-
Question 1
Correct
-
The dural venous sinuses are venous channels that drain blood from the brain. This sinuses are located between which structures?
Your Answer: Meningeal and periosteal layers of the dura mater
Explanation:The dural venous sinuses lies between the periosteal and meningeal layer of the dura mater. Dural venous sinuses is unique because it does not run parallel with arteries and allows bidirectional flow of blood intracranially as it is valve-less.
-
This question is part of the following fields:
- Anatomy
- Head & Neck
-
-
Question 2
Incorrect
-
Which of the following morphological characteristic is a salient feature of a pure apoptotic cell?
Your Answer: Plasma membrane breakdown
Correct 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.
-
This question is part of the following fields:
- Cell Injury & Wound Healing
- Pathology
-
-
Question 3
Incorrect
-
Calculate the total peripheral resistance for a patient with a blood pressure of 130/70 mm HG and cardiac output of 5 litres / min?
Your Answer: 16 mmHg × min/l
Correct Answer: 18 mmHg × min/l
Explanation:Total peripheral resistance = Mean arterial pressure/Cardiac output. And the mean arterial pressure = Diastolic pressure + 1/3 (Systolic pressure – Diastolic pressure), i.e., 70 + 1/3 (130-70) = 90 mmHg. Therefore, total peripheral resistance = 90 mmHg/5 l per min = 18 mmHg × min/l.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 4
Correct
-
Which of the following structures is not easily palpable?
Your Answer: Styloid process of the temporal bone
Explanation:The styloid process is a thin, pointed process that projects antero-inferiorly from the base of the petrous temporal bone. It can vary in length from a short, stubby process to a slender, four to five centimetre rod. It forms from the cranial elements of the second pharyngeal arch. The tympanic plate of the temporal bone ensheathes the base of this process. The pointed, projecting portion of the process provides attachment to the stylohyoid and stylomandibular ligaments, and to three muscles – the styloglossus, stylohyoid, and stylopharyngeus. As the styloid process is covered by the various muscles, it is not easily palpable in live subjects.
-
This question is part of the following fields:
- Anatomy
- Head & Neck
-
-
Question 5
Incorrect
-
A 42-year old woman presents to the doctor with jaundice. Her investigations show conjugated hyperbilirubinemia, raised urine bilirubin levels and low urine urobilinogen levels. What is the likely cause of her jaundice?
Your Answer: Haemolytic anaemia
Correct Answer: Blockage of the common bile duct
Explanation:The description of the patient here fits the diagnosis of obstructive jaundice or cholestasis, which results in conjugated hyperbilirubinemia. Cholestasis occurs due to impairment of bile flow, which can be anywhere from the liver cell canaliculus to the ampulla of Vater. Causes can be divided into intrahepatic and extrahepatic.
– Intrahepatic causes include hepatitis, drug toxicity, alcoholic liver disease, primary biliary cirrhosis, cholestasis of pregnancy and metastatic cancer.
– Extrahepatic causes include common duct stone, pancreatic cancer, benign stricture of the common duct, ductal carcinoma, pancreatitis and sclerosing cholangitis.
There is absence of bile constituents in the intestine, which causes spillage in the systemic circulation. Symptoms include pale stools, dark urine, pruritus, malabsorption leading to steatorrhea and deficiency of fat-soluble vitamins. Chronic cases can result in osteoporosis or osteomalacia due to vitamin D deficiency and Ca2+ malabsorption. Cholesterol and phospholipid retention produces hyperlipidaemia despite fat malabsorption (although increased liver synthesis and decreased plasma esterification of cholesterol also contribute); triglyceride levels are largely unaffected. The lipids circulate as a unique, low-density lipoprotein called lipoprotein X.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 6
Incorrect
-
A 5-year-old child diagnosed with nephrotic syndrome developed generalised oedema. What is the mechanism for the development of oedema in patients with nephrotic syndrome?
Your Answer: Increased capillary hydrostatic pressure
Correct Answer: Decreased colloid osmotic pressure
Explanation:The development of oedema in nephrotic syndrome has traditionally been viewed as an underfill mechanism. According to this view, urinary loss of protein results in hypoalbuminemia and decreased plasma oncotic pressure. As a result, plasma water translocates out of the intravascular space and results in a decrease in intravascular volume. In response to the underfilled circulation, effector mechanisms are then activated that signal the kidney to secondarily retain salt and water. While an underfill mechanism may be responsible for oedema formation in a minority of patients, recent clinical and experimental findings would suggest that oedema formation in most nephrotic patients is the result of primary salt retention. Direct measurements of blood and plasma volume or measurement of neurohumoral markers that indirectly reflect effective circulatory volume are mostly consistent with either euvolemia or a volume expanded state. The ability to maintain plasma volume in the setting of a decreased plasma oncotic pressure is achieved by alterations in transcapillary exchange mechanisms known to occur in the setting of hypoalbuminemia that limit excessive capillary fluid filtration.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
-
-
Question 7
Correct
-
The superior ulnar collateral artery is a direct branch of this artery:
Your Answer: Brachial
Explanation:The brachial artery gives rise to a small branch at the middle of the arm, which is the superior collateral artery. It descends accompanied by the ulnar nerve and anastomoses with the posterior ulnar recurrent and inferior ulnar collateral.
-
This question is part of the following fields:
- Anatomy
- Upper Limb
-
-
Question 8
Correct
-
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.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 9
Incorrect
-
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: 0.1
Correct 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.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 10
Correct
-
A 50 year old man on warfarin therapy following insertion of a pacemaker presented with epistaxis. Which of the following is true regarding blood coagulation?
Your Answer: Patients with haemophilia A usually have a normal bleeding time
Explanation:A prolonged bleeding time is seen in platelet disorders like thrombocytopenia. Patients with haemophilia A or B have a prolonged PTT but not a prolonged bleeding time.
Ca2+ is necessary for coagulation.
von Willebrand factor is an important part of the factor VIII complex and promotes platelet adhesion and aggregation.
DIC results in depleted coagulation factors and accumulation of fibrin.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 11
Correct
-
A 25 year-old female medical student presents with fever, lack of appetite, rashes, sore throat and lymphadenopathy. Peripheral smear shows atypical lymphocytes. Which is the most likely organism responsible for this patient's condition?
Your Answer: Epstein–Barr virus
Explanation:Epstein-Barr virus is in the herpes family of viruses and most people will become infected with EBV sometime during their lives. EBV commonly causes infectious mononucleosis, or mono, a contagious viral illness that initially attacks the lymph nodes in the neck and throat. When these tissues become less effective in fighting infection, sore throats, swelling of the nodes and fever may result.
-
This question is part of the following fields:
- Microbiology
- Pathology
-
-
Question 12
Correct
-
To reach the oral vestibule, the parotid duct must pierce this muscle:
Your Answer: Buccinator muscle
Explanation:The parotid duct or Stensen duct is a duct and the route that saliva takes from the major salivary gland, the parotid gland into the mouth. The parotid duct is formed when several interlobular ducts—the largest ducts inside the parotid gland join. It emerges from the gland and runs forward along the lateral side of the masseter muscle. In this course, the duct is surrounded by the buccal fat pad. It takes a steep turn at the border of the masseter and passes through the buccinator muscle, opening into the vestibule of the mouth, between the cheek and the gums, at the parotid papilla, which lies across the second superior molar tooth.
-
This question is part of the following fields:
- Anatomy
- Head & Neck
-
-
Question 13
Incorrect
-
A surgeon ligates the left superior suprarenal artery whilst preforming a left adrenalectomy. Where does the left superior suprarenal artery originate?
Your Answer: Abdominal aorta
Correct Answer: Left inferior phrenic artery
Explanation:The superior suprarenal arteries arises from the inferior phrenic artery on either side.
-
This question is part of the following fields:
- Abdomen
- Anatomy
-
-
Question 14
Correct
-
Arterial blood gas analysis of a man admitted with acute exacerbation of chronic obstructive pulmonary disease (COPD) showed the following: pH = 7.28, p(CO2) = 65.5 mmHg, p(O2)= 60 mmHg and standard bicarbonate = 30.5 mmol/l. This patient had:
Your Answer: Respiratory acidosis
Explanation:Acidosis with high p(CO2) and normal standard bicarbonate indicates respiratory acidosis, commonly seen in acute worsening of COPD patients. Respiratory acidosis occurs due to alveolar hypoventilation which leads to increased arterial carbon dioxide concentration (p(CO2)). This in turn decreases the HCO3 –/p(CO2) and decreases pH. In acute respiratory acidosis, the p(CO2) is raised above the upper limit of normal (over 45 mm Hg) with a low pH. However, in chronic cases, the raised p(CO2) is accompanied with a normal or near-normal pH due to renal compensation and an increased serum bicarbonate (HCO3 – > 30 mmHg).
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 15
Correct
-
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.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 16
Incorrect
-
A blood sample of a 58 year old male patient, who underwent an abdominal aortic aneurysm repair, was sent to the laboratory. The laboratory technician said that the patient’s blood agglutinates with antisera anti-A and anti-D, while the patient’s serum agglutinates cells of blood group B. What is the blood group of this patient?
Your Answer: B positive
Correct Answer: A positive
Explanation:Group A – has only the A antigen on red cells (and B antibody in the plasma)
Group B – has only the B antigen on red cells (and A antibody in the plasma)
Group AB – has both A and B antigens on red cells (but neither A nor B antibody in the plasma)
Group O – has neither A nor B antigens on red cells (but both A and B antibody are in the plasma). Many people also have a so-called Rh factor on the red blood cell’s surface. This is also an antigen and those who have it are called Rh+. A person with Rh– blood does not have Rh antibodies naturally in the blood plasma (as one can have A or B antibodies, for instance) but they can develop Rh antibodies in the blood plasma if they receive blood from a person with Rh+ blood, whose Rh antigens can trigger the production of Rh antibodies. A person with Rh+ blood can receive blood from a person with Rh– blood without any problems. In this scenario the person has blood group A+ as he has A antigen, anti B antibody and Rh antigen
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 17
Correct
-
A 42 year old women presents with end stage renal failure and is prepared to receive a kidney from her husband. HLA testing showed that they are not a 100% match and she is given immunosuppressant therapy for this. Three months later when her renal function tests were performed she showed signs of deteriorating renal function, with decreased renal output, proteinuria of +++ and RBCs in the urine. She was given antilymphocyte globulins and her condition reversed. What type of graft did this patient receive?
Your Answer: Allograft
Explanation:Allograft describes a graft between two of the same species. As the donor and the recipients are history-incompatible, rejection of the graft is common and is controlled by immunosuppressive drug therapy. Isograft and syngraft are synonymous and referred to a graft transferred between genetically identical individuals e.g. identical twins. In this case rejection is rare as they are history-compatible.
Autograft refers to transfer of one part of the body to another location.
Xenograft is transfer of tissue from another species e.g. pig to human in valve replacement surgeries and rejection is very high.
-
This question is part of the following fields:
- Inflammation & Immunology; Renal
- Pathology
-
-
Question 18
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.
-
This question is part of the following fields:
- Cell Injury & Wound Healing; Dermatology
- Pathology
-
-
Question 19
Correct
-
When at rest, which of the following will be higher in a marathon runner compared to an untrained individual?
Your Answer: Cardiac stroke volume
Explanation:Cardiac muscle hypertrophy is seen in trained athletes as compared to the normal population. This hypertrophy results in higher stroke volume at rest and increased cardiac reserve (maximum cardiac output during exercise). However, the cardiac output at rest is almost the same in both trained and untrained people. This is because in trained athletes, the heart rate is slower, even up to 40-50 beats/min. There is minimal affect of athletic training on oxygen consumption and respiratory rate.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 20
Correct
-
A neurotransmitter of the nigrostriatal pathway is:
Your Answer: Dopamine
Explanation:Dopamine acts as a neurotransmitter in the brain, activating dopamine receptors. It is also a neurohormone released from the hypothalamus. It plays an important role in the reward system. It is believed that dopamine provides a teaching signal to parts of the brain responsible for acquiring new motor sequences (behaviours), by activation of dopamine neurons when an unexpected reward is presented. Loss of dopamine neurones in the nigrostriatal pathway causes Parkinson’s disease. In the frontal lobes, dopamine controls the flow of information from other areas of the brain, and thus, dopamine disorders in this region can cause a decline in neurocognitive functions, especially memory, attention and problem solving. Reduced dopamine concentrations in the prefrontal cortex are thought to contribute to attention-deficit disorder and some symptoms of schizophrenia. Dopamine is also the primary neuroendocrine regulator of the secretion of prolactin from the anterior pituitary gland. Dopamine is also commonly associated with the pleasure system of the brain. This plays a key role in understanding the mechanism of action of drugs (such as cocaine and the amphetamines), which seem to be directly or indirectly related to the increase of dopamine.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 21
Correct
-
A 45-year-old man complains of shortness of breath, cough and chest pain. Chest X ray revealed a perihilar mass with bronchiectasis in the left mid-lung. Which of the following is most probably associated with these findings?
Your Answer: Bronchial carcinoid
Explanation:Bronchial carcinoids are neuroendocrine tumours that arise from Kulchitsky’s cells of the bronchial epithelium. Kulchitsky’s cells belong to the diffuse endocrine system. Patients affected by this tumour may be asymptomatic or may present with symptoms of airway obstruction, like dyspnoea, wheezing, and cough. Other common findings are recurrent pneumonia, haemoptysis, chest pain and paraneoplastic syndromes. Chest radiographs are abnormal in the majority of cases. Peripheral carcinoids usually present as a solitary pulmonary nodule. For central lesions common findings include hilar or perihilar masses with or without atelectasis, bronchiectasis, or consolidation. Bronchial carcinoids most commonly arise in the large bronchi causing obstruction.
-
This question is part of the following fields:
- Pathology
- Respiratory
-
-
Question 22
Incorrect
-
A patient presents with loss of fine touch and sense of proprioception in the lower part of the body (below T6). He is likely to have a lesion involving:
Your Answer: Lateral spinothalamic tract
Correct Answer: Gracile nucleus
Explanation:The gracile nucleus is located in the medulla oblongata and is one of the dorsal column nuclei involved in the sensation of fine touch and proprioception. It contains second-order neurons of the dorsal column–medial lemniscus system, that receive inputs from sensory neurones of the dorsal root ganglia and send axons that synapse in the thalamus.
The gracile nucleus and fasciculus carry epicritic, kinaesthetic and conscious proprioceptive information from the lower part of the body (below the level of T6 in the spinal cord). Similar information from the upper part of body (above T6, except for face and ear) is carried by the cuneate nucleus and fasciculus. The information from face and ear is carried by the primary sensory trigeminal nucleus.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 23
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.
-
This question is part of the following fields:
- Pathology
- Women's Health
-
-
Question 24
Correct
-
An old man was diagnosed with squamous cell carcinoma with axillary lymph node metastasis. The doctor said he will excise the tumour and remove all axillary lymph nodes lateral to the edge of the pectoralis minor muscle. One of the following axillary lymph nodes won't be removed by this procedure. Which is it?
Your Answer: Apical
Explanation:The apical lymph node group won’t be removed which include 20 to 30 lymph nodes.
Axillary lymph nodes are grouped according to location. The lateral group, the anterior to pectoral group, the posterior or subscapular group, the central group, and the medial or apical group. The lateral, pectoral, and subscapular groups are found lateral to the pectoralis minor muscle. The central group is situated directly under that muscle. Thus, if all lymph nodes lateral to the medial edge of the pectoralis minor muscle are removed, all the above four groups will be removed. The apical group won’t be removed which is situated medial to the medial edge of the pectoralis minor muscle.
-
This question is part of the following fields:
- Anatomy
- Upper Limb
-
-
Question 25
Correct
-
A 30-year-old woman known with Von Willebrand disease (vWD) has to undergo surgery. Which of these complications is most unlikely in this patient?
Your Answer: Hemarthrosis
Explanation:Von Willebrand disease (vWD) is an inherited haemorrhagic disorder characterised by the impairment of primary haemostasis. It is caused by the deficiency or dysfunction of a protein named von Willebrand factor. The most common manifestation due to the condition is abnormal bleeding. Complications include easy bruising, hematomas, epistaxis, menorrhagia, prolonged bleeding and severe haemorrhage. Hemarthrosis is a complication that is more commonly found in haemophilia.
-
This question is part of the following fields:
- Haematology
- Pathology
-
-
Question 26
Correct
-
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.
-
This question is part of the following fields:
- Cell Injury & Wound Healing; Cardiovascular
- Pathology
-
-
Question 27
Incorrect
-
Calculate the stroke volume in an adult male with the following parameters:
Heart rate 70 beats/min
Arterial [O2] 0.24 ml O2/min
Venous [O2] 0.16 ml O2/mi
Whole body O2 consumption 500 ml/min
Pulmonary diastolic pressure 15 mmHg
Pulmonary systolic pressure 25 mmHg
Wedge pressure 5 mmHg.Your Answer: 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.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 28
Incorrect
-
A patients sciatic nerve has been severed following a stab injury. What would be affected?
Your Answer: Extension of the knee would be eliminated
Correct Answer: There would still be cutaneous sensation over the anteromedial surface of the thigh
Explanation:The sciatic nerve supplies nearly all of the sensation of the skin of the leg and the muscles of the back of the thigh, leg and foot. A transection of the sciatic nerve at its exit from the pelvis will affect all the above-mentioned functions except cutaneous sensation over the anteromedial surface of the thigh, which comes from the femoral nerve.
-
This question is part of the following fields:
- Anatomy
- Lower Limb
-
-
Question 29
Correct
-
Pain in the right upper quadrant of the abdomen on ingestion of a fatty meal is seen in a condition which involves which of the following substances?
Your Answer: Cholecystokinin
Explanation:The clinical scenario described here favours the presence of gallstones. During food ingestion, vagal discharges stimulate gallbladder contraction. Moreover, presence of fat and amino acids in the intestinal lumen stimulates the release of cholecystokinin (CCK) in the duodenum. This causes sustained gallbladder contraction and relaxation of the sphincter of Oddi. If gallstones are present, there will be inflammation in the gallbladder and CCK will aggravate it due to contractions.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 30
Incorrect
-
What Is the mechanism behind rhesus incompatibility in a new born baby?
Your Answer: Type III hypersensitivity
Correct 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.
-
This question is part of the following fields:
- Inflammation & Immunology; Haematology
- Pathology
-
SESSION STATS - PERFORMANCE PER SPECIALTY
