Incorrect
-
A patient with this condition has extracellular fluid volume expansion:
Your Answer:
Correct Answer: Nephrotic syndrome
Explanation:Nephrotic syndrome is a syndrome comprising of signs of nephrosis, including proteinuria, hypoalbuminemia, and oedema. It is a component of glomerulonephritis, in which different degrees of proteinuria occur. Essentially, loss of protein through the kidneys leads to low protein levels in the blood , which causes water to be drawn into soft tissues (oedema). Severe hypoalbuminemia can also cause a variety of secondary problems, such as water in the abdominal cavity (ascites), around the heart or lung (pericardial effusion, pleural effusion), high cholesterol, loss of molecules regulating coagulation (hence increased risk of thrombosis). The most common sign is excess fluid in the body due to the serum hypoalbuminemia. Lower serum oncotic pressure causes fluid to accumulate in the interstitial tissues. Sodium and water retention aggravates the oedema.
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This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
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Question 100
Incorrect
-
A 56-year-old man undergoes tests to determine his renal function. His results over a period of 24 hours were:
Urine flow rate: 2. 0 ml/min
Urine inulin: 1.0 mg/ml
Plasma inulin: 0.01 mg/ml
Urine urea: 260 mmol/l
Plasma urea: 7 mmol/l
What is the glomerular filtration rate?Your Answer:
Correct Answer: 200 ml/min
Explanation:Glomerular filtration rate (GFR) is the volume of fluid filtered from the renal (kidney) glomerular capillaries into the Bowman’s capsule per unit time. GFR is equal to the inulin clearance because inulin is freely filtered into Bowman’s capsule but is not reabsorbed or secreted. The clearance (C) of any substance can be calculated as follows: C = (U × V)/P, where U and P are the urine and plasma concentrations of the substance, respectively and V is the urine flow rate. Thus, glomerular filtration rate = (1.0 × 2. 0)/0.01 = 200 ml/min.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 101
Incorrect
-
A 15-day old male baby was brought to the emergency department with sweating and his lips turning blue while feeding. He was born full term. On examination, his temperature was 37.9°C, blood pressure 75/45 mmHg, pulse was 175/min, and respiratory rate was 42/min. A harsh systolic ejection murmur could be heard at the left upper sternal border. X-ray chest showed small, boot-shaped heart with decreased pulmonary vascular markings. He most likely has:
Your Answer:
Correct Answer: Tetralogy of Fallot
Explanation:The most common congenital cyanotic heart disease and the most common cause of blue baby syndrome, Tetralogy of Fallot shows four cardiac malformations occurring together. These are ventricular septal defect (VSD), pulmonary stenosis (right ventricular outflow obstruction), overriding aorta (degree of which is variable), and right ventricular hypertrophy. The primary determinant of severity of disease is the degree of pulmonary stenosis. Tetralogy of Fallot is seen in 3-6 per 10,000 births and is responsible for 5-7% congenital heart defects, with slightly higher incidence in males. It has also been associated with chromosome 22 deletions and DiGeorge syndrome. It gives rise to right-to-left shunt leading to poor oxygenation of blood. Primary symptom is low oxygen saturation in the blood with or without cyanosis at birth of within first year of life. Affected children ay develop acute severe cyanosis or ‘tet spells’ (sudden, marked increase in cyanosis, with syncope, and may result in hypoxic brain injury and death). Other symptoms include heart murmur, failure to gain weight, poor development, clubbing, dyspnoea on exertion and polycythaemia. Chest X-ray reveals characteristic coeur-en-sabot (boot-shaped) appearance of the heart. Treatment consists of immediate care for cyanotic spells and Blalock–Taussig shunt (BT shunt) followed by corrective surgery.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 102
Incorrect
-
A 45-year-old man complains of shortness of breath, cough and chest pain. Chest X ray revealed a perihilar mass with bronchiectasis in the left mid-lung. Which of the following is most probably associated with these findings?
Your Answer:
Correct Answer: Bronchial carcinoid
Explanation:Bronchial carcinoids are neuroendocrine tumours that arise from Kulchitsky’s cells of the bronchial epithelium. Kulchitsky’s cells belong to the diffuse endocrine system. Patients affected by this tumour may be asymptomatic or may present with symptoms of airway obstruction, like dyspnoea, wheezing, and cough. Other common findings are recurrent pneumonia, haemoptysis, chest pain and paraneoplastic syndromes. Chest radiographs are abnormal in the majority of cases. Peripheral carcinoids usually present as a solitary pulmonary nodule. For central lesions common findings include hilar or perihilar masses with or without atelectasis, bronchiectasis, or consolidation. Bronchial carcinoids most commonly arise in the large bronchi causing obstruction.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 103
Incorrect
-
Which of the following will increase blood pressure and cause hypokalaemia?
Your Answer:
Correct Answer: Angiotensin II
Explanation:Angiotensin is a peptide that is released in response to a decrease in blood volume and blood pressure. It has multiple functions but mainly acts to cause vasoconstriction, increase BP and release aldosterone from the adrenal cortex. It is a powerful vasoconstrictor and release of aldosterone causes increased retention of sodium and excretion of potassium.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 104
Incorrect
-
A 27-year-old woman, who had been taking a combined oral contraceptive for 6 months, presented with inguinal pain and oedema of the left leg. Which of the following investigations would you recommend to help confirm the diagnosis?
Your Answer:
Correct Answer: Duplex scan
Explanation:Oral combined contraceptive pill (OCCP) is a drug used for birth control and treating a number of other conditions. Women who take the OCP have a higher risk of developing deep vein thrombosis (DVT), usually in the legs. Duplex ultrasonography is a safe and non-invasive technique which is used for diagnosing the presence of lower extremity thrombi.
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This question is part of the following fields:
- Cardiovascular
- Pathology
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Question 105
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 106
Incorrect
-
A 58-year-old woman diagnosed with deep vein thrombosis had been using warfarin for 10 days. When she presented to the doctor she had haemorrhagic bullae and necrotic lesions in her lower limbs and buttocks. Deficiency of which of the following proteins may have caused the necrotic skin lesions?
Your Answer:
Correct Answer: Protein C
Explanation:Warfarin-induced skin necrosis is a rare complication of anticoagulant therapy that requires immediate drug cessation. The most common cutaneous findings include petechiae that progress to ecchymoses and haemorrhagic bullae. Warfarin inactivates vitamin K-dependent clotting factors II, VII, IX, and X and vitamin K-dependent proteins C and S. The concentration of protein C falls more rapidly than other vitamin K-dependent factors because they have a shorter half-lives. Skin necrosis is seen mainly in patients with prior protein C deficiency.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 107
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 108
Incorrect
-
A 63-year-old woman complains of a new, persisting headache. She is diagnosed with vasculitis and the histopathological sample revealed giant-cell arteritis. What is the most probable diagnose?
Your Answer:
Correct Answer: Temporal arteritis
Explanation:Giant cell arteritis (GCA), also known as temporal arteritis, is the most common systemic inflammatory vasculitis that occurs in adults. It is of unknown aetiology and affects arteries large to small however the involvement of the superficial temporal arteries is almost always present. Other commonly affected arteries include the ophthalmic, occipital and vertebral arteries, therefore GCA can result in systemic, neurologic, and ophthalmologic complications. GCA usually is found in patients older than 50 years of age and should always be considered in the differential diagnosis of a new-onset headache accompanied by an elevated erythrocyte sedimentation rate. Diagnosis depends on the results of artery biopsy.
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This question is part of the following fields:
- Cardiovascular
- Pathology
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Question 109
Incorrect
-
Carbachol is a cholinergic agonist. In which of these cases should carbachol be administered?
Your Answer:
Correct Answer: Cataract surgery
Explanation:Carbachol (carbamylcholine) is a cholinergic agent, a choline ester and a positively charged quaternary ammonium compound. It is primarily used for various ophthalmic purposes, such as for treating glaucoma, or for use during ophthalmic surgery. It is usually administered topically to the eye or through intraocular injection. It is not well absorbed in the gastro-intestinal tract and does not cross the blood–brain barrier.
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This question is part of the following fields:
- Pathology
- Pharmacology
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Question 110
Incorrect
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What is the normal amount of oxygen that is carried in the blood?
Your Answer:
Correct Answer: 20 ml oxygen/100 ml blood
Explanation:Normally, 100 ml of blood contains 15g haemoglobin and a single gram of haemoglobin can bind to 1.34 ml oxygen when 100% saturated. Thus, 15 × 1.34 = 20 ml O2/100 ml blood. The haemoglobin in venous blood that is leaving the tissues is about 75% saturated with oxygen, and hence it carries about 15 ml O2/100 ml venous blood. This implies that for each 10 ml of blood, 5 ml oxygen is transported to the tissues. With a p(O2) > 100 mm Hg, only 3 ml of oxygen is dissolved in every one litre of plasma. By increasing the pA(O2) by breathing 100% oxygen, one can add an extra amount of oxygen in the plasma, but the amount of oxygen carried by haemoglobin will not increase significantly as it is already > 95% saturated.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 111
Incorrect
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A 50-year-old woman goes to the doctor complaining of myalgia, muscle cramps, and weakness; she is diagnosed with severe hypokalaemia. Which of the following is the most common cause of hypokalaemia?
Your Answer:
Correct Answer: Prolonged vomiting
Explanation:Potassium is one of the body’s major ions. Nearly 98% of the body’s potassium is intracellular. The ratio of intracellular to extracellular potassium is important in determining the cellular membrane potential. Small changes in the extracellular potassium level can have profound effects on the function of the cardiovascular and neuromuscular systems. Hypokalaemia may result from conditions as varied as renal or gastrointestinal (GI) losses, inadequate diet, transcellular shift (movement of potassium from serum into cells) and medications. The important causes of hypokalaemia are:
Renal losses: renal tubular acidosis, hyperaldosteronism, magnesium depletion, leukaemia (mechanism uncertain).
GI losses: vomiting or nasogastric suctioning, diarrhoea, enemas or laxative use, ileal loop.
Medication effects: diuretics (most common cause), β-adrenergic agonists, steroids, theophylline, aminoglycosides.
Transcellular shift: insulin, alkalosis.
Severe hypokalaemia, with serum potassium concentrations of 2.5–3 meq/l, may cause muscle weakness, myalgia, tremor, muscle cramps and constipation.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 112
Incorrect
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A 50-year-old gentleman was recently diagnosed with hypertension, with no other abnormalities on physical examination. Further investigations revealed the following :
Na+ 144 mmol/l
K+ 3.0 mmol/l
Cl- 107 mmol/l
Bicarbonate 25 mmol/l.
Blood glucose 5.8 mmol/l.
What is the likely diagnosis?Your Answer:
Correct Answer: Conn syndrome
Explanation:Overproduction of aldosterone (a mineralocorticoid) by the adrenal glands is known as Conn’s syndrome. It can be either due to an aldosterone-secreting adrenal adenoma (50-60% cases) or adrenal gland hyperplasia (40-50% cases). Excess aldosterone leads to sodium and water retention, along with potassium excretion. This leads to arterial (non-essential) hypertension. Conn’s syndrome is the commonest cause of primary hyperaldosteronism. Other symptoms include muscle cramps, headache (due to hypokalaemia) and metabolic alkalosis, which occurs due to increased secretion of H+ ions by the kidney. The raised pH of the blood traps calcium leading to symptoms of hypocalcaemia, which can be mimicked by liquorice ingestion and Liddle syndrome. To diagnose Conn’s syndrome, the ratio of renin and aldosterone is measured. Due to suppression of renin secretion, there is low renin to aldosterone ratio (<0.05). However, anti-hypertensives may affect the test results and should be withdrawn for 6 weeks. Computed tomography can also be done to detect the presence of adrenal adenoma. Cushing’s syndrome does not cause hypokalaemia with normal serum glucose levels. Nelson’s syndrome refers to increased ACTH secretion due to pituitary adenoma. Pheochromocytoma will not lead to hypokalaemia even though hypertension can be seen.
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This question is part of the following fields:
- Endocrine
- Pathology
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Question 113
Incorrect
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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 114
Incorrect
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Calculate the stroke volume in an adult male with the following parameters:
Heart rate 70 beats/min
Arterial [O2] 0.24 ml O2/min
Venous [O2] 0.16 ml O2/mi
Whole body O2 consumption 500 ml/min
Pulmonary diastolic pressure 15 mmHg
Pulmonary systolic pressure 25 mmHg
Wedge pressure 5 mmHg.Your Answer:
Correct Answer: 90 ml
Explanation:Fick’s principle states that, VO2 = (CO × CAO2) – (CO × CVO2) where VO2 = oxygen consumption, CO = cardiac output, CAO2 = oxygen concentration of arterial blood and CVO2 = oxygen concentration of venous blood. Thus, CO = VO2/CAO2– CVO2, CO = 500/0.24–0.16, CO = 500/0.8, CO = 6.25 l/min. Cardiac output is also given by product of stroke volume and heart rate. Thus, stroke volume = cardiac output / heart rate = 6.25/70 × 1000 stroke volume = 90 ml approximately.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 115
Incorrect
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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 116
Incorrect
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Hormones of the anterior pituitary include which of the following?
Your Answer:
Correct Answer: Prolactin
Explanation:The anterior pituitary gland (adenohypophysis or pars distalis) synthesizes and secretes:
1. FSH (follicle-stimulating hormone)
2. LH (luteinizing hormone)
3. Growth hormone
4. Prolactin
5. ACTH (adrenocorticotropic hormone)
6. TSH (thyroid-stimulating hormone).
The posterior pituitary gland (neurohypophysis) stores and secretes 2 hormones produced by the hypothalamus:
1. ADH (antidiuretic hormone or vasopressin)
2. Oxytocin
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This question is part of the following fields:
- Endocrine
- Physiology
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Question 117
Incorrect
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In which of the following conditions will the oxygen-haemoglobin dissociation curve shift to the right?
Your Answer:
Correct Answer: Exercise
Explanation:The oxygen-haemoglobin dissociation curve plots saturated haemoglobin against the oxygen tension and is usually a sigmoid plot. Each molecule of haemoglobin can bind to four molecules of oxygen reversibly. Factors that can influence the binding include: pH, concentration of 2,3-diphosphoglycerate (2,3-DPG), temperature, type of haemoglobin molecules, and presence of toxins, especially carbon monoxide. Shape of the curve is due to interaction of bound oxygen molecules with the incoming molecules. The binding of first molecule is difficult, with easier binding of the second and third molecule and increase in difficulty with the fourth molecule – partly as a result of crowding and partly as a natural tendency of oxygen to dissociate.
Left shift of curve indicates haemoglobin’s increased affinity for oxygen (seen at lungs). Right shift indicates decreased affinity and is seen with an increase in body temperature, hydrogen ions, 2,3-diphosphoglycerate (DPG), carbon dioxide concentration and exercise. Under normal resting conditions in a healthy individual, the normal position of the curve is at a pH of 7.4. A shift in the position of the curve with a change in pH is called the Bohr effect. Left shift occurs in acute alkalosis, decrease in p(CO2), decrease in temperature and decrease in 2,3-DPG. The fetal haemoglobin curve is to the left of the adult haemoglobin to allow for oxygen diffusion across the placenta. The curve for myoglobin is even further to the left. Carbon monoxide has a much higher affinity for haemoglobin than oxygen does. Thus, carbon monoxide poisoning leads to hypoxia.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 118
Incorrect
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Whilst snorkelling, a 30-year old gentleman has the respiratory rate of 10/min, tidal volume of 550 ml and an effective anatomical dead space of 250 ml. 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 119
Incorrect
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The blood investigations of a 30-year old man with jaundice revealed the following : total bilirubin 6.5 mg/dl, direct bilirubin 1.1 mg/dl, indirect bilirubin 5.4 mg/dl and haemoglobin 7.3 mg/dl. What is the most likely diagnosis out of the following?
Your Answer:
Correct Answer: Haemolysis
Explanation:Hyperbilirubinemia can be caused due to increased bilirubin production, decreased liver uptake or conjugation, or decreased biliary excretion. Normal bilirubin level is less than 1.2 mg/dl (<20 μmol/l), with most of it unconjugated. Elevated unconjugated bilirubin (indirect bilirubin fraction >85%) can occur due to haemolysis (increased bilirubin production) or defective liver uptake/conjugation (Gilbert syndrome). Such increases are less than five-fold usually (<6 mg/dl or <100 μmol/l) unless there is coexistent liver disease.
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This question is part of the following fields:
- Gastrointestinal; Hepatobiliary
- Pathology
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Question 120
Incorrect
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Calculate the cardiac stroke volume of a patient whose oxygen consumption (measured by analysis of mixed expired gas) is 300 ml/min, arterial O2 content is 20 ml/100 ml blood, pulmonary arterial O2 content is 15 ml/100 ml blood and heart rate is 60/min.
Your Answer:
Correct Answer: 100 ml
Explanation:By Fick’s principle, VO2 = Q × (CA (O2) − CV (O2)) where VO2 = O2 consumption, Q = cardiac output and CA(O2) and CV(O2) are arterial and mixed venous O2 content respectively. Thus, in the given problem, 300 ml O2/min = Q × (20−15) ml O2/100 ml. Thus, Q = 6000 ml blood/min. Then, we can calculate stroke volume by dividing the cardiac output with heart rate. Thus, stroke volume = 6000 ml/min divided by 60/min stroke volume = 100 ml.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 121
Incorrect
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A 35 year old man presented to the surgical OPD with a lump on his right forearm which appeared 3 weeks ago and was tender on examination. He gave a history was being in a car accident with pieces of glass from the windshield piercing his forearm removed manually and on further elective surgery. Which of these cells are characteristically found during inflammation in this situation?
Your Answer:
Correct Answer: Giant cell
Explanation:A foreign body reaction Is characteristic of giant cells. Glass being the foreign object initiates an inflammatory response in this condition.
Mast cells are involved in allergic reactions.
Eosinophils are characteristic of a parasitic infection and allergic inflammatory process but are not due to foreign bodies.
Plasma cells are typical of chronic inflammation.
Lymphocytes are involved in viral infections.
Macrophages combine together to form giant cells.
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This question is part of the following fields:
- Inflammation & Immunology
- Pathology
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Question 122
Incorrect
-
Epstein-Barr virus (EBV) is known to be a carcinogen for:
Your Answer:
Correct Answer: B-cell lymphoma
Explanation:EBV is known to be carcinogenic for B-cell lymphoma, nasopharyngel carcinoma and Hodgkin’s lymphoma.
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This question is part of the following fields:
- Microbiology; Neoplasia
- Pathology
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Question 123
Incorrect
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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 124
Incorrect
-
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:
Correct 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 125
Incorrect
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A 28 years old women presents with a history of chronic cough with fever for the past 2 months. A chest x ray revealed a diffuse bilateral reticulonodular pattern. A transbronchial biopsy was performed and histological examination showed focal areas of inflammation with epithelioid macrophages, Langhans cells and lymphocytes. Which of the immune reaction is responsible for this?
Your Answer:
Correct Answer: Type IV hypersensitivity
Explanation:A reactivated tuberculosis with granuloma formation is characteristic of type IV reaction. It is also called a delayed type of hypersensitivity reaction and takes around 2-8 days to deliver. It is a cell mediated response with the involvement of CD8 and CD4 cells and the release of IL-1 from macrophages that further activate these CD cells.
Granulomatous reactions are mostly cell-mediated.
Type I reactions are allergic and anaphylactic reactions and type II are complement-mediated immune reactions.
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This question is part of the following fields:
- Inflammation & Immunology; Respiratory
- Pathology
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Question 126
Incorrect
-
When does the heart rate decrease?
Your Answer:
Correct Answer: Pressure on the eyeball
Explanation:Various vagotonic manoeuvres (e.g. Valsalva manoeuvre, carotid sinus massage, pressure on eyeballs, ice-water facial immersion, swallowing of ice-cold water) result in increased parasympathetic tone through the vagus nerve which results in a decrease in heart rate. These manoeuvres may be clinically useful in terminating supraventricular arrhythmias.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 127
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 128
Incorrect
-
A 35-year-old ultra marathon runner becomes severely dehydrated and collapses. This patient most likely has:
Your Answer:
Correct Answer: Decreased baroreceptor firing rate
Explanation:Baroreceptors are sensors located in the blood vessels of all vertebrate animals. They sense the blood pressure and relay the information to the brain, so that a proper blood pressure can be maintained. Acute dehydration results in decreased plasma volume and increased plasma osmolarity, since more water than salt is lost in sweat. The decrease in plasma volume leads to an inhibition of the baroreceptors and a lower firing rate. The increase in plasma osmolarity leads to increased ADH secretion and high plasma ADH levels, which increases water permeability of collecting duct cells. Therefore more water is reabsorbed by the kidneys and renal water excretion is low.
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This question is part of the following fields:
- Physiology
- Renal
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Question 129
Incorrect
-
A cancer patient was found to have a radio resistant tumour. Which tumour does the patient most likely have?
Your Answer:
Correct Answer: Liposarcoma
Explanation:Liposarcoma is a cancer that arises in fat cells in deep soft tissue. Commonly it occurs inside the thigh or retroperitoneum. It usually affects middle-aged and older adults, over 40 years. Liposarcoma is the most common soft-tissue sarcoma. It is very radio resistant. Five-year survival rates vary from 100% to 56% based on histological subtype.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 130
Incorrect
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What percentage of the cardiac output is delivered to the brain?
Your Answer:
Correct Answer: 15%
Explanation:Among all body organs, the brain is most susceptible to ischaemia. Comprising of only 2.5% of total body weight, the brain receives 15% of the cardiac output. Oxygen extraction is also higher with venous oxygen levels approximating 13 vol%, and arteriovenous oxygen difference of 7 vol%.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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SESSION STATS - PERFORMANCE PER SPECIALTY
