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Question 1
Incorrect
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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: 0.5
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.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 2
Incorrect
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The fundus of the stomach receives blood supply from the same artery as the greater curvature of the stomach. Which of the following arteries when ligated will disrupt blood supply to the fundus of the stomach through this artery?
Your Answer: Left gastroepiploic
Correct Answer: Splenic
Explanation:The fundus of the stomach along with the greater curvature of the stomach receive blood supply from the short gastric artery. The short gastric artery arises from the end of the splenic artery. The ligation of the splenic artery therefore would cause a disruption of blood supply to the fundus of the stomach.
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This question is part of the following fields:
- Abdomen
- Anatomy
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Question 3
Correct
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Which of the following is likely to result in hematocolpos in a 12-year old girl?
Your Answer: Imperforate hymen
Explanation:Hematocolpos means accumulation of blood in vagina and hematometra is accumulation of blood in the uterus. These are most likely seen with an imperforate hymen; which is seen I 1 in 2000 females. If spontaneous resolution does not occur, treatment involves making a hole in the hymen to allow discharge of menstrual blood.
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This question is part of the following fields:
- Pathology
- Women's Health
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Question 4
Correct
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A 42-year old woman presents to the doctor with jaundice. Her investigations show conjugated hyperbilirubinemia, raised urine bilirubin levels and low urine urobilinogen levels. What is the likely cause of her jaundice?
Your Answer: Blockage of the common bile duct
Explanation:The description of the patient here fits the diagnosis of obstructive jaundice or cholestasis, which results in conjugated hyperbilirubinemia. Cholestasis occurs due to impairment of bile flow, which can be anywhere from the liver cell canaliculus to the ampulla of Vater. Causes can be divided into intrahepatic and extrahepatic.
– Intrahepatic causes include hepatitis, drug toxicity, alcoholic liver disease, primary biliary cirrhosis, cholestasis of pregnancy and metastatic cancer.
– Extrahepatic causes include common duct stone, pancreatic cancer, benign stricture of the common duct, ductal carcinoma, pancreatitis and sclerosing cholangitis.
There is absence of bile constituents in the intestine, which causes spillage in the systemic circulation. Symptoms include pale stools, dark urine, pruritus, malabsorption leading to steatorrhea and deficiency of fat-soluble vitamins. Chronic cases can result in osteoporosis or osteomalacia due to vitamin D deficiency and Ca2+ malabsorption. Cholesterol and phospholipid retention produces hyperlipidaemia despite fat malabsorption (although increased liver synthesis and decreased plasma esterification of cholesterol also contribute); triglyceride levels are largely unaffected. The lipids circulate as a unique, low-density lipoprotein called lipoprotein X.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 5
Correct
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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.
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This question is part of the following fields:
- General
- Physiology
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Question 6
Incorrect
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A 35-year-old ultra marathon runner becomes severely dehydrated and collapses. This patient most likely has:
Your Answer: Low plasma ADH levels
Correct 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.
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This question is part of the following fields:
- Physiology
- Renal
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Question 7
Correct
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Difficulty in retracting the foreskin of the penis in an uncircumcised male is known as:
Your Answer: Phimosis
Explanation:Phimosis is the inability to fully retract the foreskin of the penis in an uncircumcised male. It can be physiological in infancy, in which it could be referred to as ‘developmental non-retractility of the foreskin. However, it is almost always pathological in older children and men. Causes include chronic inflammation (e.g. balanoposthitis), multiple catheterisations, or forceful foreskin retraction. One of the causes is chronic balanitis xerotica obliterans. It leads to development of a ring of indurated tissue near the tip of the prepuce, which prevents retraction. Contributory factors include infections, hormonal and inflammatory factors. The recommended treatment includes circumcision.
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This question is part of the following fields:
- Pathology
- Urology
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Question 8
Incorrect
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What Is the mechanism behind rhesus incompatibility in a new born baby?
Your Answer: Type IV hypersensitivity
Correct Answer: Type II hypersensitivity
Explanation:In type II hypersensitivity the antibodies that are produced by the immune response bind to the patients own cell surface antigens. These antigens can be intrinsic or extrinsic. Destruction occurs due to antibody dependent cell mediated antibodies. Antibodies bind to the cell and opsonise the cell, activating phagocytes to destroy that cell e.g. autoimmune haemolytic anaemia, Goodpasture syndrome, erythroblastosis fetalis, pernicious anaemia, Graves’ disease, Myasthenia gravis and haemolytic disease of the new-born.
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This question is part of the following fields:
- Inflammation & Immunology; Haematology
- Pathology
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Question 9
Incorrect
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Anthrax is an infection caused by the bacterium Bacillus anthracis. Anthrax spores have been used as a biological warfare weapon. What is the drug of choice in treating anthrax infection?
Your Answer: Tetracycline
Correct Answer: Ciprofloxacin
Explanation:Early antibiotic treatment of anthrax is essential. A delay may significantly lessen the chances for survival of the patient. Treatment for anthrax infection include large doses of intravenous and oral antibiotics, such as fluoroquinolones (ciprofloxacin), doxycycline, erythromycin, vancomycin, or penicillin.
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This question is part of the following fields:
- Microbiology
- Pathology
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Question 10
Correct
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What forms the pelvic diaphragm?
Your Answer: Levator ani and coccygeus muscles
Explanation:The pelvic diaphragm is formed by the levator ani and the coccygeus muscles. The levator ani forms the greater part of the pelvic floor supporting the viscera in the pelvic cavity.
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This question is part of the following fields:
- Anatomy
- Pelvis
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Question 11
Incorrect
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Which of the following is not seen as a complication of wound healing?
Your Answer: Proud flesh
Correct Answer: Malignancy
Explanation:Complications in wound healing can originate due to abnormalities in the repair process. These abnormalities are :
1) Deficient scar formation: insufficient granulation tissue can lead to wound dehiscence and ulceration. Dehiscence or wound rupture is seen most commonly in abdominal surgery due to increased intraabdominal pressure. Ulcerations are common in extremity wounds due to inadequate vascularization.
2) Excessive formation of repair components: collagen being laid down may begin normally however later lead to a raised scar also called a hypertrophic scar, which can extend beyond its boundaries to form a keloid and
3) Formation of contractures.
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This question is part of the following fields:
- Cell Injury & Wound Healing
- Pathology
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Question 12
Incorrect
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A 53 year old women with a history of atrial fibrillation developed an acute abdomen. On laparoscopic examination her bowels appeared to be dusky to red-purple in colour and her mesenteric veins appeared to be patent. Which of the following is most likely to occur in this situation?
Your Answer: Coagulative necrosis
Correct Answer: Wet gangrene
Explanation:Infarction of the small bowel following a sudden and complete occlusion of the mesenteric artery can involve any portion of the bowel, whether small or a large. The splenic flexure is at most risk for infarction as it is the watershed area between the superior and inferior mesenteric vessels. Regardless of whether the arterial or the venous blood vessels are occluded, because of the blood reflow into the damaged portion, it will appear haemorrhagic. The bowel appearing congested at first and then becoming oedematous. If the artery is occluded then there will be a clear cut demarcation and in venous occlusion the dusky colour fades with the rest of the normal bowel. Wet gangrene is characteristic of ischaemic injury to the gut.
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This question is part of the following fields:
- Cell Injury & Wound Healing; Gastrointestinal
- Pathology
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Question 13
Correct
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In which of the following situations will fat necrosis occur?
Your Answer: Trauma to the breast
Explanation:Fat necrosis is necrosis of adipose tissue with subsequent deposition of calcium, giving it a white chalky appearance. It is seen characteristically in trauma to the breast and the pancreas with subsequent involvement of the peripancreatic fat. In the breast it may present as a palpable mass with is usually painless or as an incidental finding on mammogram. Fatty acids are released from the traumatic tissue which combine with calcium in a process known as saponification, this is an example of dystrophic calcification in which calcium binds to dead tissue. The central focus is surrounded by macrophages and neutrophils initially, followed by proliferation of fibroblasts, neovascularization and lymphocytic migration to the site of the insult.
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This question is part of the following fields:
- Cell Injury & Wound Healing
- Pathology
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Question 14
Correct
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An 8 year old boy presents with a history of repeated bacterial/pyogenic infections. He had a normal recovery from chickenpox and measles and shows normal antibody response. A decrease in which of the cell types can best explain this history of repeated pyogenic infections?
Your Answer: Neutrophils
Explanation:A decrease in the number of granulocytes, particularly neutrophils is known as agranulocytosis and it increases the susceptibility of an individual towards recurrent infections. Neutropenia can be either due to decreased production or increased elimination of neutrophils.
Ineffective agranulopoiesis is seen in: 1. myeloid stem cell suppression, 2. disease conditions associated with granulopoiesis such as megaloblastic anaemia and myelodysplastic syndromes, 3. rare genetic diseases, 4. splenic sequestration and 5. increased peripheral utilization.
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This question is part of the following fields:
- Inflammation & Immunology
- Pathology
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Question 15
Correct
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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.
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This question is part of the following fields:
- Anatomy
- Upper Limb
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Question 16
Incorrect
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Thalamic syndrome will most likely result in:
Your Answer: Increased temperature
Correct Answer: Hyperaesthesia
Explanation:Signs and symptoms of thalamic syndrome include contralateral hemi anaesthesia, burning or aching sensation in one half of a body (hyperaesthesia), often accompanied by mood swings.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 17
Incorrect
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Which of the following factors will not affect the wound healing process in a young women who suffered serious burns to her chest and hands?
Your Answer: Steroids
Correct Answer: Vitamin A deficiency
Explanation:Healing can be sped-up or slowed down due to various reasons: 1. blood supply, 2. infection, 3. denervation, 4. collection of blood/hematoma, 5. mechanical stress, 6. foreign body, 7. techniques used during surgery and 8. dressing of the wound. Other systemic factors include 1. nutrition e.g. deficiency of zinc, vitamin C, protein deficiency, 2. metabolic status, 3. circulatory status and 4. hormonal influence
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This question is part of the following fields:
- Cell Injury & Wound Healing
- Pathology
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Question 18
Incorrect
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Which of the following is a likely cause of jaundice?
Your Answer: Hepatic disease if plasma acid phosphatase level is raised
Correct Answer: Hepatic disease if plasma albumin is low and serum aminotransferase elevations > 500 units
Explanation:Jaundice can occur due to any of the possible causes and treatment depends upon diagnosing the correct condition. Mild hyperbilirubinemia with normal levels of aminotransferase and alkaline phosphatase is often unconjugated (e.g., due to haemolysis or Gilbert’s syndrome rather than hepatobiliary disease). Moderate or severe hyperbilirubinemia along with increased urinary bilirubin (bilirubinuria), high alkaline phosphatase or aminotransferase levels suggest hepatobiliary disease. Hyperbilirubinemia produced by any hepatobiliary disease is largely conjugated. In this case, other blood tests include hepatitis serology for suspected hepatitis, prothrombin time (PT) or international normalised ratio (INR), albumin and globulin levels, and antimitochondrial antibody levels (suspected primary biliary cirrhosis). Low albumin and high globulin levels suggest chronic rather than acute liver disease. In cases where there is only a an elevation of alkaline phosphatase, γ-glutamyl transpeptidase (GGT) levels should be checked – the levels of which will be found high in hepatobiliary disease, but not in bone disorder which can also lead to elevated alkaline phosphatase levels. In diseases of hepatobiliary origin, aminotransferase elevations > 500 units suggest a hepatocellular cause, whereas disproportionate increases of alkaline phosphatase (e.g., alkaline phosphatase > 3 times normal and aminotransferase < 200 units) suggest cholestasis. Because hepatobiliary disease alone rarely causes bilirubin levels > 30 mg/dl, higher levels are suggestive of a combination of severe hepatobiliary disease and haemolysis or renal dysfunction. Imaging is best for diagnosing infiltrative and cholestatic causes of jaundice. Liver biopsy is rarely needed, but can be of use in intrahepatic cholestasis and in some types of hepatitis.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 19
Correct
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Choose the most correct answer regarding the obturator internus muscle.
Your Answer: It emerges from the pelvis through the lesser sciatic foramen
Explanation:The obturator internus arises from the inner surface of the anterolateral wall of the pelvis and the pelvic surface of the obturator membrane. The fibres converge rapidly towards the lesser sciatic foramen and end in four or five tendinous bands and leave the pelvis through the lesser sciatic foramen.
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This question is part of the following fields:
- Anatomy
- Lower Limb
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Question 20
Correct
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What is the primary function of the Kupffer cells found in the liver?
Your Answer: Recycling of old red blood cells
Explanation:Kupffer cells found in the liver are part of the monocyte-reticular system. They are specialised macrophages and primarily function to recycle old and damaged RBCs. The RBCs are phagocytosed and the haemoglobin is broken down into haem and globin. The haem is further broken down into iron that is recycled and bilirubin that is conjugated with glucuronic acid and excreted in the bile.
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This question is part of the following fields:
- Inflammation & Immunology; Hepatobiliary
- Pathology
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SESSION STATS - PERFORMANCE PER SPECIALTY
