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Question 1
Incorrect
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Atractyloside is an inhibitor of electron transport chain. It is expected to have little or no effect on the functioning of which of the following cell types?
Your Answer: Parietal cells of the stomach
Correct Answer: Red blood cells
Explanation:Electron transport chain is a series of electron carriers that are embedded in the mitochondrial membrane. It is the place where ATP is made. Inhibiting the electron transport chain will stop production of ATP. Red blood cells are the only cell in the given option which do not contain ATP.
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This question is part of the following fields:
- General
- Physiology
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Question 2
Correct
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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.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 3
Incorrect
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Glucose is not secreted by the kidneys, and is filtered without a limit. What is the transport maximum for glucose?
Your Answer: 100 mg/dl
Correct Answer: 300 mg/dl
Explanation:Transport maximum (or Tm) refers to the point at which increases in concentration do not result in an increase in movement of a substance across a membrane. Glucose is not secreted, thus excretion = filtration – reabsorption. Both filtration and reabsorption are directly proportional to the concentration of glucose in the plasma. However, reabsorption has a transport maximum of about 300 mg/dl in healthy nephrons, while filtration has effectively no limit (within reasonable physiological ranges). So, if the concentration rises above 300 mg/dl, the body cannot retain all the glucose, leading to glucosuria. Glucosuria is nearly always caused by elevated blood glucose levels, most commonly due to untreated diabetes mellitus.
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This question is part of the following fields:
- Physiology
- Renal
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Question 4
Incorrect
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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: 10 ml
Correct 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.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 5
Incorrect
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A 35-year-old woman in her 37th week of pregnancy complains of urinary incontinence. She is most likely to have:
Your Answer: Functional incontinence
Correct Answer: Stress incontinence
Explanation:Urinary incontinence is the involuntary excretion of urine from one’s body. It is often temporary and it almost always results from an underlying medical condition. Several types include:
– Stress incontinence is the voiding of urine following increased abdominal pressure e.g. laughing, coughing, pregnancy etc. It is the most common form of incontinence in women, most commonly due to pelvic floor muscle weakness, physical changes from pregnancy, childbirth and menopause. In men it is a common problem following a prostatectomy. Most lab results such as urine analysis, cystometry and postvoid residual volume are normal.
– Urge incontinence is involuntary loss of urine occurring for no apparent reason while suddenly feeling the need or urge to urinate. The most common cause of urge incontinence are involuntary and inappropriate detrusor muscle contractions.
– Functional incontinence – occurs when a person does not recognise the need to go to the toilet, recognise where the toilet is or get to the toilet in time. The urine loss may be large. Causes of functional incontinence include confusion, dementia, poor eyesight, poor mobility, poor dexterity or unwillingness. t
– Overflow incontinence – sometimes people find that they cannot stop their bladders from constantly dribbling or continuing to dribble for some time after they have passed urine.
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This question is part of the following fields:
- Physiology
- Renal
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Question 6
Incorrect
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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:
Correct 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).
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 7
Incorrect
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A 40 year old man suffered severe trauma following an MVA. His BP is 70/33 mmhg, heart rate of 140 beats/mins and very feeble pulse. He was transfused 3 units of blood resulting in his BP returning to 100/70 and his heart rate to 90 beats/min. What decreased following transfusion?
Your Answer:
Correct Answer: Total peripheral resistance
Explanation:The patient is in hypovolemic shock, he is transfused with blood to replace the volume lost. It is important not only to replace fluids but stop active bleeding in resuscitation. Fluid replacement will result in a decreased sympathetic discharge and adequate ventricular filling thus reducing total peripheral resistance and increasing cardiac output and cardiac filling pressures.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 8
Incorrect
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What is the most likely cause of prolonged bleeding time in a 40 year old women admitted for a laparoscopic cholecystectomy?
Your Answer:
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.
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This question is part of the following fields:
- General
- Physiology
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Question 9
Incorrect
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A 30 year old female presented in the emergency with an irregular pulse. Her ECG showed absent P-waves with irregular RR interval. What is the most likely diagnosis?
Your Answer:
Correct Answer: Atrial fibrillation
Explanation:Atrial fibrillation is one of the most common cardiac arrhythmias. It is often asymptomatic but may present with symptoms of palpitations, fainting, chest pain and heart failure. Characteristic findings are: absence of P-waves, unorganised electrical activity in their place, irregularity of RR interval due to irregular conduction of impulses to the ventricles and if paroxysmal AF is suspected, episodes may be documented with the use of Holter monitoring
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 10
Incorrect
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What is the role of factor VII in coagulation?
Your Answer:
Correct 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.
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This question is part of the following fields:
- General
- Physiology
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Question 11
Incorrect
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Which of the following will be a likely sequelae of complete ileal resection?
Your Answer:
Correct Answer: Vitamin B12 deficiency
Explanation:The ileum is a part of the small intestine and has a pH of around 7-8 (neutral or slightly alkaline). Its main function is absorption of products of digestion. The ileal wall has multiple villi, which in turn have numerous microvilli. This increases the surface area available for absorption significantly. The cells lining the ileum contain multiple enzymes such as protease and carbohydrase, which aid in the final stages of digestion. Villi contain lacteals which absorb the products of fat digestion, fatty acids and glycerol. Thus, ileal resection will lead to their decreased absorption and increased fat content in the stool. The ileum is also responsible for absorption of vitamin B12.
Maximum water absorption occurs in the colon followed by the jejunum. Hence, ileal resection is less likely to lead to fluid volume deficiency. Also, most minerals (like calcium, iron etc.) are absorbed in the duodenum, and thus will not be affected by ileal resection.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 12
Incorrect
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Rapid eye movement (REM) sleep is likely to be affected by a lesion in the:
Your Answer:
Correct Answer: Pons
Explanation:Rapid eye movement (REM) sleep is also known as paradoxical sleep, as the summed activity of the brain’s neurons is quite similar to that during waking hours. Characterised by rapid movements of the eyes, most of the vividly recalled dreams occur during this stage of sleep. The total time of REM sleep for an adult is about 90–120 min per night.
Certain neurones in the brainstem, known as REM sleep-on cells, which are located in the pontine tegmentum, are particularly active during REM sleep and are probably responsible for its occurrence. The eye movements associated with REM are generated by the pontine nucleus with projections to the superior colliculus and are associated with PGO (pons, geniculate, occipital) waves.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 13
Incorrect
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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.
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This question is part of the following fields:
- Physiology
- Renal
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Question 14
Incorrect
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Renin is secreted by pericytes in the vicinity of the afferent arterioles of the kidney from the juxtaglomerular cells. Plasma renin levels are decreased in patients with:
Your Answer:
Correct Answer: Primary aldosteronism
Explanation:Primary aldosteronism, also known as primary hyperaldosteronism or Conn’s syndrome, is excess production of the hormone aldosterone by the adrenal glands resulting in low renin levels. Most patients with primary aldosteronism (Conn’s syndrome) have an adrenal adenoma. The increased plasma aldosterone concentration leads to increased renal Na+ reabsorption, which results in plasma volume expansion. The increase in plasma volume suppresses renin release from the juxtaglomerular apparatus and these patients usually have low plasma renin levels. Salt restriction and upright posture decrease renal perfusion pressure and therefore increases renin release from the juxtaglomerular apparatus. Secondary aldosteronism is due to elevated renin levels and may be caused by heart failure or renal artery stenosis.
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This question is part of the following fields:
- Physiology
- Renal
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Question 15
Incorrect
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Which of these conditions causes haematuria, hypertension and proteinuria in children, usually after a streptococcal infection?
Your Answer:
Correct Answer: Acute nephritic syndrome
Explanation:Nephritic syndrome (or acute nephritic syndrome) is a syndrome comprising of signs of nephritis. Children between 2 and 12 are most commonly affected, but it may occur at any age. Predisposing factors/causes include:
Infections with group A streptococcal bacteria (acute post-streptococcal glomerulonephritis).
Primary renal diseases: immunoglobulin A nephropathy, membranoproliferative glomerulonephritis, idiopathic rapidly progressive crescentic glomerulonephritis.
Secondary renal diseases: subacute bacterial endocarditis, infected ventriculo–peritoneal shunt, glomerulonephritis with visceral abscess, glomerulonephritis with bacterial, viral or parasitic infections.
Multisystem diseases.
By contrast, nephrotic syndrome is characterized by only proteins moving into the urine.
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This question is part of the following fields:
- Physiology
- Renal
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Question 16
Incorrect
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A 30 year old man suffered severe blood loss, approx. 20-30% of his blood volume. What changes are most likely seen in the pulmonary vascular resistance (PVR) and pulmonary artery pressure (PAP) respectively following this decrease in cardiac output?
Your Answer:
Correct Answer: Increase Decrease
Explanation:Hypovolemia will result in the activation of the sympathetic adrenal discharge resulting is a decrease pulmonary artery pressure and an elevated pulmonary vascular resistance.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 17
Incorrect
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Diffusion is the movement of molecules from a region of high concentration to a region of low concentration. Which of these changes will decrease the rate of diffusion of a substance?
Your Answer:
Correct Answer: An increase in the molecular weight of the substance
Explanation:Unless given IV, a drug must cross several semipermeable cell membranes before it reaches the systemic circulation. Drugs may cross cell membranes by diffusion, amongst other mechanisms. The rate of diffusion of a substance is proportional to the difference in the concentration of the diffusing substance between the two sides of the membrane, the temperature of the solution, the permeability of the membrane and, in the case of ions, the electrical potential difference between the two sides of the membrane.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 18
Incorrect
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Myoglobin is released as a result of rhabdomyolysis from damaged skeletal muscles. What function do they perform in the muscle?
Your Answer:
Correct 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.
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This question is part of the following fields:
- General
- Physiology
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Question 19
Incorrect
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In which situation is a stretch reflex such as knee jerk likely to be exaggerated?
Your Answer:
Correct Answer: In upper motor neuron lesion
Explanation:A stretch reflex is a monosynaptic reflex that causes muscle contraction in response to stretching within that muscle. The sensory apparatus in a muscle that are sensitive to stretch are the muscle spindles. The patellar (knee jerk) reflex is an example. In upper motor neuron lesions, the stretch reflexes tend to be brisk due to loss of inhibitory signals on gamma neurons through the lateral reticulospinal tract.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 20
Incorrect
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Intravenous diazepam was administered to a man who was brought to the emergency department with status epilepticus. He was administered 15 l/min oxygen via a reservoir bag mask. Blood investigations showed sodium = 140 mmol/l, potassium = 4 mmol/l and chloride = 98 mmol/l. His arterial blood gas analysis revealed pH 7.08, p(CO2)= 61.5 mmHg, p(O2) = 111 mmHg and standard bicarbonate = 17 mmol/l. This patient had:
Your Answer:
Correct Answer: Mixed acidosis
Explanation:Acidosis with high p(CO2) and low standard bicarbonate indicates mixed acidosis. Lower p(O2) is due to breathing of 70% oxygen. The prolonged seizures lead to lactic acidosis and the intravenous diazepam is responsible for the respiratory acidosis. Treatment includes airway manoeuvres and oxygen, assisted ventilation if needed, and treatment with fluids.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 21
Incorrect
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A 7-year-old boy is diagnosed with metabolic acidosis as a result of severe dehydration. Which of the following conditions is most likely to cause severe dehydration and metabolic acidosis?
Your Answer:
Correct Answer: Severe diarrhoea
Explanation:Diarrhoea is defined as having three or more loose or liquid stools per day, or as having more stools than is normal for that person. Severe diarrhoea, causing fluid loss and loss of bicarbonate, will result in marked dehydration and metabolic acidosis.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 22
Incorrect
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Which of these substances is secreted by pericytes in the juxtaglomerular cells?
Your Answer:
Correct Answer: Renin
Explanation:The juxtaglomerular cells synthesise, store and secrete the enzyme renin in the kidney. They are specialised smooth muscle cells in the wall of the afferent arteriole that delivers blood to the glomerulus and thus play a critical role in the renin– angiotensin system and so in renal autoregulation.
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This question is part of the following fields:
- Physiology
- Renal
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Question 23
Incorrect
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A 47 year-old woman was admitted for elective cholecystectomy, with a past history of easy bruising and heavy menstrual periods. The patient was also diagnosed with Willebrand's disease. Willebrand's disease is:
Your Answer:
Correct Answer: Autosomal dominant
Explanation:von Willebrand disease is an autosomal dominant disorder marked by the deficiency of vWF, a large protein synthesized by the endothelial cells and megakaryocytes. It mediates adhesion of platelets to the subendothelium at site of vascular injury. Disease characteristics include impaired platelet adhesion, prolonged bleeding time and a functional deficiency of factor VIII (vWF is its carrier protein).
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This question is part of the following fields:
- General
- Physiology
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Question 24
Incorrect
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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:
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.
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This question is part of the following fields:
- General
- Physiology
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Question 25
Incorrect
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During a normal respiratory exhalation, what is the recoil alveolar pressure?
Your Answer:
Correct Answer: +10 cmH2O
Explanation:To determine compliance of the respiratory system, changes in transmural pressures (in and out) immediately across the lung or chest cage (or both) are measured simultaneously with changes in lung or thoracic cavity volume. Changes in lung or thoracic cage volume are determined using a spirometer with transmural pressures measured by pressure transducers. For the lung alone, transmural pressure is calculated as the difference between alveolar (pA; inside) and intrapleural (ppl; outside) pressure. To calculate chest cage compliance, transmural pressure is ppl (inside) minus atmospheric pressure (pB; outside). For the combined lung–chest cage, transmural pressure or transpulmonary pressure is computed as pA – pB. pA pressure is determined by having the subject deeply inhale a measured volume of air from a spirometer. Under physiological conditions the transpulmonary or recoil pressure is always positive; intrapleural pressure is always negative and relatively large, while alveolar pressure moves from slightly negative to slightly positive as a person breathes.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 26
Incorrect
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The primary somatosensory cortex is located in the:
Your Answer:
Correct Answer: Postcentral gyrus
Explanation:The primary somatic sensory cortex is located in the postcentral gyrus and is the largest cortical receiving area for information from somatosensory receptors. Through corticocortical fibres, it then sends the information to other areas of the neocortex and further analysis takes place in the posterior parietal association cortex.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 27
Incorrect
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Which organ is responsible for the secretion of enzymes that aid in digestion of complex starches?
Your Answer:
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.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 28
Incorrect
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A 50 year old woman presented with excessive bleeding after an inguinal hernia repair. Labs are suggestive of a primary haemostasis defect. Deficiency of which of the following is most likely to cause it?
Your Answer:
Correct Answer: Platelets
Explanation:Primary haemostatic control means the first line of defence against immediate bleeding. This is carried out by the platelets. They immediately form a haemostatic plug at the site of injury. Coagulation starts within 20s after an injury to the blood vessel which damage the endothelial cells. Secondary haemostasis follows which includes activation of the coagulation factors to form fibrin strands which mesh together forming the platelet plug. Platelets interact with platelet collagen receptor, glycoprotein Ia/IIa and to collagen fibres in the vascular endothelium. This adhesion is mediated by von Willebrand factor (vWF), which forms links between the platelet glycoprotein Ib/IX/V and collagen fibrils. The platelets are then activated and release the contents of their granules into the plasma, in turn activating other platelets and white blood cells.
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This question is part of the following fields:
- General
- Physiology
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Question 29
Incorrect
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How are amino acids transported across the luminal surface of the small intestinal epithelium?
Your Answer:
Correct 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.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 30
Incorrect
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Which of the following associations is correctly matched with the body's defence mechanism in fighting infection?
Your Answer:
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.
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This question is part of the following fields:
- General
- Physiology
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Question 31
Incorrect
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Evaluation of a 60-year old gentleman, who has been a coal miner all his life and is suspected to have pulmonary fibrosis reveals the following: normal FEV1, arterial oxygen saturation 92%, alveolar ventilation 6000 ml/min at a tidal volume of 600 ml and a breathing rate of 12 breaths/min. There are also pathological changes in lung compliance and residual volume. Which of the following is most accurate about his residual volume?
Your Answer:
Correct Answer: Cannot be measured directly with a spirometer
Explanation:Residual volume is the air left in the lungs after maximal expiration is done. Thus, this is not a part of vital capacity and cannot be measured with a spirometer directly. It can be measured by the methods such as body plethysmography or inert gas dilution. Expiratory reserve volume is vital capacity minus inspiratory capacity. Resting volume of lungs is he sum of residual volume and expiratory reserve volume. Lungs recoil inward until the recoil pressure becomes zero, which corresponds to a volume significantly lower than residual volume.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 32
Incorrect
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Increased resistance to flow of blood in cerebral vessels is most likely seen in:
Your Answer:
Correct Answer: Elevation in systemic arterial pressure from 100 to 130 mmHg
Explanation:Constant cerebral blood flow is maintained by autoregulation in the brain, which causes an increase in local vascular resistance to offset an increase in blood pressure. There will be an increase in cerebral blood flow (and decrease in resistance to cerebral blood flow) with a decrease in arterial oxygen or an increase in arterial CO2. Similarly, a decrease in viscosity will also increase the blood flow. Due to increased brain metabolism and activity during a seizure, there will also be an increase in the cerebral blood flow.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 33
Incorrect
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An experiment was conducted in which the skeletal muscle protein (not smooth muscle) involved in contraction was selectively inhibited. Which protein was inhibited?
Your Answer:
Correct Answer: Troponin
Explanation:The mechanism of contraction of smooth muscles is different from that of skeletal muscles in which the contractile protein is troponin whilst in smooth muscle contraction is a protein called calmodulin. Calmodulin reacts with calcium ions and stimulates the formation of myosin crossbridges.
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This question is part of the following fields:
- General
- Physiology
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Question 34
Incorrect
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During cardiac catheterisation in a 20-year old man, the following data is obtained: Pressure (mmHg), O2 saturation (%) Right atrium 7 (N = 5) 90 (N = 75), Right ventricle 35/7 (N = 25/5) 90 (N = 75), Pulmonary artery 35/8 (N = 25/15), 90 (N = 75), Left atrium 7 (N = 9) 95 (N = 95), Left ventricle 110/7 (N = 110/9) 95 (N = 95), Aorta 110/75 (N = 110/75) 95 (N = 95) where N = Normal value. What is the likely diagnosis?
Your Answer:
Correct Answer: Atrial septal defect
Explanation:A congenital heart disease, ASD or atrial septal defect leads to a communication between the right and left atria due to a defect in the interatrial septum. This leads to mixing of arterial and venous blood from the right and left side of the heart. The hemodynamic significance of this defect depends on the presence of shunting of blood. Normally, the left side of the heart has higher pressure than the right as the left side has to pump blood throughout the body. A large ASD (> 9 mm) will result in a clinically significant left-to-right shunt, causing volume overload of the right atrium and ventricle, eventually leading to heart failure. Cardiac catheterization would reveal very high oxygen saturation in the right atrium, right ventricle and pulmonary artery. Eventually, the left-to-right shunt will lead to pulmonary hypertension and increased afterload in the right ventricle, along with the increased preload due to the shunted blood. This will either cause right ventricular failure, or raise the pressure in the right side of the heart to equal or more than that in the left. Elevation of right atrial pressure to that of left atrial pressure would thus lead to diminishing or complete cessation of the shunt. If left uncorrected, there will be reversal of the shunt, known as Eisenmenger syndrome, resulting in clinical signs of cyanosis as the oxygen-poor blood form right side of the heart will mix with the blood in left side and reach the peripheral vascular system.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 35
Incorrect
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A 45-year old gentleman presents with diarrhoea for two weeks. He has no history of fever and the diarrhoea stops on fasting. Which is the most likely type of diarrhoea that he is suffering from?
Your Answer:
Correct Answer: Osmotic
Explanation:The different types of diarrhoea are:
1. Secretory diarrhoea – Due to increased secretion or decreased absorption. There is minimal to no structural damage in this type. The most common cause is cholera toxin which stimulates secretion of anions (especially chloride), with sodium and water.
2. Osmotic diarrhoea – Due to increased osmotic load, there is water loss. This occurs in cases of maldigestion syndromes, such as coeliac or pancreatic disease.
3. Motility-related diarrhoea – Occurs in cases of abnormal gastrointestinal motility. Due to increased motility, there is poor absorption and this leads to diarrhoea. This is seen post-vagotomy or in diabetic neuropathy.
4. Inflammatory diarrhoea – Due to damage to the mucosa or brush border, there is a loss of protein-rich fluids and poor absorption. Features of all the above three types can be seen in this type. Aetiology includes bacterial, viral, parasitic infections or autoimmune problems including inflammatory bowel disease.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 36
Incorrect
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Which of the following can lead to haemolytic anaemia?
Your Answer:
Correct Answer: Presence of haemoglobin S
Explanation:Haemoglobin S is an abnormal type of haemoglobin seen in sickle cell anaemia. This allows for the haemoglobin to crystalize within the RBC upon exposure to low partial pressures of oxygen. This results in rupture of the RBCs as they pass through microcirculation, especially in the spleen. This can cause blockage of the vessel down stream and ischaemic death of tissues, accompanied by severe pain.
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This question is part of the following fields:
- General
- Physiology
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Question 37
Incorrect
-
Ventricular filling follows a delay caused by?
Your Answer:
Correct 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.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 38
Incorrect
-
Which of the following substances will enhance the activity of antithrombin III?
Your Answer:
Correct Answer: Heparin
Explanation:Antithrombin III is a glycoprotein that inactivates multiple enzymes involved in the coagulation system. It inactivates factor X, factor IX, factor II, factor VII, factor XI and factor XII. Its activity is greatly increased by the action of heparin.
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This question is part of the following fields:
- General
- Physiology
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Question 39
Incorrect
-
Regarding the coagulation cascade, Factor VII:
Your Answer:
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.
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This question is part of the following fields:
- General
- Physiology
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Question 40
Incorrect
-
A lesion involving the suprachiasmatic nucleus of hypothalamus is likely to affect:
Your Answer:
Correct 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.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 41
Incorrect
-
Which of the following conditions is characterized by generalised oedema due to effusion of fluid into the extracellular space?
Your Answer:
Correct Answer: Anasarca
Explanation:Anasarca (or ‘generalised oedema’) is a condition characterised by widespread swelling of the skin due to effusion of fluid into the extracellular space. It is usually caused by liver failure (cirrhosis of the liver), renal failure/disease, right-sided heart failure, as well as severe malnutrition/protein deficiency.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 42
Incorrect
-
After a severe asthma attack, a 26-year-old woman is left in a markedly hypoxic state. In which of the following organs are the arterial beds most likely to be vasoconstricted due to the hypoxia?
Your Answer:
Correct Answer: Lungs
Explanation:Hypoxic pulmonary vasoconstriction is a local response to hypoxia resulting primarily from constriction of small muscular pulmonary arteries in response to reduced alveolar oxygen tension. This unique response of pulmonary arterioles results in a local adjustment of perfusion to ventilation. This means that if a bronchiole is obstructed, the lack of oxygen causes contraction of the pulmonary vascular smooth muscle in the corresponding area, shunting blood away from the hypoxic region to better-ventilated regions. The purpose of hypoxic pulmonary vasoconstriction is to distribute blood flow regionally to increase the overall efficiency of gas exchange between air and blood.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 43
Incorrect
-
Which of the following brings about a reduction in gastric blood flow?
Your Answer:
Correct Answer: Vasopressin
Explanation:Gastric blood flow is increased by vagal stimulation, gastrin, histamine and acetylcholine as they stimulate gastric section and the production of vasodilator metabolites. Acetylcholine and histamine also have a direct action on the gastric arterioles. Similarly, gastric blood flow is reduced by inhibitors of secretion – catecholamines, secretin and vasopressin.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 44
Incorrect
-
Causes of metabolic acidosis with a normal anion gap include:
Your Answer:
Correct Answer: Diarrhoea
Explanation:Excess acid intake and excess bicarbonate loss as in diarrhoea, are causes of metabolic acidosis with a normal anion gap. The other conditions all result in an increased anion gap.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
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Question 45
Incorrect
-
Under normal conditions, what is the major source of energy of cardiac muscles?
Your Answer:
Correct 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.
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This question is part of the following fields:
- Cardiovascular
- Physiology
-
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Question 46
Incorrect
-
A 60 year old patient with a history of carcinoma of the head of the pancreas, and obstructive jaundice presents with a spontaneous nose bleed and easy bruising. What is the most likely reason for this?
Your Answer:
Correct Answer: Vitamin-K-dependent clotting factors deficiency
Explanation:Vitamin K is a fat soluble vitamin requiring fat metabolism to function properly to allow for its absorption. People with obstructive jaundice develop vitamin k deficiency as fat digestion is impaired. Vit K causes carboxylation of glutamate residue and hence regulates blood coagulation including: prothrombin (factor II), factors VII, IX, X, protein C, protein S and protein Z.
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This question is part of the following fields:
- General
- Physiology
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Question 47
Incorrect
-
A 30-year-old man is brought to the emergency department suffering from extreme dehydration, and subsequent hypotension and tachycardia. Which part of the kidney will compensate for this loss?
Your Answer:
Correct Answer: Collecting ducts
Explanation:The collecting duct system of the kidney consists of a series of tubules and ducts that physically connect nephrons to a minor calyx or directly to the renal pelvis. The collecting duct system is the last component of the kidney to influence the body’s electrolyte and fluid balance. In humans, the system accounts for 4–5% of the kidney’s reabsorption of sodium and 5% of the kidney’s reabsorption of water. At times of extreme dehydration, over 24% of the filtered water may be reabsorbed in the collecting duct system.
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This question is part of the following fields:
- Physiology
- Renal
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Question 48
Incorrect
-
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:
Correct 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.
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This question is part of the following fields:
- General
- Physiology
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Question 49
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:
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.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 50
Incorrect
-
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:
Correct 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 51
Incorrect
-
Which of the following is involved in vitamin B12 absorption?
Your Answer:
Correct Answer: Intrinsic factor
Explanation:Absorption of vitamin B12 is by an active transport process and occurs in the ileum. Most cobalamins are bound to proteins and are released in the stomach due to low pH and pepsin. The cobalamins then bind to R proteins, i.e. haptocorrin (HC) secreted from salivary glands and gastric juice. Another cobalamin binding protein is Intrinsic factor (IF) secreted from the gastric parietal cells. The cobalamin-HC complex is digested by pancreatic proteases in the intestinal lumen, and the free cobalamin then binds to IF. The complex then reaches a transmembrane receptor in the ileum and undergoes endocytosis. Cobalamin is then released intracellularly and binds to transcobalamin II (TC II). The newly formed complex then exits the ileal cell and enters the blood circulation.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 52
Incorrect
-
The gradual depolarization in-between action potentials in pacemaker tissue is a result of?
Your Answer:
Correct Answer: A combination of gradual inactivation outward IK along with the presence of an inward ‘funny’ current (If) due to opening of channels permeable to both Na+ and K+ ions
Explanation:One of the characteristic features of the pacemaker cell is the generation of a gradual diastolic depolarization also called the pacemaker potential. In phase 0, the upstroke of the action potential caused by an increase in the Ca2+ conductance, an influx of calcium occurs and a positive membrane potential is generated. The next is phase 3 which is repolarization caused by increased K+ conductance as a result of outwards K+ current. Phase 4 is a slow depolarization which accounts for the pacemaker activity, caused by increased conductance of Na+, inwards Na+ current called IF. it is turned on by repolarization.
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This question is part of the following fields:
- General
- Physiology
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Question 53
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:
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.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 54
Incorrect
-
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:
Correct 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.
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This question is part of the following fields:
- Physiology
- Renal
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Question 55
Incorrect
-
Which of the following is a likely cause of jaundice?
Your Answer:
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 56
Incorrect
-
Which of the following is responsible for the maximum increase in total peripheral resistance on sympathetic stimulation?
Your Answer:
Correct Answer: Arterioles
Explanation:Arterioles are also known as the resistance vessels as they are responsible for approximately half the resistance of the entire systemic circulation. They are richly innervated by the autonomic nervous system and hence, will bring about the maximum increase in peripheral resistance on sympathetic stimulation.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 57
Incorrect
-
A 34-year old gentleman presented with acute pancreatitis to the emergency department. On enquiry, there was found to be a history of recurrent pancreatitis, eruptive xanthomas and raised plasma triglyceride levels associated with chylomicrons. Which of the following will be found deficient in this patient?
Your Answer:
Correct Answer: Lipoprotein lipase
Explanation:The clinical features mentioned here suggest the diagnosis of hypertriglyceridemia due to lipoprotein lipase (LPL) deficiency. LPL aids in hydrolysing the lipids in lipoproteins into free fatty acids and glycerol. Apo-CII acts as a co-factor. Deficiency of this enzyme leads to hypertriglyceridemia.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 58
Incorrect
-
Destruction of the ventromedial nucleus of the hypothalamus will result in:
Your Answer:
Correct 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.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 59
Incorrect
-
A glycogen storage disorder is characterised by increased liver glycogen with a normal structure and no increase in serum glucose after oral intake of a protein-rich diet. Deficiency of which of the following enzymes is responsible for this disorder?
Your Answer:
Correct Answer: Glucose-6-phosphatase
Explanation:The most common glycogen storage disorder is von Gierke’s disease or glycogen storage disease type I. It results from a deficiency of enzyme glucose-6-phosphatase which affects the ability of liver to produce free glucose from glycogen and gluconeogenesis; leading to severe hypoglycaemia. There is also increased glycogen storage in the liver and kidneys causing enlargement and various problems in their functioning. The disease also causes lactic acidosis and hyperlipidaemia. The main treatment includes frequent or continuous feedings of corn-starch or other carbohydrates.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 60
Incorrect
-
A patient came into the emergency in a state of shock. His blood group is not known, but on testing it clotted when mixed with Type A antibodies. Which blood should be transfused?
Your Answer:
Correct Answer: B +ve
Explanation:There are two stages to determine the blood group, known as ABO typing. The first stage is called forward typing. In this method, RBCs are mixed with two separate solutions of type A or type B antibodies to see if they agglutinate. If this blood clumps, this indicates the presence of antigens within the blood sample. For example, a sample of type B blood will clump when tested with type A antibodies as it contains type B antigens. Group B – has only the B antigen on red cells (and A antibody in the plasma)
Group B – has only the B antigen on red cells (and A antibody in the plasma)
Group AB – has both A and B antigens on red cells (but neither A nor B antibody in the plasma)
Group O – has neither A nor B antigens on red cells (but both A and B antibody are in the plasma). Many people also have a Rh factor on the red blood cell’s surface. This is also an antigen and those who have it are called Rh+. Those who have not are called Rh–. A person with Rh– blood does not have Rh antibodies naturally in the blood plasma (as one can have A or B antibodies, for instance) but they can develop Rh antibodies in the blood plasma if they receive blood from a person with Rh+ blood, whose Rh antigens can trigger the production of Rh antibodies. A person with Rh+ blood can receive blood from a person with Rh– blood without any problems. The patient’s blood group is B positive as he has antigen B, antibody A and Rh antigens.
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This question is part of the following fields:
- General
- Physiology
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Question 61
Incorrect
-
Which of the following clinical signs will be demonstrated in a case of Brown-Séquard syndrome due to hemisection of the spinal cord at mid-thoracic level?
Your Answer:
Correct Answer: Ipsilateral spastic paralysis, ipsilateral loss of vibration and proprioception (position sense) and contralateral loss of pain and temperature sensation beginning one or two segments below the lesion
Explanation:Brown–Séquard syndrome results due to lateral hemisection of the spinal cord and results in a loss of motricity (paralysis and ataxia) and sensation. The hemisection of the cord results in a lesion of each of the three main neural systems: the principal upper motor neurone pathway of the corticospinal tract, one or both dorsal columns and the spinothalamic tract. As a result of the injury to these three main brain pathways the patient will present with three lesions. The corticospinal lesion produces spastic paralysis on the same side of the body (the loss of moderation by the upper motor neurons). The lesion to fasciculus gracilis or fasciculus cuneatus results in ipsilateral loss of vibration and proprioception (position sense). The loss of the spinothalamic tract leads to pain and temperature sensation being lost from the contralateral side beginning one or two segments below the lesion. At the lesion site, all sensory modalities are lost on the same side, and an ipsilateral flaccid paralysis.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 62
Incorrect
-
Production of pain is most likely associated with:
Your Answer:
Correct Answer: Substance P
Explanation:Substance P is a short-chain polypeptide that functions as a neurotransmitter and as a neuromodulator, and is thus, a neuropeptide. It has been linked with pain regulation, mood disorders, stress, reinforcement, neurogenesis, respiratory rhythm, neurotoxicity, nausea and emesis. It is also a potent vasodilator as it brings about release of nitric oxide from the endothelium. Its release can also cause hypotension.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 63
Incorrect
-
Abnormal breathing is noticed in a of victim of a road traffic accident, who sustained a head injury. The breathing pattern is characterised by alternate periods of waxing and waning tidal volumes with interspersed periods of apnoea. This breathing pattern is known as:
Your Answer:
Correct Answer: Cheyne–Stokes breathing
Explanation:Cheyne-Stokes breathing is an abnormal breathing pattern with breathing periods of gradually waxing and waning tidal volumes, with apnoeic periods interspersed. It is usually the first breathing pattern to be seen with a rise in intracranial pressure and is caused by failure of the respiratory centre in the brain to compensate quickly enough to changes in serum partial pressure of oxygen and carbon dioxide. The aetiology includes strokes, head injuries, brain tumours and congestive heart failure. It is also a sign of altitude sickness in normal people, a symptom of carbon monoxide poisoning or post-morphine administration. Biot’s respiration (cluster breathing) is characterized by cluster of quick, shallow inspirations followed by regular or irregular periods of apnoea. It is different from ataxic respiration, which has completely irregular breaths and pauses. It results due to damage to the medulla oblongata by any reason (stroke, uncal herniation, trauma) and is a poor prognostic indicator. Kussmaul breathing, also known as ‘air hunger’, is basically respiratory compensation for metabolic acidosis and is characterized by quick, deep and laboured breathing. It is most often seen in in diabetic ketoacidosis. Due to forced inspiratory rate, the patients will show a low p(CO2). Ondine’s curse is congenital central hypoventilation syndrome or primary alveolar hypoventilation, which can be fatal and leads to sleep apnoea. It involves an inborn failure to control breathing autonomically during sleep and in severe cases, can affect patients even while awake. It is known to occur in 1 in 200000 liveborn children. Treatment includes tracheostomies and life long mechanical ventilator support.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 64
Incorrect
-
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:
Correct 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
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 65
Incorrect
-
In a hypertensive patient with secondary hyperaldosteronism, aldosterone is released mainly in response to:
Your Answer:
Correct Answer: Angiotensin II
Explanation:Secondary hyperaldosteronism in hypertension is either due to primary renin overproduction by the kidneys or renin overproduction secondary to decreased renal blood flow. The main stimulus for aldosterone release are adrenocorticotrophic hormone (ACTH), angiotensin II and high plasma K+ levels. Low plasma Na+ might also stimulate the adrenal cortex. Fluid overload will reduce aldosterone secretion. Atrial natriuretic peptide is secreted under conditions of expanded extracellular volume and will not lead to aldosterone secretion.
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This question is part of the following fields:
- Endocrinology
- Physiology
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Question 66
Incorrect
-
Which of the following proteins prevents red blood cells (RBCs) from bursting when they pass through capillaries?
Your Answer:
Correct Answer: Spectrin
Explanation:Spectrin is a structural protein found in the cytoskeleton that lines the intercellular side of the membrane of cells which include RBCs. They maintain the integrity and structure of the cell. It is arranged into a hexagonal arrangement formed from tetramers of spectrin and associated with short actin filaments that form junctions allowing the RBC to distort its shape.
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This question is part of the following fields:
- General
- Physiology
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Question 67
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:
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.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 68
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:
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.
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This question is part of the following fields:
- General
- Physiology
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Question 69
Incorrect
-
A syndrome responsible for failure to absorb vitamin B12 from the GIT is called?
Your Answer:
Correct Answer: Pernicious anaemia
Explanation:Pernicious anaemia is a type of autoimmune disease in which antibodies form against the parietal cells or intrinsic factor. Intrinsic factor is required for the absorption of vitamin B12. Blood testing typically shows a macrocytic, normochromic anaemia and low levels of serum vitamin B12. A Schilling test can then be used to distinguish between pernicious anaemia, vitamin B12 malabsorption and vitamin B12 deficiency. Symptoms include shortness of breath, pallor and diarrhoea etc.
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This question is part of the following fields:
- General
- Physiology
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Question 70
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:
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).
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This question is part of the following fields:
- Physiology
- Renal
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Question 71
Incorrect
-
Pain in the right upper quadrant of the abdomen on ingestion of a fatty meal is seen in a condition which involves which of the following substances?
Your Answer:
Correct Answer: Cholecystokinin
Explanation:The clinical scenario described here favours the presence of gallstones. During food ingestion, vagal discharges stimulate gallbladder contraction. Moreover, presence of fat and amino acids in the intestinal lumen stimulates the release of cholecystokinin (CCK) in the duodenum. This causes sustained gallbladder contraction and relaxation of the sphincter of Oddi. If gallstones are present, there will be inflammation in the gallbladder and CCK will aggravate it due to contractions.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 72
Incorrect
-
A 70-year-old male who has smoked since his teens complains of progressive shortness of breath and a persistent cough. He is diagnosed with COPD. Which of the following abnormalities is most likely to be present in his pulmonary function tests?
Your Answer:
Correct Answer: Increased residual volume
Explanation:Chronic obstructive pulmonary disease (COPD) is a type of obstructive lung disease characterized by long-term poor airflow. The main symptoms include shortness of breath and cough with sputum production. The best diagnostic test for evaluating patients with suspected chronic obstructive pulmonary disease (COPD) is lung function measured with spirometry. Key spirometrical measures may be obtained with a portable office spirometer and should include forced vital capacity (FVC) and the normal forced expiratory volume in the first second of expiration (FEV1). The ratio of FEV1 to forced vital capacity (FEV1/FVC) normally exceeds 0.75. Patients with COPD typically present with obstructive airflow. Complete pulmonary function testing may show increased total lung capacity, functional residual capacity and residual volume. A substantial loss of lung surface area available for effective oxygen exchange causes diminished carbon monoxide diffusion in the lung (DLco) in patients with emphysema. Tobacco smoking is the most common cause of COPD, with factors such as air pollution and genetics playing a smaller role.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 73
Incorrect
-
Which is a feature of the action of insulin?
Your Answer:
Correct Answer: Promotes protein synthesis
Explanation:Insulin is produced by the beta-cells of the islets of Langerhans in the pancreas. Its actions include:
– promoting uptake of glucose into cells
– glycogen synthesis (glycogenesis)
– protein synthesis
– stimulation of lipogenesis (fat formation).
– driving potassium into cells – used to treat hyperkaelamia.
Parathyroid hormone and activated vitamin D are the principal hormones involved in calcium/phosphate metabolism, rather than insulin.
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This question is part of the following fields:
- Endocrine
- Physiology
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Question 74
Incorrect
-
Which type of contractions are responsible for the propulsion of chyme along the small intestine?
Your Answer:
Correct Answer: Segmentation
Explanation:Two major types of intestinal contractions are segmentation and peristalsis:
Segmentation occurs most frequently and primarily involves circular muscle. It is essentially a contraction of 2- or 3-cm long intestinal segments while the muscle on either side of it relaxes. Chyme in the segment is displaced in both directions. As the contracted segment relaxes, the previously relaxed segments on either side may contract. This efficiently mixes the chyme with the digestive secretions and exposes the mucosal absorptive surface to the luminal contents. It also serves a propulsive function and contributes to the movement of chyme.
Peristalsis is a propulsive wave of contraction that is initiated by intestinal distension. It is short lived and travels only a few centimetres before dying out. The combined effects of intestinal peristalsis and segmentation provide for both adequate mixing of the intestinal contents and slow, steady movement of chyme.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 75
Incorrect
-
A sudden loud sound is more likely to result in cochlear damage than a slowly developing loud sound. This is because:
Your Answer:
Correct Answer: There is a latent period before the attenuation reflex can occur
Explanation:On transmission of a loud sound into the central nervous system, an attenuation reflex occurs after a latent period of 40-80 ms. This reflex contracts the two muscles that pull malleus and stapes closer, developing a high degree of rigidity in the entire ossicular chain. This reduces the ossicular conduction of low frequency sounds to the cochlea by 30-40 decibels. In this way, the cochlea is protected from damage due to loud sounds (these are low frequency sounds) when they develop slowly.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 76
Incorrect
-
Which of the following organelles have the capacity to regenerate and spontaneously replicate?
Your Answer:
Correct Answer: Mitochondrion
Explanation:A mitochondria is a membrane bound organelle found in eukaryotic cells. They are called the powerhouse of the cell and are the place where ATP is formed from energy generated through metabolism. They are capable of replication as well as repair and regeneration.
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This question is part of the following fields:
- General
- Physiology
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Question 77
Incorrect
-
After surgery, a patient developed a stitch granuloma . Which leukocyte in the peripheral blood will become an activated macrophage in this granuloma?
Your Answer:
Correct 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.
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This question is part of the following fields:
- General
- Physiology
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Question 78
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:
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.
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This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
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Question 79
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) = 80–100 fl, mean corpuscular haemoglobin concentration (MCHC) = 31–37 g/dl. Which of the following is correct regarding this patient’s erythrocytes:
Your Answer:
Correct Answer: Normal MCV
Explanation:MCV is the mean corpuscular volume and it is calculated from the haematocrit and the RBC count. It is normally 90 fl. Mean corpuscular haemoglobin concentration (MCHC) [g/dl] = haemoglobin [g/dl]/haematocrit = 11/0.27 = 41 g/dl and is higher than normal range (32 to 36 g/dL).
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This question is part of the following fields:
- General
- Physiology
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Question 80
Incorrect
-
What is the basic chemical reaction that takes place in the breakdown of complex foodstuffs?
Your Answer:
Correct Answer: Hydrolysis
Explanation:Breakdown of complex food into simpler compounds is achieved by hydrolysis, with the help of different enzymes specific for different compounds.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 81
Incorrect
-
Different portions of the renal tubule have varying degrees of water permeability. Which of the following renal sites is characterised by low water permeability under normal circumstances?
Your Answer:
Correct Answer: Thick ascending limb of the loop of Henlé
Explanation:Within the nephron of the kidney, the ascending limb of the loop of Henle is a segment of the loop of Henle downstream of the descending limb, after the sharp bend of the loop. Both the thin and the thick ascending limbs of the loop of Henlé have very low permeability to water. Since there are no regulatory mechanisms to alter its permeability, it remains poorly permeable to water under all circumstances. Sodium and chloride are transported out of the luminal fluid into the surrounding interstitial spaces, where they are reabsorbed. Water must remain behind because it is not reabsorbed, so the solute concentration becomes less and less (the luminal fluid becomes more dilute). This is one of the principal mechanisms (along with diminution of ADH secretion) for the production of a dilute, hypo-osmotic urine (water diuresis).
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This question is part of the following fields:
- Physiology
- Renal
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Question 82
Incorrect
-
The most likely cause of prominent U waves on the electrocardiogram (ECG) of a patient is:
Your Answer:
Correct Answer: Hypokalaemia
Explanation:The U-wave, not always visible in ECGs, is thought to represent repolarisation of papillary muscles or Purkinje fibres. When seen, it is very small and occurs after the T-wave. Inverted U-waves indicate myocardial ischaemia or left ventricular volume overload. Prominent U-waves are most commonly seen in hypokalaemia. Other causes include hypercalcaemia, thyrotoxicosis, digitalis exposure, adrenaline and class 1A and 3 anti-arrhythmic agents. It can also be seen in congenital long-QT syndrome and in intracranial haemorrhage.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 83
Incorrect
-
In the kidney, the macula densa is an area of closely packed specialized cells lining the wall of the:
Your Answer:
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.
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This question is part of the following fields:
- Physiology
- Renal
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Question 84
Incorrect
-
A 65-year old patient with altered bowl movement experienced the worsening of shortness of breath and exertional chest pains over the course of 8 weeks. Examination shows pallor and jugular venous distension. Furthermore, a test of the stool for occult blood is positive. Laboratory studies show:
Haemoglobin 7.4 g/dl
Mean corpuscular volume 70 fl Leukocyte count 5400/mm3
Platelet count 580 000/mm3 Erythrocyte sedimentation 33 mm/h
A blood smear shows hypochromic, microcytic RBCs with moderate poikilocytosis. Which of the following is the most likely diagnosis?Your Answer:
Correct Answer: Iron deficiency anaemia
Explanation:Iron deficiency anaemia is the most common type of anaemia. It can occur due to deficiency of iron due to decreased intake or due to faulty absorption. An MCV less than 80 will indicated iron deficiency anaemia. On the smear the RBC will be microcytic hypochromic and will also show piokilocytosis. iron profiles tests are important to make a diagnosis. Clinically the patient will be pale and lethargic.
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This question is part of the following fields:
- General
- Physiology
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Question 85
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:
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.
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This question is part of the following fields:
- General
- Physiology
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Question 86
Incorrect
-
A suspected recreational drug user was brought to the Emergency department in an unconscious state, and was found to be hypoventilating. Which of the following set of arterial blood gas analysis report is most consistent with hypoventilation as the primary cause? pH, pa(CO2) (mmHg), pa(O2) (mmHg).
Your Answer:
Correct Answer: 7.28, 55, 81
Explanation:Hypoventilation (or respiratory depression) causes an increase in carbon dioxide (hypercapnia) and respiratory acidosis. It can result due to drugs such as alcohol, benzodiazepines, barbiturates, opiates, mechanical conditions or holding ones breath. Strong opioids such as heroin and fentanyl are commonly implicated and can lead to respiratory arrest. In recreational drug overdose, acute respiratory acidosis occurs with an increase in p(CO2) over 45 mm Hg and acidaemia (pH < 7.35)
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 87
Incorrect
-
A young women following a road traffic accident suffered heavy blood loss and developed subsequent anaemia. Which of the following is a consequence of this?
Your Answer:
Correct Answer: A high reticulocyte count
Explanation:Anaemia refers to a decrease in the circulating levels of haemoglobin in the blood resulting in a reduced ability of the body to transport oxygen effectively. Anaemia from blood loss results in the body further compensating by releasing stored RBCs and immature RBCs from the bone marrow. Thus resulting in a high reticulocyte count.
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This question is part of the following fields:
- General
- Physiology
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Question 88
Incorrect
-
Basal Metabolic Rate (BMR) will most likely be reduced by which of the following?
Your Answer:
Correct 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.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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Question 89
Incorrect
-
A patient presents with loss of pain and temperature sensation in the left leg. He is likely to have a lesion involving:
Your Answer:
Correct Answer: Right lateral spinothalamic tract
Explanation:The spinothalamic tract is a sensory pathway originating in the spinal cord that transmits information to the thalamus. There are two main parts of the spinothalamic tract: the lateral spinothalamic tract transmits pain and temperature and the anterior spinothalamic tract transmits touch (crude touch). The decussation of this pathway occurs at the level of the spinal cord. Hence, a unilateral lesion of the lateral spinothalamic tract causes contralateral loss of pain and temperature.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 90
Incorrect
-
If a 55-year old gentleman who has suffered a stroke, develops a tremor in his fingers that worsens on reaching for an object, what part of his brain is likely to be involved?
Your Answer:
Correct Answer: Cerebellum
Explanation:The cerebellum plays an important role in the integration of sensory perception and motor output. Multiple neural pathways link the cerebellum with the motor cortex and the spinocerebellar tract. The cerebellum uses the constant feedback on body position to fine-tune the movements and integrates these pathways. The patient described here has a characteristic cerebellar tremor that is a slow, broad tremor of the extremities and occurs at the end of a purposeful movement.
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This question is part of the following fields:
- Neurology
- Physiology
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Question 91
Incorrect
-
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:
Correct 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 92
Incorrect
-
What is the normal duration of PR interval on an electrocardiogram of a healthy individual?
Your Answer:
Correct 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.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 93
Incorrect
-
Which of the following is true regarding factor XI?
Your Answer:
Correct Answer: Deficiency causes haemophilia C
Explanation:Factor XI is also known as plasma thromboplastin and is one of the enzymes of the coagulation cascade. It is produced in the liver and is a serine protease. It is activated by factor XIIa, thrombin and by itself. Deficiency of factor XI causes the rare type of haemophilia C. Low levels of factor XI also occur in other disease states, including Noonan syndrome. High levels of factor XI have been seen in thrombosis.
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This question is part of the following fields:
- General
- Physiology
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Question 94
Incorrect
-
When at rest, which of the following will be higher in a marathon runner compared to an untrained individual?
Your Answer:
Correct Answer: Cardiac stroke volume
Explanation:Cardiac muscle hypertrophy is seen in trained athletes as compared to the normal population. This hypertrophy results in higher stroke volume at rest and increased cardiac reserve (maximum cardiac output during exercise). However, the cardiac output at rest is almost the same in both trained and untrained people. This is because in trained athletes, the heart rate is slower, even up to 40-50 beats/min. There is minimal affect of athletic training on oxygen consumption and respiratory rate.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 95
Incorrect
-
What is the normal duration of the ST segment?
Your Answer:
Correct 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.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 96
Incorrect
-
In the glomerulus of the kidney, the mesangium is a structure associated with the capillaries. It has extraglomerular mesangial cells that:
Your Answer:
Correct Answer: Form the juxtaglomerular apparatus in combination with the macula densa and juxtaglomerular cells
Explanation:The mesangium is an inner layer of the glomerulus, within the basement membrane surrounding the glomerular capillaries. The mesangial cells are phagocytic and secrete the amorphous basement membrane-like material known as the mesangial matrix. They are typically separated from the lumen of the capillaries by endothelial cells. The other type of cells in the mesangium are the extraglomerular mesangial cells which form the juxtaglomerular apparatus in combination with two other types of cells: the macula densa of the distal convoluted tubule and juxtaglomerular cells of the afferent arteriole. This apparatus controls blood pressure through the renin–angiotensin–aldosterone system.
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This question is part of the following fields:
- Physiology
- Renal
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Question 97
Incorrect
-
Which of the following will be a seen in a patient with a plasma thyroid-stimulating hormone (TSH) level of 14 mU/l (normal < 5 mU/l) and a low T3 resin uptake of 19% (normal 25–35%)?
Your Answer:
Correct Answer: Periorbital swelling and lethargy
Explanation:Low T3 resin uptake combined with raised TSH is indicative of hypothyroidism. Signs and symptoms include dull expression, facial puffiness, lethargy, periorbital swelling due to infiltration with mucopolysaccharides, bradycardia and cold intolerance. Anxiety, palpitations, tachycardia, raised body temperature, heat intolerance and weight loss are all seen in hyperthyroidism.
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This question is part of the following fields:
- Endocrinology
- Physiology
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Question 98
Incorrect
-
A 79-year-old has been bedridden for 2 months after suffering from a stroke. She suddenly developed shortness of breath and chest pain, and was diagnosed with a pulmonary embolism. Which of the following is most likely to increase in this case?
Your Answer:
Correct Answer: Ventilation/perfusion ratio
Explanation:Pulmonary embolism (PE) is a blockage of an artery in the lungs by an embolus that has travelled from elsewhere in the body through the bloodstream. The change in cardiopulmonary function is proportional to the extent of the obstruction, which varies with the size and number of emboli obstructing the pulmonary arteries. The resulting physiological changes may include pulmonary hypertension with right ventricular failure and shock, dyspnoea with tachypnoea and hyperventilation, arterial hypoxaemia and pulmonary infarction. Consequent alveolar hyperventilation is manifested by a lowered pa(CO2). After occlusion of the pulmonary artery, areas of the lung are ventilated but not perfused, resulting in wasted ventilation with an increased ventilation/perfusion ratio – the physiological hallmark of PE – contributing to a further hyperventilatory state. The risk of blood clots is increased by cancer, prolonged bed rest, smoking, stroke, certain genetic conditions, oestrogen-based medication, pregnancy, obesity, and post surgery.
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This question is part of the following fields:
- Physiology
- Respiratory
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Question 99
Incorrect
-
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:
Correct 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.
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This question is part of the following fields:
- Cardiovascular
- Physiology
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Question 100
Incorrect
-
A TRUE statement regarding abolition of the cephalic phase of pancreatic secretion is that it:
Your Answer:
Correct Answer: Will result after vagotomy
Explanation:Recognition and integration of the sight, smell and taste of food triggers the cephalic phase of pancreatic secretion, causing an increase in pancreatic HCO3- and enzyme secretion. The degree of enzyme secretion in this phase is about 50% of the maximal response seen with exogenous CCK and secretin. The vagus nerve regulates the secretion through the cholinergic fibres innervating the acinar cells of the pancreas, and through peptidergic nerve fibres, which innervate ductal cells.
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This question is part of the following fields:
- Gastroenterology
- Physiology
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SESSION STATS - PERFORMANCE PER SPECIALTY
