-
Question 1
Correct
-
A 56 year old gentleman, who is a chronic smoker presents to the clinic with dyspnoea. His Chest X-ray shows intercostal space widening with increased blackening bilaterally. What is the most likely finding on his pulmonary function test?
Your Answer: Increased total lung capacity
Explanation:In patients suspected of having chronic obstructive pulmonary disease, pulmonary function testing (PFT) is useful to confirm airway obstruction, to quantify its severity and reversibility, for following disease progression and monitoring response to treatment. These tests include:
FEV1 – volume of air forcefully expired during the first second after a full breath
Forced vital capacity (FVC) – total volume of air expired with maximal force and flow-volume loops. The hallmark of airway obstruction is reduction of FEV1, FVC and the ratio of FEV1/FVC, with a concave pattern in expiratory tracing on the flow-volume loop. FEV1 and forced vital capacity (FVC) are easily measured with office spirometry and are useful to assess the severity of disease. Other parameters include increased total lung capacity, functional residual capacity and residual volume, which can help distinguish chronic obstructive pulmonary disease (COPD) from restrictive pulmonary disease where these values are lower than normal. Along with these, other tests are decreased vital capacity; and decreased diffusing capacity for carbon monoxide (DLCO). DLCO is non-specific and can be low in other disorders that affect the pulmonary vascular bed, such as interstitial lung disease. DLCO is however useful to distinguish COPD from asthma, in which DLCO is normal or elevated.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 2
Correct
-
A 78-year-old diabetic man undergoes renal function tests. Which of the following substances will be the most accurate for measuring glomerular filtration rate (GFR)?
Your Answer: Inulin
Explanation:Glomerular filtration rate (GFR) is the volume of fluid filtered from the renal glomerular capillaries into the Bowman’s capsule per unit time. Clinically, this is often measured to determine renal function. Inulin was originally used as it is not reabsorbed by the kidney after glomerular filtration, therefore its rate of excretion is directly proportional to the rate of filtration of water and solutes across the glomerular filter. However, in clinical practice, creatinine clearance is used to measure GFR. Creatinine is an endogenous molecule, synthesised in the body, that is freely filtered by the glomerulus (but also secreted by the renal tubules in very small amounts). Creatinine clearance exceeds GFR due to creatinine secretion, and is therefore a close approximation of the GFR.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 3
Incorrect
-
Extracellular body fluid as compared with intracellular body fluid:
Your Answer: Has lower tonicity
Correct Answer: Is relatively rich in glucose
Explanation:The percentages of body water contained in various fluid compartments add up to total body water (TBW). This water makes up a significant fraction of the human body, both by weight and by volume. Ensuring the right amount of body water is part of fluid balance, an aspect of homeostasis. The extracellular fluid (ECF) includes all fluids outside the cells. This fluid can be divided into three fluid departments: interstitial (in the tissue spaces) fluid, blood plasma and lymph, and specialised compartments called transcellular fluid. The extracellular fluid surrounds all the cells in the body and is in equilibrium with the intracellular fluid. So, its composition must remain fairly constant even though substances are passing into and out of the cells. The interstitial fluid, though called a fluid, is in a reality a gel-like composition made up of: water, proteoglycan molecules and collagen. The extracellular fluid constitutes 40% of total body water, with intracellular fluid making up the remaining 60%. It is relatively rich in glucose.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 4
Correct
-
How are amino acids transported across the luminal surface of the small intestinal epithelium?
Your Answer: Co-transport with sodium ions
Explanation:Once complex peptides are broken down into amino acids by the peptidases present in the brush border of small intestine, they are ready for absorption by at least four sodium-dependent amino acid co-transporters – one each for acidic, basic, neutral and amino acids, present on the luminal plasma membrane. These transporters first bind sodium and can then bind the amino acids. Thus, amino acid absorption is totally dependent on the electrochemical gradient of sodium across the epithelium. The basolateral membrane in contrast, possesses additional transporters to carry amino acids from the cell into the blood, but these are sodium-independent.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 5
Incorrect
-
Which organ is responsible for the secretion of enzymes that aid in digestion of complex starches?
Your Answer: Stomach
Correct Answer: Pancreas
Explanation:α-amylase is secreted by the pancreas, which is responsible for hydrolysis of starch, glycogen and other carbohydrates into simpler compounds.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 6
Incorrect
-
A 25 year old man presented with a history of headache and peripheral cyanosis. He had been living in the Himalayas for 6 months prior to this. What is the reason for his condition?
Your Answer: Methemoglobinemia
Correct Answer: Physiological polycythaemia
Explanation:Polycythaemia is a condition that results in an increase in the total number of red blood cells (RBCs) in the blood. It can be due to myeloproliferative syndrome or due to chronically low oxygen levels or rarely malignancy. In primary polycythaemia/polycythaemia vera the increase is due to an abnormality in the bone marrow, resulting in increases RBCs, white blood cells (WBCs) and platelets. In secondary polycythaemia the increase occurs due to high levels of erythropoietin either artificially or naturally. The increase is about 6-8 million/cm3 of blood. A type of secondary polycythaemia is physiological polycythaemia where people living in high altitudes who are exposed to hypoxic conditions produce more erythropoietin as a compensatory mechanism for thin oxygen and low oxygen partial pressure.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 7
Incorrect
-
Regarding the coagulation cascade, Factor VII:
Your Answer: Is activated by von Willebrand factor
Correct Answer: Is a serine protease
Explanation:Factor VII (FVII) is a zymogen for a vitamin K-dependent serine protease essential for the initiation of blood coagulation. It is synthesized primarily in the liver and circulates in plasma. Within the liver, hepatocytes are involved in the synthesis of most blood coagulation factors, such as fibrinogen, prothrombin, factor V, VII, IX, X, XI, XII, as well as protein C and S, and antithrombin, whereas liver sinusoidal endothelial cells produce factor VIII and von Willebrand factor.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 8
Correct
-
What is the normal glomerular filtration rate?
Your Answer: 125 mL/min
Explanation:The normal glomerular filtration rate (GFR) in humans is 125 mL/min. After the age of 40, GFR decreases progressively by about 0.4–1.2 mL/min per year.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 9
Correct
-
A medical student is asked to calculate the net pressure difference in a capillary wall, considering: Interstitial fluid hydrostatic pressure = –3 mmHg, Plasma colloid osmotic pressure = 28 mmHg, Capillary hydrostatic pressure = 17 mmHg, Interstitial fluid colloid osmotic pressure = 8 mmHg, and Filtration coefficient = 1. Which is the correct answer?
Your Answer: 0 mmHg
Explanation:The rate of filtration at any point along a capillary depends on a balance of forces sometimes called Starling’s forces after the physiologist who first described their operation in detail. The Starling principle of fluid exchange is key to understanding how plasma fluid (solvent) within the bloodstream (intravascular fluid) moves to the space outside the bloodstream (extravascular space). Fluid movement = k[(pc– pi)–(Πc– Πi)] where k = capillary filtration coefficient, pc = capillary hydrostatic pressure, pi= interstitial hydrostatic pressure, Πc = capillary colloid osmotic pressure, Πi = interstitial colloid osmotic pressure. Therefore: 1 × [capillary hydrostatic pressure (17) – interstitial fluid hydrostatic pressure (–3)] – [plasma colloid osmotic pressure (28) – interstitial fluid colloid osmotic pressure (8)] = 0 mmHg
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 10
Correct
-
Lung compliance is increased by:
Your Answer: Emphysema
Explanation:Lung compliance is increased by emphysema, acute asthma and increasing age and decreased by alveolar oedema, pulmonary hypertension, atelectasis and pulmonary fibrosis.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 11
Incorrect
-
Which of the following is NOT a nutritional factor involved in wound healing:
Your Answer: Copper
Correct Answer: Vitamin B3
Explanation:Vitamin B6 is required for collagen cross-links.
Vitamin A is required for epithelial cell proliferation.
Zinc is required for RNA and DNA synthesis.
Copper is required for cross-linking of collagen.
-
This question is part of the following fields:
- Cell Injury & Wound Healing
- Physiology
-
-
Question 12
Correct
-
A 30 year old female suffered from mismatched transfusion induced haemolysis. Which substance will be raised in the plasma of this patient?
Your Answer: Bilirubin
Explanation:Bilirubin is a yellow pigment that is formed due to the break down of RBCs. Haemolysis results in haemoglobin that is broken down into a haem portion and globin which is converted into amino acids and used again. Haem is converted into unconjugated bilirubin in the macrophages and shunted to the liver. In the liver it is conjugated with glucuronic acid making it water soluble and thus excreted in the urine. Its normal levels are from 0.2-1 mg/dl. Increased bilirubin causes jaundice and yellowish discoloration of the skin.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 13
Correct
-
The majority of gallstones are mainly composed of:
Your Answer: Cholesterol
Explanation:Bile salts are formed out of cholesterol in the liver cells. Occasionally, precipitation of cholesterol occurs resulting into cholesterol stones developing in the gall bladder.
These cholesterol gallstones are the most common type and account for 80% of all gallstones. Another type, accounting for 20% gallstones is pigment stones which are composed of bilirubin and calcium salts. Occasionally, stones of mixed origin are also seen.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 14
Correct
-
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.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 15
Correct
-
Question 16
Incorrect
-
A 59-year-old woman with hyperaldosteronism is prescribed a diuretic. Which of the following diuretics promotes diuresis by opposing the action of aldosterone?
Your Answer: Loop diuretic
Correct Answer: Potassium-sparing diuretic
Explanation:The term potassium-sparing refers to an effect rather than a mechanism or location. Potassium-sparing diuretics act by either antagonising the action of aldosterone (spironolactone) or inhibiting Na+ reabsorption in the distal tubules (amiloride). This group of drugs is often used as adjunctive therapy, in combination with other drugs, for the management of chronic heart failure. Spironolactone, the first member of the class, is also used in the management of hyperaldosteronism (including Conn’s syndrome) and female hirsutism (due to additional antiandrogen actions).
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 17
Correct
-
What is the most likely cause of prolonged bleeding time in a 40 year old women admitted for a laparoscopic cholecystectomy?
Your 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 18
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 19
Correct
-
The Henderson–Hasselbalch equation describes the derivation of pH as a measure of acidity. According to this equation, the buffering capacity of the system is at maximum when the number of free anions compared with undissociated acid is:
Your Answer: Equal
Explanation:In 1908, Lawrence Joseph Henderson wrote an equation describing the use of carbonic acid as a buffer solution. Later, Karl Albert Hasselbalch re-expressed that formula in logarithmic terms, resulting in the Henderson–Hasselbalch equation. The equation is also useful for estimating the pH of a buffer solution and finding the equilibrium pH in acid–base reactions. Two equivalent forms of the equation are: pH = pKa + log10 [A–]/[HA] or pH = pKa + log10 [base]/[acid]. Here, pKa is − log10(Ka) where Ka is the acid dissociation constant, that is: pKa = –log10(Ka) = –log10 ([H3 O+][A–]/[HA]) for the reaction: HA + H2 O ≈ A– + H3 O+ In these equations, A– denotes the ionic form of the relevant acid. Bracketed quantities such as [base] and [acid] denote the molar concentration of the quantity enclosed. Maximum buffering capacity is found when pH = pKa or when the number of free anions to undissociated acid is equal and buffer range is considered to be at a pH = pKa ± 1.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 20
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 21
Incorrect
-
A 15-day old male baby was brought to the emergency department with sweating and his lips turning blue while feeding. He was born full term. On examination, his temperature was 37.9°C, blood pressure 75/45 mmHg, pulse was 175/min, and respiratory rate was 42/min. A harsh systolic ejection murmur could be heard at the left upper sternal border. X-ray chest showed small, boot-shaped heart with decreased pulmonary vascular markings. He most likely has:
Your Answer: Atrial septal defect
Correct Answer: Tetralogy of Fallot
Explanation:The most common congenital cyanotic heart disease and the most common cause of blue baby syndrome, Tetralogy of Fallot shows four cardiac malformations occurring together. These are ventricular septal defect (VSD), pulmonary stenosis (right ventricular outflow obstruction), overriding aorta (degree of which is variable), and right ventricular hypertrophy. The primary determinant of severity of disease is the degree of pulmonary stenosis. Tetralogy of Fallot is seen in 3-6 per 10,000 births and is responsible for 5-7% congenital heart defects, with slightly higher incidence in males. It has also been associated with chromosome 22 deletions and DiGeorge syndrome. It gives rise to right-to-left shunt leading to poor oxygenation of blood. Primary symptom is low oxygen saturation in the blood with or without cyanosis at birth of within first year of life. Affected children ay develop acute severe cyanosis or ‘tet spells’ (sudden, marked increase in cyanosis, with syncope, and may result in hypoxic brain injury and death). Other symptoms include heart murmur, failure to gain weight, poor development, clubbing, dyspnoea on exertion and polycythaemia. Chest X-ray reveals characteristic coeur-en-sabot (boot-shaped) appearance of the heart. Treatment consists of immediate care for cyanotic spells and Blalock–Taussig shunt (BT shunt) followed by corrective surgery.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 22
Incorrect
-
A medical student is told a substance is freely filtered but is not metabolised, secreted, or stored in the kidney. It has a plasma concentration of 1000 mg/l and its urine excretion rate is 25 mg/min, and the inulin clearance is 100 ml/min. What is the rate of tubular reabsorption of the substance?
Your Answer: 250 mg/min
Correct Answer: 75 mg/min
Explanation:Reabsorption or tubular reabsorption is the process by which the nephron removes water and solutes from the tubular fluid (pre-urine) and returns them to the circulating blood. To calculate the reabsorption rate of substance Z we use the following equation: excretion = (filtration + secretion) – reabsorption. As this substance is freely filtered, its filtration rate is equal to that of inulin. So 25 = (100 + 0) – reabsorption. Reabsorption = 100 – 25 therefore reabsorption = 75 mg/min.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 23
Incorrect
-
A 26-year-old female patient had the following blood report: RBC count = 4. 0 × 106/μl, haematocrit = 27% and haemoglobin = 11 g/dl, mean corpuscular volume (MCV) = 90 fl, mean corpuscular haemoglobin concentration (MCHC) = 41 g/dl. Further examination of blood sample revealed increased osmotic fragility of the erythrocytes. Which of the following is the most likely cause of this patient’s findings?
Your Answer: Liver disease
Correct Answer: Spherocytosis
Explanation:Spherocytes are small rounded RBCs. It is due to an inherited defect of the RBC cytoskeleton membrane tethering proteins. Membrane blebs form that are lost over time and cells become round instead of biconcave. As it is a normochromic anaemia, the MCV is normal. it is diagnosed by osmotic fragility test which reveals increased fragility in a hypotonic solution.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 24
Correct
-
Which of the following causes the maximum increase in the secretion of antidiuretic hormone (ADH)?
Your Answer: Increased plasma osmolarity
Explanation:The most potent stimulus for ADH release is increased plasma osmolarity. Decreased plasma volume is a less potent stimulus in comparison. However, decrease blood volume and arterial pressure due to severe haemorrhage does lead to ADH secretion. Hypothalamic releasing factors do not control the release of posterior pituitary hormones ADH and oxytocin.
-
This question is part of the following fields:
- Endocrinology
- Physiology
-
-
Question 25
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 26
Incorrect
-
Which of the following physiological changes will you see in a young man who has been trekking in the Himalayas for 3 years?
Your Answer: Pulmonary vasodilatation
Correct Answer: Increased renal excretion of HCO3 –
Explanation:The atmospheric pressure is lower at high altitudes as compared with sea level. This leads to a decrease in the partial pressure of oxygen. Once 2100 m (7000 feet) of altitude is reached, there is a drop in saturation of oxyhaemoglobin. The oxygen saturation of haemoglobin determines the oxygen content in the blood. The body physiological tries to adapt to high altitude by acclimatization. Immediate effects include hyperventilation, fluid loss (due to a decreased thirst drive), increase in heart rate and slightly lowered stroke volume. Long term effects include lower lactate production, compensatory alkali loss in urine, decrease in plasma volume, increased erythropoietin release and red cell mass, increased haematocrit, higher concentration of capillaries in striated muscle tissue, increase in myoglobin, increase in mitochondria, increase in aerobic enzyme concentration such as 2,3-DPG and pulmonary vasoconstriction.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 27
Incorrect
-
Which of the following associations is correctly matched with the body's defence mechanism in fighting infection?
Your Answer: First line of defence → neutrophils
Correct Answer: Specific cellular mechanism → cytotoxic T cells
Explanation:The immune system has certain levels of defence against pathogens. First line includes simple barriers such as skin, mucosa and stomach acid that prevent the pathogen from entering into the body. If this barrier is breached then the innate immune system is activated which includes leukocytes (macrophages, neutrophils, mast cells, eosinophils, basophils, natural killer cells). If the pathogens invade the second layer of defence then the third layer, adaptive immunity is activated, which includes B and T lymphocytes. B cells provide a humoral response whereas cytotoxic T cells have specific cellular mechanisms. They maintain a memory of past infections and are activated faster following a recurrence.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 28
Correct
-
A 40-year old gentleman, known with a history of peptic ulcer disease, was brought to the clinic in a dehydrated state with persistent vomiting. His blood investigations revealed:
- sodium = 142 mmol/l
- potassium = 2.6 mmol/l
- chloride = 85 mmol/l
- pH = 7.55
- p(CO2) = 50 mmHg
- p(O2) = 107 mmHg
- standard bicarbonate = 40 mmol/l
Your Answer: Metabolic alkalosis
Explanation:High pH with high standard bicarbonate indicates metabolic alkalosis. The pa(CO2) was appropriately low in compensation. This is hypokalaemic hypochloraemic metabolic acidosis due to prolonged vomiting. Treatment includes treating the cause and intravenous sodium chloride with potassium.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 29
Correct
-
What is a major source of fuel being oxidised by the skeletal muscles of a man who has undergone starvation for 7 days?
Your Answer: Serum fatty acids
Explanation:Starvation is the most extreme form of malnutrition. Prolonged starvation can lead to permanent organ damage and can be fatal. Starved individuals eventually lose significant fat and muscle mass as the body uses these for energy.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 30
Incorrect
-
A patient admitted for esophagectomy showed low levels of the lightest plasma protein in terms of weight. Which of the following is the lightest plasma protein:
Your Answer: Gamma-globulin
Correct Answer: Albumin
Explanation:Albumin is the most abundant and the lightest of all the plasma proteins. It maintains osmotic pressure, transports unconjugated bilirubin, thyroid hormones, fatty acids, drugs and acts as a buffer for pH.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 31
Incorrect
-
Which of the following organs is most likely to have dendritic cells?
Your Answer: Spleen
Correct Answer: Skin
Explanation:Dendritic cells are part of the immune system and they function mainly as antigen presenting cells. They are present in small quantities in tissues which are in contact in the external environment. Mainly in the skin and to a lesser extent in the lining of the nose, lungs, stomach and intestines. In the skin they are known as Langerhans cells.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 32
Correct
-
A 23-year-old woman decides to donate a kidney through a kidney chain. Which of the following indices would be expected to be decreased in the donor after full recovery from the operation?
Your Answer: Creatinine clearance
Explanation:Since medication to prevent rejection is so effective, donors do not need to be similar to their recipient. Most donated kidneys come from deceased donors; however, the utilisation of living donors is on the rise. Most problems encountered with live donation are associated with the donor. Firstly, there are the potentially harmful investigative procedures carried out in the assessment phase, the most hazardous being renal angiography, where there is cannulation of the artery and injection of a radio-opaque dye to determine the blood supply to the kidney. Secondly, there are the short-term risks of nephrectomy surgery. According to the literature, there is a mortality rate of between 1 in 1600 and 1 in 3000, but this is no more than is associated with any anaesthetic. In the initial postoperative period creatinine clearance may be decreased but this recovers fully over a few weeks to months. Long-term complications include prolonged wound pain.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 33
Correct
-
What is the role of factor VII in coagulation?
Your Answer: Initiates the process of coagulation in conjunction with tissue factor
Explanation:The main role of factor VII is to initiate the process of coagulation along with tissue factor (TF). TF is found in the blood vessels and is not normally exposed to the bloodstream. When a vessel is injured tissue factor is exposed to blood and circulating factor VII. Factor VII is converted to VIIa by TF.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 34
Correct
-
The most likely cause of a low p(O2) in arterial blood is:
Your Answer: Hypoxic hypoxia
Explanation:Hypoxia is when the whole body or a region is deprived of adequate oxygen supply. Different types of hypoxia include the following:
– Hypoxic hypoxia, which occurs due to poor oxygen supply, as a result of low partial pressure of oxygen in arterial blood. This could be due to low partial pressure of atmospheric oxygen (e.g., at high altitude), sleep apnoea, poor ventilation because of chronic obstructive pulmonary disease or respiratory arrest, or shunts. The other types of hypoxia have a normal partial pressure of oxygen.
– Anaemic hypoxia occurs due to low total oxygen content of the blood, with a normal arterial oxygen pressure.
– Hyperaemic hypoxia occurs due to poor delivery of oxygen to target tissues, such as in carbon monoxide poisoning or methemoglobinemia.
– Histotoxic hypoxia results due to inability of the cells to use the delivered oxygen due to disabled oxidative phosphorylation enzymes.
– Ischaemic (or stagnant) hypoxia occurs due to local flow restriction of well-oxygenated blood, seen in cases like cerebral ischaemia, ischaemic heart disease and intrauterine hypoxia.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 35
Incorrect
-
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: Basal lamina
Correct 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 36
Correct
-
One sensitive indicator of heavy alcohol dependence is:
Your Answer: Elevated serum gamma-glutamyl transpeptidase
Explanation:Elevated serum gamma-glutamyl transpeptidase (GGT) may be the only laboratory abnormality in patients who are dependent on alcohol. Heavy drinkers may also have an increased MCV.
-
This question is part of the following fields:
- Hepatobiliary
- Physiology
-
-
Question 37
Correct
-
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: 0.2
Explanation:Flow in any vessel = Effective perfusion pressure divided by resistance, where effective perfusion pressure is the mean intraluminal pressure at the arterial end minus the mean pressure at the venous end. Thus, in the given problem, resistance = (60 − 20)/200 = 0.2 mmHg/ml per min.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 38
Incorrect
-
The blood-brain barrier is a membrane that separates the circulating blood from the brain extracellular fluid in the central nervous system (CNS). Which of the following statements regarding the blood– brain barrier is CORRECT?
Your Answer: It permits carbon dioxide to pass via facilitated diffusion
Correct Answer: It breaks down in areas of brain that are infected
Explanation:The blood–brain barrier is a membrane that controls the passage of substances from the blood into the central nervous system. It is a physical barrier between the local blood vessels and most parts of the central nervous system and stops many substances from travelling across it. During meningitis, the blood–brain barrier may be disrupted. This disruption may increase the penetration of various substances (including either toxins or antibiotics) into the brain. A few regions in the brain, including the circumventricular organs, do not have a blood–brain barrier.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 39
Correct
-
What is the normal amount of oxygen that is carried in the blood?
Your Answer: 20 ml oxygen/100 ml blood
Explanation:Normally, 100 ml of blood contains 15g haemoglobin and a single gram of haemoglobin can bind to 1.34 ml oxygen when 100% saturated. Thus, 15 × 1.34 = 20 ml O2/100 ml blood. The haemoglobin in venous blood that is leaving the tissues is about 75% saturated with oxygen, and hence it carries about 15 ml O2/100 ml venous blood. This implies that for each 10 ml of blood, 5 ml oxygen is transported to the tissues. With a p(O2) > 100 mm Hg, only 3 ml of oxygen is dissolved in every one litre of plasma. By increasing the pA(O2) by breathing 100% oxygen, one can add an extra amount of oxygen in the plasma, but the amount of oxygen carried by haemoglobin will not increase significantly as it is already > 95% saturated.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 40
Correct
-
Calculate the pulmonary vascular resistance in an adult male with the following parameters:
Heart rate 70 beats/min
Arterial [O2] 0.24 ml O2/min
Venous [O2] 0.16 ml O2/mi
Whole body O2 consumption 500 ml/min
Pulmonary diastolic pressure 15 mmHg
Pulmonary systolic pressure 25 mmHg
Wedge pressure 5 mmHg.Your Answer: 2.0 resistance units (mmHg/l per min)
Explanation:Pulmonary vascular resistance (PVR) = (Mean pulmonary artery pressure – Pulmonary capillary wedge pressure) divided by Cardiac output. To get cardiac output, Fick’s principle needs to be applied which states that VO2 = (CO × CAO2) – (CO × CVO2) where VO2 = oxygen consumption, CO = cardiac output, CA = oxygen concentration of arterial blood and CVO2 = oxygen concentration of venous blood. Thus, CO = VO2/CAO2– CVO2, CO = 500/0.24–0.16, CO = 500/0.8, CO = 6.25 l/min. To calculate mean pulmonary artery pressure, we use the formula: Mean pulmonary artery pressure = Diastolic pressure + 1/3(Systolic pressure – Diastolic pressure). Thus, Mean pulmonary artery pressure = 15 + 1/3(25 – 15) = 15 + 3. 33 = 18.33. Substituting these values in the first formula, PVR = 18.3–5/6.25 = 13.5/5.25 = 2.0 resistance units (mmHg/l per min) approximately.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 41
Correct
-
Basal Metabolic Rate (BMR) will most likely be reduced by which of the following?
Your Answer: Decrease in body temperature
Explanation:The basal metabolic rate (BMR) is defined as the rate of calorie consumption after an overnight fast, in the absence of any muscular activity, with the patient in a restful state. Various factors affect the BMR including weight, body surface area and age. The BMR is 30 kcal/m2 per hour at birth; at age 2, the rate is 57 kcal/m2 per hour; and at age 20, 41 kcal/m2 per hour. After this, the BMR decreases by 10% between 20-60 years of age. Women are known to have a 10% lower BMR than men (due to higher fat content). A one-degree change in body temperature leads to a 10% change in BMR in the same direction. However, shivering and increasing ambient temperature brings about a rise in BMR, and so does stress, physical activity, caffeine, theophylline and hyperthyroidism. Also, thermogenesis induced by diet results in increased metabolic rate and hence, BMR should be ideally measured after overnight fasting.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 42
Incorrect
-
Bile salt reuptake principally occurs where?
Your Answer: In the colon
Correct Answer: In the ileum
Explanation:90 – 95% of the bile salts are absorbed from the small intestine (mostly terminal ileum and then excreted again from the liver. This is known as the enterohepatic circulation. The entire pool recycles twice per meal and approximately 6 to 8 times per day.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 43
Correct
-
Calculate the stroke volume in an adult male with the following parameters:
Heart rate 70 beats/min
Arterial [O2] 0.24 ml O2/min
Venous [O2] 0.16 ml O2/mi
Whole body O2 consumption 500 ml/min
Pulmonary diastolic pressure 15 mmHg
Pulmonary systolic pressure 25 mmHg
Wedge pressure 5 mmHg.Your Answer: 90 ml
Explanation:Fick’s principle states that, VO2 = (CO × CAO2) – (CO × CVO2) where VO2 = oxygen consumption, CO = cardiac output, CAO2 = oxygen concentration of arterial blood and CVO2 = oxygen concentration of venous blood. Thus, CO = VO2/CAO2– CVO2, CO = 500/0.24–0.16, CO = 500/0.8, CO = 6.25 l/min. Cardiac output is also given by product of stroke volume and heart rate. Thus, stroke volume = cardiac output / heart rate = 6.25/70 × 1000 stroke volume = 90 ml approximately.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 44
Correct
-
Which organ is most vulnerable to haemorrhagic shock?
Your Answer: Kidneys
Explanation:At rest, the brain receives 15% cardiac output, muscles 15%, gastrointestinal tract 30% and kidneys receive 20%. However, if normalised by weight, the largest specific blood flow is received by the kidneys at rest (400 ml/min x 100g), making them highly vulnerable in the case of a haemorrhagic shock.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 45
Correct
-
Myoglobin is released as a result of rhabdomyolysis from damaged skeletal muscles. What function do they perform in the muscle?
Your Answer: Acts like haemoglobin and binds with O2
Explanation:Myoglobin is a pigmented globular protein made up of 153 amino acids with a prosthetic group containing haem around which the apoprotein folds. It is the primary oxygen carrying protein of the muscles. The binding of oxygen to myoglobin is unaffected by the oxygen pressure as it has an instant tendency to bind given its hyperbolic oxygen curve. It releases oxygen at very low pO2 levels.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 46
Correct
-
Which of the following conditions causes an elevation of the pH in the tissues with elevated arterial CO2 content?
Your Answer: Metabolic alkalosis
Explanation:Metabolic alkalosis is a metabolic condition in which the pH of tissue is elevated beyond the normal range (7.35-7.45). This is the result of decreased hydrogen ion concentration, leading to increased bicarbonate, or alternatively a direct result of increased bicarbonate concentrations. Normally, arterial pa(CO2) increases by 0.5–0.7 mmHg for every 1 mEq/l increase in plasma bicarbonate concentration, a compensatory response that is very quick. If the change in pa(CO2) is not within this range, then a mixed acid–base disturbance occurs.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 47
Incorrect
-
what is the cause of a prolonged PT(prothrombin time)?
Your Answer: von Willebrand factor deficiency
Correct Answer: Liver disease
Explanation:PT measure the intrinsic pathway of coagulation. It determines the measure of the warfarin dose regime, liver disease and vit K deficiency status along with the clotting tendency of blood. PT measured factors are II,V,VII,X and fibrinogen. It is used along with aPTT which measure the intrinsic pathway.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 48
Correct
-
Which of the following has the highest content of triglycerides?
Your Answer: Chylomicron
Explanation:Created by the small intestinal cells, chylomicrons are large lipoprotein molecules which transport lipids to the liver, adipose, cardiac and skeletal tissue. Chylomicrons are mainly composed of triglycerides (,85%) along with some cholesterol and cholesteryl esters. Apo B-48 is the main apolipoprotein content.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 49
Correct
-
A victim of road traffic accident presented to the emergency department with a blood pressure of 120/90 mm Hg, with a drop in systolic pressure to 100 mm Hg on inhalation. This is known as:
Your Answer: Pulsus paradoxus
Explanation:Weakening of pulse with inhalation and strengthening with exhalation is known as pulsus paradoxus. This represents an exaggeration of the normal variation of the pulse in relation to respiration. It indicates conditions such as cardiac tamponade and lung disease. The paradox refers to the auscultation of extra cardiac beats on inspiration, as compared to the pulse. Due to a decrease in blood pressure, the radial pulse becomes impalpable along with an increase in jugular venous pressure height (Kussmaul sign). Normal systolic blood pressure variation (with respiration) is considered to be >10 mmHg. It is >100 mmHg in Pulsus paradoxus. It is also predictive of the severity of cardiac tamponade.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 50
Correct
-
After having donated a unit of blood. The blood bank will prefer to use which of the following anticoagulants to store the blood?
Your Answer: Citrate
Explanation:Calcium is necessary for coagulation to occur. Citrate being a chelator and combining with calcium ions to form un-ionised compound will prevent coagulation. Following transfusion the citrate is removed by the liver with in a few minutes. Oxalate also works on the same principle but it is toxic to the body.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 51
Correct
-
A neonate with failure to pass meconium is being evaluated. His abdomen is distended and X-ray films of the abdomen show markedly dilated small bowel and colon loops. The likely diagnosis is:
Your Answer: Aganglionosis in the rectum
Explanation:Hirschsprung’s disease (also known as aganglionic megacolon) leads to colon enlargement due to bowel obstruction by an aganglionic section of bowel that starts at the anus. A blockage is created by a lack of ganglion cells needed for peristalsis that move the stool. 1 in 5000 children suffer from this disease, with boys affected four times more commonly than girls. It develops in the fetus in early stages of pregnancy. Symptoms include not having a first bowel movement (meconium) within 48 hours of birth, repeated vomiting and a swollen abdomen. Two-third of cases are diagnosed within 3 months of birth. Some children may present with delayed toilet training and some might not show symptoms till early childhood. Diagnosis is by barium enema and rectal biopsy (showing lack of ganglion cells).
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 52
Incorrect
-
C5a (a complement component) is a potent?
Your Answer: Cytokine
Correct Answer: Anaphylotoxin
Explanation:C5a is a strong chemoattractant as well as an anaphylotoxin and is involved in the recruitment of inflammatory cells such as neutrophils, eosinophils, monocytes, and T lymphocytes. It is also involved in activation of phagocytic cells, release of granule-based enzymes and generation of oxidants. All of which contribute to innate immune functions.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 53
Correct
-
Ventricular filling follows a delay caused by?
Your Answer: AV node
Explanation:The AV node is a conducting tissue found between the atria and the ventricles of the heart. It conducts electrical signal from the atria to the ventricles and acts a delaying mechanism preventing the atria and the ventricles from contracting at the same time. This decremental conduction prevents premature ventricular contraction in cases such as atrial fibrillation. A delay in the AV node is the reason for the PR segment seen on the ECG. In certain types of supraventricular tachycardia, a person could have two AV nodes; this will cause a loop in electrical current and uncontrollably rapid heart beat. When this electricity catches up with itself, it will dissipate and return to a normal heart rate.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
SESSION STATS - PERFORMANCE PER SPECIALTY
