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Question 1
Correct
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Diffusion is the movement of molecules from a region of high concentration to a region of low concentration. Which of these changes will decrease the rate of diffusion of a substance?
Your Answer: An increase in the molecular weight of the substance
Explanation:Unless given IV, a drug must cross several semipermeable cell membranes before it reaches the systemic circulation. Drugs may cross cell membranes by diffusion, amongst other mechanisms. The rate of diffusion of a substance is proportional to the difference in the concentration of the diffusing substance between the two sides of the membrane, the temperature of the solution, the permeability of the membrane and, in the case of ions, the electrical potential difference between the two sides of the membrane.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 2
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 3
Incorrect
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A patient came into the emergency in a state of shock. His blood group is not known, but on testing it clotted when mixed with Type A antibodies. Which blood should be transfused?
Your Answer: A +ve
Correct Answer: B +ve
Explanation:There are two stages to determine the blood group, known as ABO typing. The first stage is called forward typing. In this method, RBCs are mixed with two separate solutions of type A or type B antibodies to see if they agglutinate. If this blood clumps, this indicates the presence of antigens within the blood sample. For example, a sample of type B blood will clump when tested with type A antibodies as it contains type B antigens. Group B – has only the B antigen on red cells (and A 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 Rh factor on the red blood cell’s surface. This is also an antigen and those who have it are called Rh+. Those who have not 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. The patient’s blood group is B positive as he has antigen B, antibody A and Rh antigens.
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This question is part of the following fields:
- General
- Physiology
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Question 4
Incorrect
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A 65-year old gentleman presents to the clinic with chronic back pain and weight loss. His blood count shows a white blood cell count of 10 × 109/l, with a differential count of 66 polymorphonuclear leukocytes, 7 bands, 3 metamyelocytes, 3 myelocytes, 14 lymphocytes, 7 monocytes, and 5 nucleated red blood cells. The haemoglobin is 13 g/dl with a haematocrit of 38.1%, a mean corpuscular volume of 82 fl, and a platelet count of 126 × 109/l. What is the likely diagnosis?
Your Answer: Chronic lymphocytic leukaemia
Correct Answer: Metastatic carcinoma
Explanation:The peripheral blood findings suggest a leucoerythroblastic picture, the common causes of which in a 65-year old gentleman includes prostatic or lung malignancy.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 5
Incorrect
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A drug abuser developed an infection which spread from the dorsum of the hand to the medial side of the arm along the course of the large cutaneous vein. Which vein is involved?
Your Answer: Cephalic
Correct Answer: Basilic
Explanation:The basilic vein is one of two veins found in the forearm, the other is the cephalic vein. These veins originate from the deep venous arch of the hand. The cephalic vein ascends along the lateral side of the forearm, and the basilic vein runs up the medial side of the forearm.
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This question is part of the following fields:
- Anatomy
- Upper Limb
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Question 6
Correct
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A 14 year-old girl is found to have haemophilia B. What pathological problem does she have?
Your Answer: Deficiency of factor IX
Explanation:Haemophilia B (also known as Christmas disease) is due to a deficiency in factor IX. Haemophilia A is due to a deficiency in factor VIII.
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This question is part of the following fields:
- Haematology
- Pathology
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Question 7
Correct
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Passing through the lesser sciatic foramen are the:
Your Answer: Pudendal nerve
Explanation:Structures that pass through the lesser sciatic foramen include:
– the pudendal nerve
– the nerve to obturator internus
– internal pudendal artery
– the tendon of obturator internus
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This question is part of the following fields:
- Anatomy
- Pelvis
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Question 8
Correct
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The specimen sent to the pathologist for examination was found to be benign. Which one of the following is most likely a benign tumour?
Your Answer: Warthin’s tumour
Explanation:Warthin’s tumour is also known as papillary cystadenoma lymphomatosum. It is a benign cystic tumour of the salivary glands containing abundant lymphocytes and germinal centres. It has a slightly higher incidence in males and most likely occur in older adults aged between 60 to 70 years. This tumour is also associated with smoking. Smokers have an eight-fold greater risk in developing the tumour compared to non-smokers.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 9
Correct
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A patient under went repair of a lingual artery aneurysm in the floor of the mouth. During surgical dissection from the inside of the mouth which muscle would you have to pass through to reach the main portion of the lingual artery?
Your Answer: Hyoglossus
Explanation:The lingual artery first runs obliquely upward and medialward to the greater horns of the hyoid bone. It then curves downward and forward, forming a loop which is crossed by the hypoglossal nerve, and passing beneath the digastric muscle and stylohyoid muscle it runs horizontally forward, beneath the hyoglossus, and finally, ascending almost perpendicularly to the tongue, turns forward on its lower surface as far as the tip, to become the deep lingual artery.
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This question is part of the following fields:
- Anatomy
- Head & Neck
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Question 10
Correct
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A 44-year old man, who was euthyroid underwent thyroidectomy following neoplastic cells found on fine-needle aspiration. Frozen section of multiple thyroid masses showed malignant neoplasm of polygonal cells in nests. The neoplasm also showed presence of amyloid which was positive with Congo-red staining. Immunoperoxidase staining for calcitonin was also positive. Chest X-ray revealed no abnormality. However, his blood pressure was found to be raised, and his serum ionised calcium was high. What is the likely diagnosis?
Your Answer: Multiple endocrine neoplasia type IIA
Explanation:MEN (Multiple Endocrine Neoplasia) syndromes are a group of three separate familial disease which consists of adenomatous hyperplasia and neoplasia in several endocrine glands. All three conditions are inherited as an autosomal dominant trait, with a single gene producing multiple effects. MEN IIA is characterized by medullary carcinoma of the thyroid, pheochromocytoma and hyperparathyroidism. It should be suspected in patients with bilateral pheochromocytoma, a familial history of MEN, or at least two characteristic endocrine manifestations. Genetic testing is used to confirm the diagnosis. Early diagnosis is crucial to aid in complete excision of the localized tumour. Pheochromocytomas can be detected by plasma free metanephrines and fractionated urinary catecholamines, particularly adrenaline (epinephrine).
Imaging studies such as computed tomography or magnetic resonance imaging might also prove useful. Hyperparathyroidism is diagnosed by the standard finding of hypercalcaemia, hypophosphatemia and an increased parathyroid hormone level. Once MEN IIA syndrome is identified in any patient, it is recommended that his or her first-degree relatives and any other symptomatic also undergo genetic testing. Relatives should be subjected to annual screening for hyperparathyroidism and pheochromocytoma beginning in early childhood and continue indefinitely. Serum calcium levels help in screening for hyperparathyroidism. Similarly, screening for pheochromocytoma is by history, measurement of the blood pressure and laboratory testing.
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This question is part of the following fields:
- Endocrine
- Pathology
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Question 11
Correct
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A 45-year-old man presents to the emergency department with an irregular pulse and shortness of breath. Electrocardiography findings show no P waves, normal QRS complexes and an irregularly irregular rhythm. The patient most probably has:
Your Answer: Atrial fibrillation
Explanation:Atrial fibrillation is one of the most common arrhythmias, characterised by an irregular and rapid heart rate. Due to the decreased cardiac output, atrial fibrillation increases the risk of heart failure. It can also lead to thrombus formation which may lead to thromboembolic events. Clinical findings include palpitations, shortness of breath, fatigue, chest pain and confusion. The diagnosis is made by electrocardiographic findings which include absent P wave, fibrillatory (f) waves between QRS complexes and irregularly irregular R-R intervals.
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This question is part of the following fields:
- Cardiovascular
- Pathology
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Question 12
Correct
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A 57-year-old male smoker noted a lump on his inner lip. Upon physical examination the lump measured more than 2 cm but less than 4 cm in its greatest dimension. He is diagnosed with squamous cell carcinoma of the lip. What is the stage of the patient's cancer according to the TNM staging for head and neck cancers?
Your Answer: T2
Explanation:Head and neck cancer is a group of cancers that starts within the mouth, nose, throat, larynx, sinuses, or salivary glands. The TNM staging system used for head and neck cancers is a clinical staging system that allows physicians to compare results across patients, assess prognosis, and design appropriate treatment regimens. The staging is as follows; Primary tumour (T): Tis: pre-invasive cancer (carcinoma in situ), T0: no evidence of primary tumour, T1: tumour 2 cm or less in its greatest dimension, T2: tumour more than 2 cm but not more than 4 cm, T3: tumour larger than 4 cm, T4: tumour with extension to bone, muscle, skin, antrum, neck, etc and TX: minimum requirements to assess primary tumour cannot be met. Regional lymph node involvement (N): N0: no evidence of regional lymph node involvement, N1: evidence of involvement of movable homolateral regional lymph nodes, N2: evidence of involvement of movable contralateral or bilateral regional lymph nodes, N3: evidence of involvement of fixed regional lymph nodes and NX: Minimum requirements to assess the regional nodes cannot be met. Distant metastases (M): M0: no evidence of distant metastases, M1: evidence of distant metastases and MX: minimum requirements to assess the presence of distant metastases cannot be met. Staging: Stage I: T1 N0 M0, Stage II: T2 N0 M0, Stage III: T2NOMO and T3N1MO, Stage IV: T4N1M0, any TN2M0, any TN3M0, any T and any NM1. The depth of infiltration is predictive of the prognosis. With increasing depth of invasion of the primary tumour, the risk of nodal metastasis increases and survival decreases. The patient in this scenario therefore has a T2 tumour.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 13
Correct
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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 14
Incorrect
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Which of the following is a likely consequence of severe diarrhoea?
Your Answer: An increase in the bicarbonate content of the body
Correct Answer: A decrease in the sodium content of the body
Explanation:Diarrhoea can occur due to any of the numerous aetiologies, which include infectious, drug-induced, food related, surgical, inflammatory, transit-related or malabsorption. Four mechanisms have been implicated in diarrhoea: increased osmotic load, increased secretion, inflammation and decreased absorption time. Diarrhoea can result in fluid loss with consequent dehydration, electrolyte loss (Na+, K+, Mg2+, Cl–) and even vascular collapse. Loss of bicarbonate ions can lead to a metabolic acidosis.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 15
Incorrect
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What is the action of the muscle of the orbit that originates on the lesser wing of the sphenoid bone, just above the optic foramen?
Your Answer: Elevation of the eyeball
Correct Answer: Elevation of the upper eyelid
Explanation:The levator palpebrae superioris is the muscle in the orbit that elevates the superior (upper) eyelid. The levator palpebrae superioris originates on the lesser wing of the sphenoid bone, just above the optic foramen and receives somatic motor input from the ipsilateral superior division of the oculomotor nerve.
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This question is part of the following fields:
- Anatomy
- Head & Neck
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Question 16
Correct
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You are asked to help a junior medical student studying anatomy to identify the left lung. Which of the following features found only in the left lung will you use the identify it?
Your Answer: Cardiac notch
Explanation:Oblique fissure: is found on both the left and the right lungs. It separates the upper from the lower lobes in both lungs and the middle lobe from the lower lobe in the right lung(which has three lobes.)
The superior lobar bronchus is found in both lungs.
Cardiac notch: found only on the left lung.
Horizontal fissure: a deep groove separating the middle lobe from the upper lobe of the right lung is absent on the left lung.
Diaphragmatic surface: refers to the part of the lung, both the left and the right, that is in contact with the diaphragm.
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This question is part of the following fields:
- Anatomy
- Thorax
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Question 17
Correct
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Which of these foramen is located at the base of the skull and transmits the accessory meningeal artery?
Your Answer: Foramen ovale
Explanation:At the base of the skull the foramen ovale is one of the larger of the several holes that transmit nerves through the skull. The following structures pass through foramen ovale: mandibular nerve, motor root of the trigeminal nerve, accessory meningeal artery, lesser petrosal nerve, a branch of the glossopharyngeal nerve, emissary vein connecting the cavernous sinus with the pterygoid plexus of veins and occasionally the anterior trunk of the middle meningeal vein.
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This question is part of the following fields:
- Anatomy
- Head & Neck
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Question 18
Correct
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During the fetal stage, the mesonephric tubules gives rise to the?
Your Answer: Wolffian duct
Explanation:The development of the kidney proceeds through a series of successive phases, each marked by the development of a more advanced kidney: the pronephros, mesonephros, and metanephros. The development of the pronephric duct proceeds in a cranial-to-caudal direction. As it elongates caudally, the pronephric duct induces nearby intermediate mesoderm in the thoracolumbar area to become epithelial tubules called mesonephric tubules. Each mesonephric tubule receives a blood supply from a branch of the aorta, ending in a capillary tuft analogous to the glomerulus of the definitive nephron. The mesonephric tubule forms a capsule around the capillary tuft, allowing for filtration of blood. This filtrate flows through the mesonephric tubule and is drained into the continuation of the pronephric duct, now called the mesonephric duct or Wolffian duct. The nephrotomes of the pronephros degenerate while the mesonephric duct extends towards the most caudal end of the embryo, ultimately attaching to the cloaca.
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This question is part of the following fields:
- Anatomy
- Embryology
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Question 19
Correct
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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 20
Incorrect
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An excised lesion is found to be a premalignant during examination by the pathologist. What is the most likely histopathology report of this lesion?
Your Answer: Familial polyposis
Correct Answer: Solar keratosis
Explanation:Premalignant condition is a state of disordered morphology of cells that is associated with an increased risk of cancer. If this condition is left untreated, it may lead to the development of cancer. The following are examples of pre-malignant lesions: actinic keratosis, Barret’s oesophagitis, atrophic gastritis, ductal carcinoma in situ, dyskeratosis congenita, sideropenic dysphagia, lichen planus, oral submucous fibrosis, solar elastosis, cervical dysplasia, leucoplakia and erythroplakia.
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This question is part of the following fields:
- Neoplasia
- Pathology
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Question 21
Correct
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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 22
Correct
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A 26-year-old female patient had the following blood report: RBC count = 4. 0 × 106/μl, haematocrit = 27% and haemoglobin = 11 g/dl, mean corpuscular volume (MCV) = 80–100 fl, mean corpuscular haemoglobin concentration (MCHC) = 31–37 g/dl. Which of the following is correct regarding this patient’s erythrocytes:
Your Answer: Normal MCV
Explanation:MCV is the mean corpuscular volume and it is calculated from the haematocrit and the RBC count. It is normally 90 fl. Mean corpuscular haemoglobin concentration (MCHC) [g/dl] = haemoglobin [g/dl]/haematocrit = 11/0.27 = 41 g/dl and is higher than normal range (32 to 36 g/dL).
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This question is part of the following fields:
- General
- Physiology
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Question 23
Correct
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Which nuclei of the posterior grey column of the spinal cord are likely affected in a patient who has lost the sensation of pain and temperature?
Your Answer: Substantia gelatinosa
Explanation:Substantia gelatinosa is one of the nuclei in the posterior grey column along side other posterior grey column nuclei like the nucleus dorsalis, nucleus proprius, and posteromarginal nucleus. These nuclei are a collection of cells in the posterior grey area found in throughout the spinal cord. The substantia gelatinosa receives direct input from the dorsal nerve roots (sensory), especially from thermoreceptors and nociceptors (receptors for temperature and pain).
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This question is part of the following fields:
- Anatomy
- Neurology
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Question 24
Correct
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A medical student is told a substance is freely filtered but is not metabolised, secreted, or stored in the kidney. It has a plasma concentration of 1000 mg/l and its urine excretion rate is 25 mg/min, and the inulin clearance is 100 ml/min. What is the rate of tubular reabsorption of the substance?
Your Answer: 75 mg/min
Explanation:Reabsorption or tubular reabsorption is the process by which the nephron removes water and solutes from the tubular fluid (pre-urine) and returns them to the circulating blood. To calculate the reabsorption rate of substance Z we use the following equation: excretion = (filtration + secretion) – reabsorption. As this substance is freely filtered, its filtration rate is equal to that of inulin. So 25 = (100 + 0) – reabsorption. Reabsorption = 100 – 25 therefore reabsorption = 75 mg/min.
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This question is part of the following fields:
- Physiology
- Renal
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Question 25
Correct
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A 50-year old gentleman who suffered a stroke was brought to the emergency department by his relatives. The patient however denied the presence of paralysis of his left upper and lower limbs. What is the most likely site of the lesion in this patient?
Your Answer: Right posterior parietal cortex
Explanation:A large injury to the non-dominant parietal cortex can make the patient neglect or refuse to acknowledge the presence of paralysis on the contralateral side. This can also involve the perception of the external world. Smaller injuries in this area which involve the precentral gyrus (primary motor cortex) or postcentral gyrus (primary sensory cortex) cause contralateral spastic paralysis or contralateral loss of tactile sensation respectively. A lesion in posterior inferior gyrus of the dominant frontal lobe results in motor aphasia. Involvement of the posterior superior gyrus of the dominant frontal lobe produces sensory aphasia.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 26
Correct
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A 56-year-old man undergoes tests to determine his renal function. His results over a period of 24 hours were:
Urine flow rate: 2. 0 ml/min
Urine inulin: 1.0 mg/ml
Plasma inulin: 0.01 mg/ml
Urine urea: 260 mmol/l
Plasma urea: 7 mmol/l
What is the glomerular filtration rate?Your Answer: 200 ml/min
Explanation:Glomerular filtration rate (GFR) is the volume of fluid filtered from the renal (kidney) glomerular capillaries into the Bowman’s capsule per unit time. GFR is equal to the inulin clearance because inulin is freely filtered into Bowman’s capsule but is not reabsorbed or secreted. The clearance (C) of any substance can be calculated as follows: C = (U × V)/P, where U and P are the urine and plasma concentrations of the substance, respectively and V is the urine flow rate. Thus, glomerular filtration rate = (1.0 × 2. 0)/0.01 = 200 ml/min.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 27
Correct
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A 27 year-old male patient was admitted to the hospital due to recurrent fever for the past 2 weeks. The patient claimed that he is an intravenous drug user. Following work up, the patient was diagnosed with infective endocarditis. Which is the most likely organism responsible for this diagnosis?
Your Answer: Staphylococcus aureus
Explanation:Acute bacterial endocarditis is a fulminant illness lasting over days to weeks (<2weeks). It is most likely due to Staphylococcus aureus especially in intravenous drug abusers.
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This question is part of the following fields:
- Microbiology
- Pathology
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Question 28
Correct
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A 65 year old man with a history of diabetes and hypertension presented with a stroke a few months ago severely affecting his speech and movement in the right arm and leg. A cerebral angiogram revealed a middle cerebral artery occlusion. A recent CT scan was done which revealed a 5 cm cystic space in his left parietal lobe. This lesion is a result of which of the following forms of resolution?
Your Answer: Liquefactive necrosis
Explanation:Characteristically, the brain will undergo liquefactive necrosis following ischaemic injury. This leaves a cystic space in that region which would show up on a CT scan. Atrophy would result in a generalized decrease in the brain size. Coagulative necrosis typically occurs in parenchymal organs e.g. the spleen or kidney which have a lower lipid content. Caseous necrosis is typical in granulomatous tuberculosis infection. Apoptosis will not form a cystic area as it is programmed cell death involving a individual cells. Gangrenous necrosis is characteristic of ischaemic injury of the lower limb and GI tract. Fibrinous necrosis results from necrotic damage to the blood vessels with the leaking of proteins into the vessel, appearing bright pink on H & E staining.
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This question is part of the following fields:
- Cell Injury & Wound Healing; Neurology
- Pathology
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Question 29
Incorrect
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A 7-year-old boy with facial oedema was brought to the hospital by his parents. Renal function is normal and urinalysis revealed the presence of a profound proteinuria. Which of the following is the most probable cause of these findings?
Your Answer: Membranoproliferative glomerulonephritis
Correct Answer: Minimal-change disease
Explanation:Minimal-change disease (MCD) refers to a histopathologic glomerular lesion, typically found in children, that is almost always associated with nephrotic syndrome. The most noticeable symptom of MCD is oedema, which can develop very rapidly. Due to the renal loss of proteins muscle wasting and growth failure may be seen in children. Renal function is usually not affected and a proteinuria of more than 40 mg/h/m2 is the only abnormal finding in urinalysis.
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This question is part of the following fields:
- Pathology
- Renal
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Question 30
Correct
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A 15-year-old girl was diagnosed with bacterial meningitis. Gram staining of the spinal fluid shows numerous polymorphonuclear neutrophils and Gram-positive cocci. Which is the empiric drug of choice to be given to the patient until the antibiotic sensitivity report is available?
Your Answer: Ceftriaxone
Explanation:Acute meningitis is a medical emergency. All suspects should receive their first dose of antibiotics immediately and be transferred to hospital as soon as possible. If facilities for blood culture and/or lumbar puncture (LP) are immediately available, they should be performed before administration
of the first dose of antibiotics (see contraindications to LP below). Neither procedure should lead to a significant delay in antibiotic administration.
Administer ceftriaxone 80-100 mg/kg (maximum 2 g, 12 hourly) intravenously. The intramuscular or intraosseous route can be used if there is no vascular access. Penicillin allergy is not a contraindication to ceftriaxone in acute meningitis (C-1). Omit ceftriaxone only if there has been documented ceftriaxone anaphylaxis. Give chloramphenicol 25 mg/kg (maximum 500 mg) intravenously instead, if available. Administer adequate analgesia and transfer the patient immediately to hospital, detailing all administered
medication in the referral letter. -
This question is part of the following fields:
- Microbiology
- Pathology
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Question 31
Correct
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During an operation to repair an indirect inguinal hernia, you are asked to indicate the position of the deep inguinal ring. You indicate this as being:
Your Answer: Above the midpoint of the inguinal ligament
Explanation:The deep inguinal ring is near the midpoint of the inguinal ligament, below the anterior superior iliac spine. It is lateral to the inferior epigastric artery. The superficial ring, however, is found above the pubic tubercle. The supravesical fossa is the space between the median and medial umbilical folds.
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This question is part of the following fields:
- Abdomen
- Anatomy
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Question 32
Correct
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What is correct regarding the obturator artery?
Your Answer: It is found in the medial compartment of the thigh
Explanation:The obturator artery is a branch of the internal iliac artery, which passes antero-inferiorly on the lateral wall of the pelvis, to the upper part of the obturator foramen. The posterior branch follows the posterior margin of the foramen and turns forward on the inferior ramus of the ischium. It also supplies an articular branch, which enters the hip joint through the acetabular notch, sending a branch along the ligamentum teres to the head of the femur. It is the main source of arterial supply to the medial compartment of the thigh
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This question is part of the following fields:
- Anatomy
- Lower Limb
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Question 33
Correct
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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 34
Correct
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A 31 -year-old female patient had a blood gas done on presentation to the emergency department. She was found to have a metabolic acidosis and decreased anion gap. The most likely cause of these findings in this patient would be?
Your Answer: Hypoalbuminemia
Explanation:A low anion gap might be caused by alterations in serum protein levels, primarily albumin (hypoalbuminemia), increased levels of calcium (hypercalcaemia) and magnesium (hypermagnesemia) or bromide and lithium intoxication. However, the commonest cause is hypoalbuminemia, thus if the albumin concentration falls, the anion gap will also be lower. The anion gap should be corrected upwards by 2.5 mmol/l for every 10g/l fall in the serum albumin.
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This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
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Question 35
Correct
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What occurs during cellular atrophy?
Your Answer: Cell size decreases
Explanation:Atrophy is the decrease in the size of cells, tissues, or organs. There are several causes including inadequate nutrition, poor circulation, loss of hormonal support or nerve supply, disuse, lack of exercise, or disease. An increase in cell size is termed hypertrophy which is distinguished from hyperplasia, in which the cells remain approximately the same size but increase in number.
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This question is part of the following fields:
- Cell Injury & Wound Healing; Urology
- Pathology
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Question 36
Incorrect
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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: Ciprofloxacin
Correct Answer: Oral vancomycin
Explanation:Three antibiotics are effective against Clostridium difficile:
Metronidazole 500 mg orally three times daily is the drug of choice, because of superior tolerability, lower price and comparable efficacy.
Oral vancomycin 125 mg four times daily is second-line therapy in particular cases of relapse or where the infection is unresponsive to metronidazole treatment.
Thirdly, the use of linezolid might also be considered.
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This question is part of the following fields:
- Pathology
- Pharmacology
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Question 37
Incorrect
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A young man came to the emergency room after an accident. The anterior surface of his wrist was lacerated with loss of sensation over the thumb side of his palm. Which nerves have been damaged?
Your Answer: Radial
Correct Answer: Median
Explanation:The median nerve provides cutaneous innervation to the skin of the palmar radial three and a half fingers. Also the site of injury indicates that the medial nerve may have been injured as it passes into the hand by crossing over the anterior wrist.
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This question is part of the following fields:
- Anatomy
- Upper Limb
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Question 38
Incorrect
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What's the nodal stage of a testicular seminoma if several lymph nodes between 2cm and 5cm are found?
Your Answer: N3
Correct Answer: N2
Explanation:According to the American Joint Committee on Cancer (AJCC) 2002 guidelines, the nodal staging of testicular seminoma is the following:
N0: no regional lymph node metastases
N1: metastasis with lymph nodes 2 cm or less in their greatest dimension or multiple lymph nodes, none more than 2 cm
N2: metastasis with lymph nodes greater than 2 cm but not greater than 5 cm in their greatest dimension, or multiple lymph nodes, any one mass greater than 2 cm, but not more than 5 cm
N3: metastasis with lymph nodes greater than 5 cm in their greatest dimension.
The patient in this case has N2 testicular seminoma. This TNM staging is extremely important because treatment options are decided depending on this classification.
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This question is part of the following fields:
- Pathology
- Urology
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Question 39
Incorrect
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A 49-year-old woman with acute renal failure has a total plasma [Ca2+] = 2. 5 mmol/l and a glomerular filtration rate of 160 l/day. What is the estimated daily filtered load of calcium?
Your Answer: 550 mmol/day
Correct Answer: 240 mmol/day
Explanation:Calcium is the most abundant mineral in the human body. The average adult body contains in total approximately 1 kg of calcium of which 99% is in the skeleton in the form of calcium phosphate salts. The extracellular fluid (ECF) contains approximately 22 mmol, of which about 9 mmol is in the plasma. About 40% of total plasma Ca2+ is bound to proteins and not filtered at the glomerular basement membrane. Therefore, the estimated daily filtered load is 1.5 mmol/l × 160 l/day = 240 mmol/day. The exact amount of free versus total Ca2+ depends on the blood pH: free Ca2+ increases during acidosis and decreases during alkalosis.
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This question is part of the following fields:
- Physiology
- Renal
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Question 40
Incorrect
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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: Squamous cell carcinoma
Correct 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 41
Incorrect
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A 54-year-old woman with amyotrophic lateral sclerosis is diagnosed with respiratory acidosis. The patient’s renal excretion of potassium would be expected to:
Your Answer: Rise, since acid and potassium excretion are coupled
Correct Answer: Fall, since tubular secretion of potassium is inversely coupled to acid secretion
Explanation:Respiratory acidosis is a medical emergency in which decreased ventilation (hypoventilation) increases the concentration of carbon dioxide in the blood and decreases the blood’s pH (a condition generally called acidosis). Secretion of acid and potassium by the renal tubule are inversely related. So, increased excretion of H+ during renal compensation for respiratory acidosis will result in decreased secretion (or increased retention) of potassium ions, with the result that the body’s potassium store rises. An increase in K+ excretion would be associated with renal compensation for respiratory alkalosis. The filtered load of K+depends only on K+ plasma concentration and glomerular filtration rate, not on plasma pH.
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This question is part of the following fields:
- Physiology
- Renal
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Question 42
Correct
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The blood investigations of a 30-year old man with jaundice revealed the following : total bilirubin 6.5 mg/dl, direct bilirubin 1.1 mg/dl, indirect bilirubin 5.4 mg/dl and haemoglobin 7.3 mg/dl. What is the most likely diagnosis out of the following?
Your Answer: Haemolysis
Explanation:Hyperbilirubinemia can be caused due to increased bilirubin production, decreased liver uptake or conjugation, or decreased biliary excretion. Normal bilirubin level is less than 1.2 mg/dl (<20 μmol/l), with most of it unconjugated. Elevated unconjugated bilirubin (indirect bilirubin fraction >85%) can occur due to haemolysis (increased bilirubin production) or defective liver uptake/conjugation (Gilbert syndrome). Such increases are less than five-fold usually (<6 mg/dl or <100 μmol/l) unless there is coexistent liver disease.
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This question is part of the following fields:
- Gastrointestinal; Hepatobiliary
- Pathology
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Question 43
Correct
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During a normal respiratory exhalation, what is the recoil alveolar pressure?
Your 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 44
Incorrect
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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: Primary osteogenic sarcoma
Correct 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 45
Correct
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Staphylococcus aureus can be identified in the laboratory based on the clotting of plasma. Which microbial product is responsible for this activity?
Your Answer: Coagulase
Explanation:Staphylococcus aureus is the most pathogenic species and is implicated in a variety of infections. S. aureus can be identified due to its production of coagulase. The staphylococcal enzyme coagulase will cause inoculated citrated rabbit plasma to gel or coagulate. The coagulase converts soluble fibrinogen in the plasma into insoluble fibrin.
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This question is part of the following fields:
- Microbiology
- Pathology
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Question 46
Correct
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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 47
Correct
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The superior rectal artery is a continuation of the:
Your Answer: Inferior mesenteric artery
Explanation:The superior rectal artery or superior haemorrhoidal artery is the continuation of the inferior mesenteric artery. It descends into the pelvis between the layers of the mesentery of the sigmoid colon, crossing the left common iliac artery and vein.
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This question is part of the following fields:
- Abdomen
- Anatomy
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Question 48
Correct
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Normally, the O2 transfer in the lungs from alveolar to capillary is perfusion-limited. In which of the following situations does it become a diffusion-limited process?
Your Answer: Pulmonary oedema
Explanation:Normally, the transfer of oxygen from air spaces to blood takes place across the alveolar-capillary membrane by simple diffusion and depends entirely on the amount of blood flow (perfusion-limited process). Diseases that affect this diffusion will transform the normal process to a diffusion limited process. Thus, the diseases which cause a thickened barrier (such as pulmonary oedema due to increased extravascular lung water or asbestosis) will limit the diffusion of oxygen. Chronic obstructive lung diseases will have little effect on diffusion. Inhaling hyperbaric gas mixtures might overcome the diffusion limitation in patients with mild asbestosis or interstitial oedema, by increasing the driving force. Strenuous (not mild) exercise might also favour diffusion limitation and decrease passage time. Increasing the rate of ventilation will not have this affect but will only maintain a high oxygen gradient from air to blood.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 49
Incorrect
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In an anatomy demonstration, the instructor asked one of the medical students to pass his index finger inferior to the root of the left lung. The student notices that his finger is blocked by a structure. Which structure do you think is responsible for this?
Your Answer: Left pulmonary vein
Correct Answer: Pulmonary ligament
Explanation:The pulmonary ligament is dual layer of pleura stretching from the inferior part of the hilar reflection toward the diaphragm.
The costodiaphragmatic recess is the cavity at the inferior border of the lung where the costal pleura becomes the diaphragmatic pleura.
The cupola: is part of the pleura that extends superiorly above the first rib and has no association with the root of the lung.
Inferior vena cava is located in the mediastinum, not near the root of the lung.
Left pulmonary veins being part of the root of the lung, would not block access to behind the lung. Costomediastinal recess is the part of the pleura where the costal pleura become the mediastinal pleura.
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This question is part of the following fields:
- Anatomy
- Thorax
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Question 50
Correct
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A 30 year old female suffered from mismatched transfusion induced haemolysis. Which substance will be raised in the plasma of this patient?
Your Answer: Bilirubin
Explanation:Bilirubin is a yellow pigment that is formed due to the break down of RBCs. Haemolysis results in haemoglobin that is broken down into a haem portion and globin which is converted into amino acids and used again. Haem is converted into unconjugated bilirubin in the macrophages and shunted to the liver. In the liver it is conjugated with glucuronic acid making it water soluble and thus excreted in the urine. Its normal levels are from 0.2-1 mg/dl. Increased bilirubin causes jaundice and yellowish discoloration of the skin.
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This question is part of the following fields:
- General
- Physiology
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SESSION STATS - PERFORMANCE PER SPECIALTY
