-
Question 1
Correct
-
A 36-year-old woman suddenly suffers from a generalized seizure. She was previously healthy. An emergency CT scan reveals a mass in the posterior fossa, with distortion of the lateral ventricles. After removing the tumour, the biopsy reveals it contains glial fibrillary acidic protein (GEAP). What's the most likely diagnosis?
Your Answer: Astrocytoma
Explanation:Astrocytomas are primary intracranial tumours derived from astrocyte cells of the brain. They can arise in the cerebral hemispheres, in the posterior fossa, in the optic nerve and, rarely, in the spinal cord. These tumours express glial fibrillary acidic protein (GFAP). In almost half of cases, the first symptom of an astrocytoma is the onset of a focal or generalised seizure. Between 60% and 75% of patients will have recurrent seizures during the course of their illness. Secondary clinical sequelae may be caused by elevated intracranial pressure (ICP) cause by the direct mass effect, increased blood volume, or increased cerebrospinal fluid (CSF) volume. CT will usually show distortion of the third and lateral ventricles, with displacement of the anterior and middle cerebral arteries. Histological diagnosis with tissue biopsy will normally reveal an infiltrative character suggestive of the slow growing nature of the tumour.
-
This question is part of the following fields:
- Neurology
- Pathology
-
-
Question 2
Correct
-
Regarding the extensor retinaculum of the wrist, which of these is CORRECT?
Your Answer: It prevents the tendons of the posterior compartment of the forearm from ‘bowstringing’ when the hand is extended at the wrist
Explanation:This extensor retinaculum, as the name indicates, holds the tendons of the extensors against the dorsal surface of the distal radius and ulna. Therefore, the correct answer is that it prevents bowstringing of the extensor tendons with wrist extension. It forms compartments between it and its bony attachment, and these compartments guide and hold the tendons.
-
This question is part of the following fields:
- Anatomy
- Upper Limb
-
-
Question 3
Correct
-
During a splenectomy procedure of a 45-year old male patient with a bleeding ruptured spleen, the surgeon decided to clamp the splenic artery near the coeliac trunk to stop the bleeding. Which of the following organ/s parts will be least affected by the clamping?
Your Answer: Duodenum
Explanation:The duodenum is the only organ in the list that would not be affect by the clumping of the splenic artery as it is supplied by common hepatic artery (through the gastroduodenal artery) and the superior mesenteric artery (through the inferior pancreaticoduodenal artery). The splenic artery is the artery that supplies the spleen with oxygenated blood. The splenic artery before reaching the spleen also gives off branches that supply the stomach and the pancreas. The greater curvature and the fundus of the stomach is supplied the short gastric artery which branches off from the splenic artery. The left portion of the greater curvature of the stomach together with the greater omentum is supplied by the left gastro-omental artery of the splenic artery. The body and tail of the pancreas is supplied by branches of the splenic artery through the dorsal and superior pancreatic arteries and the caudal pancreatic arteries respectively.
-
This question is part of the following fields:
- Abdomen
- Anatomy
-
-
Question 4
Correct
-
A 30-year-old woman is diagnosed with Hodgkin's lymphoma. Which of the following chemotherapy regimens would be used in this case?
Your Answer: ABVD
Explanation:ABVD is a chemotherapy regimen used in the first-line treatment of Hodgkin’s lymphoma. It consists of concurrent treatment with the chemotherapy drugs, adriamycin, bleomycin, vinblastine and dacarbazine. It supplanted the older MOPP protocol.
-
This question is part of the following fields:
- Pathology
- Pharmacology
-
-
Question 5
Correct
-
A 23 year old woman is Rh -ve and she delivered a baby with a Rh+ blood group. What measure can be performed to prevent Rh incompatibility in the next pregnancy?
Your Answer: Immunoglobulin D
Explanation:Rh disease is also known as erythroblastosis fetalis and is a disease of the new-born. In mild states it can cause anaemia with reticulocytosis and in severe forms causes severe anaemia, morbus hemolytcus new-born and hydrops fetalis. RBCs of the Rh+ baby can cross the placenta and enter into the maternal blood. As she is Rh- her body will form antibodies against the D antigen which will pass through the placenta in subsequent pregnancies.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 6
Correct
-
Which of the following structures is affected or damaged when the 'anterior drawer test' is positive?
Your Answer: Anterior cruciate ligament
Explanation:The anterior drawer test is an important orthopaedic test that is used to test weakness of the anterior cruciate ligament. The test is done by having the patient sit in a supine position with his/her knees flexed at 90 degrees. An examiner stabilises the patients feet and softly pulls or pushes on the proximal tibia. The test is positive if there is translational movement of the tibia in relation to the femur, indicating a weakened anterior cruciate ligament. The anterior cruciate ligament prevents the tibia from sliding out in front of the femur.
-
This question is part of the following fields:
- Anatomy
- Lower Limb
-
-
Question 7
Correct
-
Question 8
Correct
-
The muscle that forms the posterior wall of the axilla along with the scapula, subscapularis muscle and teres major muscle is the?
Your Answer: Latissimus dorsi
Explanation:The latissimus dorsi forms the posterior wall of the axilla along with the scapula. It is responsible for extension, adduction, transverse extension also known as horizontal abduction, flexion from an extended position, and (medial) internal rotation of the shoulder joint. It also has a synergistic role in extension and lateral flexion of the lumbar spine.
-
This question is part of the following fields:
- Anatomy
- Upper Limb
-
-
Question 9
Correct
-
Signals pass through neuromuscular junctions via the neurotransmitter acetylcholine. After release from the skeletal neuromuscular junction, acetylcholine:
Your Answer: Causes postsynaptic depolarisation
Explanation:Acetylcholine is released from the presynaptic membrane into the cleft where it binds to the ion gated channels on the post synaptic membrane, causing them to open. This results in sodium entering into the fibre and further depolarizing it, creating an action potential.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 10
Incorrect
-
Following nerve injury, paralysis of the quadriceps femoris muscle occurs. Which of the following movements will be affected?
Your Answer: Flexion of the thigh
Correct Answer: Extension of the leg
Explanation:The quadriceps muscle is a great extensor of the thigh. Therefore, following nerve injury or cutting nerve supply to the quadriceps will affect extension of the thigh
-
This question is part of the following fields:
- Anatomy
- Lower Limb
-
-
Question 11
Correct
-
During thymectomy the surgeon accidentally nicks a vein that lies just posterior to the thymus. Which of the following vessels is likely to be injured?
Your Answer: Left brachiocephalic vein
Explanation:The thymus is located superficially in the anterior mediastinum. The left brachiocephalic vein courses through the mediastinum to join the right brachiocephalic vein and form the superior vena cava on the right side of the thorax. The left brachiocephalic vein, being superficial, courses just deep to the thymus so that it may be susceptible to compression by an adjacent tumour. The left pulmonary vein, left bronchial vein and right pulmonary artery are deep, and enter and exit the lung at its root, thus are not near the thymus. The right superior intercostal vein drains the 2nd to the 4th intercostal spaces and drains into the arch of the azygos vein. It is not, therefore, closely related with the thymus.
-
This question is part of the following fields:
- Anatomy
- Thorax
-
-
Question 12
Incorrect
-
The inferior palpebral nerve ascends behind the orbicularis oculi. What is the terminal branch of the inferior palpebral nerve?
Your Answer: Facial nerve
Correct Answer: Infraorbital nerve
Explanation:The inferior palpebral nerve is a branch of the maxillary nerve. It supplies the skin and conjunctiva of the lower eyelid.
-
This question is part of the following fields:
- Anatomy
- Head & Neck
-
-
Question 13
Correct
-
The most important difference between interstitial fluid and plasma is the:
Your Answer: Protein concentration
Explanation:Interstitial fluid (or tissue fluid or intercellular fluid) is a solution that surrounds the cells of multicellular animals. It is the main component of the extracellular fluid, which also includes plasma, lymph and transcellular fluid. Plasma, the major component in blood, communicates freely with interstitial fluid through pores and intercellular clefts in capillary endothelium. Interstitial fluid consists of a water solvent containing amino acids, sugars, fatty acids, coenzymes, hormones, neurotransmitters, salts, as well as waste products from the cells. Red blood cells, platelets and plasma proteins cannot pass through the walls of the capillaries. The resulting mixture that does pass through is essentially blood plasma without the plasma proteins. Tissue fluid also contains certain types of white blood cells. Once the extracellular fluid collects into small vessels it is considered to be lymph, and the vessels that carry it back to the blood are called the lymphatic vessels. The lymphatic system returns protein and excess interstitial fluid to the circulation.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 14
Correct
-
Under normal conditions, what is the major source of energy of cardiac muscles?
Your Answer: Fatty acids
Explanation:Under basal conditions, most of the energy needed by cardiac muscle for metabolism is derived from fats (60%), 35% by carbohydrates, and 5% by ketones and amino acids. However, after intake of large amounts of glucose, lactate and pyruvate are mainly used. During prolonged starvation, fat acts as the primary source. 50% of the used lipids are sourced from circulating fatty acids.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 15
Incorrect
-
A 12-year old girl was brought to the hospital with recurrent headaches for 6 months. Her physical examination revealed no abnormality. A CT scan of the head revealed a suprasellar mass with calcifications, eroding the surrounding sella turcica. The lesion is likely to represent:
Your Answer: Prolactinoma
Correct Answer: Craniopharyngioma
Explanation:Craniopharyngiomas (also known as Rathke pouch tumours, adamantinomas or hypophyseal duct tumours) affect children mainly between the age of 5 and 10 years. It constitutes 9% of brain tumours affecting the paediatric population. These are slow-growing tumours which can also be cystic, and arise from the pituitary stalk, specifically the nests of epithelium derived from Rathke’s pouch. Histologically, this tumour shows nests of squamous epithelium which is lined on the outside by radially arranged cells. Calcium deposition is often seen with a papillary type of architecture.
ACTH-secreting pituitary adenomas are rare and mostly microadenomas. Paediatric astrocytoma’s usually occur in the posterior fossa. Although null cell adenomas can cause mass effect and give rise to the described symptoms, they are not suprasellar. Prolactinomas can also show symptoms of headache and disturbances in the visual field, however they are known to be small and slow-growing.
-
This question is part of the following fields:
- Endocrine
- Pathology
-
-
Question 16
Incorrect
-
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 I
Correct 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.
-
This question is part of the following fields:
- Endocrine
- Pathology
-
-
Question 17
Correct
-
A 76-year-old man with a urinary tract obstruction due to prostatic hyperplasia develops acute renal failure. Which of the following physiological abnormalities of acute renal failure will be most life threatening for this patient?
Your Answer: Acidosis
Explanation:Acute renal failure (ARF) is a rapid loss of renal function due to damage to the kidneys, resulting in retention of nitrogenous (urea and creatinine) and non-nitrogenous waste products that are normally excreted by the kidney. This accumulation may be accompanied by metabolic disturbances, such as metabolic acidosis and hyperkalaemia, changes in body fluid balance and effects on many other organ systems. Metabolic acidosis and hyperkalaemia are the two most serious biochemical manifestations of acute renal failure and may require medical treatment with sodium bicarbonate administration and antihyperkalaemic measures. If not appropriately treated these can be life-threatening. ARF is diagnosed on the basis of characteristic laboratory findings, such as elevated blood urea nitrogen and creatinine, or inability of the kidneys to produce sufficient amounts of urine.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 18
Correct
-
Which of the following tumours has the best prognosis following surgery?
Your Answer: Schwannoma
Explanation:Schwannomas are histologically benign neoplasms that arise from nerve sheaths, most commonly from sensory nerve roots. They are generally contained within a capsule, and so can be surgically removed without injuring the associated nerve root. They are histologically benign and less than 1% become malignant.
-
This question is part of the following fields:
- Neurology
- Pathology
-
-
Question 19
Correct
-
A chest x ray is ordered for a 39 year old man who presents with a history of a cough and weight loss for over a month. It shows a rounded opacity in the pleural cavity near the cardiac notch. The opacity is most likely to be in the:
Your Answer: Costomediastinal recess
Explanation:The costomediastinal recess is the point where the costal pleura becomes the mediastinal pleura, located right next to the cardiac notch.
The cupola: part of the parietal pleura that extends above the first rib to the root of the lung.
Hilum: located on the medial surface of the lung where neurovascular structures enter and leave the lung.
Pulmonary ligament: pleural fold found below the root of the lung, is a point of continuity between the visceral and mediastinal pleura.
Costodiaphragmatic recess: the lowest extent of the pleural cavity.
Superior mediastinum: part of the mediastinum that contains the great vessels leaving and entering the heart.
The cardiac notch is in the inferior mediastinum.
-
This question is part of the following fields:
- Anatomy
- Thorax
-
-
Question 20
Incorrect
-
Both the arytenoid muscles and the lateral cricoarytenoid muscles perform this action on the glottis:
Your Answer: Abduction
Correct Answer: Adduction
Explanation:Both the arytenoid and the cricoartenoid muscles close the glottis. The lateral cricoarytenoid muscles extend from the lateral cricoid cartilage to the muscular process of the arytenoid cartilage. By rotating the arytenoid cartilages medially, these muscles adduct the vocal cords and thereby close the rima glottidis. The arytenoid muscle adducts or approximates the arytenoid cartilages, and thus closes the aperture of the glottis.
-
This question is part of the following fields:
- Anatomy
- Head & Neck
-
-
Question 21
Correct
-
Which lymph nodes are likely to be enlarged in a patient who has malignant growth involving the anus?
Your Answer: Superficial inguinal
Explanation:The lymphatics from the anus, skin of the perineum and the scrotum end in the superficial inguinal nodes. In case of a malignant growth of the anus, the superficial inguinal lymph nodes would most likely be enlarge.
-
This question is part of the following fields:
- Anatomy
- Pelvis
-
-
Question 22
Correct
-
Which of the following muscles aid in inspiration?
Your Answer: Diaphragm and external intercostals
Explanation:The diaphragm and external intercostals are muscles of inspiration as they increase the volume of thoracic cavity and reduce the intrathoracic pressure. Muscles of expiration include abdominal muscles and internal intercostals.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 23
Incorrect
-
A 5-year-old child is rushed to the emergency department because of fever, erythema and neck stiffness. The child is also hypotensive upon physical examination. Which is the most likely toxin responsible for this child's condition?
Your Answer: Exotoxin
Correct Answer: Endotoxin
Explanation: An endotoxin is part of the bacterial outer membrane, and it is not released until the bacterium is killed by the immune system. The body’s response to an endotoxin can involve severe inflammation. In general, the inflammation process is usually considered beneficial to the infected host, but if the reaction is severe enough, it can lead to sepsis.
-
This question is part of the following fields:
- Microbiology
- Pathology
-
-
Question 24
Correct
-
QT interval in the electrocardiogram of a healthy individual is normally:
Your Answer: 0.40 s
Explanation:QT interval extends from beginning of the QRS complex till the end of he T-wave and normally lasts for 0.40 s. It is important in the diagnosis of long-QT and short-QT syndrome. The QT interval varies on the basis of heart rate and may need to be corrected.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 25
Correct
-
A 50-year-old man is diagnosed with emphysema and cirrhosis of the liver. Which of the following condition may be the cause of both cirrhosis and emphysema in this patient?
Your Answer: Alpha1-antitrypsin deficiency
Explanation:Alpha-1 antitrypsin (A1AT) deficiency is a condition characterised by the lack of a protein that protects the lungs and liver from damage, called alpha1-antytripsin. The main complications of this condition are liver diseases such as cirrhosis and chronic hepatitis, due to accumulation of abnormal alpha 1-antytripsin and emphysema due to loss of the proteolytic protection of the lungs.
-
This question is part of the following fields:
- Pathology
- Respiratory
-
-
Question 26
Correct
-
What is the normal duration of the ST segment?
Your Answer: 0.08 s
Explanation:The ST segment lies between the QRS complex and the T-wave. The normal duration of the ST segment is 0.08 s. ST-segment elevation or depression may indicate myocardial ischaemia or infarction.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 27
Incorrect
-
Which of the following muscles is innervated by the inferior branch of the right recurrent laryngeal nerve?
Your Answer: Thyroarytenoid
Correct Answer: Posterior cricoarytenoid
Explanation:The posterior cricoarytenoid muscle, which is the sole abductor of the vocal folds, receives its innervation from the inferior laryngeal nerve which is a continuation of the recurrent laryngeal nerve.
-
This question is part of the following fields:
- Anatomy
- Head & Neck
-
-
Question 28
Correct
-
Following a bee sting, a women develops a 2cm red, raised, swollen lesion at the site of the sting . Which of the following findings is likely to be seen in this lesion?
Your Answer: Vasodilation
Explanation:Inflammation is the immediate response of the body towards infections or irritations. The cardinal signs of inflammation are 1. redness/rubor, 2. tumour/swelling, 3.dolar/pain, 4.calor/heat and organ dysfunction (function laesa). Inflammation has 2 components; vascular and cellular. Blood vessels dilate upstream of the inflamed area leading to the rubor and calor and constrict downstream, increasing pressure and causing fluid to leak out of the capillary, resulting in swelling. The cellular component includes infiltration by neutrophils. Leukocyte arrival and functions include; 1. margination: cells marginated from the centre to the periphery of the vessel, 2. rolling: selectins are upregulated on the vessel walls, 3. adhesion: upregulation of the adhesion molecules ICAM and VCAM on the endothelium interact with integrins on the leukocytes resulting in adhesion, 4. diapedesis and chemotaxis: diapedesis is the transmigration of the leukocyte across the endothelium of the capillary and towards a chemotactic product and 5. phagocytosis: engulfing the offending substance/cell.
-
This question is part of the following fields:
- Inflammation & Immunology
- Pathology
-
-
Question 29
Incorrect
-
In order to arrest sudden bleeding during pericardiectomy that started after accidental injury to a major vasculature in the pericardium; the surgeon inserted his left index finger through the transverse pericardial sinus, pulled forward on the two large vessels lying ventral to his finger, and compressed these vessels with his thumb to control the bleeding. Which vessels were these?
Your Answer: Pulmonary trunk and brachiocephalic trunk
Correct Answer: Pulmonary trunk and aorta
Explanation:Transverse pericardial sinus: located behind two great vessels (aorta and pulmonary trunk) and in front of the superior vena cava and is accessed from above as in this case. The brachiocephalic trunk is located above the pericardium and the right pulmonary artery is above the pericardial reflections.
-
This question is part of the following fields:
- Anatomy
- Thorax
-
-
Question 30
Incorrect
-
A 50 year old man on warfarin therapy following insertion of a pacemaker presented with epistaxis. Which of the following is true regarding blood coagulation?
Your Answer: Disseminated intravascular coagulation (DIC) results in depletion of fibrin split products
Correct Answer: Patients with haemophilia A usually have a normal bleeding time
Explanation:A prolonged bleeding time is seen in platelet disorders like thrombocytopenia. Patients with haemophilia A or B have a prolonged PTT but not a prolonged bleeding time.
Ca2+ is necessary for coagulation.
von Willebrand factor is an important part of the factor VIII complex and promotes platelet adhesion and aggregation.
DIC results in depleted coagulation factors and accumulation of fibrin.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 31
Correct
-
A 70-year old man who is suspected to have a perforated colonic diverticulum is explored in theatre through a midline incision. This incision will be through the:
Your Answer: Linea alba
Explanation:The linea alba is the point where this incision was made. It is a tendinous raphe in the midline of the abdomen extending between the xiphoid process and the symphysis pubis. It is placed between the medial borders of the recti and is formed by the blending of the aponeuroses of the external and internal obliques and transversi.
The linea aspera is a vertical ridge on the posterior surface of the femur.
The arcuate line is the inferior border of the posterior rectus sheath behind the rectus abdominis muscle.
The semilunar line is the lateral margin of the rectus abdominis.
The iliopectineal line is a line on the pelvic bones formed by the arcuate line of the ilium and the pectineal line of the pubis.
-
This question is part of the following fields:
- Abdomen
- Anatomy
-
-
Question 32
Correct
-
During pregnancy the uterus enlarges however after delivery it regresses to its original size. Which of the following organelles is responsible for this regression?
Your Answer: Lysosomes
Explanation:Lysosomes are formed by budding of the Golgi apparatus and contain enzymes which digest macromolecules. They are found in both plants and animals and are active in autophagic cell death, digestion after phagocytosis and for the cells own recycling process. They fuse with the molecules and release their content resulting in digestion.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 33
Correct
-
Where is factor VIII predominantly synthesised?
Your Answer: Vascular endothelium
Explanation:Factor VIII is an important part of the coagulation cascade. Deficiency causes haemophilia A. It is synthesised predominantly by the vascular endothelium and is not affected by liver disease. In the circulation it is bound to von Willebrand factor and it forms a stable complex with it. It is activated by thrombin or factor Xa and acts as a co factor to factor IXa to activate factor X which is a co factor to factor Va. Thrombin cleaves fibrinogen in fibrin and forms a meshwork to trap RBC and platelets to form a clot.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 34
Incorrect
-
A 36-year old gentleman with a history of cough for 4 weeks came to the hospital. Examination revealed multiple lymphadenopathy with splenomegaly. Investigations revealed haemoglobin 11 g/dl, haematocrit 32.4%, mean corpuscular volume (MCV) 93 fl, white blood cell count 63 × 109/l, and platelet count 39 × 109/l; along with characteristic Auer rods on peripheral blood smear. What is the likely diagnosis?
Your Answer: Plasma cell leukaemia
Correct Answer: Acute myelogenous leukaemia (AML)
Explanation:AML, or acute myeloid leukaemia is the commonest acute leukaemia affecting adults. increasing in incidence with age. It is a malignancy of the myeloid line of white blood cells. It results in rapid proliferation of abnormal cells, which accumulate in the marrow. Interference with normal cell production leads to a drop in red blood cells, white blood cells and platelets. This causes symptoms such as tiredness, shortness of breath, tendency to bleed or bruise easily and recurrent infections. AML is known to progress quickly and can lead to death in weeks and months if not treated. Leukemic blasts of AML show presence of Auer rods. These are clumps of azurophilic granular material that form needles in the cytoplasm. Composed of fused lysosomes, these contain peroxidase, lysosomal enzymes and crystalline inclusions. Auer rods are classically present in myeloid blasts of M1, M2, M3 and M4 acute leukaemia. They also help to distinguish the preleukemia myelodysplastic syndromes.
-
This question is part of the following fields:
- Haematology
- Pathology
-
-
Question 35
Incorrect
-
What is the most likely cause of prolonged bleeding time in a 40 year old women admitted for a laparoscopic cholecystectomy?
Your Answer: Protein S deficiency
Correct Answer: Thrombocytopaenia
Explanation:Bleeding time is related to platelet function, thus a decrease in platelet function, as seen in thrombocytopenia, DIC and von Willebrand disease in which platelet aggregation is defective, leads to an increase in bleeding time. It is not affected by a decrease or deficiency of any other clotting factors. Aspirin and other COX inhibitors prolong bleeding time along with warfarin and heparin.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 36
Incorrect
-
Which of these nerves controls adduction of hand?
Your Answer: Musculocutaneous nerve
Correct Answer: Ulnar nerve
Explanation:The adductors of the fingers are the palmer interossei. They are supplied by the ulnar nerve, which is a branch of the medical cord of the brachial plexus.
-
This question is part of the following fields:
- Anatomy
- Upper Limb
-
-
Question 37
Incorrect
-
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: Neutrophils
Correct 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.
-
This question is part of the following fields:
- Cell Injury & Wound Healing
- Pathology
-
-
Question 38
Correct
-
Which of the following structures is not easily palpable?
Your Answer: Styloid process of the temporal bone
Explanation:The styloid process is a thin, pointed process that projects antero-inferiorly from the base of the petrous temporal bone. It can vary in length from a short, stubby process to a slender, four to five centimetre rod. It forms from the cranial elements of the second pharyngeal arch. The tympanic plate of the temporal bone ensheathes the base of this process. The pointed, projecting portion of the process provides attachment to the stylohyoid and stylomandibular ligaments, and to three muscles – the styloglossus, stylohyoid, and stylopharyngeus. As the styloid process is covered by the various muscles, it is not easily palpable in live subjects.
-
This question is part of the following fields:
- Anatomy
- Head & Neck
-
-
Question 39
Incorrect
-
Which of the following structures lying posterior to the ovary are at risk of injury in excision of a malignant tumour in the right ovary?
Your Answer: Uterine tube
Correct Answer: Ureter
Explanation:The ovaries are two nodular structures situated one on either side of the uterus in relation to the lateral wall of the pelvis and attached to the back of the broad ligament of the uterus, lying posteroinferiorly to the fallopian tubes. Each ovary has a lateral and medial surface. The ureter is at greater risk of iatrogenic injury at this location.
-
This question is part of the following fields:
- Anatomy
- Pelvis
-
-
Question 40
Correct
-
A 51-year old woman after undergoing a barium swallowing was discovered to be suffering from an oesophageal hiatal hernia. Which muscle fibres of the diaphragm border this hernia directly if the stomach herniates through an enlarged oesophageal hiatus?
Your Answer: Right crus
Explanation:The oesophageal hiatus is a natural fissure on the thoracic diaphragm that allows passage of the oesophagus and the vagal nerve. The oesophageal hiatus is located in one of the tendinous structures of the diaphragm that connect it to the spine which is known as the right crus. In case of an hiatal hernia, this diaphragmatic structure would be the one bordering the hernia as it is the structure that encircles the oesophageal hiatus.
-
This question is part of the following fields:
- Abdomen
- Anatomy
-
-
Question 41
Correct
-
A 25-year-old woman complains of generalised swelling and particularly puffiness around the eyes which is worst in the morning. Laboratory studies showed:
Blood urea nitrogen (BUN) = 30 mg/dl
Creatinine = 2. 8 mg/dl
Albumin = 2. 0 mg/dl
Alanine transaminase (ALT) = 25 U/l
Bilirubin = 1 mg/dl
Urine analysis shows 3+ albumin and no cells.
Which of the following is the most likely diagnosis?Your Answer: Nephrotic syndrome
Explanation:Nephrotic syndrome is a disorder in which the glomeruli have been damaged, characterized by:
– Proteinuria (>3.5 g per 1.73 m2 body surface area per day, or > 40 mg per square meter body surface area per hour in children)
– Hypoalbuminemia (< 2,5 g/dl) – Hyperlipidaemia, and oedema (generalized anasarca).
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 42
Correct
-
A lesion involving the suprachiasmatic nucleus of hypothalamus is likely to affect:
Your Answer: Regulation of circadian rhythm
Explanation:The suprachiasmatic nucleus (SCN) in the hypothalamus is responsible for controlling endogenous circadian rhythms and destruction of the SCN leads to a loss of circadian rhythm.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 43
Incorrect
-
The tensor villi palatini muscle is a broad thin, ribbon-like muscle in the head that tenses the soft palate. Which of the following structures is associated with the tensor villi palatini muscle?
Your Answer: The middle pharyngeal constrictor
Correct Answer: The hamulus of the medial pterygoid plate
Explanation:The pterygoid hamulus is a hook-like process at the lower extremity of the medial pterygoid plate of the sphenoid bone around which the tendon of the tensor veli palatini passes.
-
This question is part of the following fields:
- Anatomy
- Head & Neck
-
-
Question 44
Correct
-
A lesion involving the lateral portion of the dorsal columns at the level of the nape of the neck will most likely affect:
Your Answer: Vibratory sensations from the ipsilateral arm
Explanation:At the level mentioned in the question, the lateral portion of dorsal columns comprises of the fasciculus cuneatus. Axons carrying the sensations of touch, vibration and proprioception from the ipsilateral arm enter the spinal cord and ascend in the fasciculus cuneatus, synapsing in the nucleus cuneatus of the caudal medulla. Secondary neurons from this nucleus give rise to internal arcuate fibres, which decussate and ascend to the thalamus as the medial lemniscus. Tertiary neurons from there project to the ipsilateral somatosensory cortex. Thus, any damage to the fasciculus cuneatus will result in a deficit in tactile, proprioceptive and vibratory sensations in the ipsilateral arm, and not the contralateral arm.
Fine motor control of the fingers is mainly carried by the ipsilateral lateral corticospinal tract in the lateral funiculus of the cord. Motor control of the contralateral foot is carried by the ipsilateral corticospinal tract in the lateral funiculus of the cord. Lack of sweating of the face could be produced by interruption of sympathetic innervation. Proprioception from the ipsilateral leg is carried by the fasciculus gracilis in the medial part of the dorsal columns.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 45
Correct
-
A 70 year old women, died suddenly. She had a history of hypertension and aortic stenosis. On autopsy her heart weighed 550g. What is the most likely cause of this pathology?
Your Answer: Hypertrophy
Explanation:Due to increased pressure on the heart as a result of hypertension and aortic stenosis, the myocardial fibres hypertrophied to adapt to the increased pressure and to effectively circulate blood around the body. Hyperplasia could not occur, as myocardial fibres are stable cells and cannot divide further.
Fat does not deposit in the heart due to volume overload.
Myocardial oedema is not characteristic of a myocardial injury.
Metaplasia is a change in the type of epithelium.
Atrophy would result in a decreased heart size and inability to function.
-
This question is part of the following fields:
- Cell Injury & Wound Healing; Cardiovascular
- Pathology
-
-
Question 46
Correct
-
A 46 -year old patient diagnosed with chronic rhinosinusitis, was to undergo surgery to improve drainage from his frontal sinus to the nose. Which is a route that one would take to enter into the frontal sinus through the nasal cavity?
Your Answer: Middle meatus
Explanation:The middle meatus is a nasal passageway located inferior to the middle concha and superior to the inferior concha. On the superior aspect of this meatus is a bulge produced by the middle ethmoidal cells known as the bulla ethmoidalis. Below this bulge is a curved fissure, the hiatus semilunaris, which is also bordered inferiorly by the edge of the uncinate process of the ethmoid. It is through this curved fissure, hiatus semilunaris, that the middle meatus communicates with the frontal sinus. It first forms a communication with a curved passage way known as the infundibulum. The infundibulum anteriorly communicates with the anterior ethmoidal cells and continues upward as the frontonasal duct into the frontal sinus.
-
This question is part of the following fields:
- Anatomy
- Head & Neck
-
-
Question 47
Correct
-
The glossopharyngeal nerve provides the parasympathetic innervation of the:
Your Answer: Parotid salivary gland
Explanation:The glossopharyngeal nerve provides parasympathetic innervation for the parotid salivary gland via the auriculotemporal nerve. The facial nerve supplies the parasympathetic innervation of the lacrimal, nasal, sublingual and submandibular glands.
-
This question is part of the following fields:
- Anatomy
- Head & Neck; Neurology
-
-
Question 48
Correct
-
What is the result of maltase deficiency in the brush border of the small intestine?
Your Answer: Results in increased passage of maltose in stool
Explanation:Maltase is an enzyme produced from the surface cells of the villi, lining the small intestine and aids in hydrolysing the disaccharide maltose, which splits into two molecules of α-glucose. It is done by breaking the glycosidic bond between the ‘first’ carbon of one glucose and the ‘fourth’ carbon of the other (a 1–4 bond). Hence, a deficiency of enzyme maltase will result in the increased passage of maltose in the stool.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 49
Incorrect
-
A surgeon performing a Whipple's procedure, that involves mobilizing the head of the pancreases, accidentally injured a structure immediately posterior to the neck of the pancrease which bled out. Which structure is most likely to have been injured?
Your Answer: Hepatic artery
Correct Answer: Superior mesenteric artery
Explanation:The structure immediately posterior to the neck of the pancreas is the superior mesenteric vein. The neck of the pancreas lies anterior to the superior mesenteric vein, which joins with the splenic vein to form the portal vein.
The superior mesenteric artery (SMA) is also located in proximity to the neck of the pancreas. Specifically, the SMA emerges from the abdominal aorta just below the level of the pancreas and passes posterior to the neck of the pancreas before it descends into the mesentery to supply the intestines. Thus, both the superior mesenteric vein and the superior mesenteric artery are key vascular structures related to the posterior aspect of the neck of the pancreas.
Bleeding out would suggest an arterial injury.
-
This question is part of the following fields:
- Abdomen
- Anatomy
-
-
Question 50
Correct
-
Which one of the following groups of lymph nodes is most likely to be inflamed due to paronychia involving the big toe?
Your Answer: Vertical group of superficial inguinal lymph nodes
Explanation:Paronychia affecting the big toe will result in inflammation of the superficial inguinal lymph nodes as it drains lymph from the big toe.
-
This question is part of the following fields:
- Anatomy
- Lower Limb
-
-
Question 51
Incorrect
-
Intracellular shifting of hydrogen ions can generate a metabolic alkalosis. In which of the following conditions is metabolic alkalosis caused by this mechanism ?
Your Answer: Vomiting
Correct Answer: Hypokalaemia
Explanation:Metabolic alkalosis is characterized by a primary increase in the concentration of serum bicarbonate ions. This may occur as a consequence of a loss of hydrogen ions or a gain in bicarbonate. Hydrogen ions may be lost through the kidneys or the GI tract, as for example during vomiting, nasogastric suction or use of diuretics. Intracellular shifting of hydrogen ions develops mainly during hypokalaemia to maintain neutrality. Gain in bicarbonate ions may develop during administration of sodium bicarbonate in high amounts or in amounts that exceed the capacity of excretion of the kidneys, as seen in renal failure. Fluid losses may be another cause of metabolic alkalosis, causing the reduction of extracellular fluid volume.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
-
-
Question 52
Correct
-
A CT-scan report of a patient with gastric carcinoma shows that the lymph nodes around the coeliac trunk are enlarged. The coeliac trunk:
Your Answer: Gives rise to the splenic, left gastric and common hepatic arteries
Explanation:The coeliac trunk is a branch of the aorta arising just below the aortic hiatus of the diaphragm to pass nearly horizontally forward to divide into 3 large branches i.e. the left gastric, the hepatic and the splenic arteries. Occasionally it may give off one of the inferior phrenic arteries. It is covered by the lesser omentum and on the right side bordered by the right coeliac ganglion and the cardiac end of the stomach. Inferiorly it is bordered by the upper border of the pancreas and splenic vein. The embryonic midgut is supplied by the superior mesenteric artery.
-
This question is part of the following fields:
- Abdomen
- Anatomy
-
-
Question 53
Correct
-
In a study, breast lumps were analysed to determine the characteristic of malignant neoplasm on biopsy. What microscopic findings are suggestive of malignancy?
Your Answer: Invasion
Explanation:Invasion is suggestive of malignancy and an even better option would have been metastasis. Pleomorphism is found in both benign and malignant neoplasms along with atypia and anaplasia. A height nuclear/cytoplasmic ratio is suggestive of malignancy but not the best indicator. Malignant tumours are aggressive and growth rapidly. Necrosis can be seen in benign tumours if they deplete their blood supply.
-
This question is part of the following fields:
- Neoplasia
- Pathology
-
-
Question 54
Correct
-
Severe abdominal pain radiating to the back, along with increased serum amylase levels, is seen in which of the following conditions?
Your Answer: Pancreatitis
Explanation:The primary test for diagnosis and monitoring of pancreatitis is amylase. Increased plasma levels of amylase can be found in: salivary trauma (including anaesthetic intubation), mumps, pancreatitis and renal failure. However, a rise in the total amylase levels over 10 times the upper limit of normal (ULN) is suggestive of pancreatitis; 5–10 times the ULN may indicate ileus or duodenal disease or renal failure. Lower levels are commonly found in salivary gland disease.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 55
Correct
-
A 45-year old farmer was referred to the surgical clinic with complaints of pain in his right hypochondrium. Investigations confirmed the diagnosis of hepatocellular carcinoma with malignant ascites. According to you, what is the most likely cause of HCC in this patient?
Your Answer: Aflatoxin
Explanation:Aflatoxins are naturally occurring toxins produced by the Aspergillus fungus (most often, A. flavus and A. parasiticus). These organisms are common and their native habitat is soil, decaying vegetation and grains. They can contaminate the grain before harvest or after storage, more likely in high-humidity (at least 7%) or high temperature environment of after stressful conditions like drought. Aflatoxins are mycotoxins and also carcinogenic. They get metabolized in the liver to an epoxide, aflatoxin M1. High exposure can lead to acute necrosis, cirrhosis or liver carcinoma. These substances can cause haemorrhage, acute liver damage, oedema, and alteration in digestion, absorption and/or metabolism of nutrients. Although humans are susceptible to these toxins like all other animals, they have a high tolerance level and hence, rarely develop acute aflatoxicosis. However, children are particularly susceptible to exposure leading to growth impairment and delayed development. Chronic exposure carries a high risk of hepatic cancer, due to intercalation of its metabolite aflatoxin M1 into the DNA and alkylation of the bases because of its epoxide moiety.
-
This question is part of the following fields:
- Gastrointestinal; Hepatobiliary
- Pathology
-
-
Question 56
Correct
-
Calculate the total peripheral resistance for a patient with a blood pressure of 130/70 mm HG and cardiac output of 5 litres / min?
Your Answer: 18 mmHg × min/l
Explanation:Total peripheral resistance = Mean arterial pressure/Cardiac output. And the mean arterial pressure = Diastolic pressure + 1/3 (Systolic pressure – Diastolic pressure), i.e., 70 + 1/3 (130-70) = 90 mmHg. Therefore, total peripheral resistance = 90 mmHg/5 l per min = 18 mmHg × min/l.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 57
Incorrect
-
The vascular structure found on the right side of the fifth lumbar vertebra is?
Your Answer: Right external iliac vein
Correct Answer: Inferior vena cava
Explanation:The most likely vascular structure is the inferior vena cava. The inferior vena cava is formed by the joining of the two common iliac arteries, the right and the left iliac artery, at the level of the fifth lumbar vertebra( L5). The inferior vena cava passes along the right side of the vertebral column. It enters the thoracic cavity into the underside of the heart through the caval opening of the diaphragm at the level of the eight thoracic vertebra (T8).
-
This question is part of the following fields:
- Abdomen
- Anatomy
-
-
Question 58
Correct
-
Which Statement is true of the brachial plexus?
Your Answer: The posterior cord continues as the axillary nerve
Explanation:The lateral cord continues as the musculocutaeous nerve.
The medial cord continues as the ulnar nerve.
The posterior cord continues as the radial nerve and the axillary nerve.
The nerve to subclavius muscle is a branch of the C6 root.
The suprascapular nerve is a branch from the upper trunk.
-
This question is part of the following fields:
- Anatomy
- Upper Limb
-
-
Question 59
Correct
-
The prostatic plexus of nerves contains nerve fibres that innervate penile tissue allowing for erection. From which of the following nerves do these fibres originate?
Your Answer: Pelvic splanchnics
Explanation:Pelvic splanchnic nerves are examples of the parasympathetic nerves that innervate the smooth muscle and glands of the pelvic viscera. They are also the nerves contributing fibres to the prostatic plexus which innervate penile erectile tissue.
-
This question is part of the following fields:
- Anatomy
- Pelvis
-
-
Question 60
Correct
-
During cardiac catheterisation, if the blood sample from the catheter shows an oxygen saturation of 70%, and the pressure ranging from 12 to 24 mm Hg, it implies that the catheter tip is located in the:
Your Answer: Pulmonary artery
Explanation:Normal values for various parameters are as follows:
Systolic arterial blood pressure (SBP): 90–140 mmHg.
Diastolic arterial blood pressure: 60–90 mmHg.
Mean arterial blood pressure (MAP): SBP + (2 × DBP)/3 (normal range: 70-105 mmHg).
Right atrial pressure (RAP): 2–6 mmHg.
Systolic right ventricular pressure (RVSP): 15–25 mmHg.
Diastolic right ventricular pressure (RVDP): 0–8 mmHg.
Pulmonary artery pressure (PAP): Systolic (PASP) is 15-25 mmHg and Diastolic (PADP) is 8–15 mmHg.
Pulmonary artery wedge pressure (PAWP): 6–12 mmHg.
Left atrial pressure (LAP): 6–12 mmHg.
Thus, the given value indicates that the position of catheter tip is likely to be in the pulmonary artery.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 61
Incorrect
-
The mandibular nerve, which is the largest of the 3 divisions of the trigeminal nerve, exits the cranial cavity through which foramen?
Your Answer: Foramen rotundum
Correct 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.
-
This question is part of the following fields:
- Anatomy
- Head & Neck
-
-
Question 62
Correct
-
The rotator cuff surrounds the shoulder joint and consists of the supraspinatus, infraspinatus, subscapularis and which one other muscle?
Your Answer: Teres minor
Explanation:The correct answer is the teres minor muscle. These group of muscles play an important role in protecting the shoulder joint and keeping the head of the humerus in the glenoid fossa of the scapula. This fossa is somehow shallow and needs support to allow for the full mobility that the shoulder joint has.
-
This question is part of the following fields:
- Anatomy
- Upper Limb
-
-
Question 63
Correct
-
A textbook of cardiac surgery explains that while doing a left pneumonectomy, a surgeon must be careful to avoid injury to a vital structure that leaves an impression on the mediastinal surface of the left lung. Which structure is it referring to?
Your Answer: Aortic arch
Explanation:Structures that leave an impression on the mediastinal surface of the left lung include: the oesophagus, subclavian artery, brachiocephalic vein, first rib, thymus, the heart, the diaphragm, descending aorta and arch of the aorta. The other structures form an impression on the mediastinal aspect of the right lung.
-
This question is part of the following fields:
- Anatomy
- Thorax
-
-
Question 64
Correct
-
Destruction of the ventromedial nucleus of the hypothalamus will result in:
Your Answer: Loss of satiety
Explanation:The ventromedial nucleus of the hypothalamus is divided into an anterior and a superior part. The anterior part controls the female sexual drive, whereas the superior part is responsible for satiety. Destruction of the superior part of the nucleus will result in overeating, as no signal tells the body that it is satisfied.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 65
Incorrect
-
In the kidney, the macula densa is an area of closely packed specialized cells lining the wall of the:
Your Answer: Juxtaglomerular cells
Correct Answer: Distal convoluted tubule
Explanation:In the kidney, the macula densa is an area of closely packed specialised cells lining the region of the distal convoluted tubule (DCT) lying next to the glomerular vascular pole. The cells of the macula densa are sensitive to the ionic content and water volume of the fluid in the DCT, producing signals that promote renin secretion by other cells of the juxtaglomerular apparatus.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 66
Correct
-
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.
-
This question is part of the following fields:
- Neoplasia
- Pathology
-
-
Question 67
Incorrect
-
Calculate the cardiac stroke volume of a patient whose oxygen consumption (measured by analysis of mixed expired gas) is 300 ml/min, arterial O2 content is 20 ml/100 ml blood, pulmonary arterial O2 content is 15 ml/100 ml blood and heart rate is 60/min.
Your Answer: 50 ml
Correct Answer: 100 ml
Explanation:By Fick’s principle, VO2 = Q × (CA (O2) − CV (O2)) where VO2 = O2 consumption, Q = cardiac output and CA(O2) and CV(O2) are arterial and mixed venous O2 content respectively. Thus, in the given problem, 300 ml O2/min = Q × (20−15) ml O2/100 ml. Thus, Q = 6000 ml blood/min. Then, we can calculate stroke volume by dividing the cardiac output with heart rate. Thus, stroke volume = 6000 ml/min divided by 60/min stroke volume = 100 ml.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 68
Correct
-
A 20-year-old woman had profuse watery diarrhoea for 2 days. She felt dizzy and weak, and thus decided to seek medical attention. At the emergency room her BP was 80/60 mmHg with a pulse of 118/min. What is the most appropriate intravenous treatment that should be given?
Your Answer: Isotonic saline
Explanation:Normal saline is typically the first fluid used when hypovolemia is severe enough to threaten the adequacy of blood circulation. It is isotonic and has long been believed to be the safest fluid to give quickly in large volumes.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Pathology
-
-
Question 69
Correct
-
Calculate the stroke volume in a patient admitted for coronary bypass surgery, with the following parameters pre-operatively:
Oxygen consumption = 300 ml/min
Arterial oxygen content = 20 ml/100 ml blood
Pulmonary arterial oxygen content = 15 ml/100 ml blood and Heart rate = 100 beats/min.Your Answer: 60 ml
Explanation:By Fick’s principle, cardiac output can be calculated as follows: VO2 = CO × (CAO2– CVO2) where VO2= oxygen consumption, CO = cardiac output, CAO2 = arterial oxygen content and CvO2 = mixed venous oxygen content. Thus, in the given problem, 300 ml/min = CO × (20 – 15) ml/100 ml CO = 300 × 100/5 ml/min CO = 6000 ml/min. Also, cardiac output = stroke volume × heart rate. Thus, 6000 ml/min = stroke volume × 100 beats/min. Hence, stroke volume = 6000/100 ml/min which is 60 ml/min.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 70
Correct
-
A 55-year-old woman died 3 years after a cardiac transplant due to worsening congestive heart failure. Autopsy revealed diffuse hyperplasia of the vascular intima involving the entire length of the coronary arteries. The most probable cause of deterioration of the cardiac function is:
Your Answer: Coronary atherosclerosis
Explanation:Allograft coronary artery disease (CAD) can begin right after the transplant and is the major cause of later death in cardiac transplant recipients. This form of atherosclerosis progresses quickly resulting in allograft failure. Due to lack of premonitory symptoms CAD may lead to sudden death.
-
This question is part of the following fields:
- Cardiovascular
- Pathology
-
-
Question 71
Incorrect
-
Which of the following arteries, that runs on the superior aspect of the first part of the duodenum, forms the lower boundary of the epiploic foramen (also known as the foramen of Winslow)?
Your Answer: Superior pancreaticoduodenal
Correct Answer: Hepatic
Explanation:The epiploic foramen is an important anatomical opening that allows for the communication between the greater peritoneal sac and the lesser peritoneal sac. Its boundaries are formed; superiorly by the caudate lobe of the liver, anteriorly by the hepatoduodenal ligament (containing the components of the portal triad), inferiorly by the first part of the duodenum and posteriorly by the peritoneum covering the inferior vena cava. The superior aspect of the first part of the duodenum, which forms the inferior boundary of the foramen of Winslow, forms the course of the hepatic artery before it ascends to the porta hepatis where it divides into its right and left branches.
-
This question is part of the following fields:
- Abdomen
- Anatomy
-
-
Question 72
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 73
Correct
-
A 7-year old child from a rural setting complains of recurrent abdominal pain. The child is found to have a heavy parasitic infestation and anaemia. Which type of anaemia is most likely seen in this patient?
Your Answer: Iron deficiency anaemia
Explanation:The most common cause of iron deficiency anaemia in children in developing countries is parasitic infection (hookworm, amoebiasis, schistosomiasis and whipworm).
-
This question is part of the following fields:
- Microbiology
- Pathology
-
-
Question 74
Correct
-
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.
-
This question is part of the following fields:
- Anatomy
- Embryology
-
-
Question 75
Correct
-
A diabetic 58-year-old man, after the injection of radiographic contrast, has a decreased urine output and decreased level of consciousness. Which of the following conditions has he most likely developed
Your Answer: Acute tubular necrosis
Explanation:Acute tubular necrosis (ATN) involves damage to the tubule cells of the kidneys and is the most common cause of acute kidney injury. ATN in the majority of the cases is caused by ischaemia of the kidneys due to lack of perfusion and oxygenation but it may also occur due to poison or harmful substance. Contrast used for radiology may cause ATN in patients with several risk factors e.g. diabetic nephropathy. Symptoms may include oliguria, nausea, fluid retention, fatigue and decreased consciousness.
-
This question is part of the following fields:
- Pathology
- Renal
-
-
Question 76
Correct
-
A 15-day old baby was brought to the emergency department with constipation for 4 days. On examination, the abdomen of the baby was found to be distended and tender all over. No bowel sounds were heard. A sigmoid colon biopsy was carried out, which showed absent ganglion cells. What is the diagnosis?
Your Answer: Hirschsprung’s disease
Explanation:Hirschsprung’s disease is characterized by congenital absence of the autonomic plexus (Meissner’s and Auerbach’s plexus) in the intestinal wall. Usually limited to the distal colon, it can occasionally involve the entire colon or even the small bowel. There is abnormal or absent peristalsis in the affected segment, resulting in continuous spasm of smooth muscle and partial/complete obstruction. This causes accumulation of intestinal contents and dilatation of proximal segment. Skip lesions are highly uncommon. This disease is seen early in life with 15% patients presenting in first month, 60% by 1 year of age and 85% by the age of 4 years. Symptoms include severe and complete constipation, abdominal distension and vomiting. Patients with involvement of ultra-short segments might have mild constipation with intervening diarrhoea. In older children, symptoms include failure to thrive, anorexia, and lack of an urge to defecate. On examination, an empty rectum is revealed with stool palpable high up in the colon. If not diagnosed in time, it can lead to Hirschsprung’s enterocolitis (toxic megacolon), which can be fulminant and lead to death. Diagnosis involves a barium enema or a rectal suction biopsy. Barium enema shows a transition in diameter between the dilated, normal colon proximal to the narrowed, affected distal segment. It is to be noted that barium enema should be done without prior preparation, which can dilate the abnormal segment, leading to a false-negative result. A 24-hour post-evacuation film can be obtained in the neonatal period – if the colon is still filled with barium, there is a high likelihood of Hirschsprung’s disease. Full-thickness rectal biopsy is diagnostic by showing the absence of ganglion cells. Acetylcholinesterase staining can be done to highlight the enlarged nerve trunks. Abnormal innervation can also be demonstrated by rectal manometry.
-
This question is part of the following fields:
- Gastrointestinal; Hepatobiliary
- Pathology
-
-
Question 77
Correct
-
Which of the following variables are needed to calculate inspiratory reserve volume of a patient?
Your Answer: Tidal volume, vital capacity and expiratory reserve volume
Explanation:Vital capacity = inspiratory reserve volume + tidal volume + expiratory reserve volume. Thus, inspiratory reserve volume can be calculated if tidal volume, vital capacity and expiratory reserve volume are known.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 78
Correct
-
Which nerve mediates the sensation to itch from the skin that is just over the base of the spine of your scapula?
Your Answer: Dorsal primary ramus of C7
Explanation:The first branches off spinal nerves are called the dorsal and ventral rami. The dorsal rami mediate sensation of the skin over the back and motor supply to the true muscles of the back whilst the ventral rami gives sensation to the skin over the limbs and the skin that is over the ventral side of the trunk. It also gives motor supply to the skeletal muscles of the neck, the trunk and extremities. Hence, itchiness of the part of the skin that is over the spine of the scapula would be mediated by the primary ramus of C7. Accessory nerve doesn’t have any sensory innervation.
-
This question is part of the following fields:
- Anatomy
- Neurology
-
-
Question 79
Correct
-
Which of the following is true about a patient who has undergone total colectomy and ileostomy?
Your Answer: Following total colectomy and ileostomy, the volume and water content of ileal discharge decreases over time
Explanation:After a patient has undergone total colectomy and ileostomy, the volume of ileal discharge, along with its water content gradually decreases over time. Post surgery, most patients can live a normal life. Iron and vitamin B12 absorption do not take place in the colon and hence are not affected significantly by a colectomy.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 80
Correct
-
A surgeon trainee is assisting in an operation to ligate the ductus arteriosus. The consultant supervising explains that caution is required when placing a clamp on the ductus to avoid injury to an important structure immediately dorsal to it. To which structure is the consultant referring?
Your Answer: Left recurrent laryngeal nerve
Explanation:The left recurrent laryngeal nerve branches off the vagus and wraps around the aorta, posterior to the ductus arteriosus/ligamentum arteriosum from whence it courses superiorly to innervate the laryngeal muscles.
Accessory Hemiazygous vein is on the left side of the body draining the posterolateral chest wall and emptying blood into the azygos vein.
The left internal thoracic artery is branch of the left subclavian artery supplying blood to the anterior wall of the thorax.
Left phrenic nerve is lateral to the vagus nerve.
Thoracic duct: is behind the oesophagus, coursing between the aorta and the azygos vein in the posterior chest.
Right recurrent laryngeal nerve: loops around the right subclavian artery and is not in danger in this procedure.
-
This question is part of the following fields:
- Anatomy
- Thorax
-
-
Question 81
Correct
-
Rapid Eye Movement (REM) sleep is typically associated with:
Your Answer: Penile erections
Explanation:Normal sleep comprises of alternate cycles between slow-wave sleep (non-REM sleep) and REM sleep. REM sleep is characterized by increased metabolic brain activity and EEG desynchronization. Somnambulism (sleepwalking), enuresis (bedwetting) and night terrors all occur during slow-wave sleep or during arousal from slow-wave sleep. In comparison, REM sleep is characterized by hypotonia of major muscle groups (excluding ocular muscles), dreams, nightmares and penile erection.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 82
Correct
-
the action of the semimembranosus muscle is:
Your 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
-
-
Question 83
Correct
-
Which of these illnesses is most likely to precede Guillain-Barré syndrome?
Your Answer: Viral pneumonia
Explanation:Guillain–Barré syndrome (GBS) is characterized by a rapid-onset muscle weakness caused by the immune system damaging the peripheral nervous system. In about two-thirds of patients, the syndrome begins 5 days to 3 weeks after an infectious disease, surgery or vaccination. Infection is the trigger in over 50% of patients; common pathogens include Campylobacter jejuni, enteric viruses, herpesviruses (including cytomegalovirus and those causing infectious mononucleosis) and Mycoplasma species. The underlying mechanism involves an autoimmune disorder in which the body’s immune system mistakenly attacks the peripheral nerves and damages their myelin insulation, although the cause for this is still unknown.
-
This question is part of the following fields:
- Neurology
- Pathology
-
-
Question 84
Correct
-
The posterior cord contains nerve fibres from which of the following levels of the spinal cord?
Your Answer: C5, C6, C7, C8 and T1
Explanation:THE correct answer is A. The posterior cord derives its fibres from the spinal nerves C5,C6,C7,C8,T1. This cord is formed from the fusion of the posterior divisions of the upper, lower, and middle trunks.
-
This question is part of the following fields:
- Anatomy
- Upper Limb
-
-
Question 85
Correct
-
A 50 year old man was admitted to the surgical ICU following a hemicolectomy for carcinoma of the caecum. A full blood count revealed: haematocrit = 30%, erythrocytes = 4 × 106/μ, haemoglobin level = 8 g/dl. To determine the likely cause of his anaemia, red blood cell indices were calculated. Which RBC indices are correct?
Your Answer: MCHC = haemoglobin concentration/haematocrit
Explanation:Mean corpuscular haemoglobin concentration (MCHC) is calculated simply by dividing the haemoglobin concentration (8 g/dl) by the haematocrit (0.3). The normal range is 31–36 g/dl. This patient has a hypochromic anaemia (MCHC = 8/0.3 = 26.7 g/dl). Dividing the haemoglobin concentration × 10 by erythrocyte number yields mean corpuscular haemoglobin (MCH). Normal range is 25.4–34.6 pg/cell and this patient has a significantly reduced cellular haemoglobin content (MCH = 8 × 10/4 = 20 pg/cell). Mean corpuscular volume (MCV) is calculated by dividing haematocrit × 1000 by erythrocyte number (4 × 106/μl). Normal range is 80–100 fl and this patient has a microcytic anaemia (MCV = 0.3 × 1000/4 = 75 fl). Microcytic, hypochromic anaemia is characteristic for iron-deficiency.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 86
Correct
-
Arterial blood gas analysis of a man admitted with acute exacerbation of chronic obstructive pulmonary disease (COPD) showed the following: pH = 7.28, p(CO2) = 65.5 mmHg, p(O2)= 60 mmHg and standard bicarbonate = 30.5 mmol/l. This patient had:
Your Answer: Respiratory acidosis
Explanation:Acidosis with high p(CO2) and normal standard bicarbonate indicates respiratory acidosis, commonly seen in acute worsening of COPD patients. Respiratory acidosis occurs due to alveolar hypoventilation which leads to increased arterial carbon dioxide concentration (p(CO2)). This in turn decreases the HCO3 –/p(CO2) and decreases pH. In acute respiratory acidosis, the p(CO2) is raised above the upper limit of normal (over 45 mm Hg) with a low pH. However, in chronic cases, the raised p(CO2) is accompanied with a normal or near-normal pH due to renal compensation and an increased serum bicarbonate (HCO3 – > 30 mmHg).
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 87
Correct
-
A 55-year-old man underwent CT scan of the whole abdomen. The result showed renal cell carcinoma with a tumour size of 7cm and extension into the regional lymph. What is the clinical stage of his renal cell cancer?
Your Answer: Stage III
Explanation:Renal cell carcinoma is a kidney cancer that originates in the lining of the proximal convoluted tubule. It is the most common type of kidney cancer in adults, responsible for approximately 90–95% of cases. Renal cell carcinomas can be staged by using the American Joint Committee on Cancer (AJCC) TNM (tumour-node-metastasis) classification, as follows: Stage I: tumours that are 7 cm or smaller and confined to the kidney, Stage II: tumours that are larger than 7 cm but still confined to the kidney, Stage III: tumours extending into the renal vein or vena cava, involving the ipsilateral adrenal gland and/or perinephric fat, or which have spread to one local lymph node and Stage IV: tumours extending beyond Gerota’s fascia, to more than one local node, or with distant metastases Recent literature has questioned whether the cut-off in size between stage I and stage II tumours should be 5 cm instead of 7 cm. The patient’s cancer in this case is stage III.
-
This question is part of the following fields:
- Neoplasia
- Pathology
-
-
Question 88
Correct
-
During strenuous exercise, what else occurs besides tachycardia?
Your Answer: Increased stroke volume
Explanation:During strenuous exercise there is an increase in:
– Heart rate, stroke volume and therefore cardiac output. (CO = HR x SV)
– Respiratory rate (hyperventilation) which will lead to a reduction in Paco2.
– Oxygen demand of skeletal muscle, therefore leading to a reduction in mixed venous blood oxygen concentration.
Renal blood flow is autoregulated, so renal blood flow is preserved and will tend to remain the same. Mean arterial blood pressure is a function of cardiac output and total peripheral resistance and will increase with exercise, mainly as a result of the increase in cardiac output that occurs.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 89
Correct
-
A 30 year old gym coach presented to the surgical out patient clinic with a lump in the inguinal region. He was booked for surgery, during which, the surgeon opened the inguinal region and found a hernial sac with a small segment of intestine projecting through the abdominal wall. It was located just above the inguinal ligament and lateral to the inferior epigastric vessels. What type of hernia was this?
Your Answer: An indirect inguinal hernia
Explanation:An indirect inguinal hernia exits the abdominal cavity lateral to the inferior epigastric vessels and enters the inguinal canal through the deep inguinal ring. These are the most common types of hernias often caused by heavy weigh lifting. Direct inguinal hernias exit the abdominal cavity medial to the inferior epigastric vessels through weak fascia.
Congenital inguinal hernias are indirect hernias that occur due to persistence of the processus vaginalis.
Femoral hernias occur when abdominal viscera push through the femoral ring in the femoral canal.
Incisional hernia occurs after surgery when the omentum or organ protrudes through a previous site of incision.
Obturator hernia are a very rare type of hernia where the pelvic or abdominal contents protrude through the obturator foramen.
-
This question is part of the following fields:
- Abdomen
- Anatomy
-
-
Question 90
Correct
-
Which best describes the sartorius muscle?
Your Answer: Will flex the leg at the knee joint
Explanation:The sartorius muscle arises from tendinous fibres from the superior iliac spine. It passes obliquely across the thigh from lateral to medial and is inserted into the upper part of the medial side of the tibia. When the sartorius muscle contracts it will flex the leg at the knee joint.
-
This question is part of the following fields:
- Anatomy
- Lower Limb
-
-
Question 91
Correct
-
A 45 year old women has a 4 cm non tender mass in her right breast which is fixed to the chest wall. Another 2cm non-tender mass is palpable in the left axilla. Chest x ray reveals multiple nodules ranging from 0.5 - 0.2 cm in both the lungs. What is the stage of her disease?
Your Answer: T4 N1 M1
Explanation:This woman has an invasive primary tumour mass with axillary node and lung metastases, making this stage T4 N1 M1. Looking at the other stems, T1 N1 M0 signifies a small primary cancer with nodal metastases but no distant metastases; T1 N0 M1 signifies a small primary cancer with no lymph node metastases but with distant metastases; T2 N1 M0 signifies a larger primary cancer with nodal metastases but no distant metastases; and T3 N0 M0 indicates a larger primary cancer with no metastases to either lymph nodes or to distant sites.
-
This question is part of the following fields:
- Neoplasia; Female Health
- Pathology
-
-
Question 92
Correct
-
A 41 year old women presents with a history of carcinoma involving the right breast with enlarged axillary nodes on the same side. She underwent mastectomy and axillary node clearance. These were sent for histopathological examination. They showed no signs of metastasis. What could be cause of this enlargement in the lymph nodes?
Your Answer: Sinus histiocytosis
Explanation:Sinus histiocytosis also referred to as reticular hyperplasia, refers to the enlargement, distention and prominence of the sinusoids of the lymph nodes. This is a non-specific form of hyperplasia characteristically seen in lymph nodes that drain tumours. The endothelial lining of the lymph node becomes markedly hypertrophied, along with an increase in the number of macrophages which results in the distortion, distention and enlargement of the sinus. In this scenario there is no evidence that an infection or another malignancy could account for the enlargement. Paracortical lymphoid hyperplasia is caused by an immune response.
-
This question is part of the following fields:
- Inflammation & Immunology; Female Health
- Pathology
-
-
Question 93
Incorrect
-
In a neurological exam on a robbery with violence victim, it was discovered that the victim had lost sense of touch to the skin over her cheek and chin (maxilla and mandible region). Where are the cell bodies of the nerve that is responsible for touch sensations of this region located?
Your Answer: Cranial nerve VII ganglion
Correct Answer: Cranial nerve V ganglion
Explanation:The skin over the cheek and the maxilla are innervated by the trigeminal nerve (CN V). The trigeminal nerve has three major branches and it is the largest cranial nerve. The three branches of the trigeminal nerve are; the ophthalmic nerve, the maxillary nerve and the mandibular nerve. The trigeminal nerves ganglion is a sensory nerve ganglion know as the trigeminal ganglion (also referred to as the Gasser’s ganglion or the semilunar ganglion). It is contained in the dura matter in a cavity known as the Meckel’s cave, which covers the trigeminal impression near the apex of the petrous part of the temporal bone.
-
This question is part of the following fields:
- Anatomy
- Head & Neck
-
-
Question 94
Correct
-
Purkinje fibres in the heart conduct action potentials at the rate of:
Your Answer: 1.5–4.0 m/s
Explanation:Purkinje fibres control the heart rate along with the sinoatrial node (SA node) and the atrioventricular node (AV node). The QRS complex is associated with the impulse passing through the Purkinje fibres. These fibres conduct action potential about six times faster than the velocity in normal cardiac muscle.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 95
Correct
-
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: 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.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 96
Correct
-
Question 97
Correct
-
Multiple cells were labelled using a fluorescent dye that doesn’t cross the cell membrane. One cell in the middle was bleached with a light that destroys the dye, but the cell soon recovers its stain. The presence of which structures best explains this?
Your Answer: Gap junctions
Explanation:Gap junctions are attachments between cells that permit intercellular communication e.g. they permit current flow and electrical coupling between myocardial cells. They allow direct electrical transmission among cells and also permit certain substance to pass through as well. They are either homotypic, formed by two identical hemichannels or heterotypic, formed by different hemichannels.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 98
Correct
-
After surgery, a patient developed a stitch granuloma . Which leukocyte in the peripheral blood will become an activated macrophage in this granuloma?
Your Answer: Monocyte
Explanation:Monocytes are leukocytes that protect the body against infections and move to the site of infection within 8-12 hours to deal with it. They are produced in the bone marrow and shortly after being produced are released into the blood stream where they circulate until an infection is detected. When called upon they leave the circulation and transform into macrophages within the tissue fluid and thus gain the capability to phagocytose the offending substance. Monocyte count is part of a complete blood picture. Monocytosis is the state of excess monocytes in the peripheral blood and may be indicative of various disease states. Examples of processes that can increase a monocyte count include: • chronic inflammation • stress response • hyperadrenocorticism • immune-mediated disease • pyogranulomatous disease • necrosis • red cell regeneration.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 99
Correct
-
What is the normal duration of PR interval on an electrocardiogram of a healthy individual?
Your Answer: 0.12–0.20 s
Explanation:PR interval extends from the beginning of the P-wave until the beginning of the QRS complex. The normal duration of the PR interval is 0.12-0.20 s. It can be prolonged in first degree heart block, and reduced in Wolff-Parkinson-White syndrome.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 100
Incorrect
-
Which part of the nephron would have to be damaged to stop the reabsorption of the majority of salt and water?
Your Answer:
Correct Answer: Proximal tubule
Explanation:The proximal tubule is the portion of the duct system of the nephron of the kidney which leads from Bowman’s capsule to the loop of Henle. It is conventionally divided into the proximal convoluted tubule (PCT) and the proximal straight tubule (PST). The proximal tubule reabsorbs the majority (about two-thirds) of filtered salt and water. This is done in an essentially iso-osmotic manner. Both the luminal salt concentration and the luminal osmolality remain constant (and equal to plasma values) along the entire length of the proximal tubule. Water and salt are reabsorbed proportionally because the water is dependent on and coupled with the active reabsorption of Na+. The water permeability of the proximal tubule is high and therefore a significant transepithelial osmotic gradient is not possible.
-
This question is part of the following fields:
- Physiology
- Renal
-
SESSION STATS - PERFORMANCE PER SPECIALTY
