-
Question 1
Correct
-
Which portion of the renal tubule absorbs amino acids and glucose?
Your Answer: Proximal convoluted tubule
Explanation:In relation to the morphology of the kidney as a whole, the convoluted segments of the proximal tubules are confined entirely to the renal cortex. Glucose, amino acids, inorganic phosphate and some other solutes are reabsorbed via secondary active transport in the proximal renal tubule through co-transport channels driven by the sodium gradient.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 2
Correct
-
A 35-year-old woman is in a comatose state following a traumatic head injury, and is receiving intravenous (IV) antibiotics and IV fluids containing saline and 5% dextrose. A serum biochemistry analysis is performed five days later which shows a low serum potassium level. This is most likely to be due to:
Your Answer: Nothing per oral regimen
Explanation:In this patient the cause for hypokalaemia is insufficient consumption of potassium as she is nil-per mouth with no intravenous supplementation. Parenteral nutrition has been used for comatose patients, although enteral feeding is usually preferable, and less prone to complications.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 3
Correct
-
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: 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.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 4
Correct
-
If the blood flow is constant, oxygen extraction by tissues will show the greatest decrease due to which of the following interventions?
Your Answer: Tissue cooling
Explanation:With a constant blood flow to a given tissue bed, there will be an increase in oxygen extraction by the tissue with the following; an increase in tissue metabolism and oxygen requirements: warming (or fever), exercise, catecholamines and thyroxine. With cooling, the demand for oxygen decreases, leading to decreased oxygen extraction.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 5
Correct
-
Whilst snorkelling, a 30-year old gentleman has the respiratory rate of 10/min, tidal volume of 550 ml and an effective anatomical dead space of 250 ml. Which of the following will bring about a maximum increase in his alveolar ventilation?
Your Answer: A 2x increase in tidal volume and a shorter snorkel
Explanation:Alveolar ventilation = respiratory rate × (tidal volume − anatomical dead space volume). Increase in respiratory rate simply causes movement of air in the anatomical dead space, with no contribution to the alveolar ventilation. By use of a shorter snorkel, the effective anatomical dead space will decrease and will cause a maximum rise in alveolar ventilation along with doubling of tidal volume.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 6
Correct
-
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: 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).
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 7
Correct
-
Which of the following cytokines produced by the T cell induce MHC-II proteins?
Your Answer: γ-Interferon
Explanation:Interferons elicit a non-specific antiviral activity by inducing specific RNA synthesis and expression of proteins in neighbouring cells. Common interferon inducers are viruses, double-stranded RNA and micro-organisms. INF-γ is produced mainly by CD4+, CD8+ T cells and less commonly by B cells and natural killer cells. INF-γ has antiviral and antiparasitic activity but its main biological activity appears to be immunomodulatory. Among its many functions are activation of macrophages and enhanced expression of MHC-II proteins or macrophages.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 8
Correct
-
Arterial blood gas analysis of a man admitted with acute exacerbation of chronic obstructive pulmonary disease (COPD) showed the following: pH = 7.28, p(CO2) = 65.5 mmHg, p(O2)= 60 mmHg and standard bicarbonate = 30.5 mmol/l. This patient had:
Your Answer: Respiratory acidosis
Explanation:Acidosis with high p(CO2) and normal standard bicarbonate indicates respiratory acidosis, commonly seen in acute worsening of COPD patients. Respiratory acidosis occurs due to alveolar hypoventilation which leads to increased arterial carbon dioxide concentration (p(CO2)). This in turn decreases the HCO3 –/p(CO2) and decreases pH. In acute respiratory acidosis, the p(CO2) is raised above the upper limit of normal (over 45 mm Hg) with a low pH. However, in chronic cases, the raised p(CO2) is accompanied with a normal or near-normal pH due to renal compensation and an increased serum bicarbonate (HCO3 – > 30 mmHg).
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 9
Correct
-
What is the result of maltase deficiency in the brush border of the small intestine?
Your Answer: Results in increased passage of maltose in stool
Explanation:Maltase is an enzyme produced from the surface cells of the villi, lining the small intestine and aids in hydrolysing the disaccharide maltose, which splits into two molecules of α-glucose. It is done by breaking the glycosidic bond between the ‘first’ carbon of one glucose and the ‘fourth’ carbon of the other (a 1–4 bond). Hence, a deficiency of enzyme maltase will result in the increased passage of maltose in the stool.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 10
Correct
-
The Henderson–Hasselbalch equation describes the derivation of pH as a measure of acidity. According to this equation, the buffering capacity of the system is at maximum when the number of free anions compared with undissociated acid is:
Your Answer: Equal
Explanation:In 1908, Lawrence Joseph Henderson wrote an equation describing the use of carbonic acid as a buffer solution. Later, Karl Albert Hasselbalch re-expressed that formula in logarithmic terms, resulting in the Henderson–Hasselbalch equation. The equation is also useful for estimating the pH of a buffer solution and finding the equilibrium pH in acid–base reactions. Two equivalent forms of the equation are: pH = pKa + log10 [A–]/[HA] or pH = pKa + log10 [base]/[acid]. Here, pKa is − log10(Ka) where Ka is the acid dissociation constant, that is: pKa = –log10(Ka) = –log10 ([H3 O+][A–]/[HA]) for the reaction: HA + H2 O ≈ A– + H3 O+ In these equations, A– denotes the ionic form of the relevant acid. Bracketed quantities such as [base] and [acid] denote the molar concentration of the quantity enclosed. Maximum buffering capacity is found when pH = pKa or when the number of free anions to undissociated acid is equal and buffer range is considered to be at a pH = pKa ± 1.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 11
Correct
-
Which is the site of action of the drug omeprazole?
Your Answer: H+/K+ ATPase
Explanation:H+/K+-ATPase or ‘proton pump’ located in the canalicular membrane plays a major role in acid secretion. The ATPase here is magnesium-dependent. Omeprazole is a proton pump inhibitor and blocks H+/K+- ATPase.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 12
Correct
-
During a normal respiratory exhalation, what is the recoil alveolar pressure?
Your 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.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 13
Incorrect
-
For calculation of cardiac output by Fick's principle, which of the following vessels is the best source of venous blood to determine the arterial-to-venous oxygen tension difference?
Your Answer: Pulmonary vein
Correct Answer: Pulmonary artery
Explanation:Fick’s principle states that the total uptake (or release) of a substance by peripheral tissues is equal to the product of the blood flow to the peripheral tissues and the arterial– venous concentration difference (gradient) of the substance. It is used to measure the cardiac output, and the formula is Cardiac output = oxygen consumption divided by arteriovenous oxygen difference. Assuming there are no shunts across the pulmonary system, the pulmonary blood flow equals the systemic blood flow. The arterial and venous blood oxygen content is measured by sampling from the pulmonary artery (low oxygen content) and pulmonary vein (high oxygen content). Peripheral arterial blood is used as a surrogate for the pulmonary vein.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 14
Correct
-
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: 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.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 15
Correct
-
Diuretics that act on the ascending limb of the loop of Henle produce:
Your Answer: Reduced active transport of sodium
Explanation:The loop of Henlé leads from the proximal convoluted tubule to the distal convoluted tubule. Its primary function uses a counter current multiplier mechanism in the medulla to reabsorb water and ions from the urine. It can be divided into four parts:
1. Descending limb of loop of Henlé – low permeability to ions and urea, while being highly permeable to water
2. Thin ascending limb of loop of Henlé – not permeable to water, but it is permeable to ions
3. Medullary thick ascending limb of loop of Henlé – sodium (Na+), potassium (K+) and chloride (Cl–) ions are reabsorbed by active transport. K+ is passively transported along its concentration gradient through a K+ channel in the basolateral aspect of the cells, back into the lumen of the ascending limb.
4. The cortical thick ascending limb – the site of action where loop diuretics such as furosemide block the K+/Na+/2Cl− co-transporters = reduced active transport.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 16
Correct
-
When the pitch of a sound increases, what is the physiological response seen in the listener?
Your Answer: The location of maximal basilar membrane displacement moves toward the base of the cochlea
Explanation:An increase in the frequency of sound waves results in a change in the position of maximal displacement of the basilar membrane in the cochlea. Low pitch sound produces maximal displacement towards the cochlear apex and greatest activation of hair cells there. With an increasing pitch, the site of greatest displacement moves towards the cochlear base. However, increased amplitude of displacement, increase in the number of activated hair cells, increased frequency of discharge of units in the auditory nerve and increase in the range of frequencies to which such units respond, are all seen in increases in the intensity or a sound stimulus.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 17
Correct
-
What is the normal amount of oxygen that is carried in the blood?
Your Answer: 20 ml oxygen/100 ml blood
Explanation:Normally, 100 ml of blood contains 15g haemoglobin and a single gram of haemoglobin can bind to 1.34 ml oxygen when 100% saturated. Thus, 15 × 1.34 = 20 ml O2/100 ml blood. The haemoglobin in venous blood that is leaving the tissues is about 75% saturated with oxygen, and hence it carries about 15 ml O2/100 ml venous blood. This implies that for each 10 ml of blood, 5 ml oxygen is transported to the tissues. With a p(O2) > 100 mm Hg, only 3 ml of oxygen is dissolved in every one litre of plasma. By increasing the pA(O2) by breathing 100% oxygen, one can add an extra amount of oxygen in the plasma, but the amount of oxygen carried by haemoglobin will not increase significantly as it is already > 95% saturated.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 18
Correct
-
The gradual depolarization in-between action potentials in pacemaker tissue is a result of?
Your 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.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 19
Correct
-
Production of pain is most likely associated with:
Your 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.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 20
Correct
-
A 31-year-old woman is diagnosed with adrenal hyperplasia, and laboratory samples are taken to measure serum aldosterone and another substance. Which is most likely to be the other test that was prescribed to this patient?
Your Answer: Plasma renin
Explanation:The evaluation of a patient in whom hyperaldosteronism is first to determine that hyperaldosteronism is present (serum aldosterone) and, if it is present, to differentiate primary from secondary causes of hyperaldosteronism. The aldosterone-to-renin ratio (ARR) is the most sensitive means of differentiating primary from secondary causes of hyperaldosteronism as it is abnormally increased in primary hyperaldosteronism, and decreased or normal but with high renin levels in secondary hyperaldosteronism.
-
This question is part of the following fields:
- Physiology
- Renal
-
-
Question 21
Correct
-
Signals pass through neuromuscular junctions via the neurotransmitter acetylcholine. After release from the skeletal neuromuscular junction, acetylcholine:
Your Answer: Causes postsynaptic depolarisation
Explanation:Acetylcholine is released from the presynaptic membrane into the cleft where it binds to the ion gated channels on the post synaptic membrane, causing them to open. This results in sodium entering into the fibre and further depolarizing it, creating an action potential.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 22
Correct
-
Extracellular body fluid as compared with intracellular body fluid:
Your Answer: Is relatively rich in glucose
Explanation:The percentages of body water contained in various fluid compartments add up to total body water (TBW). This water makes up a significant fraction of the human body, both by weight and by volume. Ensuring the right amount of body water is part of fluid balance, an aspect of homeostasis. The extracellular fluid (ECF) includes all fluids outside the cells. This fluid can be divided into three fluid departments: interstitial (in the tissue spaces) fluid, blood plasma and lymph, and specialised compartments called transcellular fluid. The extracellular fluid surrounds all the cells in the body and is in equilibrium with the intracellular fluid. So, its composition must remain fairly constant even though substances are passing into and out of the cells. The interstitial fluid, though called a fluid, is in a reality a gel-like composition made up of: water, proteoglycan molecules and collagen. The extracellular fluid constitutes 40% of total body water, with intracellular fluid making up the remaining 60%. It is relatively rich in glucose.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 23
Correct
-
A blood sample of a 58 year old male patient, who underwent an abdominal aortic aneurysm repair, was sent to the laboratory. The laboratory technician said that the patient’s blood agglutinates with antisera anti-A and anti-D, while the patient’s serum agglutinates cells of blood group B. What is the blood group of this patient?
Your Answer: A positive
Explanation:Group A – has only the A antigen on red cells (and B 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 so-called Rh factor on the red blood cell’s surface. This is also an antigen and those who have it 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. In this scenario the person has blood group A+ as he has A antigen, anti B antibody and Rh antigen
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 24
Correct
-
The mechanism of action of streptokinase involves:
Your Answer: Direct conversion of plasminogen to plasmin
Explanation:Streptokinase is an enzyme that is produced by group A beta haemolytic streptococcus and is an effective and cost efficient method for the dissolution of a clot used in cases of MI and pulmonary embolism. It works by directly converting plasminogen to plasmin which breaks down the blood components in the clot and fibrin, dissolving the clot. Streptokinase is a bacterial product and thus the body will develop immunity against it.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 25
Correct
-
A child defecates a few minutes after being fed by the mother. This is most likely due to:
Your Answer: Gastrocolic reflex
Explanation:The gastrocolic reflex is a physiological reflex that involves increase in colonic motility in response to stretch in the stomach and by-products of digestion in the small intestine. It is shown to be uneven in its distribution throughout the colon, with the sigmoid colon affected more than the right side of the colon in terms of a phasic response. Various neuropeptides have been proposed as mediators of this reflex, such as serotonin, neurotensin, cholecystokinin and gastrin.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 26
Correct
-
Which of the following associations is correctly matched with the body's defence mechanism in fighting infection?
Your Answer: Specific cellular mechanism → cytotoxic T cells
Explanation:The immune system has certain levels of defence against pathogens. First line includes simple barriers such as skin, mucosa and stomach acid that prevent the pathogen from entering into the body. If this barrier is breached then the innate immune system is activated which includes leukocytes (macrophages, neutrophils, mast cells, eosinophils, basophils, natural killer cells). If the pathogens invade the second layer of defence then the third layer, adaptive immunity is activated, which includes B and T lymphocytes. B cells provide a humoral response whereas cytotoxic T cells have specific cellular mechanisms. They maintain a memory of past infections and are activated faster following a recurrence.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 27
Correct
-
What is the normal duration of PR interval on an electrocardiogram of a healthy individual?
Your Answer: 0.12–0.20 s
Explanation:PR interval extends from the beginning of the P-wave until the beginning of the QRS complex. The normal duration of the PR interval is 0.12-0.20 s. It can be prolonged in first degree heart block, and reduced in Wolff-Parkinson-White syndrome.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 28
Correct
-
Which of the following will be a likely sequelae of complete ileal resection?
Your 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.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 29
Incorrect
-
The anatomical dead space in a patient with low oxygen saturation, is 125 ml, with a tidal volume of 500 ml and pa(CO2) of 40 mm Hg. The dead space was determined by Fowler's method. If we assume that the patient's lungs are healthy, what will his mixed expired CO2 tension [pE(CO2)] be?
Your Answer: 0 mmHg
Correct Answer: 30 mmHg
Explanation:According to Bohr’s equation, VD/VT = (pA(CO2) − pE(CO2))/pA(CO2), where pE(CO2) is mixed expired CO2 and pA(CO2) is alveolar CO2pressure. Normally, the pa(CO2) is virtually identical to pA(CO2). Thus, VD/VT = (pa(CO2)) − pE(CO2)/pa(CO2). By Fowler’s method, VD/VT= 0.25. In the given problem, (pa(CO2) − pE(CO2)/pa(CO2) = (40 − pE(CO2)/40 = 0.25. Thus, pE(CO2) = 30 mmHg. If there is a great perfusion/ventilation inequality, pE(CO2) could be significantly lower than 30 mm Hg, and the patient’s physiological dead space would exceed the anatomical dead space.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 30
Correct
-
Which of the following is NOT a nutritional factor involved in wound healing:
Your Answer: Vitamin B3
Explanation:Vitamin B6 is required for collagen cross-links.
Vitamin A is required for epithelial cell proliferation.
Zinc is required for RNA and DNA synthesis.
Copper is required for cross-linking of collagen.
-
This question is part of the following fields:
- Cell Injury & Wound Healing
- Physiology
-
-
Question 31
Incorrect
-
A 26-year-old female patient had the following blood report: RBC count = 4. 0 × 106/μl, haematocrit = 27% and haemoglobin = 11 g/dl, mean corpuscular volume (MCV) = 90 fl, mean corpuscular haemoglobin concentration (MCHC) = 41 g/dl. Further examination of blood sample revealed increased osmotic fragility of the erythrocytes. Which of the following is the most likely cause of this patient’s findings?
Your Answer: Iron-deficiency anaemia
Correct Answer: Spherocytosis
Explanation:Spherocytes are small rounded RBCs. It is due to an inherited defect of the RBC cytoskeleton membrane tethering proteins. Membrane blebs form that are lost over time and cells become round instead of biconcave. As it is a normochromic anaemia, the MCV is normal. it is diagnosed by osmotic fragility test which reveals increased fragility in a hypotonic solution.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 32
Incorrect
-
A recognised side-effect of prefrontal leukotomy is:
Your Answer: Receptive aphasia
Correct Answer: Confusion
Explanation:Used previously as a treatment for psychiatric disorders, prefrontal leucotomy severs the connection between the prefrontal cortical association area and the thalamus. This leads to functional isolation of the prefrontal and orbitofrontal association cortex. Thus, along with the desired reduction in anger and frustration, undesirable side effects included changes in mood and affect, as well as confusion.
-
This question is part of the following fields:
- Neurology
- Physiology
-
-
Question 33
Correct
-
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: Increased renal excretion of HCO3 –
Explanation:The atmospheric pressure is lower at high altitudes as compared with sea level. This leads to a decrease in the partial pressure of oxygen. Once 2100 m (7000 feet) of altitude is reached, there is a drop in saturation of oxyhaemoglobin. The oxygen saturation of haemoglobin determines the oxygen content in the blood. The body physiological tries to adapt to high altitude by acclimatization. Immediate effects include hyperventilation, fluid loss (due to a decreased thirst drive), increase in heart rate and slightly lowered stroke volume. Long term effects include lower lactate production, compensatory alkali loss in urine, decrease in plasma volume, increased erythropoietin release and red cell mass, increased haematocrit, higher concentration of capillaries in striated muscle tissue, increase in myoglobin, increase in mitochondria, increase in aerobic enzyme concentration such as 2,3-DPG and pulmonary vasoconstriction.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 34
Correct
-
Calculate the cardiac output of a patient with the following measurements: oxygen uptake 200 ml/min, oxygen concentration in the peripheral vein 7 vol%, oxygen concentration in the pulmonary artery 10 vol% and oxygen concentration in the aorta 15 vol%.
Your Answer: 4000 ml/min
Explanation:The Fick’s principle states that the uptake of a substance by an organ equals the arteriovenous difference of the substance multiplied by the blood flowing through the organ. We can thus calculate the pulmonary blood flow with pulmonary arterial (i.e., mixed venous) oxygen content, aortic oxygen content and oxygen uptake. The pulmonary blood flow, systemic blood flow and cardiac output can be considered the same assuming there are no intracardiac shunts. Thus, we can calculate the cardiac output. Cardiac output = oxygen uptake/(aortic − mixed venous oxygen content) = 200 ml/min/(15 ml O2/100 ml − 10 ml O2/100 ml) = 200 ml/min/(5 ml O2/100 ml) = 200 ml/min/0.05 = 4000 ml/min.
It is crucial to remember to use pulmonary arterial oxygen content and not peripheral vein oxygen content, when calculating the cardiac output by Fick’s method.
-
This question is part of the following fields:
- Cardiovascular
- Physiology
-
-
Question 35
Correct
-
Which of the following can occur even in the absence of brainstem co-ordination?
Your Answer: Gastric emptying
Explanation:Although gastric emptying is under both neural and hormonal control, it does not require brainstem co-ordination. Increased motility of the orad stomach (decreased distensibility) or of the distal stomach (increased peristalsis), decreased pyloric tone, decreased duodenal motility or a combination of these, all increase the rate of gastric emptying. The major control mechanism for gastric emptying is through duodenal gastric feedback. The duodenum has receptors for the presence of acid, carbohydrate, fat and protein digestion products, osmolarity different from that of plasma, and distension. Activating these receptors decreases the rate of gastric emptying. Neural mechanisms involve both enteric and vagal pathways and a vagotomy impairs the gastric emptying regulation. CCK (cholecystokinin) slows gastric emptying at physiological levels of the hormone. Gastrin, secretin and glucose-1-phosphate also slow gastric emptying, but require higher doses.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 36
Correct
-
Question 37
Correct
-
What is the role of factor VII in coagulation?
Your Answer: Initiates the process of coagulation in conjunction with tissue factor
Explanation:The main role of factor VII is to initiate the process of coagulation along with tissue factor (TF). TF is found in the blood vessels and is not normally exposed to the bloodstream. When a vessel is injured tissue factor is exposed to blood and circulating factor VII. Factor VII is converted to VIIa by TF.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 38
Correct
-
A 77-year-old woman's renal function is tested. The following results were obtained during a 24-h period:
Urine flow rate: 2. 0 ml/min
Urine inulin: 0.5 mg/ml
Plasma inulin: 0.02 mg/ml
Urine urea: 220 mmol/l
Plasma urea: 5 mmol/l.
What is the urea clearance?Your Answer: 88 ml/min
Explanation:Urea is reabsorbed in the inner medullary collecting ducts of the nephrons. The clearance (C) of any substance can be calculated as follows: C = (U × V)/P, where U and P are the urine and plasma concentrations of the substance, respectively and V is the urine flow rate. So, glomerular filtration rate = (0.220 × 2. 0)/0.005 = 88 ml/min.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 39
Correct
-
Which of the following variables are needed to calculate inspiratory reserve volume of a patient?
Your Answer: Tidal volume, vital capacity and expiratory reserve volume
Explanation:Vital capacity = inspiratory reserve volume + tidal volume + expiratory reserve volume. Thus, inspiratory reserve volume can be calculated if tidal volume, vital capacity and expiratory reserve volume are known.
-
This question is part of the following fields:
- Physiology
- Respiratory
-
-
Question 40
Correct
-
A medical student is asked to calculate the net pressure difference in a capillary wall, considering: Interstitial fluid hydrostatic pressure = –3 mmHg, Plasma colloid osmotic pressure = 28 mmHg, Capillary hydrostatic pressure = 17 mmHg, Interstitial fluid colloid osmotic pressure = 8 mmHg, and Filtration coefficient = 1. Which is the correct answer?
Your Answer: 0 mmHg
Explanation:The rate of filtration at any point along a capillary depends on a balance of forces sometimes called Starling’s forces after the physiologist who first described their operation in detail. The Starling principle of fluid exchange is key to understanding how plasma fluid (solvent) within the bloodstream (intravascular fluid) moves to the space outside the bloodstream (extravascular space). Fluid movement = k[(pc– pi)–(Πc– Πi)] where k = capillary filtration coefficient, pc = capillary hydrostatic pressure, pi= interstitial hydrostatic pressure, Πc = capillary colloid osmotic pressure, Πi = interstitial colloid osmotic pressure. Therefore: 1 × [capillary hydrostatic pressure (17) – interstitial fluid hydrostatic pressure (–3)] – [plasma colloid osmotic pressure (28) – interstitial fluid colloid osmotic pressure (8)] = 0 mmHg
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
-
Question 41
Correct
-
Which is a feature of the action of insulin?
Your 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.
-
This question is part of the following fields:
- Endocrine
- Physiology
-
-
Question 42
Correct
-
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: Physiological polycythaemia
Explanation:Polycythaemia is a condition that results in an increase in the total number of red blood cells (RBCs) in the blood. It can be due to myeloproliferative syndrome or due to chronically low oxygen levels or rarely malignancy. In primary polycythaemia/polycythaemia vera the increase is due to an abnormality in the bone marrow, resulting in increases RBCs, white blood cells (WBCs) and platelets. In secondary polycythaemia the increase occurs due to high levels of erythropoietin either artificially or naturally. The increase is about 6-8 million/cm3 of blood. A type of secondary polycythaemia is physiological polycythaemia where people living in high altitudes who are exposed to hypoxic conditions produce more erythropoietin as a compensatory mechanism for thin oxygen and low oxygen partial pressure.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 43
Correct
-
Post-total gastrectomy, there will be a decreased production of which of the following enzymes?
Your Answer: Pepsin
Explanation:Pepsin is a protease that is released from the gastric chief cells and acts to degrade proteins into peptides. Released as pepsinogen, it is activated by hydrochloric acid and into pepsin itself. Gastrin and the vagus nerve trigger the release of pepsinogen and HCl when a meal is ingested. Pepsin functions optimally in an acidic environment, especially at a pH of 2.
-
This question is part of the following fields:
- Gastroenterology
- Physiology
-
-
Question 44
Correct
-
Selective destruction of which of the following cells will affect antibody synthesis?
Your Answer: Plasma cells
Explanation:Plasma cell are memory cells. After the antigen Is engulfed by the B cells it is presented to the CD4+ helper cells via the MCH II receptor and this leads to their activation which in turn stimulates the B cells to form antibodies against that specific antigen. Some B cells differentiate into plasma cells also called memory cells that get activated after subsequent infection.
-
This question is part of the following fields:
- General
- Physiology
-
-
Question 45
Correct
-
The blood-brain barrier is a membrane that separates the circulating blood from the brain extracellular fluid in the central nervous system (CNS). Which of the following statements regarding the blood– brain barrier is CORRECT?
Your Answer: It breaks down in areas of brain that are infected
Explanation:The blood–brain barrier is a membrane that controls the passage of substances from the blood into the central nervous system. It is a physical barrier between the local blood vessels and most parts of the central nervous system and stops many substances from travelling across it. During meningitis, the blood–brain barrier may be disrupted. This disruption may increase the penetration of various substances (including either toxins or antibiotics) into the brain. A few regions in the brain, including the circumventricular organs, do not have a blood–brain barrier.
-
This question is part of the following fields:
- Fluids & Electrolytes
- Physiology
-
00
Correct
00
Incorrect
00
:
00
:
0
00
Session Time
00
:
00
Average Question Time (
Secs)