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Question 1
Incorrect
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Which of the following is the most common cause of hyperthyroidism:
Your Answer: Toxic multinodular goitre
Correct Answer: Graves disease
Explanation:Graves disease is the most common cause of hyperthyroidism. It is an autoimmune disease in which autoantibodies against TSH receptors are produced. These antibodies bind to and stimulate these TSH receptors leading to an excess production of thyroid hormones. Therefore, the signs and symptoms of Graves disease are the same as those of hyperthyroidism, reflecting the actions of increased circulating levels of thyroid hormones: increased heat production, weight loss, increased 02 consumption and cardiac output and exophthalmos (bulging eyes, not drooping eyelids). TSH levels will be decreased (not increased) as a result of the negative feedback effect of increased T3 levels on the anterior pituitary.
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This question is part of the following fields:
- Endocrine
- Physiology
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Question 2
Incorrect
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A 64-year-old woman with a history of chronic breathlessness is referred for lung function testing.Which of the following statements regarding lung function testing is FALSE? Select ONE answer only.
Your Answer:
Correct Answer: In restrictive lung disease, the FVC is increased
Explanation:In restrictive lung disorders there is a reduction in the forced vital capacity (FVC) and the forced expiratory volume in one second (FEV1). The decline in the FVC is greater than that of the FEV1, resulting in preservation of the FEV1/FVC ratio (>0.7%).In obstructive lung disease, FEV1is reduced to <80% of normal and FVC is usually reduced but to a lesser extent than FEV1. The FEV1/FVC ratio is reduced to 80% in the presence of symptomsModerate airflow obstruction = FEV1 of 50-79%Severe airflow obstruction = FEV1 of 30-49%Very severe airflow obstruction = FEV1<30%.Spirometry is a poor predictor of durability and quality of life in COPD but can be used as part of the assessment of severity.COPD can only be diagnosed on spirometry if the FEV1 is <80% and FEV1/FVC ratio is < 0.7.
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This question is part of the following fields:
- Physiology
- Respiratory Physiology
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Question 3
Incorrect
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Aside from the inability to extend the leg above the knee, which of the following clinical symptoms should you anticipate seeing in a patient who had a pelvic and right leg injury as well as femoral nerve damage?
Your Answer:
Correct Answer: Loss of sensation over the anterior thigh
Explanation:The femoral nerve runs down the front of the leg from the pelvis. It gives the front of the thigh and a portion of the lower leg sensation. Extension of the leg at the knee joint, flexion of the thigh at the hip are produced by muscles that is primarily innervated by the femoral nerve.
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This question is part of the following fields:
- Anatomy
- Lower Limb
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Question 4
Incorrect
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An analytical cohort study aimed to determine a relationship between intake of dietary calcium and incidence of hip fractures among post-menopausal women. The following are the data obtained from the study:No. of post-menopausal women who took Calcium: 500No. of post-menopausal women who took Calcium and suffered a hip fracture: 10No. of post-menopausal women who took placebo: 500No. of post-menopausal women who took placebo and suffered a hip fracture: 25Compute for the absolute risk reduction of a hip fracture.
Your Answer:
Correct Answer: 0.03
Explanation:Absolute risk reduction (ARR) is computed as the difference between the absolute risk in the control group (ARC) and the absolute risk in the treatment group (ART).ARR = ARC-ARTARR = (25/500) – (10/500)ARR = 0.03
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This question is part of the following fields:
- Evidence Based Medicine
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Question 5
Incorrect
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A patient with diplopia is found to have eye deviation downwards and outwards. The likely nerves that are affected are:
Your Answer:
Correct Answer: Oculomotor nerve
Explanation:The results of an oculomotor (CN III) nerve palsy are a depressed and abducted (down and out) eye, ptosis, diplopia, and a fixed and dilated pupil.
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This question is part of the following fields:
- Anatomy
- Cranial Nerve Lesions
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Question 6
Incorrect
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Which of the following anatomic structures will gallstones most likely lodge into, and cause cholestasis?
Your Answer:
Correct Answer: Hartmann’s pouch
Explanation:Hartmann’s pouch is a diverticulum that can occur at the neck of the gallbladder. It is one of the rarest congenital anomalies of the gallbladder. Hartmann’s gallbladder pouch is a frequent but inconsistent feature of normal and pathologic human gallbladders. It is caused by adhesions between the cystic duct and the neck of the gallbladder. As a result, it is classified as a morphologic rather than an anatomic entity.There is a significant association between the presence of Hartmann’s pouch and gallbladder stones. It is the most common location for gallstones to become lodged and cause cholestasis.
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This question is part of the following fields:
- Abdomen And Pelvis
- Anatomy
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Question 7
Incorrect
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A 39-year-old woman's son with meningococcal meningitis was recently admitted to the Paediatric Intensive Care Unit. She is currently 22 weeks pregnant and is concerned about the possibility of her also contracting the disease as she cared closely for her son during his admission. Which antibiotic would be the MOST appropriate choice for chemoprophylaxis in this case?
Your Answer:
Correct Answer: Ciprofloxacin
Explanation:Ciprofloxacin is recommended for use as meningococcal chemoprophylaxis in all age groups and in pregnancy, and is the most appropriate for this patient.However, rifampicin is the drug of choice for meningococcal chemoprophylaxis because it is licensed for chemoprophylaxis, but multiple doses are necessary and it is not readily available in community pharmacies. It also interacts with oral contraceptives.
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This question is part of the following fields:
- Microbiology
- Specific Pathogen Groups
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Question 8
Incorrect
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A 29-year-old woman presents with night sweats, fever, and haemoptysis. A diagnosis of tuberculosis was suspected.Which of the following statements regarding the diagnosis of tuberculosis is considered correct?
Your Answer:
Correct Answer: Mycobacteria tuberculosis can be typed using a RFLP method
Explanation:Although a variety of clinical specimens may be submitted to thelaboratory to recover MTB and NTM, respiratory secretions suchas sputum and bronchial aspirates are the most common. Anearly-morning specimen should be collected on three consecutivedays, although recent studies have suggested that the addition ofa third specimen does not significantly increase the sensitivityof detecting Mycobacteria.Mycobacterium tuberculosis appear red on acid-fast staining because they take up the primary stain, which is carbolfuchsin, and is not decolorized by the acid alcohol anymore.Culture on Lowenstein-Jensen medium should be read within 5 to 7 days after inoculation and once a week thereafter for up to 8 weeks.Nucleic acid amplification assays designed to detect M. tuberculosis complexbacilli directly from patient specimens can be performed in as little as 6 to 8 hours on processed specimens.
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This question is part of the following fields:
- Microbiology
- Specific Pathogen Groups
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Question 9
Incorrect
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A 18 year old with known asthma presents himself to ED with acute breathlessness and wheeze for the past 20 minutes. On examination he is tachypneic and tachycardic. His oxygen saturations are 96% on air. What is the first line treatment for acute asthma:
Your Answer:
Correct Answer: Salbutamol
Explanation:High-dose inhaled short-acting beta2-agonists are the first line treatment for acute asthma(salbutamol or terbutaline). Oxygen should only been given to hypoxaemic patients (to maintain oxygen saturations of 94 – 98%). A pressurised metered dose inhaler with spacer device is preferred in patients with moderate to severe asthma (4 puffs initially, followed by 2 puffs every 2 minutes according to response, up to 10 puffs, whole process repeated every 10 – 20 minutes if necessary). The oxygen-driven nebuliser route is recommended for patients with life-threatening features or poorly responsive severe asthma (salbutamol 5 mg at 15 – 30 minute intervals). Continuous nebulisation should be considered in patients with severe acute asthma that is poorly responsive to initial bolus dose (salbutamol at 5 – 10 mg/hour) The intravenous route should be reserved for those in whom inhaled therapy cannot be used reliably.
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This question is part of the following fields:
- Pharmacology
- Respiratory
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Question 10
Incorrect
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Antidiuretic hormone (ADH) is synthesised by which of the following:
Your Answer:
Correct Answer: Hypothalamus
Explanation:Antidiuretic hormone is synthesised in the hypothalamus and transported to the posterior pituitary within nerve fibres where it is stored in secretory granules. ADH binds V2 receptors on renal principal cells in the late distal tubule and collecting ducts, raising cAMP levels and causing intracellular vesicles to fuse with the apical membrane. In their membrane these vesicles have water channels called aquaporins, which increase the water permeability allowing greater water reabsorption and concentration of urine. Excess levels of ADH results in syndrome of inappropriate ADH secretion (SIADH) characterised by hyponatremia with concomitant hypo-osmolality and high urine osmolality.
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This question is part of the following fields:
- Endocrine
- Physiology
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Question 11
Incorrect
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A CT pulmonary angiography of a patient with a massive pulmonary embolus will most likely show which of the following signs?
Your Answer:
Correct Answer: Increased alveolar dead space
Explanation:A CT pulmonary angiogram is an angiogram of the blood vessels of the lungs. It is a diagnostic imaging test used to check for pulmonary embolism. A pulmonary embolism is caused by a blood clot or thrombus that has become lodged in an artery in the lung and blocks blood flow to the lung. A patient with pulmonary embolism may feel an abrupt onset of pleuritic chest pain, shortness of breath, and hypoxia. Also, pulmonary embolism can result in alveolar dead space. Dead space represents the volume of ventilated air that does not participate in gas exchange. The alveolar dead space is caused by ventilation/perfusion imbalances in the alveoli. It is defined as the sum of the volumes of alveoli that are ventilated but not perfused. Aside from pulmonary embolism, smoking, bronchitis, emphysema, and asthma are among the other causes of alveolar dead space. The other types of dead space are the following: Anatomical dead space is the portion of the airways that conducts gas to the alveoli. This is usually around 150 mL, and there is no possibility of gas exchange in these areas. Physiological dead space is the sum of anatomical and alveolar dead spaces. Physiological dead space can account for up to 30% of the tidal volume.
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This question is part of the following fields:
- Physiology
- Respiratory Physiology
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Question 12
Incorrect
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Which statement accurately describes the osmolality in the various parts of the Henle Loop?
Your Answer:
Correct Answer: The osmolality of fluid in the descending loop equals that of the peritubular fluid
Explanation:The Loop of Henle connects the proximal tubule to the distal convoluted tubule and lies parallel to the collecting ducts. It consists of three major segments, including the descending thin limb, the ascending thin limb, and the ascending thick limb. These segments are differentiated based on structure, anatomic location, and function. The main function of the loop of Henle is to recover water and sodium chloride from urine. When fluid enters the loop of Henle, it has an osmolality of approximately 300 mOsm, and the main solute is sodium. The thin descending limb has a high water permeability but a low ion permeability. Because it lacks solute transporters, it cannot reabsorb sodium. Aquaporin 1 (AQP1) channels are used to passively absorb water in this area. The peritubular fluid becomes increasingly concentrated as the loop descends into the medulla, causing water to osmose out of the tubule. The tubular fluid in this area now equalizes to the osmolality of the peritubular fluid, to a maximum of approximately 1200 mOsm in a long medullary loop of Henle and 600 mOsm in a short cortical loop of Henle. The thin ascending limb is highly permeable to ions and impermeable to water. It allows the passive movement of sodium, chloride, and urea down their concentration gradients, so urea enters the tubule and sodium and chloride leave. Reabsorption occurs paracellularly due to the difference in osmolarity between the tubule and the interstitium. The thick ascending limb is also impermeable to water but actively transports sodium, potassium, and chloride out of the tubular fluid. The osmolality of the tubular fluid is lower compared to the surrounding peritubular fluid. This area is water impermeable. This results in tubular fluid leaving the loop of Henle with an osmolality of approximately 100 mOsm, which is lower than the osmolality of the fluid entering the loop, and urea being the solute.
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This question is part of the following fields:
- Physiology
- Renal Physiology
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Question 13
Incorrect
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A patent has weakness of humeral flexion and extension. A CT scan reveals that he has suffered damage to the nerve that innervates pectoralis major.Pectoralis major receives its innervation from which of the following nerves? Select ONE answer only.
Your Answer:
Correct Answer: Lateral and medial pectoral nerves
Explanation:Pectoralis major is a thick, fan-shaped muscle situated in the chest. It makes up the bulk of the chest musculature in the male and lies underneath the breast in the female. It overlies the thinner pectoralis minor muscle.Superficial muscles of the chest and arm showing pectoralis major (from Gray’s Anatomy)Pectoralis major has two heads; the clavicular head and the sternocostal head. The clavicular head originates from the anterior border and medial half of the clavicle. The sternocostal head originates from the anterior surface of the sternum, the superior six costal cartilages and the aponeurosis of the external oblique muscle. It inserts into the lateral lip of the bicipital groove of the humerus.Pectoralis major receives dual innervation from the medial pectoral nerve and the lateral pectoral nerve.Its main actions are as follows:Flexes humerus (clavicular head)Extends humerus (sternocostal head)Adducts and medially rotates the humerusDraws scapula anteriorly and inferiorly
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This question is part of the following fields:
- Anatomy
- Upper Limb
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Question 14
Incorrect
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Which of the following clotting factors is NOT vitamin K-dependent:
Your Answer:
Correct Answer: V
Explanation:Fat-soluble vitamin K is obtained from green vegetables and bacterial synthesis in the gut. Deficiency may present in the newborn (haemorrhagic disease of the newborn) or in later life. Deficiency may be caused by an inadequate diet, malabsorption or inhibition of vitamin K by drugs such as warfarin. The activity of factors II, VII, IX and X are vitamin K dependent as well as that of protein C and protein S. Both PT and APTT are prolonged.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 15
Incorrect
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Which of the following is typically a cause of a normal anion gap metabolic acidosis:
Your Answer:
Correct Answer: Diarrhoea
Explanation:FUSEDCARS can be used to remember some of the causes of a normal anion gap acidosis:Fistula (pancreaticoduodenal)Ureteroenteric conduitSaline administrationEndocrine (hyperparathyroidism)DiarrhoeaCarbonic anhydrase inhibitors (e.g. acetazolamide)Ammonium chlorideRenal tubular acidosisSpironolactone
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This question is part of the following fields:
- Physiology
- Renal
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Question 16
Incorrect
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A 18 year old student presents to ED with a headache, fever and photophobia. You suspect meningitis and agree to observe your junior performing a lumbar puncture. What is the highest safest vertebral level to perform lumbar puncture in adults:
Your Answer:
Correct Answer: L3/L4
Explanation:In adults, the spinal cord typically ends between L1/L2 whereas the subarachnoid space extends to approximately the lower border of vertebra S2. Lumbar puncture is performed in the intervertebral space L4/L5 or L3/L4.
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This question is part of the following fields:
- Anatomy
- Head And Neck
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Question 17
Incorrect
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Regarding probability distribution, which of the following statements is CORRECT:
Your Answer:
Correct Answer: In a negative skew, the mass of distribution is concentrated on the right.
Explanation:Distribution of data is usually unimodal (one peak) but may be bimodal (two peaks) or uniform (no peaks, each value equally likely). The normal distribution is a symmetrical bell-shaped curve. The mean, median, and mode of a normal distribution are equal. In a positive skew, the right tail is longer and the mass of distribution is concentrated on the left; mean > median > mode. In a negative skew, the left tail is longer and the mass of distribution is concentrated on the right; mean < median < mode.
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This question is part of the following fields:
- Evidence Based Medicine
- Statistics
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Question 18
Incorrect
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Which of the following laboratory findings are indicative of von Willebrand disease (VWD):
Your Answer:
Correct Answer: Prolonged APTT
Explanation:Laboratory results often show that:PFA-100 test results are abnormal.Low levels of factor VIII (if a factor VIII/VWF binding assay is conducted)APTT is Prolonged (or normal)PT is normalVWF values are low.Defective Platelet aggregationThe platelet count is normal.
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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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Which of the following medications can block the effect of adrenaline?
Your Answer:
Correct Answer: Beta-blockers
Explanation:Beta-blockers may reduce the response to adrenaline in the treatment of anaphylactic reactions. Noncardioselective beta-blockers in particular can antagonize the broncho dilating and cardio stimulatory effects of adrenaline by blocking beta-2 adrenergic receptors in the smooth muscles of the bronchial tree and in the heart muscles.
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This question is part of the following fields:
- Pharmacology
- Respiratory
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Question 20
Incorrect
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Regarding hypertensive crises, which of the following statements is CORRECT:
Your Answer:
Correct Answer: In a hypertensive emergency, blood pressure should be reduced by 20 - 25% within 2 hours.
Explanation:A hypertensive emergency is defined as severe hypertension (blood pressure ≥ 180/110 mmHg) with acute damage to the target organs. Prompt treatment with intravenous antihypertensive therapy is generally required; over the first few minutes or within 2 hours, blood pressure should be reduced by 20 – 25%. Severe hypertension without acute target organ damage is defined as hypertensive urgency.; blood pressure should be reduced gradually over 24 – 48 hours with oral antihypertensive therapy. If blood pressure is reduced too quickly in the management of hypertensive crises, there is a risk of reduced organ perfusion leading to cerebral infarction, blindness, deterioration in renal function, and myocardial ischaemia.
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This question is part of the following fields:
- Cardiovascular
- Pharmacology
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Question 21
Incorrect
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Which of the following statements regarding the pituitary gland is CORRECT:
Your Answer:
Correct Answer: Cortisol inhibits release of adrenocorticotropic hormone (ACTH) from the anterior pituitary.
Explanation:Anterior pituitary hormones are released under the control of hypothalamic releasing or inhibiting hormones originating from small neurons with their cell bodies in the hypothalamus and released into the blood at the median eminence. These hypothalamic hormones are transported directly to the anterior pituitary via hypophyseal portal vessels. The anterior pituitary hormones (and the hormones released by their target organs) inhibit further release of hypothalamic and anterior pituitary hormones by negative feedback mechanisms e.g. cortisol inhibits the release of ACTH. Prolactin release from the anterior pituitary is inhibited by dopamine.The posterior pituitary is really a direct extension of the hypothalamus. Oxytocin and ADH are manufactured in the cell bodies of large neurons in the hypothalamus and are transported down the axons of these cells to their terminals on capillaries originating from the inferior hypophyseal artery within the posterior pituitary gland. ADH release is controlled by negative feedback mechanisms based on plasma osmolality and blood volume, oxytocin however is involved in positive feedback mechanisms.
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This question is part of the following fields:
- Endocrine
- Physiology
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Question 22
Incorrect
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Which of the following statements is incorrect regarding potassium replacement?
Your Answer:
Correct Answer: Oral potassium supplements are often required for patients taking spironolactone.
Explanation:It is very seldom that potassium supplements are required with the small doses of diuretics given to treat hypertension. Potassium-sparing diuretics like spironolactone (rather than potassium supplements), are recommended for hypokalaemia prevention when diuretics are given to eliminate oedema, such as furosemide or the thiazides.
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This question is part of the following fields:
- Fluids And Electrolytes
- Pharmacology
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Question 23
Incorrect
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A 65-year-old man complains of severe vertigo, nausea, and tinnitus. Upon presenting himself to the emergency room, it was observed that he is exhibiting ataxia, right-sided loss of pain and temperature sense on the face, and left-sided sensory loss to the body. An MRI and CT scan was ordered and the results showed that he is suffering from a right-sided stroke. Branches of which of the following arteries are most likely implicated in the case?
Your Answer:
Correct Answer: Basilar artery
Explanation:The lateral pontine syndrome occurs due to occlusion of perforating branches of the basilar and anterior inferior cerebellar (AICA) arteries. It is also known as Marie-Foix syndrome or Marie-Foix-Alajouanine syndrome. It is considered one of the brainstem stroke syndromes of the lateral aspect of the pons.It is characterized by ipsilateral limb ataxia, loss of pain and temperature sensation of the face, facial weakness, hearing loss, vertigo and nystagmus, hemiplegia/hemiparesis, and loss of pain and temperature sensation.
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This question is part of the following fields:
- Anatomy
- Central Nervous System
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Question 24
Incorrect
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A patient presents with a rash for dermatological examination. A large area of purplish discolouration of the skin that measures 2 cm in diameter and does not blanch when pressure is applied is seen .What is the best description of this rash that you have found on examination?
Your Answer:
Correct Answer: Ecchymoses
Explanation:Ecchymosis are discolouration of the skin or mucous membranes caused by extravasation of blood. They are usually red or purple in colour and measure greater than 1 cm in diameter and do not blanch on applying pressure. A macule is a flat, well circumscribed area of discoloured skin less than 1 cm in diameter with no changes in the thickness or texture of the skin. Petechiae are discolouration of the skin measuring less than 3 mm in diameterPurpura are discolouration of the skin measuring between 0.3 cm and 1 cm in diameter.Erythema is redness of the skin or mucous membranes caused by hyperaemia of superficial capillaries caused by skin injury, infection or inflammation. Erythema blanches when pressure is applied whereas ecchymosis, purpura and petechiae do not.
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This question is part of the following fields:
- General Pathology
- Pathology
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Question 25
Incorrect
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Which of the following drug classes may cause bronchoconstriction:
Your Answer:
Correct Answer: Beta-blockers
Explanation:Beta-blockers, including those considered to be cardioselective, should usually be avoided in patients with a history of asthma, bronchospasm or a history of obstructive airways disease. However, when there is no alternative, a cardioselective beta-blocker can be given to these patients with caution and under specialist supervision. In such cases the risk of inducing bronchospasm should be appreciated and appropriate precautions taken.
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This question is part of the following fields:
- Cardiovascular
- Pharmacology
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Question 26
Incorrect
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Which of the following is NOT one of the cardinal features of acute inflammation:
Your Answer:
Correct Answer: Discharge
Explanation:Classic signs:Rubor (redness)Calor (heat)Dolour (pain)Tumour (swelling)Functio laesa (loss of function)These classic signs are produced by a rapid vascular response and cellular events. The main function of these events is to bring elements of the immune system to the site of injury and prevent further tissue damage.
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This question is part of the following fields:
- Inflammatory Responses
- Pathology
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Question 27
Incorrect
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C-reactive protein (CRP) synthesis is predominantly stimulated by which of the following cytokines:
Your Answer:
Correct Answer: IL-6
Explanation:Activated leukocytes, adipocytes, and endothelial cells all release interleukin 6 (IL-6), a significant proinflammatory cytokine. The main downstream mediator of the acute phase response is C-reactive protein, which is predominantly produced by IL-6–dependent hepatic biosynthesis.
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This question is part of the following fields:
- Inflammatory Responses
- Pathology
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Question 28
Incorrect
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Question 29
Incorrect
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What proportion of peripheral blood leukocytes are monocytes?
Your Answer:
Correct Answer: 5 - 10%
Explanation:Monocytes account for around 5 to 10% of peripheral white cells. Monocytes in peripheral blood are generally bigger than other leukocytes and feature a large central oval or indented nucleus with clumped chromatin. The abundant cytoplasm staining blue and containing numerous fine vacuoles gives the appearance of ground glass. Cytoplasmic granules are another type of granule. Monocytes evolve from the granulocyte-macrophage progenitor to become monoblasts, promonocytes, monocytes, and tissue macrophages (in increasing order of maturity). Monocytes only stay in the bone marrow for a short time before exiting to circulate in the bloodstream for 20-40 hours before becoming macrophages.
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This question is part of the following fields:
- Immune Responses
- Pathology
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Question 30
Incorrect
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An ambulance transports a 23-year-old woman who has taken a witnessed overdose of her mother's diazepam tablets. She has no significant medical history and does not take any medications on a regular basis.In this case, what is the SINGLE MOST APPROPRIATE FIRST DRUG TREATMENT?
Your Answer:
Correct Answer: Flumazenil IV 200 μg
Explanation:Flumazenil is a benzodiazepine antagonist that can be helpful in some overdose situations. It works quickly (in less than a minute), but the effects are fleeting, lasting less than an hour. The dose is 200 micrograms every 1-2 minutes with a maximum dose of 3 milligrams per hour.Flumazenil should be avoided by patients who are addicted to benzodiazepines or who take tricyclic antidepressants because it can cause withdrawal symptoms. It can cause seizures or cardiac arrest in these situations.
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This question is part of the following fields:
- CNS Pharmacology
- Pharmacology
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SESSION STATS - PERFORMANCE PER SPECIALTY
