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Question 1
Correct
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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 2
Correct
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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 3
Correct
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A 40-year old lady presented to the hospital with fever and mental confusion for 1 week. On examination, she was found to have multiple petechiae all over her skin and mucosal surfaces. Blood investigations revealed low platelet count and raised urea and creatinine. A platelet transfusion was carried out, following which she succumbed to death. Autopsy revealed pink hyaline thrombi in myocardial arteries. What is the likely diagnosis?
Your Answer: Thrombotic thrombocytopenic purpura
Explanation:Hyaline thrombi are typically associated with thrombotic thrombocytopenic purpura (TTP), which is caused by non-immunological destruction of platelets. Platelet transfusion is contraindicated in TTP. Platelets and red blood cells also get damaged by loose strands of fibrin deposited in small vessels. Multiple organs start developing platelet-fibrin thrombi (bland thrombi with no vasculitis) typically at arteriocapillary junctions. This is known as ‘thrombotic microangiopathy’. Treatment consists of plasma exchange.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 4
Incorrect
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The anatomical dead space in a patient with low oxygen saturation, is 125 ml, with a tidal volume of 500 ml and pa(CO2) of 40 mm Hg. The dead space was determined by Fowler's method. If we assume that the patient's lungs are healthy, what will his mixed expired CO2 tension [pE(CO2)] be?
Your Answer: 0 mmHg
Correct Answer: 30 mmHg
Explanation:According to Bohr’s equation, VD/VT = (pA(CO2) − pE(CO2))/pA(CO2), where pE(CO2) is mixed expired CO2 and pA(CO2) is alveolar CO2pressure. Normally, the pa(CO2) is virtually identical to pA(CO2). Thus, VD/VT = (pa(CO2)) − pE(CO2)/pa(CO2). By Fowler’s method, VD/VT= 0.25. In the given problem, (pa(CO2) − pE(CO2)/pa(CO2) = (40 − pE(CO2)/40 = 0.25. Thus, pE(CO2) = 30 mmHg. If there is a great perfusion/ventilation inequality, pE(CO2) could be significantly lower than 30 mm Hg, and the patient’s physiological dead space would exceed the anatomical dead space.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 5
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: Klebsiella 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 6
Correct
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A 22-year old man presented with a mass in his left scrotum which was more prominent when standing and felt like a 'bag of worms'. Examination revealed a non-tender mass along the spermatic cord. Also, the right testis was larger than the left testis. What is the likely diagnosis?
Your Answer: Varicocele
Explanation:Varicocele refers to dilatation and increased tortuosity of the pampiniform plexus – which is a network of veins found in spermatic cord that drain the testicle. Defective valves or extrinsic compression can result in outflow obstruction and cause dilatation near the testis. Normal diameter of the small vessels ranges from 0.5 – 1.5mm. A varicocele is a dilatation more than 2mm.
The plexus travels from the posterior aspect of testis into the inguinal canal with other structures forming the spermatic cord. They then form the testicular veins out of which the right testicular vein drains into the inferior vena cava and the left into the left renal vein.
It affects 15-20% men, and 40% of infertile males. Usually diagnosed in 15-25 years of age, they are rarely seen after 40 years of age. Because of the vertical path taken by the left testicular vein to drain into left renal vein, 98% idiopathic varicoceles occur on the left side. It is bilateral in 70% cases. Right-sided varicoceles are rare.
Symptoms include pain or heaviness in the testis, infertility, testicular atrophy, a palpable mass, which is non-tender and along the spermatic cord (resembling a ‘bag of worms’). The testis on the affected side might be smaller.
Diagnosis can be made by ultrasound. Provocative measures such as Valsalva manoeuvre or making the patient stand up to increase the dilatation by increasing the intra-abdominal venous pressure.
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This question is part of the following fields:
- Pathology
- Urology
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Question 7
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: T1a
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 8
Correct
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In which of the following conditions will the oxygen-haemoglobin dissociation curve shift to the right?
Your Answer: Exercise
Explanation:The oxygen-haemoglobin dissociation curve plots saturated haemoglobin against the oxygen tension and is usually a sigmoid plot. Each molecule of haemoglobin can bind to four molecules of oxygen reversibly. Factors that can influence the binding include: pH, concentration of 2,3-diphosphoglycerate (2,3-DPG), temperature, type of haemoglobin molecules, and presence of toxins, especially carbon monoxide. Shape of the curve is due to interaction of bound oxygen molecules with the incoming molecules. The binding of first molecule is difficult, with easier binding of the second and third molecule and increase in difficulty with the fourth molecule – partly as a result of crowding and partly as a natural tendency of oxygen to dissociate.
Left shift of curve indicates haemoglobin’s increased affinity for oxygen (seen at lungs). Right shift indicates decreased affinity and is seen with an increase in body temperature, hydrogen ions, 2,3-diphosphoglycerate (DPG), carbon dioxide concentration and exercise. Under normal resting conditions in a healthy individual, the normal position of the curve is at a pH of 7.4. A shift in the position of the curve with a change in pH is called the Bohr effect. Left shift occurs in acute alkalosis, decrease in p(CO2), decrease in temperature and decrease in 2,3-DPG. The fetal haemoglobin curve is to the left of the adult haemoglobin to allow for oxygen diffusion across the placenta. The curve for myoglobin is even further to the left. Carbon monoxide has a much higher affinity for haemoglobin than oxygen does. Thus, carbon monoxide poisoning leads to hypoxia.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 9
Correct
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As per the Poiseuille-Hagen formula, doubling the diameter of a vessel will change the resistance of the vessel from 16 peripheral resistance units (PRU) to:
Your Answer: 1 PRU
Explanation:Poiseuille-Hagen formula for flow in along narrow tube states that F = (PA– PB) × (Π/8) × (1/η) × (r4/l) where F = flow, PA– PB = pressure difference between the two ends of the tube, η = viscosity, r = radius of tube and L = length of tube. Also, flow is given by pressure difference divided by resistance. Hence, R = 8ηL ÷ Πr4. Hence, the resistance of the vessel changes in inverse proportion to the fourth power of the diameter. So, if the diameter of the vessel is increased to twice the original, it will lead to decrease in resistance to one-sixteenth its initial value.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 10
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 11
Correct
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Which of the following is the most accurate test for the diagnosis of primary syphilis?
Your Answer: Dark-field microscopy
Explanation:Primary syphilis is transmitted via sexual contact. Lesions on genitalia, called a chancre occur after an asymptomatic incubation period of 10-90 days (average 21 days) after exposure. This chancre is a typically solitary (can be multiple), firm, painless, ulceration over the skin at the point of exposure to spirochete, seen on penis, vagina or rectum. It heals spontaneously after 4-6 weeks. Local lymphadenopathy can be seen.
Diagnosis is made by microscopy of fluid from lesion using dark-field illumination, taking care to not confuse with other treponemal disease. Screening tests include rapid plasma regain (RPR) and Venereal Diseases Research Laboratory (VDRL) tests. False positives are known to occur with these tests and can be seen in viral infections like hepatitis, varicella, Epstein-Barr virus, tuberculosis, lymphoma, pregnancy and IV drug use. More specific tests should therefore be carried out in case these screening tests are positive.
The Treponema pallidum hemagglutination assay (TPHA) and the fluorescent treponemal antibody absorption (FTAABS) test are based on monoclonal antibodies and immunofluorescence and are more specific. However, they can too show false positives with other treponemal diseases like yaws or pinta. Other confirmatory tests include those based on enzyme-linked immunoassays.
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This question is part of the following fields:
- Pathology
- Urology
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Question 12
Correct
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A 65-year-old man with no history of smoking complains of shortness of breath and persistent cough over the past 8 months. He reveals that in the 1960s he worked for several years as a boiler operator. Chest X-ray shows diffuse lung infiltrates. Which of the following is the most probable cause of these findings?
Your Answer: Asbestosis
Explanation:Asbestosis is a chronic lung disease which leads to long-term respiratory complications and is caused by the inhalation of asbestos fibres. Symptoms due to long exposure to asbestos usually appear 10 to 40 years after initial exposure and include shortness of breath, cough, weight loss, clubbing of the fingers and chest pain. Typical chest X-ray findings include diffuse lung infiltrates that cause the appearance of shaggy heart borders.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 13
Correct
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What is the most likely cause of bitemporal hemianopia?
Your Answer: Prolactinoma
Explanation:Prolactinoma is the mot common pituitary adenoma; leading to hyperprolactinaemia. By virtue of their size, macroprolactinomas press on the adjacent structures leading to headaches and loss of vision due to the pressure effect on optic chiasm. Women notice a change in their menstrual cycle due to raised prolactin levels. In comparison, the problem goes unnoticed in men in the initial stages. Craniopharyngioma is a less common space-occupying lesions affecting children and young adults.
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This question is part of the following fields:
- Endocrine
- Pathology
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Question 14
Correct
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A 56 year old gentleman, who is a chronic smoker presents to the clinic with dyspnoea. His Chest X-ray shows intercostal space widening with increased blackening bilaterally. What is the most likely finding on his pulmonary function test?
Your Answer: Increased total lung capacity
Explanation:In patients suspected of having chronic obstructive pulmonary disease, pulmonary function testing (PFT) is useful to confirm airway obstruction, to quantify its severity and reversibility, for following disease progression and monitoring response to treatment. These tests include:
FEV1 – volume of air forcefully expired during the first second after a full breath
Forced vital capacity (FVC) – total volume of air expired with maximal force and flow-volume loops. The hallmark of airway obstruction is reduction of FEV1, FVC and the ratio of FEV1/FVC, with a concave pattern in expiratory tracing on the flow-volume loop. FEV1 and forced vital capacity (FVC) are easily measured with office spirometry and are useful to assess the severity of disease. Other parameters include increased total lung capacity, functional residual capacity and residual volume, which can help distinguish chronic obstructive pulmonary disease (COPD) from restrictive pulmonary disease where these values are lower than normal. Along with these, other tests are decreased vital capacity; and decreased diffusing capacity for carbon monoxide (DLCO). DLCO is non-specific and can be low in other disorders that affect the pulmonary vascular bed, such as interstitial lung disease. DLCO is however useful to distinguish COPD from asthma, in which DLCO is normal or elevated.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 15
Correct
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Etoposide is a chemotherapeutic agent used in the treatment of different types of cancer. Which of the following is the correct indication for this drug?
Your Answer: Lung cancer
Explanation:Etoposide phosphate is an inhibitor of the enzyme topoisomerase II. It is used as a form of chemotherapy for malignancies such as lung cancer, testicular cancer, lymphoma, non-lymphocytic leukaemia and glioblastoma multiforme. Side effects are very common and can include low blood cell counts, vomiting, loss of appetite, diarrhoea, hair loss, and fever.
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This question is part of the following fields:
- Pathology
- Pharmacology
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Question 16
Correct
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A 12-year old girl was brought to the hospital with recurrent headaches for 6 months. Her physical examination revealed no abnormality. A CT scan of the head revealed a suprasellar mass with calcifications, eroding the surrounding sella turcica. The lesion is likely to represent:
Your Answer: Craniopharyngioma
Explanation:Craniopharyngiomas (also known as Rathke pouch tumours, adamantinomas or hypophyseal duct tumours) affect children mainly between the age of 5 and 10 years. It constitutes 9% of brain tumours affecting the paediatric population. These are slow-growing tumours which can also be cystic, and arise from the pituitary stalk, specifically the nests of epithelium derived from Rathke’s pouch. Histologically, this tumour shows nests of squamous epithelium which is lined on the outside by radially arranged cells. Calcium deposition is often seen with a papillary type of architecture.
ACTH-secreting pituitary adenomas are rare and mostly microadenomas. Paediatric astrocytoma’s usually occur in the posterior fossa. Although null cell adenomas can cause mass effect and give rise to the described symptoms, they are not suprasellar. Prolactinomas can also show symptoms of headache and disturbances in the visual field, however they are known to be small and slow-growing.
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This question is part of the following fields:
- Endocrine
- Pathology
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Question 17
Correct
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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 18
Correct
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The absence of which of the following components characterizes the grey platelet syndrome (GPS)?
Your Answer: Alpha granules
Explanation:Grey platelet syndrome (GPS) is a rare inherited bleeding disorder associated with an almost total absence of α-granules and their contents. The syndrome is characterised by thrombocytopenia, enlarged platelets that have a grey appearance, myelofibrosis, and splenomegaly. Alpha granules store proteins and growth factors that promote platelet adhesiveness and wound healing. Patients with GPS develop symptoms and signs such as easy bruising, prolonged bleeding, and nose bleeds.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 19
Incorrect
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A 77-year-old woman's renal function is tested. The following results were obtained during a 24-h period:
Urine flow rate: 2. 0 ml/min
Urine inulin: 0.5 mg/ml
Plasma inulin: 0.02 mg/ml
Urine urea: 220 mmol/l
Plasma urea: 5 mmol/l.
What is the urea clearance?Your Answer: 4.4 ml/min
Correct Answer: 88 ml/min
Explanation:Urea is reabsorbed in the inner medullary collecting ducts of the nephrons. The clearance (C) of any substance can be calculated as follows: C = (U × V)/P, where U and P are the urine and plasma concentrations of the substance, respectively and V is the urine flow rate. So, glomerular filtration rate = (0.220 × 2. 0)/0.005 = 88 ml/min.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 20
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 21
Correct
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Out of the following options, which malignancy has the highest potential for multicentricity?
Your Answer: Transitional cell carcinoma
Explanation:Transitional cell carcinomas can arise anywhere in the urothelium lining the urinary tract; and hence are known to be multicentric and recur commonly. Prostatic adenocarcinoma most commonly involves the posterior lobe of the prostate gland. Although renal cell carcinomas occasionally show multicentricity, it is not common. Penile carcinomas are usually locally infiltrative lesions. Wilm’s tumours are usually solitary, but can be bilateral or multicentric in 10% cases. Small cell carcinoma of lung and teratomas are usually solitary.
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This question is part of the following fields:
- Pathology
- Renal
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Question 22
Correct
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A 68-year-old woman complains of headaches, dizziness, and memory loss. About a month ago, she fell from a staircase but only suffered mild head trauma. What is the most likely diagnosis in this case?
Your Answer: Chronic subdural haematoma
Explanation:A quarter to a half of patients with chronic subdural haematoma have no identifiable history of head trauma. If a patient does have a history of head trauma, it usually is mild. The average time between head trauma and chronic subdural haematoma diagnosis is 4–5 weeks. Symptoms include decreased level of consciousness, balance problems, cognitive dysfunction and memory loss, motor deficit (e.g. hemiparesis), headache or aphasia. Some patients present acutely. They usually result from tears in bridging veins which cross the subdural space, and may cause an increase in intracranial pressure (ICP).
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This question is part of the following fields:
- Neurology
- Pathology
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Question 23
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: 80 ml
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 24
Correct
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If a 70-year-old man with known atrial fibrillation dies suddenly, which of these is the most likely cause of death?
Your Answer: Thromboembolism
Explanation:In atrial fibrillation, the abnormal atrial contraction can cause blood to stagnate in the left atrium and form a thrombus, which may then embolize. The patient’s history of AF suggest an embolic disease, which lead to his death.
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This question is part of the following fields:
- Cardiovascular
- Pathology
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Question 25
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 26
Correct
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A 70 year old women, died suddenly. She had a history of hypertension and aortic stenosis. On autopsy her heart weighed 550g. What is the most likely cause of this pathology?
Your Answer: Hypertrophy
Explanation:Due to increased pressure on the heart as a result of hypertension and aortic stenosis, the myocardial fibres hypertrophied to adapt to the increased pressure and to effectively circulate blood around the body. Hyperplasia could not occur, as myocardial fibres are stable cells and cannot divide further.
Fat does not deposit in the heart due to volume overload.
Myocardial oedema is not characteristic of a myocardial injury.
Metaplasia is a change in the type of epithelium.
Atrophy would result in a decreased heart size and inability to function.
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This question is part of the following fields:
- Cell Injury & Wound Healing; Cardiovascular
- Pathology
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Question 27
Correct
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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 28
Correct
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A medical student is asked to calculate the net pressure difference in a capillary wall, considering: Interstitial fluid hydrostatic pressure = –3 mmHg, Plasma colloid osmotic pressure = 28 mmHg, Capillary hydrostatic pressure = 17 mmHg, Interstitial fluid colloid osmotic pressure = 8 mmHg, and Filtration coefficient = 1. Which is the correct answer?
Your Answer: 0 mmHg
Explanation:The rate of filtration at any point along a capillary depends on a balance of forces sometimes called Starling’s forces after the physiologist who first described their operation in detail. The Starling principle of fluid exchange is key to understanding how plasma fluid (solvent) within the bloodstream (intravascular fluid) moves to the space outside the bloodstream (extravascular space). Fluid movement = k[(pc– pi)–(Πc– Πi)] where k = capillary filtration coefficient, pc = capillary hydrostatic pressure, pi= interstitial hydrostatic pressure, Πc = capillary colloid osmotic pressure, Πi = interstitial colloid osmotic pressure. Therefore: 1 × [capillary hydrostatic pressure (17) – interstitial fluid hydrostatic pressure (–3)] – [plasma colloid osmotic pressure (28) – interstitial fluid colloid osmotic pressure (8)] = 0 mmHg
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 29
Incorrect
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Perforin are present in the granules of which cell?
Your Answer: Mast cell
Correct Answer: Natural killer cell
Explanation:Perforins are characteristically found In the granules of CD8+ T cells and natural killer cells. They are cytolytic proteins that insert into the target plasma membrane forming a hole and resulting in lysis. They along with granzyme B induce apoptosis in the target cell.
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This question is part of the following fields:
- Inflammation & Immunology
- Pathology
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Question 30
Correct
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A 39-year-old man, after radiological evaluation and thoracentesis, was found to have chylothorax. What is the most probable cause of this diagnose?
Your Answer: Mediastinal malignant lymphoma
Explanation:Chylothorax is a potentially lethal condition characterized by fluid (chyle) accumulation in the pleural cavity, resulting from disruption of lymphatic drainage in the thoracic duct. Chyle is a fluid rich in triglycerides and chylomicrons and can originate from the thorax, the abdomen or both. Malignant tumours, especially lymphoma, are the most common causes of nontraumatic chylothorax. Bronchogenic carcinoma and trauma are the most common causes after lymphomas. Other rare causes of chylothorax are; granulomatous diseases, tuberculosis, congenital malformations, nephrotic syndrome, hypothyroidism, cirrhosis, decompensated heart failure and idiopathic chylothorax.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 31
Correct
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A suspected recreational drug user was brought to the Emergency department in an unconscious state, and was found to be hypoventilating. Which of the following set of arterial blood gas analysis report is most consistent with hypoventilation as the primary cause? pH, pa(CO2) (mmHg), pa(O2) (mmHg).
Your Answer: 7.28, 55, 81
Explanation:Hypoventilation (or respiratory depression) causes an increase in carbon dioxide (hypercapnia) and respiratory acidosis. It can result due to drugs such as alcohol, benzodiazepines, barbiturates, opiates, mechanical conditions or holding ones breath. Strong opioids such as heroin and fentanyl are commonly implicated and can lead to respiratory arrest. In recreational drug overdose, acute respiratory acidosis occurs with an increase in p(CO2) over 45 mm Hg and acidaemia (pH < 7.35)
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 32
Correct
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Calculate the pulmonary vascular resistance in an adult male with the following parameters:
Heart rate 70 beats/min
Arterial [O2] 0.24 ml O2/min
Venous [O2] 0.16 ml O2/mi
Whole body O2 consumption 500 ml/min
Pulmonary diastolic pressure 15 mmHg
Pulmonary systolic pressure 25 mmHg
Wedge pressure 5 mmHg.Your Answer: 2.0 resistance units (mmHg/l per min)
Explanation:Pulmonary vascular resistance (PVR) = (Mean pulmonary artery pressure – Pulmonary capillary wedge pressure) divided by Cardiac output. To get cardiac output, Fick’s principle needs to be applied which states that VO2 = (CO × CAO2) – (CO × CVO2) where VO2 = oxygen consumption, CO = cardiac output, CA = oxygen concentration of arterial blood and CVO2 = oxygen concentration of venous blood. Thus, CO = VO2/CAO2– CVO2, CO = 500/0.24–0.16, CO = 500/0.8, CO = 6.25 l/min. To calculate mean pulmonary artery pressure, we use the formula: Mean pulmonary artery pressure = Diastolic pressure + 1/3(Systolic pressure – Diastolic pressure). Thus, Mean pulmonary artery pressure = 15 + 1/3(25 – 15) = 15 + 3. 33 = 18.33. Substituting these values in the first formula, PVR = 18.3–5/6.25 = 13.5/5.25 = 2.0 resistance units (mmHg/l per min) approximately.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 33
Correct
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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 34
Incorrect
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Whilst snorkelling, a 30-year old gentleman has the respiratory rate of 10/min, tidal volume of 550 ml and an effective anatomical dead space of 250 ml. What is his alveolar ventilation?
Your Answer: 5000 ml/min
Correct Answer: 3000 ml/min
Explanation:Alveolar ventilation is the amount of air reaching the alveoli per minute. Alveolar ventilation = respiratory rate × (tidal volume – anatomical dead space volume). Thus, alveolar ventilation = 10 × (550 − 250) = 3000 ml/min.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 35
Correct
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A 60-year-old woman complains of left sided headaches which have been recurring for several years. She recently suffered from a focal seizure for the first time a few days ago. A CT scan shows a mass in the left hemisphere of the brain. The most likely diagnosis is:
Your Answer: Meningioma
Explanation:Meningiomas are a common benign intracranial tumour, and their incidence is higher in women between the ages of 40-60 years old. Many of these tumours are asymptomatic and are diagnosed incidentally, although some of them may have malignant presentations (less than 2% of cases). These benign tumours can develop wherever there is dura, over the convexities near the venous sinuses, along the base of the skull, in the posterior fossa and, within the ventricles.
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This question is part of the following fields:
- Neurology
- Pathology
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Question 36
Correct
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A previously healthy 40-year-old housewife suddenly complains of a headache and loses consciousness. A CT scan reveals subarachnoid haemorrhage. Which of the following is the most probable cause?
Your Answer: Ruptured berry aneurysm
Explanation:Saccular aneurysms, also known as berry aneurysms, appear as a round outpouching and are the most common form of cerebral aneurysm. They are a congenital intracranial defect, and haemorrhage can occur at any age, but is most common between the ages of 40-65 years. A second rupture (rebleeding) sometimes occurs, most often within about 7 days of the first bleed.
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This question is part of the following fields:
- Neurology
- Pathology
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Question 37
Correct
-
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 38
Correct
-
A 30-year-old woman feels thirsty. This thirst is probably due to:
Your Answer: Increased level of angiotensin II
Explanation:Thirst is the basic need or instinct to drink. It arises from a lack of fluids and/or an increase in the concentration of certain osmolites such as salt. If the water volume of the body falls below a certain threshold or the osmolite concentration becomes too high, the brain signals thirst. Excessive thirst, known as polydipsia, along with excessive urination, known as polyuria, may be an indication of diabetes. Angiotensin II is a hormone that is a powerful dipsogen (i.e. it stimulates thirst) that acts via the subfornical organ. It increases secretion of ADH in the posterior pituitary and secretion of ACTH in the anterior pituitary.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 39
Correct
-
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 40
Correct
-
A patient is admitted to the ICU, and is prescribed tazobactam, amongst other drugs. What is the mechanism of action of tazobactam?
Your Answer: Inhibits beta-lactamase
Explanation:Tazobactam is a compound which inhibits the action of bacterial beta-lactamases. It is added to the extended-spectrum beta-lactam antibiotic piperacillin to broaden the spectrum of piperacillin by making it effective against organisms that express beta-lactamase and would normally degrade piperacillin.
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This question is part of the following fields:
- Pathology
- Pharmacology
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Question 41
Correct
-
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: 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 42
Correct
-
A 42 year old man presents with end stage renal failure and is prepared to receive a kidney from his best friend. HLA testing showed that they are not a 100% match and he is given immunosuppressant therapy for this. Three months later when his renal function is assessed, he showed signs of deteriorating renal function, with decreased renal output, proteinuria of +++ and RBCs in the urine. He was given antilymphocyte globulins and his condition reversed. During the crisis period the patient is likely to be suffering from?
Your Answer: Acute rejection
Explanation:This patients is most likely experiencing an acute rejection. It is a cell mediated attack against the organ that has been transplanted. Antigens are either presented by blood borne cells with in the graft or antigen presenting cells in the body may be presenting class I and class II molecules that have been shed by the graft. Class I will activate CD8 and class II, CD4 cells, both of which will attack the graft.
Chronic rejection is a slow process which occurs months to years after the transplant. The exact mechanism is not very well understood but it probably involves a combination of Type III and Type IV hypersensitivity directed against the foreign MHC molecules which look like self-MHC presenting a foreign antigen.
Hyperacute Transplant Rejection occurs almost immediately and is often evident while you are still in surgery. It is caused by accidental ABO Blood type mismatching of the donor and recipient which almost never happens anymore. This means the host has preformed antibodies against the donated tissue.
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This question is part of the following fields:
- Inflammation & Immunology; Renal
- Pathology
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Question 43
Correct
-
A woman that presented with dyspnoea, chest pain and cough was found to have a serous pleural effusion. This finding is most likely to be associated with which of the following conditions?
Your Answer: Congestive heart failure
Explanation:A pleural effusion is defined as an abnormal collection of fluid in the pleural space. Pleural effusion can result from excess fluid production or decreased absorption or both. Thoracentesis and laboratory testing help determine the origin of the accumulated fluid. Serous fluid accumulation in the pleural space indicates the presence of a hydrothorax and is most likely to develop secondary to congestive heart failure.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 44
Correct
-
A 40-year old gentleman, known with a history of peptic ulcer disease, was brought to the clinic in a dehydrated state with persistent vomiting. His blood investigations revealed:
- sodium = 142 mmol/l
- potassium = 2.6 mmol/l
- chloride = 85 mmol/l
- pH = 7.55
- p(CO2) = 50 mmHg
- p(O2) = 107 mmHg
- standard bicarbonate = 40 mmol/l
Your Answer: Metabolic alkalosis
Explanation:High pH with high standard bicarbonate indicates metabolic alkalosis. The pa(CO2) was appropriately low in compensation. This is hypokalaemic hypochloraemic metabolic acidosis due to prolonged vomiting. Treatment includes treating the cause and intravenous sodium chloride with potassium.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 45
Correct
-
An abnormal opening of the urethra on the under surface of the penis (ventral surface) is known as:
Your Answer: Hypospadias
Explanation:Hypospadias is the condition where the urethra opens along the underside or ventral aspect of penile shaft. First-degree hypospadias is seen in 50-75% cases, where the urethra open on the glans penis. Second-degree hypospadias is seen in 20% cases where the urethra opens on the shaft, and third-degree in 30% cases with the urethra opening on the perineum. The severe cases are usually associated with undescended testis (cryptorchidism) or chordee, where the penis is tethered downwards and not completely separated from the perineum.
It is a common male genital birth defect but varying incidences are noted in different countries. There is no obvious inheritance pattern noted. No exact cause has been determined, however several hypotheses include poor response to androgen, or interference by environmental factors.
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This question is part of the following fields:
- Pathology
- Urology
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Question 46
Incorrect
-
Which of these structures is most likely to be damaged if a patient loses consciousness days or weeks after an otherwise insignificant head trauma, especially in elderly patients?
Your Answer: Middle meningeal artery
Correct Answer: Dural bridging vein
Explanation:A subdural haematoma is a type of hematoma, usually associated with traumatic brain injury, in which blood collects between the dura mater and the pia-arachnoid mater. Symptoms of subdural haemorrhage have a slower onset than those of epidural haemorrhages because the lower pressure veins bleed more slowly than arteries. These injuries are more common in elderly patients, especially those taking antiplatelet or anticoagulant drugs. Oedema and increased intracranial pressure are unusual.
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This question is part of the following fields:
- Neurology
- Pathology
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Question 47
Correct
-
The passage of leukocytes through the wall of the blood vessels is best described by which of the following terms?
Your Answer: Diapedesis
Explanation:The steps involved in leukocyte arrival and function are:
1. margination: cells migrate from the centre to the periphery of the vessel
2. rolling: selectins are upregulated on the vessel walls
3. adhesion: upregulation of the adhesion molecules ICAM and VCAM on the endothelium interact with integrins on the leukocytes, interaction of these results in adhesion
4. diapedesis and chemotaxis: diapedesis is the transmigration of the leukocyte across the endothelium of the capillary and towards a chemotactic product
5. phagocytosis: engulfing the offending substance/cell.
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This question is part of the following fields:
- Cell Injury & Wound Healing
- Pathology
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Question 48
Correct
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Which of the following diseases is caused by intra-articular and/or extra-articular deposition of calcium pyrophosphate dihydrate (CPPD) crystals, due to unknown causes?
Your Answer: Pseudogout
Explanation:Pseudogout or chondrocalcinosis is a rheumatological disease caused by the accumulation of crystals of calcium pyrophosphate dihydrate (CPPD) in the connective tissues. It is frequently associated with other conditions, such as trauma, amyloidosis, gout, hyperparathyroidism and old age, which suggests that it is secondary to degenerative or metabolic changes in the tissues. The knee is the most commonly affected joint. It causes symptoms similar to those of rheumatoid arthritis or osteoarthritis.
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This question is part of the following fields:
- Orthopaedics
- Pathology
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Question 49
Correct
-
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 50
Incorrect
-
In the kidney, the macula densa is an area of closely packed specialized cells lining the wall of the:
Your Answer: Juxtaglomerular cells
Correct Answer: Distal convoluted tubule
Explanation:In the kidney, the macula densa is an area of closely packed specialised cells lining the region of the distal convoluted tubule (DCT) lying next to the glomerular vascular pole. The cells of the macula densa are sensitive to the ionic content and water volume of the fluid in the DCT, producing signals that promote renin secretion by other cells of the juxtaglomerular apparatus.
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This question is part of the following fields:
- Physiology
- Renal
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Question 51
Incorrect
-
An elderly, diabetic man has firm, tender nodules at the base of his left middle and ring fingers, which he can't extend fully. What's the most likely diagnosis?
Your Answer: Dystrophic calcification
Correct Answer: Fibromatosis
Explanation:This case is suggestive of Dupuytren’s contracture due to palmar fibromatosis. Its incidence is higher in men over the age of 45 years, and it increases in patients with diabetes, alcoholism, or epilepsy. These nodules are benign, usually appearing as a tender nodule in the palm which becomes painless. The disease has an aggressive clinical behaviour and recurs frequently.
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This question is part of the following fields:
- Orthopaedics
- Pathology
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Question 52
Correct
-
Lack of findings in the bladder but presence of atypical epithelial cells in urinalysis is most often associated with which of the following conditions?
Your Answer: Transitional cell carcinoma of renal pelvis
Explanation:The presence of atypical cells in urinalysis without findings in the bladder suggests a lesion located higher up, most probably in ureters or renal pelvis. Transitional cell cancer of the renal pelvis is a disease in which malignant cells form in the renal pelvis and is characterised by the presence of abnormal cells in urine cytology.
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This question is part of the following fields:
- Pathology
- Renal
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Question 53
Correct
-
What Is the mechanism behind rhesus incompatibility in a new born baby?
Your Answer: Type II hypersensitivity
Explanation:In type II hypersensitivity the antibodies that are produced by the immune response bind to the patients own cell surface antigens. These antigens can be intrinsic or extrinsic. Destruction occurs due to antibody dependent cell mediated antibodies. Antibodies bind to the cell and opsonise the cell, activating phagocytes to destroy that cell e.g. autoimmune haemolytic anaemia, Goodpasture syndrome, erythroblastosis fetalis, pernicious anaemia, Graves’ disease, Myasthenia gravis and haemolytic disease of the new-born.
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This question is part of the following fields:
- Inflammation & Immunology; Haematology
- Pathology
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Question 54
Correct
-
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: 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 55
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 56
Correct
-
13 year old girl developed sun burnt cheeks after spending the day playing on the beach. What is the underlying mechanism to her injury?
Your Answer: Free radical injury
Explanation:Free radicals are a by-product of chemical reactions with an unpaired electron in their outer most shell. They are capable of causing wide spread damage to cells. They can cause autolytic reactions thereby converting the reactants into free radicals. By absorbing sun light, the energy is used to hydrolyse water into hydroxyl (OH) and hydrogen (H) free radicals which can cause injury by lipid peroxidation of membranes, oxidative modification of proteins and damage to the DNA structure.
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This question is part of the following fields:
- Cell Injury & Wound Healing; Dermatology
- Pathology
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Question 57
Incorrect
-
A 30-year-old woman is diagnosed with Hodgkin's lymphoma. Which of the following chemotherapy regimens would be used in this case?
Your Answer: CHOP
Correct Answer: ABVD
Explanation:ABVD is a chemotherapy regimen used in the first-line treatment of Hodgkin’s lymphoma. It consists of concurrent treatment with the chemotherapy drugs, adriamycin, bleomycin, vinblastine and dacarbazine. It supplanted the older MOPP protocol.
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This question is part of the following fields:
- Pathology
- Pharmacology
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Question 58
Correct
-
What is the role of ICAM-1 and VCAM-1 in the inflammatory process?
Your Answer: Leukocyte adhesion
Explanation:Steps involved in leukocyte arrival and function include:
1. margination: cells migrate from the centre to the periphery of the vessel.
2. rolling: selectins are upregulated on the vessel walls.
3. adhesion: upregulation of the adhesion molecules ICAM and VCAM on the endothelium interact with integrins on the leukocytes. Interaction of these results in adhesion.
4. diapedesis and chemotaxis: diapedesis is the transmigration of the leukocyte across the endothelium of the capillary and towards a chemotactic product.
5. phagocytosis: engulfing the offending substance/cell.
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This question is part of the following fields:
- Inflammation & Immunology
- Pathology
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Question 59
Correct
-
A patient is diagnosed with Conn’s syndrome. Aldosterone is secreted from where?
Your Answer: Zona glomerulosa of the adrenal cortex
Explanation:The adrenal gland comprises an outer cortex and an inner medulla, which represent two developmentally and functionally independent endocrine glands.
The adrenal medulla secretes adrenaline (70%) and noradrenaline (30%)
The adrenal cortex consists of three layers (remembered by the mnemonic GFR):
G = zona glomerulosa – secretes aldosterone
F = zona fasciculata – secretes cortisol and sex steroids
R = zona reticularis – secretes cortisol and sex steroids.
Aldosterone facilitates the reabsorption of sodium and water and the excretion of potassium and hydrogen ions from the distal convoluted tubule and collecting ducts. Conn’s syndrome is characterized by increased aldosterone secretion from the adrenal glands.
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This question is part of the following fields:
- Endocrine
- Physiology
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Question 60
Correct
-
In the glomerulus of the kidney, the mesangium is a structure associated with the capillaries. It has extraglomerular mesangial cells that:
Your Answer: Form the juxtaglomerular apparatus in combination with the macula densa and juxtaglomerular cells
Explanation:The mesangium is an inner layer of the glomerulus, within the basement membrane surrounding the glomerular capillaries. The mesangial cells are phagocytic and secrete the amorphous basement membrane-like material known as the mesangial matrix. They are typically separated from the lumen of the capillaries by endothelial cells. The other type of cells in the mesangium are the extraglomerular mesangial cells which form the juxtaglomerular apparatus in combination with two other types of cells: the macula densa of the distal convoluted tubule and juxtaglomerular cells of the afferent arteriole. This apparatus controls blood pressure through the renin–angiotensin–aldosterone system.
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This question is part of the following fields:
- Physiology
- Renal
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Question 61
Correct
-
How much blood can the pulmonary vessels of a 45-year-old healthy man accommodate when he is at rest?
Your Answer: 500 ml
Explanation:Pulmonary circulation is the portion of the cardiovascular system which carries deoxygenated blood away from the heart, to the lungs, and returns oxygenated blood back to the heart. The vessels of the pulmonary circulation are very compliant (easily distensible) and so typically accommodate about 500 ml of blood in an adult man. This large lung blood volume can serve as a reservoir for the left ventricle, particularly during periods when left ventricular output momentarily exceeds venous return.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 62
Correct
-
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: 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 63
Correct
-
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: 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 64
Correct
-
A 60-year-old male is suspected to have pancreatic cancer. What particular tumour marker should be requested to aid in the confirmation of the diagnosis of pancreatic cancer?
Your Answer: Carcinoembryonic antigen (CEA)
Explanation:Carcinoembryonic antigen (CEA) is used as a tumour marker. CEA test measures the amount of this protein that may appear in the blood of some people who have certain types of cancers especially cancer of the colon and rectal cancer. It may also be present in the pancreas, breast, ovary or lung.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 65
Correct
-
An ECG of a 30 year old woman revealed low voltage QRS complexes. This patient is most probably suffering from?
Your Answer: Pericardial effusion
Explanation:The QRS complex is associated with current that results in the contraction of both the ventricles. As ventricles have more muscle mass than the atria, they result in a greater deflection on the ECG. The normal duration of a QRS complex is 10s. A wide and deep Q wave depicts myocardial infarction. Abnormalities in the QRS complex maybe indicative of a bundle block, ventricular tachycardia or hypertrophy of the ventricles. Low voltage QRS complexes are characteristic of pericarditis or a pericardial effusion.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 66
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 67
Incorrect
-
During a normal respiratory exhalation, what is the recoil alveolar pressure?
Your Answer: Unchanged
Correct Answer: +10 cmH2O
Explanation:To determine compliance of the respiratory system, changes in transmural pressures (in and out) immediately across the lung or chest cage (or both) are measured simultaneously with changes in lung or thoracic cavity volume. Changes in lung or thoracic cage volume are determined using a spirometer with transmural pressures measured by pressure transducers. For the lung alone, transmural pressure is calculated as the difference between alveolar (pA; inside) and intrapleural (ppl; outside) pressure. To calculate chest cage compliance, transmural pressure is ppl (inside) minus atmospheric pressure (pB; outside). For the combined lung–chest cage, transmural pressure or transpulmonary pressure is computed as pA – pB. pA pressure is determined by having the subject deeply inhale a measured volume of air from a spirometer. Under physiological conditions the transpulmonary or recoil pressure is always positive; intrapleural pressure is always negative and relatively large, while alveolar pressure moves from slightly negative to slightly positive as a person breathes.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 68
Incorrect
-
T lymphocytes that express the MCH type II antigen are most likely to produce which of the following cytokines?
Your Answer: Tumour necrosis factor
Correct Answer: Gamma interferon
Explanation:Interferon gamma is a soluble cytokine previously known as the macrophage activating factor. It is the only member belonging to the type II class of the interferons. It is secreted by a number of cells taking part in the immune reaction including: T-helper cells (CD-4), cells with immunological memory (CD45PA), killer cells (CD8), dendrite cells (CD23,35), natural killer cells (CD16) and B lymphocytes (CD22,CD23). It has both a defending as well as a pathological effect. It induces differentiation in the myeloid cell in the bone marrow. If macrophages are infected by parasites it activates the macrophages to destroy them. IFN-γ strengthens the anti-tumour activities of the cytotoxic lymphocytes. Together with CD4 or CD8 toxins, produced by lymphocytes, it suppresses the growth of the tumour cells. along with these functions it increases the non specific response of the natural killer cells, causing changes in the cell membrane surface to prevent adhesion and penetration of a virus. It can either increase or decrease B cell response and it activates osteoclasts which increases bone resorption.
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This question is part of the following fields:
- Inflammation & Immunology
- Pathology
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Question 69
Correct
-
Which of these HLA alleles is most likely to be present in ankylosing spondylitis?
Your Answer: HLA-B27
Explanation:Ankylosing spondylitis usually appears between the ages of 20-40 years old and is more frequent in men. It is strongly associated with HLA-B27, along with other spondyloarthropathies, which can be remembered through the mnemonic PAIR (Psoriasis, Ankylosing spondylitis, Inflammatory bowel disease, and Reactive arthritis).
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This question is part of the following fields:
- Orthopaedics
- Pathology
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Question 70
Correct
-
A histology report of a cervical biopsy taken from a patient with tuberculosis revealed the presence of epithelioid cells. What are these cells formed from?
Your Answer: Macrophages
Explanation:Granulomas formed in tuberculosis are called tubercles and are made up polynuclear phagocytes, Langhans cells and epithelioid cells. Macrophages when enlarged, consist of abundant cytoplasm and have a tendency of arranging themselves very closely to each other representing epithelial cells. These enlarged macrophages are therefore termed as epithelioid cells.
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This question is part of the following fields:
- Cell Injury & Wound Healing
- Pathology
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Question 71
Correct
-
What percentage of the cardiac output is delivered to the brain?
Your 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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Question 72
Correct
-
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: 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 73
Correct
-
Coagulation in the body (in vivo) is a process in which several proteins known as coagulation factors are activated in a cascade effect to stop bleeding. Which of the following initiates this cascade effect?
Your Answer: Tissue factor
Explanation:Tissue factor (TF), also known as ‘factor III’ or ‘thromboplastin’, is an anti-coagulation protein that initiates the extrinsic coagulation. TF acts as a transmembrane receptor for Factor VII/VIIa . It is expressed by endothelial cells but also certain tissues, such as the heart and brain.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 74
Correct
-
The likely cause of a tender and swollen breast in a lactating mother is:
Your Answer: Acute mastitis
Explanation:Acute mastitis results due to bacterial infection of the breast and results in signs of inflammation. It commonly occurs 2-3 weeks postpartum and common causative microorganisms are Staphylococcus aureus, Streptococcus species, and Escherichia coli. Complications like an abscess can be avoided by prompt treatment, which includes antibiotics and rest along with continued lactation.
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This question is part of the following fields:
- Pathology
- Women's Health
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Question 75
Correct
-
A 24 year old mother is breastfeeding her first child. Which of the following cellular adaptations occurred in her breast tissue to allow her to do this?
Your Answer: Lobular hyperplasia
Explanation:Under the influence of oestrogen in pregnancy, there is an increase in the number of lobules which will facilitate lactation.
Steatocytes occur due to loss of weight and nutritional deficit.
Metaplasia is a normal physiological process which is due to a change in normal epithelium with another type.
Lobular atrophy will result in a decreased capacity to provide milk.
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This question is part of the following fields:
- Cell Injury & Wound Healing; Female Health
- Pathology
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Question 76
Correct
-
Hormones of the anterior pituitary include which of the following?
Your 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 77
Correct
-
A 50-year old, obese gentleman with a compression fracture of T11 vertebra was admitted in the hospital. Examination revealed a raised blood pressure 165/112 mmHg and blood glucose 8.5 mmol/l. His abdomen had the presence of purplish striae. What condition is he likely to be suffering from?
Your Answer: Adrenal cortical carcinoma
Explanation:Adrenocortical carcinomas are rare tumours with reported incidence being only two in a million. However, they have a poor prognosis. These are large tumours and range from 4-10 cm in diameter. They arise from the adrenal cortex and 10% cases are bilateral. 50-80% are known to be functional, leading to Cushing syndrome. Even though the tumour affects both sexes equally, functional tumours are slightly commoner in women and non-functional tumours are commoner in men. As compared to women, men also develop this tumour at an older age and seem to have a poorer prognosis.
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This question is part of the following fields:
- Endocrine
- Pathology
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Question 78
Incorrect
-
Cancer of the testis most likely metastases to which set of lymph nodes?
Your Answer: Internal iliac
Correct Answer: Aortic
Explanation:The lymphatic drainage of an organ is related to its blood supply. The lymphatic drainage of the testis drains along the testicular artery to reach the lymph nodes along the aorta.
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This question is part of the following fields:
- Endocrine
- Pathology
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Question 79
Incorrect
-
In a cardiac cycle, what event does the opening of the atrioventricular (AV) valves coincide with?
Your Answer: End of diastole
Correct Answer: Beginning of diastole
Explanation:Cardiac diastole refers to the time period when the heart is relaxed after contraction and is preparing to refill with blood. Both ventricular and atrial diastole are together known as complete cardiac diastole. At its beginning, the ventricles relax, causing a drop in the ventricular pressure. As soon as the left ventricular pressure drops below that in left atrium, the mitral valve opens and there is ventricular filling of blood. Similarly, the tricuspid valve opens filling the right atrium.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 80
Correct
-
A 25 year old women is pregnant with her second child. She is A- blood group. Her first child was Rh+ and the father is also Rh+. The second child is at a risk of developing which condition?
Your Answer: Haemolytic disease of the new-born
Explanation:This infant is at risk for haemolytic disease of the new born also known as erythroblastosis fetalis. In the pregnancy, Rh-positive RBC’s cross the placenta and enter the mothers blood system. She then becomes sensitised and forms IgG antibodies/anti-Rh antibodies against them. The second child is at a greater risk for this disease than the first child with Rh-positive blood group as during the second pregnancy, a more powerful response is produced. IgG has the ability to cross the placenta and bind to the fetal RBCs (type II hypersensitivity reaction) which are phagocytosed by the macrophages.
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This question is part of the following fields:
- Inflammation & Immunology; Haematology
- Pathology
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Question 81
Correct
-
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: 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 82
Correct
-
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: 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 83
Correct
-
A 70-year-old man suffers an ischaemic stroke and develops a left homonymous hemianopia. Where is the likely infarct?
Your Answer: Right occipital lobe
Explanation:The man has a left homonymous hemianopia which means he is unable to view objects in the left visual field. This information is processed by the right primary visual cortex which lies in the right occipital lobe.
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This question is part of the following fields:
- Neurology
- Pathology
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Question 84
Correct
-
Question 85
Correct
-
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: 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 86
Correct
-
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: 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 87
Correct
-
Calculate the stroke volume in a patient admitted for coronary bypass surgery, with the following parameters pre-operatively:
Oxygen consumption = 300 ml/min
Arterial oxygen content = 20 ml/100 ml blood
Pulmonary arterial oxygen content = 15 ml/100 ml blood and Heart rate = 100 beats/min.Your Answer: 60 ml
Explanation:By Fick’s principle, cardiac output can be calculated as follows: VO2 = CO × (CAO2– CVO2) where VO2= oxygen consumption, CO = cardiac output, CAO2 = arterial oxygen content and CvO2 = mixed venous oxygen content. Thus, in the given problem, 300 ml/min = CO × (20 – 15) ml/100 ml CO = 300 × 100/5 ml/min CO = 6000 ml/min. Also, cardiac output = stroke volume × heart rate. Thus, 6000 ml/min = stroke volume × 100 beats/min. Hence, stroke volume = 6000/100 ml/min which is 60 ml/min.
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This question is part of the following fields:
- Cardiovascular
- Physiology
-
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Question 88
Correct
-
Tumours derived from all three germ-cell layers in new-borns usually occur in which of the following sites?
Your Answer: Sacrococcygeal area
Explanation:A teratoma is a tumour with tissue or organ components resembling normal derivatives of more than one germ layer. It is derived from all three cell layers. The most common location of teratoma in new-born infants is in the sacrococcygeal area.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 89
Incorrect
-
When a penile tumour invades the subepithelial connective tissue of the penis, what is its stage?
Your Answer: T2
Correct Answer: T1
Explanation:The TNM staging used for penile cancer is as follows:
TX: primary tumour cannot be assessed
T0: primary tumour is not evident
Tis: carcinoma in situ is present
Ta: non-invasive verrucous carcinoma is present
T1: tumour is invading subepithelial connective tissue
T2: tumour is invading the corpora spongiosum or cavernosum
T3: tumour invading the urethra or prostate
T4: tumour invading other adjacent structures.
In this case, the patient has a T1 tumour.
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This question is part of the following fields:
- Pathology
- Urology
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Question 90
Correct
-
The extent of cancer growth can be described through staging. What is taken into consideration when staging a cancer?
Your Answer: Local invasion
Explanation:Cancer stage is based on four characteristics: the size of cancer, whether the cancer is invasive or non-invasive, whether the cancer has spread to the lymph nodes, and whether the cancer has spread to other parts of the body, in this case beyond the breast. Staging is important as it is often a good predictor of outcomes and treatment is adjusted accordingly.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 91
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 92
Correct
-
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: 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 93
Correct
-
Which of the following is responsible for the maximum increase in total peripheral resistance on sympathetic stimulation?
Your Answer: Arterioles
Explanation:Arterioles are also known as the resistance vessels as they are responsible for approximately half the resistance of the entire systemic circulation. They are richly innervated by the autonomic nervous system and hence, will bring about the maximum increase in peripheral resistance on sympathetic stimulation.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 94
Correct
-
Which of these antibiotics is the first choice treatment for infections caused by Pseudomonas aeruginosa?
Your Answer: Piperacillin
Explanation:Piperacillin is an extended-spectrum beta-lactam antibiotic of the ureidopenicillin class. It is normally used with a beta-lactamase inhibitor such as tazobactam. The combination has activity against many Gram-positive and Gram-negative pathogens and anaerobes, including Pseudomonas aeruginosa. Piperacillin is sometimes referred to as an anti-pseudomonal penicillin.
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This question is part of the following fields:
- Pathology
- Pharmacology
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Question 95
Incorrect
-
A 41 year old women presents with a history of carcinoma involving the right breast with enlarged axillary nodes on the same side. She underwent mastectomy and axillary node clearance. These were sent for histopathological examination. They showed no signs of metastasis. What could be cause of this enlargement in the lymph nodes?
Your Answer: Paracortical lymphoid hyperplasia
Correct Answer: Sinus histiocytosis
Explanation:Sinus histiocytosis also referred to as reticular hyperplasia, refers to the enlargement, distention and prominence of the sinusoids of the lymph nodes. This is a non-specific form of hyperplasia characteristically seen in lymph nodes that drain tumours. The endothelial lining of the lymph node becomes markedly hypertrophied, along with an increase in the number of macrophages which results in the distortion, distention and enlargement of the sinus. In this scenario there is no evidence that an infection or another malignancy could account for the enlargement. Paracortical lymphoid hyperplasia is caused by an immune response.
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This question is part of the following fields:
- Inflammation & Immunology; Female Health
- Pathology
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Question 96
Correct
-
Which of the following is the most likely cause of massive splenomegaly in a 35-year old gentleman?
Your Answer: Myelofibrosis
Explanation:Causes of massive splenomegaly include chronic myelogenous leukaemia, chronic lymphocytic leukaemia, lymphoma, hairy cell leukaemia, myelofibrosis, polycythaemia vera, sarcoidosis, Gaucher’s disease and malaria.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 97
Correct
-
After a prolonged coronary artery bypass surgery, a 60-year old gentleman was transfused 3 units of fresh-frozen plasma and 2 units of packed red cells. Two days later, the nurse noticed that he was tachypnoeic and chest X-ray showed signs consistent with adult respiratory distress syndrome. Which of the following variables will be low in this patient?
Your Answer: Compliance of the lung
Explanation:Acute or adult respiratory distress syndrome (ARDS) is a reaction to several forms of lung injuries and is commonly associated with sepsis and SIRS (systemic inflammatory response syndrome), severe traumatic injury, severe head injury, narcotics overdose, drowning, pulmonary contusion, and multiple blood transfusions. There is an increase in risk due to pre-existing liver disease or coagulation abnormalities. It results due to indirect toxic effects of neutrophil-derived inflammatory mediators in the lungs. ARDS is defined by the 1994 American–European Consensus Committee as the acute onset of bilateral infiltrates on chest X-ray, a partial pressure of arterial oxygen (pa(O2)) to fraction of inspired oxygen Fi(O2) ratio of less than 200 mmHg and a pulmonary artery occlusion pressure of less than 18 or the absence of clinical evidence of left arterial hypertension. ARDS is basically pulmonary oedema in the absence of volume overload or poor left ventricular function. This is different from acute lung injury, which shows a pa(O2)/Fi(O2) ratio of less than 300 mmHg. Pathogenesis of ARDS starts from damage to alveolar epithelium and vascular endothelium, causing increased permeability. Damage to surfactant-producing type II cells disrupts the production and function of pulmonary surfactant, causing micro atelectasis and poor gas exchange. There is a decrease in lung compliance and increase in work of breathing. Eventually, there is resorption of alveolar oedema, regeneration of epithelial cells, proliferation and differentiation of type II alveolar cells and alveolar remodelling. Some show resolution and some progress to fibrosing alveolitis, which involves the deposition of collagen in alveolar, vascular and interstitial spaces.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 98
Correct
-
Which of the following morphological features is most characteristic of hyaline degeneration?
Your Answer: Homogeneous, ground-glass, pink-staining appearance of cells
Explanation:The characteristic morphological features of hyaline degeneration is ground-glass, pinking staining cytoplasm with an intact cell membrane. The accumulation of lipids, calcium salts, lipofuscin and an amorphous cytoplasm with an intact cell membrane are all characteristically found in different situations.
Pyknotic nucleus and orphan Annie eye nucleus are not seen in hyaline degeneration.
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This question is part of the following fields:
- Cell Injury & Wound Healing
- Pathology
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Question 99
Incorrect
-
What is the primary function of the Kupffer cells found in the liver?
Your Answer: Synthesis of intrinsic factor
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 100
Correct
-
A 27-year old lady presented with dull, abdominal pain and some pain in her lower limbs. On enquiry, it was revealed that she has been suffering from depression for a few months. Physical examination and chest X-ray were normal. Further investigations revealed serum calcium 3.5 mmol/l, albumin 3.8 g/dl and phosphate 0.65 mmol/l. What is the diagnosis?
Your Answer: Parathyroid adenoma
Explanation:Hypercalcaemia with hypophosphatemia indicates parathyroid disorder and adenomas are more common than hyperplasia. In this young age group, metastatic disease is unlikely. Solitary adenomas are responsible for 80-85% cases of primary hyperparathyroidism. 10-15% cases are due to parathyroid hyperplasia and carcinomas account for 2-3% cases. Symptoms include bone pain (bones), nephrolithiasis (stones), muscular aches, peptic ulcer disease, pancreatitis (groans), depression (moans), anxiety and other mental disturbances.
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This question is part of the following fields:
- Endocrine
- Pathology
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Question 101
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 102
Correct
-
Digital rectal examination of a 75-year old gentleman who presented to the surgical clinic with urinary retention revealed an enlarged, nodular prostate. PSA was found to be elevated, favouring the diagnosis of prostatic malignancy. Which of the given options is the most common malignant lesion affecting the prostate gland?
Your Answer: Adenocarcinoma
Explanation:Prostatic adenocarcinoma is the commonest solid malignancy and non-dermatological cancer in men above 50 years age. Increasing in incidence with age and the highest risk seen in the black population. About 75% of cases are seen in men over 65 years. Other tumours affecting the prostate include undifferentiated prostate cancer, squamous cell carcinoma, and ductal transitional carcinoma, but these occur less commonly. Sarcomas usually affect children. Hormones play a role in the aetiology of prostate adenocarcinoma unlike the other types. Intraepithelial neoplasia is considered a precursor of invasive malignancy.
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This question is part of the following fields:
- Pathology
- Urology
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Question 103
Incorrect
-
Which cells are most commonly seen in a granulomatous lesion that suggests an underlying chronic inflammation?
Your Answer: Mast cells
Correct Answer: Lymphocytes
Explanation:Lymphocytes and monocytes are commonly and characteristically recognised in a case of chronic inflammation.
Eosinophils and neutrophils are seen with acute inflammation.
Mast cells release histamine in early inflammation.
Basophils are seen with allergies.
Plasma cells are seen with viral infection.
Platelets are not characteristic of any type of inflammation.
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This question is part of the following fields:
- Inflammation & Immunology
- Pathology
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Question 104
Correct
-
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: 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 105
Correct
-
Causes of metabolic acidosis with a normal anion gap include:
Your Answer: Diarrhoea
Explanation:Excess acid intake and excess bicarbonate loss as in diarrhoea, are causes of metabolic acidosis with a normal anion gap. The other conditions all result in an increased anion gap.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
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Question 106
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: Collecting duct
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 107
Correct
-
Bloody discharge from the nipple of a 40-year old woman with no obvious lump or abnormality on mammography is suggestive of:
Your Answer: Intraductal papilloma
Explanation:A small benign tumour, namely intraductal papilloma is most common in women between 35-55 years of age. It is also the commonest cause of spontaneous discharge from a single duct. A lump below the nipple may be sometimes palpable. Ultrasound and ductography are useful investigations., along with cytology of discharge to assess the presence of malignant cells. Confirmation is by breast biopsy.
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This question is part of the following fields:
- Pathology
- Women's Health
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Question 108
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 109
Incorrect
-
Vincristine is a chemotherapy agent used to treat a number of types of cancer. Which of the following is a recognised major side-effect of vincristine?
Your Answer: Myelosuppression
Correct Answer: Peripheral neuropathy
Explanation:Vincristine is an alkaloid chemotherapeutic agent. It is used to treat a number of types of cancer including acute lymphocytic leukaemia, acute myeloid leukaemia, Hodgkin’s disease, neuroblastoma, and small cell lung cancer among others. The main side-effects of vincristine are peripheral neuropathy and constipation.
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This question is part of the following fields:
- Pathology
- Pharmacology
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Question 110
Correct
-
After being admitted to the hospital, a 60-year-old man is administered sodium nitroprusside. Which class of drugs does nitroprusside belong to?
Your Answer: Vasodilators
Explanation:Sodium nitroprusside is a potent peripheral vasodilator that affects both arterioles and venules. It is often administered intravenously to patients who are experiencing a hypertensive emergency. It reduces both total peripheral resistance as well as venous return, so decreasing both preload and afterload. For this reason it can be used in severe cardiogenic heart failure where this combination of effects can act to increase cardiac output. It is administered by intravenous infusion. Onset is typically immediate and effects last for up to ten minutes. The duration of treatment should not exceed 72 hours.
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This question is part of the following fields:
- Pathology
- Pharmacology
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Question 111
Incorrect
-
A 73-year-old woman goes to the doctor complaining of pain and stiffness in her shoulders and hips for 4 months, which is worst in the mornings. She has also been suffering from fatigue, weight loss and depression. There were no abnormal findings on physical examination. The erythrocyte sedimentation rate was 110 mm/hour, and serum rheumatoid factor and antinuclear antibody assays were negative. Mild normochromic normocytic anaemia was also found. What is the most likely diagnosis in this case?
Your Answer: Fibromyositis
Correct Answer: Polymyalgia rheumatica
Explanation:Polymyalgia rheumatica (PMR) affects older adults, with an acute or subacute onset. Symptoms include severe pain and stiffness of the neck and pectoral or pelvic girdles, which is worse in the morning or after a period of inactivity and is usually bilateral. Other symptoms can include fatigue, weight loss, depression and fever. The erythrocyte sedimentation rate is usually elevated, and normochromic normocytic anaemia can occur. Other tests are usually normal in this disease.
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This question is part of the following fields:
- Orthopaedics
- Pathology
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Question 112
Correct
-
The neurotransmitters adrenaline, noradrenaline and dopamine are derived from which amino acid?
Your Answer: Tyrosine
Explanation:Tyrosine is the precursor to adrenaline, noradrenaline and dopamine. Tyrosine hydroxylase converts tyrosine to DOPA, which is in turn converted to dopamine, then to noradrenaline and finally adrenaline.
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This question is part of the following fields:
- Endocrine
- Physiology
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Question 113
Incorrect
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An organ transplant patient may be at risk of developing which type of cancer?
Your Answer: Thyroid cancer
Correct Answer: Skin cancer
Explanation:The most common malignancies encountered in the post–solid organ transplant setting are non-melanoma skin cancers, post-transplant lymphoproliferative disorders and Kaposi’s sarcoma (KS). The pathogenesis of these tumours is likely related to the immunosuppressive drugs used post-transplantation and subsequent viral infection.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 114
Incorrect
-
A chloride sweat test was performed on a 13-year-old boy. Results indicated a high likelihood of cystic fibrosis. This diagnosis is associated with a higher risk of developing which of the following?
Your Answer: Adenocarcinoma of the lung
Correct Answer: Bronchiectasis
Explanation:Cystic fibrosis is a life-threatening disorder that causes the build up of thick mucus in the lungs, digestive tract, and other areas of the body. It is a hereditary autosomal-recessive disease caused by mutations of the CFTR gene. Cystic fibrosis eventually results in bronchiectasis which is defined as a permanent dilatation and obstruction of bronchi or bronchioles.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 115
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: Collecting duct
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 116
Correct
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A young female in the 15th week of pregnancy presented to the emergency department with the passage of grape-like masses per vagina. Dilatation and curettage was done and microscopy revealed large avascular villi with trophoblastic proliferation. Which one of the following investigations is best recommended for her follow-up?
Your Answer: Serum β-hCG
Explanation:Trophoblast is the layer of cells surrounding the blastocyst and that later develops into the chorion and amnion. Gestational trophoblastic disease is a tumour arising from this trophoblast. It can occur during or after either an intrauterine or ectopic pregnancy. If it occurs in a pregnant woman, it usually leads to spontaneous abortion, eclampsia or fetal death. It can be either malignant or benign.
In suspected cases, investigations include measurement of serum beta subunit of human chorionic gonadotrophin (β-hCG) and pelvic ultrasound. Confirmatory test is a biopsy. Post-removal, the disease is classified clinically to assess further treatment. To assess the presence of metastases, further work-up includes computed tomography of the brain, chest, abdomen and pelvis. Chemotherapy is usually needed for persistent disease. If at least three consecutive, weekly serum β-hCG measurements are normal, treatment is considered successful. Follow-up is also done by measuring β-hCG.
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This question is part of the following fields:
- Pathology
- Women's Health
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Question 117
Correct
-
A 62-year-old woman presented to the doctor complaining of spine pain, fatigue and oliguria. She is diagnosed with chronic renal failure. Dipstick testing shows no protein, glucose, nitrite or ketones but a semi-quantitative sulphosalicylic acid test for urine protein is positive. Which of the following is the most probable cause of chronic renal failure in this patient.
Your Answer: Multiple myeloma
Explanation:Dipstick results are negative because the proteins found in the urine of this patient are not albumin but Bence Jones proteins. A Bence Jones protein is a monoclonal globulin protein commonly detected in patients affected by multiple myeloma. Multiple myeloma is a malignancy of plasma cells characterised by the production of monoclonal immunoglobulin. Symptoms include bone pain, bone fractures, bleeding, neurologic symptoms, fatigue, frequent infections and weight loss.
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This question is part of the following fields:
- Pathology
- Renal
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Question 118
Correct
-
A 55 year old man presented with a 4 day history of cough and fever. His sputum culture showed the presence of Strep pneumoniae. Which of the following substances produced by the inflammatory cells will result in effective clearance of this organism from the lung parenchyma?
Your Answer: Hydrogen peroxide
Explanation:Hydrogen peroxide is produced by myeloperoxidase to form a potent oxidant that eliminates bacteria, but is not effective in chronic granulomatous diseases.
Platelet activating factor will lead to the activation, adhesion and aggregation of platelets but will not directly kill bacteria.
Prostaglandins cause vasodilation but do not activate neutrophils.
Kallikrein promotes formation of bradykinin that leads to vasodilation.
Leukreines increase vascular permeability.
Cytokines are communicating molecules between immune cells but directly will not kill bacteria.
Interleukins will regulate the immune response.
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This question is part of the following fields:
- Inflammation & Immunology; Respiratory
- Pathology
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Question 119
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 120
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: Aneurysmal bone cyst
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 121
Correct
-
Which of the following conditions causes an elevation of the pH in the tissues with elevated arterial CO2 content?
Your Answer: Metabolic alkalosis
Explanation:Metabolic alkalosis is a metabolic condition in which the pH of tissue is elevated beyond the normal range (7.35-7.45). This is the result of decreased hydrogen ion concentration, leading to increased bicarbonate, or alternatively a direct result of increased bicarbonate concentrations. Normally, arterial pa(CO2) increases by 0.5–0.7 mmHg for every 1 mEq/l increase in plasma bicarbonate concentration, a compensatory response that is very quick. If the change in pa(CO2) is not within this range, then a mixed acid–base disturbance occurs.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 122
Correct
-
Which of these conditions causes haematuria, hypertension and proteinuria in children, usually after a streptococcal infection?
Your Answer: Acute nephritic syndrome
Explanation:Nephritic syndrome (or acute nephritic syndrome) is a syndrome comprising of signs of nephritis. Children between 2 and 12 are most commonly affected, but it may occur at any age. Predisposing factors/causes include:
Infections with group A streptococcal bacteria (acute post-streptococcal glomerulonephritis).
Primary renal diseases: immunoglobulin A nephropathy, membranoproliferative glomerulonephritis, idiopathic rapidly progressive crescentic glomerulonephritis.
Secondary renal diseases: subacute bacterial endocarditis, infected ventriculo–peritoneal shunt, glomerulonephritis with visceral abscess, glomerulonephritis with bacterial, viral or parasitic infections.
Multisystem diseases.
By contrast, nephrotic syndrome is characterized by only proteins moving into the urine.
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This question is part of the following fields:
- Physiology
- Renal
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Question 123
Correct
-
Which of the following morphological characteristic is a salient feature of a pure apoptotic cell?
Your Answer: Chromatin condensation
Explanation:Apoptosis is the programmed death of cells which occurs as a normal and controlled part of an organism’s growth or development. The changes which occur in this process include blebbing, cell shrinkage, nuclear fragmentation, chromatin condensation, chromosomal DNA fragmentation, and global mRNA decay. The cell membrane however remains intact and the dead cells are phagocytosed prior to any content leakage and thus inflammatory response.
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This question is part of the following fields:
- Cell Injury & Wound Healing
- Pathology
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Question 124
Correct
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A 38-year old lady presented to the hospital with abnormal passing of blood per vagina. On examination, she was found to have an endocervical polypoidal mass. On enquiry, she gave history of oral contraceptive usage for 3 years. What finding is expected on the histopathology report of biopsy of the mass?
Your Answer: Microglandular hyperplasia
Explanation:Endocervical polyps or microglandular hyperplasia are benign growths occurring in the endocervical canal, in about 2-5% women and occur secondary to use of oral contraceptives. They are usually < 1cm in size, friable and reddish-pink. Usually asymptomatic, they can cause bleeding or become infected, leading to leucorrhoea (purulent vaginal discharge). They are usually benign but need to be differentiated from adenocarcinomas by histology.
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This question is part of the following fields:
- Pathology
- Women's Health
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Question 125
Correct
-
If the blood flow is constant, oxygen extraction by tissues will show the greatest decrease due to which of the following interventions?
Your Answer: Tissue cooling
Explanation:With a constant blood flow to a given tissue bed, there will be an increase in oxygen extraction by the tissue with the following; an increase in tissue metabolism and oxygen requirements: warming (or fever), exercise, catecholamines and thyroxine. With cooling, the demand for oxygen decreases, leading to decreased oxygen extraction.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 126
Correct
-
If a tumour is found in both lobes of the prostate, without nodal involvement or metastases, a histological grade of G2 and elevated PSA, what is the overall prostatic cancer stage?
Your Answer: Stage II
Explanation:The AJCC uses the TNM, Gleason score and PSA levels to determine the overall stage of prostatic cancer. This staging is as follows:
Stage I: T1, N0, M0, Gleason score 6 or less, PSA less than 10; or T2a, N0, M0, Gleason score 6 or less, PSA less than 10
Stage IIa: T1, N0, M0, Gleason score of 7, PSA less than 20; or T1, N0, M0, Gleason score of 6 or less, PSA at least 10 but less than 20; or T2a or T2b, N0, M0, Gleason score of 7 or less, PSA less than 20
Stage IIb: T2c, N0, M0, any Gleason score, any PSA; or T1 or T2, N0, M0, any Gleason score PSA of 20 or more; or T1 or T2, N0, M0, Gleason score of 8 or higher, any PSA
Stage III: T3, N0, M0, any Gleason score, any PSA Stage IV: T4, N0, M0,any Gleason score, any PSA; or any T, N1, M0,any Gleason score, any PSA; or Any T, any N, M1, any Gleason score, any PSA.
The patient in this case has a T2 N0 M0 G2 tumour, meaning it belongs in stage II
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This question is part of the following fields:
- Pathology
- Urology
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Question 127
Correct
-
A young girl who presented with a clinical picture of type I hypersensitivity reaction with eosinophilia is most likely to have?
Your Answer: Liver flukes
Explanation:Usually a parasitic infection will be associated with a type I hypersensitivity reaction.
Amyloid deposition will not cause an immune reaction.
Organic dust will lead to a type III hypersensitivity reaction.
Cell mediated as well as humoral immune mechanism play a part in syphilis, but they are do not specifically cause a type I reaction.
Malaria is cause by plasmodium and is not cause of a hypersensitivity reaction.
Atopic dermatitis will not be accompanied by eosinophilia.
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This question is part of the following fields:
- Inflammation & Immunology
- Pathology
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Question 128
Correct
-
If a 68-year-old man is diagnosed with a testicular seminoma that reaches the tunica albuginea and involves the tunica vaginalis, with retroperitoneal lymph nodes greater than 5cm, LDH 1.4 times the reference levels, β-hCG 4250 mIU/ml and AFP 780 ng/ml, what's the clinical stage in this case?
Your Answer: Stage IIC
Explanation:According to the AJCC, the clinical staging for testicular seminoma is:
Stage IA: T1 N0 M0 S0
Stage IB: T2/3/4 N0 M0 S0
Stage IC: any T N0 M0 S1/2/3
Stage IIA: any T N1 M0 S0/1
Stage IIB: any T N2 M0 S0/1
Stage IIC: any T N3 M0 S0/1
Stage IIIA: any T any N M1a S0/1
Stage IIIB: any T any N M0/1a S2
Stage IIIC: any T any N M1a/1b S3.
The patient in this case has IIC stage -
This question is part of the following fields:
- Pathology
- Urology
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Question 129
Correct
-
For calculation of cardiac output by Fick's principle, which of the following vessels is the best source of venous blood to determine the arterial-to-venous oxygen tension difference?
Your Answer: Pulmonary artery
Explanation:Fick’s principle states that the total uptake (or release) of a substance by peripheral tissues is equal to the product of the blood flow to the peripheral tissues and the arterial– venous concentration difference (gradient) of the substance. It is used to measure the cardiac output, and the formula is Cardiac output = oxygen consumption divided by arteriovenous oxygen difference. Assuming there are no shunts across the pulmonary system, the pulmonary blood flow equals the systemic blood flow. The arterial and venous blood oxygen content is measured by sampling from the pulmonary artery (low oxygen content) and pulmonary vein (high oxygen content). Peripheral arterial blood is used as a surrogate for the pulmonary vein.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 130
Incorrect
-
Gastric acid secretion is stimulated by which of the following?
Your Answer: The glossopharyngeal nerve
Correct Answer: Gastrin
Explanation:Gastric acid secretion is stimulated by three factors:
– Acetylcholine, from parasympathetic neurones of the vagus nerve that innervate parietal cells directly
– Gastrin, produced by pyloric G-cells
– Histamine, produced by mast cells.
Gastric acid is inhibited by three factors:
– Somatostatin
– Secretin
– Cholecystokinin
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This question is part of the following fields:
- Endocrine
- Physiology
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Question 131
Correct
-
A patient is suspected to have a chromosomal abnormality. Which tumour and chromosomal association is correct?
Your Answer: Neuroblastoma – chromosome 1
Explanation:Neuroblastoma is associated with a deletion on chromosome 1 and inactivation of a suppressor gene. Neurofibromas and osteogenic sarcoma are associated with an abnormality on chromosome 17. Retinoblastoma (Rb) is associated with an abnormality on chromosome 13. Wilms’ tumours of the kidney are associated with an abnormality on chromosome 11.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 132
Correct
-
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 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 133
Correct
-
Routine evaluation of a 38 year old gentleman showed a slightly lower arterial oxygen [pa(O2)] than the alveolar oxygen [pA(O2)]. This difference is:
Your Answer: Is normal and due to shunted blood
Explanation:Blood that bypasses the ventilated parts of lung and enters the arterial circulation directly is known as shunted blood. It happens in normal people due to mixing of arterial blood with bronchial and some myocardial venous blood (which drains into the left heart). Diffusion limitation and reaction velocity with haemoglobin are immeasurably small. CO2 unloading will not affect the difference between alveolar and arterial p(O2). A large VSD will result in much lower arterial O2 as compared to alveolar O2.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 134
Correct
-
Cervical intraepithelial neoplasia on Pap smear of a 34-year old lady is most likely associated with which of the following?
Your Answer: Human papillomavirus infection
Explanation:CIN (Cervical intraepithelial neoplasia) is considered a precursor of cervical cancer and is likely caused due to infection with human papillomavirus (HPV) types 16, 18, 31, 33, 35 or 39. The risk factors for cervical cancer include multiple sex partners, young age at the time of first intercourse, intercourse with men whose previous partners had cervical cancer. Also, smoking and immunodeficient states are considered contributory. CIN is graded as mild (grade I), moderate (grade II) and severe dysplasia or carcinoma in situ (grade III). CIN III rarely regresses spontaneously and can lead to invasive carcinoma by invading the basement membrane. Squamous cell carcinomas are the commonest cervical cancer seen in 80-85% of all cases. Others are commonly adenocarcinomas. Cervical cancer can spread by direct extension, lymphatic spread to pelvic and para-aortic nodes or by hematogenous route.
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This question is part of the following fields:
- Pathology
- Women's Health
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Question 135
Correct
-
A 13 year old girl presented with signs of shortness of breath, chest pain, non-productive cough, oedema of the lower extremities and cyanosis of the fingertips. She has a history of a ventricular septal defect not surgically corrected. The most probable cause of these symptoms is:
Your Answer: Shunt reversal
Explanation:A ventricular septal defect (VSD) is a common form of congenital heart defects and is characterised by the presence of a hole in the wall that separates the right from the left ventricle. Medium or large defects can cause many complications. One of these complication is Eisenmenger syndrome, characterised by reversal of the shunt (from left-to-right shunt into a right-to-left) ,cyanosis and pulmonary hypertension.
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This question is part of the following fields:
- Cardiovascular
- Pathology
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Question 136
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 137
Correct
-
A new-born was found to have an undeveloped spiral septum in the heart. This is characteristic of which of the following?
Your Answer: Persistent truncus arteriosus
Explanation:Persistent truncus arteriosus is a congenital heart disease that occurs when the primitive truncus does not divide into the pulmonary artery and aorta, resulting in a single arterial trunk. The spiral septum is created by fusion of a truncal septum and the aorticopulmonary spiral septum. Incomplete development of these septa results in incomplete separation of the common tube of the truncus arteriosus and the aorticopulmonary trunk.
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This question is part of the following fields:
- Cardiovascular
- Pathology
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Question 138
Correct
-
Which of the following is NOT a nutritional factor involved in wound healing:
Your Answer: Vitamin B3
Explanation:Vitamin B6 is required for collagen cross-links.
Vitamin A is required for epithelial cell proliferation.
Zinc is required for RNA and DNA synthesis.
Copper is required for cross-linking of collagen.
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This question is part of the following fields:
- Cell Injury & Wound Healing
- Physiology
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Question 139
Incorrect
-
A 70-year-old man who was previously healthy develops a rapidly progressing neurological deficit. A MRI reveals a large, poorly defined mass with central necrosis in his left temporal lobe. What is the most likely diagnosis?
Your Answer: Low-grade astrocytoma
Correct Answer: Glioblastoma multiforme
Explanation:Glioblastoma multiforme, or grade 4 astrocytoma, is the most common and aggressive primary brain tumour. They are formed by small areas of necrotising tissue surrounded by highly anaplastic cells. Most of them arise from the deep white matter of the brain and quickly infiltrate it, becoming very large before they are symptomatic. The most common symptom is progressive memory, personality or neurological deficit due to temporal and frontal lobe involvement. It is most common in men, and risk factors include: neurofibromatosis, tuberous sclerosis, Von Hippel-Lindau disease, Li-Fraumeni syndrome, and Turcot syndrome.
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This question is part of the following fields:
- Neurology
- Pathology
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Question 140
Correct
-
The chest X-ray of a 72 year old patient reveals the presence of a round lesion containing an air-fluid level in the left lung. These findings are most probably suggestive of:
Your Answer: Lung abscess
Explanation:Lung abscesses are collections of pus within the lung that arise most commonly as a complication of aspiration pneumonia caused by oral anaerobes. Older patients are more at risk due to poor oral hygiene, gingivitis an inability to handle their oral secretions due to other diseases. Chest X-ray most commonly reveals the appearance of an irregularly shaped cavity with an air-fluid level.
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 141
Correct
-
Which of the following features is indicative of poor prognosis in a case of breast carcinoma?
Your Answer: Axillary lymph node metastases
Explanation:Lymphatic spread indicates poor prognosis. Presence of family history is not a prognostic factor despite being linked to higher incidence. Aneuploidy is a poor prognostic factor. A breast tumour positive for oestrogen receptors is a good prognostic factor as it increases the responsiveness of the tumour to certain therapies. In-situ tumours carry the best prognosis.
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This question is part of the following fields:
- Pathology
- Women's Health
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Question 142
Correct
-
Difficulty in retracting the foreskin of the penis in an uncircumcised male is known as:
Your Answer: Phimosis
Explanation:Phimosis is the inability to fully retract the foreskin of the penis in an uncircumcised male. It can be physiological in infancy, in which it could be referred to as ‘developmental non-retractility of the foreskin. However, it is almost always pathological in older children and men. Causes include chronic inflammation (e.g. balanoposthitis), multiple catheterisations, or forceful foreskin retraction. One of the causes is chronic balanitis xerotica obliterans. It leads to development of a ring of indurated tissue near the tip of the prepuce, which prevents retraction. Contributory factors include infections, hormonal and inflammatory factors. The recommended treatment includes circumcision.
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This question is part of the following fields:
- Pathology
- Urology
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Question 143
Correct
-
A 45-year-old man with short bowel syndrome requires parenteral nutrition. The solution of choice for parenteral nutrition is:
Your Answer: Crystalline amino acids
Explanation:Total parenteral nutrition (TPN), is the practice of feeding a person intravenously, circumventing the gut. It is normally used in the following situations: surgery, when feeding by mouth is not possible, when a person’s digestive system cannot absorb nutrients due to chronic disease or if a person’s nutrient requirement cannot be met by enteral feeding and supplementation. A sterile bag of nutrient solution, between 500 ml and 4L, is provided. The pump infuses a small amount (0.1–10 ml/h) continuously to keep the vein open. The nutrient solution consists of water, glucose, salts, amino acids, vitamins and sometimes emulsified fats. Ideally each patient is assessed individually before commencing on parenteral nutrition, and a team consisting of doctors, nurses, clinical pharmacists and dietitians evaluate the patient’s individual data and decide what formula to use and at what rate.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 144
Correct
-
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: 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 145
Correct
-
Which of the following cells would be increased in a patient suffering from a hydatid cyst in the liver?
Your Answer: Eosinophils
Explanation:Eosinophils are granulocytes that respond to parasitic infections. They are also involved in allergy response and asthma. They contain granules which stain red with Romanowsky’s method and contain peroxidase, Rnase, Dnase, histamine, lipase and major basic proteins that are toxic to the parasite as well as the hosts tissue. They are about 1-5% of the total WBC population and persist in the blood for 6-12 hours.
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This question is part of the following fields:
- Inflammation & Immunology
- Pathology
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Question 146
Correct
-
Which tumour occurs in young adults, affecting the epiphyses of the bones and sometimes extending to the soft tissues?
Your Answer: Benign giant-cell tumour
Explanation:Benign giant-cell tumours tend to affect adults in their twenties and thirties, occur in the epiphyses and can erode the bone and extend into the soft tissues. These tumours have a strong tendency to recur.
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This question is part of the following fields:
- Orthopaedics
- Pathology
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Question 147
Correct
-
Which of the following diseases affects young adults, causing pain in any bone -particularly long bones- which worsens at night, and is typically relieved by common analgesics, such as aspirin?
Your Answer: Osteoid osteoma
Explanation:Osteoid osteoma, which tends to affect young adults, can occur in any bone but is most common in long bones. It can cause pain (usually worse at night) that is typically relieved by mild analgesics, such as non-steroidal anti-inflammatory drugs. X-ray findings include a small radiolucent zone surrounded by a larger sclerotic zone.
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This question is part of the following fields:
- Orthopaedics
- Pathology
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Question 148
Correct
-
The most likely cause of prominent U waves on the electrocardiogram (ECG) of a patient is:
Your Answer: Hypokalaemia
Explanation:The U-wave, not always visible in ECGs, is thought to represent repolarisation of papillary muscles or Purkinje fibres. When seen, it is very small and occurs after the T-wave. Inverted U-waves indicate myocardial ischaemia or left ventricular volume overload. Prominent U-waves are most commonly seen in hypokalaemia. Other causes include hypercalcaemia, thyrotoxicosis, digitalis exposure, adrenaline and class 1A and 3 anti-arrhythmic agents. It can also be seen in congenital long-QT syndrome and in intracranial haemorrhage.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 149
Correct
-
A 55-year-old male has a mass on the right lung with involvement of the right mediastinal lymph nodes. What is the nodal staging according to the TNM staging?
Your Answer: N2
Explanation:The N stages for lung cancer are from NO to N3 : NO, there is no lung cancer in any lymph nodes: N1, there is lung cancer in the proximal lymph nodes: N2, there is lung cancer in the mediastinal hilar lymph nodes, but on the same side as the affected lung or there is lung cancer in the carinal lymph nodes: N3, there is metastatic lung cancer in lymph nodes on the opposite side of the chest, in the cervical or apical lymph nodes. In this patient the ipsilateral mediastinal node is involved, thus it is classified as N2.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 150
Correct
-
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: 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 151
Incorrect
-
During routine laboratory tests, a 66-year-old man is found to be suffering from hypercholesterolaemia and is prescribed atorvastatin. What is the mechanism of action of atorvastatin?
Your Answer: Decreases cholesterol absorption in the small intestine
Correct Answer: Inhibits cholesterol synthesis
Explanation:Atorvastatin is a member of the drug class of statins, used for lowering cholesterol. The mode of action of statins is inhibition of 3-hydroxy-3-methylglutaryl-coenzyme A (HMG-CoA) reductase. This enzyme is needed by the body to make cholesterol. The primary uses of atorvastatin is for the treatment of dyslipidaemia and the prevention of cardiovascular disease.
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This question is part of the following fields:
- Pathology
- Pharmacology
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Question 152
Correct
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A cancer patient was found to have a radio resistant tumour. Which tumour does the patient most likely have?
Your 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 153
Correct
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Calculate the resistance of the artery if the pressure at one end is 60 mmHg, pressure at the other end is 20 mm Hg and the flow rate in the artery is 200 ml/min.
Your Answer: 0.2
Explanation:Flow in any vessel = Effective perfusion pressure divided by resistance, where effective perfusion pressure is the mean intraluminal pressure at the arterial end minus the mean pressure at the venous end. Thus, in the given problem, resistance = (60 − 20)/200 = 0.2 mmHg/ml per min.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 154
Correct
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Fine-needle aspiration is a type of biopsy procedure. When performing a fine-needle aspiration of the lungs, which is the most common complication of the procedure?
Your Answer: Pneumothorax
Explanation:Pneumothorax is the most common complication of a fine-needle aspiration procedure. Various factors, such as lesion size, have been associated with increased risk of pneumothorax .
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This question is part of the following fields:
- Pathology
- Respiratory
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Question 155
Correct
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Work of breathing (WOB) is the energy expended to inhale and exhale a breathing gas. Normally, maximal amount of work of breathing is required to overcome:
Your Answer: Elastic lung compliance
Explanation:The forces of elastance (compliance), frictional resistance and inertia have been identified as the forces that oppose lung inflation and deflation. The normal relaxed state of the lung and chest is partially empty. Further exhalation requires muscular work. Inhalation is an active process requiring work. About 60–66% of the total work performed by the respiratory muscles is used to overcome the elastic or compliance characteristics of the lung–chest cage, 30–35% is used to overcome frictional resistance and only 2–5% of the work is used for inertia.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 156
Correct
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Which organ is most vulnerable to haemorrhagic shock?
Your Answer: Kidneys
Explanation:At rest, the brain receives 15% cardiac output, muscles 15%, gastrointestinal tract 30% and kidneys receive 20%. However, if normalised by weight, the largest specific blood flow is received by the kidneys at rest (400 ml/min x 100g), making them highly vulnerable in the case of a haemorrhagic shock.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 157
Correct
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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: 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 158
Correct
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Congenital anomalies of genitourinary tract are more common than any other system. Which of the following anomalies carries the greatest risk of morbidity?
Your Answer: Bladder exstrophy
Explanation:Bladder exstrophy is the condition where the urinary bladder opens from the anterior aspect suprapubically. The mucosa of the bladder is continuous with the abdominal skin and there is separation of the pubic bones. The function of the upper urinary tract remains normal usually. Treatment consists of surgical reconstruction of the bladder and returning it to the pelvis. There can be a need for continent urinary diversion along with reconstruction of the genitals.
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This question is part of the following fields:
- Pathology
- Renal
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Question 159
Correct
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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 160
Incorrect
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Evaluation of a 60-year old gentleman, who has been a coal miner all his life and is suspected to have pulmonary fibrosis reveals the following: FEV1 of 75% (normal > 65%), arterial oxygen saturation 92%, alveolar ventilation 6000 ml/min at a tidal volume of 600 ml and a breathing rate of 12 breaths/min. There are also pathological changes in lung compliance and residual volume. Calculate his anatomical dead space.
Your Answer: 150 ml
Correct Answer: 100 ml
Explanation:Dead space refers to inhaled air that does not take part in gas exchange. Because of this dead space, taking deep breaths slowly is more effective for gas exchange than taking quick, shallow breaths where a large proportion is dead space. Use of a snorkel by a diver increases the dead space marginally. Anatomical dead space refers to the gas in conducting areas such as mouth and trachea, and is roughly 150 ml (2.2 ml/kg body weight). This corresponds to a third of the tidal volume (400-500 ml). It can be measured by Fowler’s method, a nitrogen wash-out technique. It is posture-dependent and increases with increase in tidal volume. Physiological dead space is equal to the anatomical dead space plus the alveolar dead space, where alveolar dead space is the area in the alveoli where no effective exchange takes place due to poor blood flow in capillaries. This physiological dead space is very small normally (< 5 ml) but can increase in lung diseases. Physiological dead space can be measured by Bohr’s method. Total ventilation per minute (minute ventilation) is given by the product of tidal volume and the breathing rate. Here, the total ventilation is 600 ml times 12 breaths/min = 7200 ml/min. The problem mentions alveolar ventilation to be 6000 ml/min. Thus, the difference between the alveolar ventilation and total ventilation is 7200 – 6000 ml/min = 1200 ml/min, or 100 ml per breath at 12 breaths per min. This 100 ml is the dead space volume.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 161
Correct
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A 60-year-old man complains of pain in his left thigh. An X-ray reveals bowing of the affected femur, increased bone density, bony enlargement, abnormal bone architecture with coarse cortical trabeculations, and stress microfractures. Which is the most likely diagnosis in this case?
Your Answer: Paget’s disease of bone
Explanation:Paget’s disease of bone is a chronic disorder of the adult skeleton in which bone turnover is accelerated in localised areas, replacing normal matrix with softened and enlarged bone and causing gradual pain and deformity in some cases. It is more predominant in men over the age of 40. Characteristic X-ray findings include increased bone density, abnormal architecture with coarse cortical trabeculation or cortical thickening, bowing and bony enlargement; there might also be stress microfractures of the tibia or femur.
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This question is part of the following fields:
- Orthopaedics
- Pathology
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Question 162
Incorrect
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A patient with testicular seminoma has the following tumour markers: LDH 1.3 times the reference levels, β-hCG 4500 mIU/ml and AFP 875 ng/ml. What's the serum tumour marker stage in this case?
Your Answer: S4
Correct Answer: S1
Explanation:According to AJCC guidelines, the serum tumour marker staging is the following:
S0: marker studies within normal limits
S1: lactate dehydrogenase (LDH) less than 1.5 times the reference range, beta-human chorionic gonadotrophin (β-hCG) <5000 mIU/ml, and alpha-fetoprotein (AFP) <1000 ng/ml S2: LDH 1.5–10 times the reference range, β-hCG 5000–50,000 mIU/ml or AFP 1000–10,000 ng/ml S3: LDH greater than 10 times the reference range, β-hCG >50,000 mIU/ml or AFP >10,000 ng/ml.
According to this, the patient’s tumour belongs to the S1 stage.
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This question is part of the following fields:
- Pathology
- Urology
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Question 163
Correct
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A 55-year-old woman died 3 years after a cardiac transplant due to worsening congestive heart failure. Autopsy revealed diffuse hyperplasia of the vascular intima involving the entire length of the coronary arteries. The most probable cause of deterioration of the cardiac function is:
Your Answer: Coronary atherosclerosis
Explanation:Allograft coronary artery disease (CAD) can begin right after the transplant and is the major cause of later death in cardiac transplant recipients. This form of atherosclerosis progresses quickly resulting in allograft failure. Due to lack of premonitory symptoms CAD may lead to sudden death.
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This question is part of the following fields:
- Cardiovascular
- Pathology
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Question 164
Correct
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Which of the following is likely to result in hematocolpos in a 12-year old girl?
Your Answer: Imperforate hymen
Explanation:Hematocolpos means accumulation of blood in vagina and hematometra is accumulation of blood in the uterus. These are most likely seen with an imperforate hymen; which is seen I 1 in 2000 females. If spontaneous resolution does not occur, treatment involves making a hole in the hymen to allow discharge of menstrual blood.
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This question is part of the following fields:
- Pathology
- Women's Health
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Question 165
Correct
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A histopathological analysis of a specimen showed loss of individual cell structure with karyorrhexis and fragmentation. The overall integrity of the tissue structure is preserved. This is typical of which of the following pathologies?
Your Answer: Viral hepatitis
Explanation:Viral infections will cause necrosis of the hepatocytes with characteristic changes of karyorrhexis and cell fragmentation.
Brown atrophy of the heart is due to accumulation of lipofuscin in the myocardium.
Tissue destruction associated with transplant rejection leads to widespread loss of structural integrity.
Single cell necrosis is not characteristically seen in chronic alcoholic liver.
Barbiturate overdose will result in hypertrophy of the smooth endoplasmic reticulum.
Carcinoma insitu will cause dysplastic cells without the overall structural integrity being disrupted.
Atrophy is due to apoptosis with ordered cellular fragmentation and phagocytosis and will not induce an inflammatory process unlike necrosis.
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This question is part of the following fields:
- Cell Injury & Wound Healing
- Pathology
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Question 166
Correct
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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: 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 167
Correct
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A 44-year old gentleman presented to the hospital in congestive cardiac failure. On further investigation, he was found to have a right-sided retroperitoneal mass. On enquiry, he gave a history of intermittent hypertensive attacks. 24-hour urine specimen revealed raised metanephrine and vanillylmandelic acid levels. What is the likely diagnosis?
Your Answer: Pheochromocytoma
Explanation:A neuroendocrine tumour of the chromaffin cells in adrenal medulla, pheochromocytoma secretes excessive catecholamines – adrenaline (epinephrine) and noradrenaline (norepinephrine). Pheochromocytomas are also known as the ‘10% tumour’. This is because 10% of them are bilateral, 10% are malignant, and 10% are extra-adrenal. Extra-adrenal paragangliomas (also known as extra-adrenal pheochromocytomas) are less common than pheochromocytomas and arise in the ganglia of sympathetic nervous system. Around 25% of pheochromocytomas are familial. Symptoms are described as feeling of ‘impending doom’, and include tachycardia, hypertension, palpitations, anxiety, headaches and pallor. Orthostatic hypertension is typically seen where there is a more than 100 mmHg fall in systolic pressure when the patient stands up. Pheochromocytomas can also lead to malignant hypertension. Diagnosis is by measurement of catecholamines and metanephrines in blood or in 24-hour urine, after exclusion of other possible causes such as stress, hypoglycaemia and drugs (methyldopa, dopamine agonists, ganglion-blocking antihypertensive). Imaging is also needed to localize the tumour. Localization of the tumour can also be done by Iodine-131 meta-iodobenzylguanidine (I131-MIBG) imaging.
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This question is part of the following fields:
- Endocrine
- Pathology
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Question 168
Correct
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A 72 year old man suffered a MI. What is the approximate time needed by the scar tissue of the MI to recover and attain full strength?
Your Answer: Several months
Explanation:A week following a MI attack, a little collagen starts to form and deposit. By the end of the 2nd week, neovascularisation of the scar occurs, with some collagen being laid down in a haphazard fashion. By this time the scar attains some strength. During the next 6 months, collagen is constantly being laid down and is rearranged in order to shrink the scar. Most of the blood vessels by this time have regenerated, decreasing vascularity of the scar reaching full maturity.
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This question is part of the following fields:
- Cell Injury & Wound Healing; Cardiovascular
- Pathology
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Question 169
Correct
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A 40 year old man suffered severe trauma following an MVA. His BP is 70/33 mmhg, heart rate of 140 beats/mins and very feeble pulse. He was transfused 3 units of blood resulting in his BP returning to 100/70 and his heart rate to 90 beats/min. What decreased following transfusion?
Your Answer: Total peripheral resistance
Explanation:The patient is in hypovolemic shock, he is transfused with blood to replace the volume lost. It is important not only to replace fluids but stop active bleeding in resuscitation. Fluid replacement will result in a decreased sympathetic discharge and adequate ventricular filling thus reducing total peripheral resistance and increasing cardiac output and cardiac filling pressures.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 170
Incorrect
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A 56-year-old woman weighs 75 kg. In this patient, total body water, intracellular fluid and extracellular fluid are respectively:
Your Answer: 45 l, 35 l, 10 l
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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SESSION STATS - PERFORMANCE PER SPECIALTY
