-
Question 1
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 2
Incorrect
-
What is the pH of freshly formed saliva at ultimate stimulation?
Your Answer: 4.4
Correct Answer: 8
Explanation:Saliva has four major components: mucus (lubricant), α-amylase (enzyme that initiates digestion of starch), lingual lipase (enzyme that begins fat digestion), and a slightly alkaline electrolyte solution for moistening food. As the secretion rate of saliva increases, its osmolality increases. Moreover, the pH changes from slightly acidic (at rest) to basic (pH 8) at ultimate stimulation. This occurs due to increase of HCO3-. Amylase and mucus also increase in concentration after stimulation.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 3
Correct
-
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) = 90 fl, mean corpuscular haemoglobin concentration (MCHC) = 41 g/dl. Further examination of blood sample revealed increased osmotic fragility of the erythrocytes. Which of the following is the most likely cause of this patient’s findings?
Your Answer: Spherocytosis
Explanation:Spherocytes are small rounded RBCs. It is due to an inherited defect of the RBC cytoskeleton membrane tethering proteins. Membrane blebs form that are lost over time and cells become round instead of biconcave. As it is a normochromic anaemia, the MCV is normal. it is diagnosed by osmotic fragility test which reveals increased fragility in a hypotonic solution.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 4
Incorrect
-
Selective destruction of which of the following cells will affect antibody synthesis?
Your Answer: T lymphocytes
Correct Answer: Plasma cells
Explanation:Plasma cell are memory cells. After the antigen Is engulfed by the B cells it is presented to the CD4+ helper cells via the MCH II receptor and this leads to their activation which in turn stimulates the B cells to form antibodies against that specific antigen. Some B cells differentiate into plasma cells also called memory cells that get activated after subsequent infection.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 5
Incorrect
-
A 13-year-old girl has complained of pain in her left arm for 4 months. An X-ray reveals a mass along with erosion of the affected humerus. Histologically, the tumour is found to be formed by small, round, blue cells. What is the most likely diagnosis?
Your Answer: Neuroblastoma
Correct Answer: Ewing’s sarcoma
Explanation:Ewing’s sarcoma is formed by small, round, blue cells, and is common in children. The usually develop in limbs, and clinical findings include pain and inflammation, with lytic destruction showing up on X-rays.
-
This question is part of the following fields:
- Orthopaedics
- Pathology
-
-
Question 6
Incorrect
-
Calculate the cardiac output in an adult male with the following parameters:
Heart rate 70 beats/min
Arterial [O2] 0.24 ml O2/min
Venous [O2] 0.16 ml O2/mi
Whole body O2 consumption 500 ml/min
Pulmonary diastolic pressure 15 mmHg
Pulmonary systolic pressure 25 mmHg
Wedge pressure 5 mmHg.Your Answer: 3.00 l/min
Correct Answer: 6.25 l/min
Explanation:As per Fick’s principle, VO2 = (CO × CAO2) – (CO × CVO2) where VO2 = oxygen consumption, CO = cardiac output, CAO2 = oxygen concentration of arterial blood and CVO2 = oxygen concentration of venous blood. Thus, CO = VO2/CAO2– CVO2, CO = 500/0.24 – 0.16, CO = 500/0.8, CO = 6.25 l/min.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 7
Incorrect
-
A 54 year old man fell off his bicycle. He felt a severe pain in his hand and wasn't able to move it. It was found that a carpal bone in the distal row was fractured. Which is the most likely bone?
Your Answer: Scaphoid
Correct Answer: Trapezium
Explanation:There are eight carpal bones which are arranged in two rows, proximal and distal; scaphoid, lunate, triquetral and pisiform are of proximal row. Trapezium, trapezoid, capitate and hamate are of the distal row.
-
This question is part of the following fields:
- Anatomy
- Upper Limb
-
-
Question 8
Correct
-
What is the likely course of a pulmonary embolism arising from the leg veins and ending in the apical segmental pulmonary artery that supplies the superior lobe of left lung?.
Your Answer: Inferior vena cava – right atrium – tricuspid valve – right ventricle – pulmonary trunk – left pulmonary artery – left superior lobar artery – left apical segmental artery
Explanation:A clot originating in the leg vein will go to the inferior vena cava, into the right atrium, through the tricuspid valve, into the right ventricle, through the pulmonary trunk, into the left pulmonary artery, into the left superior lobar artery and then finally reach the left apical segmental artery.
-
This question is part of the following fields:
- Anatomy
- Thorax
-
-
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 20-year old cyclist falls off a speeding bike and fractures a structure that articulates with the tubercle of the 7th rib. Which structure is fractured?
Your Answer: Transverse process of vertebra T7
Explanation:A rib tubercle is a projection that is postero-inferior and lateral to the neck of a rib which articulates with the transverse process of the corresponding vertebra i.e. of the same number. Therefore, the tubercle of rib 7 articulates with the transverse process of T7 vertebra. The head of the rib 7, on the other hand, articulates with the 6th vertebra superiorly and the 7th vertebra inferiorly.
-
This question is part of the following fields:
- Anatomy
- Thorax
-
-
Question 11
Correct
-
When the pitch of a sound increases, what is the physiological response seen in the listener?
Your Answer: The location of maximal basilar membrane displacement moves toward the base of the cochlea
Explanation:An increase in the frequency of sound waves results in a change in the position of maximal displacement of the basilar membrane in the cochlea. Low pitch sound produces maximal displacement towards the cochlear apex and greatest activation of hair cells there. With an increasing pitch, the site of greatest displacement moves towards the cochlear base. However, increased amplitude of displacement, increase in the number of activated hair cells, increased frequency of discharge of units in the auditory nerve and increase in the range of frequencies to which such units respond, are all seen in increases in the intensity or a sound stimulus.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 12
Incorrect
-
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: Medial to the inferior epigastric artery
Correct 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.
-
This question is part of the following fields:
- Abdomen
- Anatomy
-
-
Question 13
Correct
-
A 44-year old gentleman presented to the hospital in congestive cardiac failure. On further investigation, he was found to have a right-sided retroperitoneal mass. On enquiry, he gave a history of intermittent hypertensive attacks. 24-hour urine specimen revealed raised metanephrine and vanillylmandelic acid levels. What is the likely diagnosis?
Your Answer: Pheochromocytoma
Explanation:A neuroendocrine tumour of the chromaffin cells in adrenal medulla, pheochromocytoma secretes excessive catecholamines – adrenaline (epinephrine) and noradrenaline (norepinephrine). Pheochromocytomas are also known as the ‘10% tumour’. This is because 10% of them are bilateral, 10% are malignant, and 10% are extra-adrenal. Extra-adrenal paragangliomas (also known as extra-adrenal pheochromocytomas) are less common than pheochromocytomas and arise in the ganglia of sympathetic nervous system. Around 25% of pheochromocytomas are familial. Symptoms are described as feeling of ‘impending doom’, and include tachycardia, hypertension, palpitations, anxiety, headaches and pallor. Orthostatic hypertension is typically seen where there is a more than 100 mmHg fall in systolic pressure when the patient stands up. Pheochromocytomas can also lead to malignant hypertension. Diagnosis is by measurement of catecholamines and metanephrines in blood or in 24-hour urine, after exclusion of other possible causes such as stress, hypoglycaemia and drugs (methyldopa, dopamine agonists, ganglion-blocking antihypertensive). Imaging is also needed to localize the tumour. Localization of the tumour can also be done by Iodine-131 meta-iodobenzylguanidine (I131-MIBG) imaging.
-
This question is part of the following fields:
- Endocrine
- Pathology
-
-
Question 14
Correct
-
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.
-
This question is part of the following fields:
- Cell Injury & Wound Healing; Urology
- Pathology
-
-
Question 15
Correct
-
A 20-year old gentleman was brought to the emergency department with headache and nausea for 2 days. He also complained of intolerance to bright light and loud sounds. Lumbar puncture showed glucose < 45 mg/dl, protein > 5 mg/dl and neutrophil leucocytosis. The likely diagnosis is:
Your Answer: Meningitis
Explanation:Diagnosis of meningitis can be carried out with examination of cerebrospinal fluid (CSF) with a lumbar puncture (LP). In a case of bacterial meningitis, the CSF analysis will show:
– Opening pressure: > 180 mmH2O
– White blood cell count: 10–10 000/μl with neutrophil predominance
– Glucose: < 40 mg/dl – CSF glucose to serum glucose ratio: < 0.4 – Protein: > 4.5 mg/dl
– Gram stain: positive in > 60%
– Culture: positive in > 80%
– Latex agglutination: may be positive in meningitis due to Streptococcus pneumoniae, Neisseria meningitidis, Haemophilus influenzae, Escherichia coli and group B streptococci
– Limulus, lysates: positive in Gram-negative meningitis
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 16
Correct
-
A 45-year-old pregnant woman develops high blood pressure at 20 weeks. She complains of headaches and swollen feet, and a test reveals proteinuria (350 mg/day). Which of the following is the most likely diagnosis?
Your Answer: Pre-eclampsia
Explanation:Pre-eclampsia (PE) is a disorder of pregnancy characterized by the onset of high blood pressure (two separate readings taken at least 6 h apart of 140/90 or more) and often a significant amount of protein in the urine (>300 mg of protein in a 24-h urine sample). While blood pressure elevation is the most visible sign of the disease, it involves generalised damage to the maternal endothelium of the kidneys and liver, with the release of vasopressive factors only secondary to the original damage. Pre-eclampsia may develop at varying times within pregnancy and its progress differs among patients; most cases present pre-term. It has no known cure apart from ending the pregnancy (induction of labour or abortion). It may also present up to 6 weeks post partum. Risk factors for pre-eclampsia include obesity, prior hypertension, older age, and diabetes mellitus.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 17
Incorrect
-
A 45-year-old man with short bowel syndrome requires parenteral nutrition. The solution of choice for parenteral nutrition is:
Your Answer: Darrow’s solution
Correct Answer: Crystalline amino acids
Explanation:Total parenteral nutrition (TPN), is the practice of feeding a person intravenously, circumventing the gut. It is normally used in the following situations: surgery, when feeding by mouth is not possible, when a person’s digestive system cannot absorb nutrients due to chronic disease or if a person’s nutrient requirement cannot be met by enteral feeding and supplementation. A sterile bag of nutrient solution, between 500 ml and 4L, is provided. The pump infuses a small amount (0.1–10 ml/h) continuously to keep the vein open. The nutrient solution consists of water, glucose, salts, amino acids, vitamins and sometimes emulsified fats. Ideally each patient is assessed individually before commencing on parenteral nutrition, and a team consisting of doctors, nurses, clinical pharmacists and dietitians evaluate the patient’s individual data and decide what formula to use and at what rate.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 18
Incorrect
-
A 49-year-old man, smoker, complains of a persisting and worsening cough over the past few months. He also has noted blood in his sputum. The patient has no other major health conditions. Which of the following investigative procedures should be done first?
Your Answer: Bronchoalveolar lavage
Correct Answer: Sputum cytology
Explanation:Sputum cytology is a diagnostic test used for the examination of sputum under a microscope to determine if abnormal cells are present. It may be used as the first diagnostic procedure to help detect a suspected lung cancer or certain non-cancerous lung conditions.
-
This question is part of the following fields:
- Pathology
- Respiratory
-
-
Question 19
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 20
Incorrect
-
A tumour growing in the posterior mediastinum is found in a 40-year-old man who presented to the out patient clinic with chest pain. Such a tumour is likely to compress the following structure:
Your Answer: Arch of the aorta
Correct Answer: Oesophagus
Explanation:The boundaries of the posterior mediastinum are: the superiorly through the sternal angle and T4/5, inferiorly, the diaphragm, anteriorly, by the middle mediastinal structures and posteriorly by the spinal cord. Structures in the posterior mediastinum include the descending thoracic aorta, the azygos system, oesophagus, thoracic duct and lymph nodes. The great vessels and structures at the root of the lung are part of the middle mediastinum. The oesophagus is the only structure in the posterior mediastinum among the choices.
-
This question is part of the following fields:
- Anatomy
- Thorax
-
-
Question 21
Incorrect
-
Which of the following causes the maximum increase in the secretion of antidiuretic hormone (ADH)?
Your Answer: Decreased plasma volume
Correct Answer: Increased plasma osmolarity
Explanation:The most potent stimulus for ADH release is increased plasma osmolarity. Decreased plasma volume is a less potent stimulus in comparison. However, decrease blood volume and arterial pressure due to severe haemorrhage does lead to ADH secretion. Hypothalamic releasing factors do not control the release of posterior pituitary hormones ADH and oxytocin.
-
This question is part of the following fields:
- Endocrinology
- Physiology
-
-
Question 22
Incorrect
-
Elevated mean corpuscular volume with hypersegmented neutrophils and low reticulocyte index is seen in on the blood count of a middle-aged lady about to undergo elective surgery. On enquiry, she mentions feeling tired for a few months. Which of the following investigations should be carried out in her to reach a diagnosis?
Your Answer: Serum ferritin
Correct Answer: Serum vitamin B12 and folate
Explanation:Elevated levels of MCV indicates megaloblastic anaemia, which are associated with hypersegmented neutrophils. Likely causes include vitamin B12 or folate deficiency. Megaloblastic anaemia results from defective synthesis of DNA. As RNA production continues, the cells enlarge with a large nucleus. The cytoplasmic maturity becomes greater than nuclear maturity. Megaloblasts are produced initially in the marrow, before blood. Dyspoiesis makes erythropoiesis ineffective, causing direct hyperbilirubinemia and hyperuricemia. As all cell lines are affected, reticulocytopenia, thrombocytopenia and leukopenia develop. Large, oval blood cells (macro-ovalocytes) are released in the circulation, along with presence of hypersegmented neutrophils.
-
This question is part of the following fields:
- Haematology
- Pathology
-
-
Question 23
Incorrect
-
A patient with a long standing lower motor neuron lesion will have:
Your Answer: Hyper-reflexia
Correct Answer: Muscle wasting
Explanation:Lower motor neurons (LMNs) connect the brainstem and spinal cord to muscle fibres. Damage to lower motor neurons is indicated by abnormal electromyographic potentials, fasciculations, paralysis, weakening and wasting of skeletal muscles.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 24
Incorrect
-
During uterogrowth, the second pharyngeal arch gives rise to which structures?
Your Answer: Incus
Correct Answer: Stylohyoid muscle
Explanation:The second pharyngeal arch or hyoid arch, is the second of six pharyngeal arches that develops in fetal life during the fourth week of development and assists in forming the side and front of the neck. Derivatives:
Skeletal – From the cartilage of the second arch arises:
Stapes,
Temporal styloid process,
Stylohyoid ligament, and
Lesser cornu of the hyoid bone.
Muscles:
Muscles of face
Occipitofrontalis muscle
Platysma
Stylohyoid muscle
Posterior belly of Digastric
Stapedius muscle
Auricular muscles
Nerve supply: Facial nerve
-
This question is part of the following fields:
- Anatomy
- Embryology
-
-
Question 25
Incorrect
-
Hepatomegaly with greatly increased serum alpha-fetoprotein is seen in which of the following conditions?
Your Answer: Pancreatic carcinoma
Correct Answer: Hepatocellular carcinoma
Explanation:Hepatocellular carcinoma or hepatoma affects people with pre-existing cirrhosis and is more common in areas with higher prevalence of hepatitis B and C. Diagnosis include raise alpha-fetoprotein levels, imaging and liver biopsy if needed. Patients at high-risk for developing this disease can undergo screening by periodic AFP measurement and abdominal ultrasonography. The malignancy carries poor prognosis (see also Answer to 10.4).
-
This question is part of the following fields:
- Gastrointestinal; Hepatobiliary
- Pathology
-
-
Question 26
Incorrect
-
A 70-year-old female who underwent coronary artery bypass graft developed post-operative acute renal failure. Her urinary catheter was left in place to monitor urine output. 6 days later she developed fever and chills. She also complained of suprapubic and left flank pain. She is found to have developed acute ascending pyelonephritis. Which of the following organism was most likely isolated during urine culture?
Your Answer: Staphylococcus aureus
Correct Answer: Escherichia coli with pili
Explanation:The most common cause of urinary tract infection is Escherichia coli. Pilated strains of E. coli ascend the urethra to infect the kidney and the bladder. Catheters have been associated with an increased risk of UTIs.
-
This question is part of the following fields:
- Microbiology
- Pathology
-
-
Question 27
Incorrect
-
A 25 year old women is pregnant with her second child. She is A- blood group. Her first child was Rh+ and the father is also Rh+. The second child is at a risk of developing which condition?
Your Answer: An autoimmune disease
Correct Answer: Haemolytic disease of the new-born
Explanation:This infant is at risk for haemolytic disease of the new born also known as erythroblastosis fetalis. In the pregnancy, Rh-positive RBC’s cross the placenta and enter the mothers blood system. She then becomes sensitised and forms IgG antibodies/anti-Rh antibodies against them. The second child is at a greater risk for this disease than the first child with Rh-positive blood group as during the second pregnancy, a more powerful response is produced. IgG has the ability to cross the placenta and bind to the fetal RBCs (type II hypersensitivity reaction) which are phagocytosed by the macrophages.
-
This question is part of the following fields:
- Inflammation & Immunology; Haematology
- Pathology
-
-
Question 28
Correct
-
A 35-year-old ultra marathon runner becomes severely dehydrated and collapses. This patient most likely has:
Your Answer: Decreased baroreceptor firing rate
Explanation:Baroreceptors are sensors located in the blood vessels of all vertebrate animals. They sense the blood pressure and relay the information to the brain, so that a proper blood pressure can be maintained. Acute dehydration results in decreased plasma volume and increased plasma osmolarity, since more water than salt is lost in sweat. The decrease in plasma volume leads to an inhibition of the baroreceptors and a lower firing rate. The increase in plasma osmolarity leads to increased ADH secretion and high plasma ADH levels, which increases water permeability of collecting duct cells. Therefore more water is reabsorbed by the kidneys and renal water excretion is low.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 29
Incorrect
-
The dural venous sinuses are venous channels that drain blood from the brain. This sinuses are located between which structures?
Your Answer: Periosteal and meningeal layers of the arachnoid mater
Correct 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 30
Incorrect
-
the action of the semimembranosus muscle is:
Your Answer:
Correct Answer: Extension of the hip and flexion of the knee
Explanation:the semimembranosus is situated at the back and medial side of the thigh. It arises from the upper and outer impression on the tuberosity of the ischium, above and lateral to the biceps femoris and semitendinosus. It is inserted mainly into the horizontal groove on the posterior medial aspect of the medial condyle of the tibia. it flexes the knee and assists in extension of the hip.
-
This question is part of the following fields:
- Anatomy
- Lower Limb
-
SESSION STATS - PERFORMANCE PER SPECIALTY
