Physio 50q

USMLE WORLD STEP 1 PHYSIOLOGYQuestion List Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: 1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 2 0 2 1 2 2 2 3 2 4 2 5 2 6 2 7 2 8 2 9 Renal Renal Pulmonology Blood vessels Neurology Pulmonology Musculoskeletal Cardiology Cardiology Cardiology Pulmonology Pulmonology Cardiology Gastrointestinal system Pulmonology Renal Endocrinology Neurology Cardiology Musculoskeletal Pulmonology Renal Renal Cardiology Endocrinology Pulmonology Pulmonology Renal Pulmonology Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: 4 3 4 4 4 5 4 6 4 7 4 8 4 9 5 0 5 1 5 2 5 3 5 4 5 5 5 6 5 7 5 8 5 9 6 0 6 1 6 2 6 3 6 4 6 5 6 6 6 7 6 8 6 9 7 0 7 1 Cardiology Neurology Musculoskeletal Pulmonology Cardiology Cardiology Pulmonology Cardiology Cardiology Renal Pulmonology Renal Pulmonology Blood vessels Pulmonology Pulmonology Renal Endocrinology Neurology Pulmonology Cardiology Endocrinology Endocrinology Dermatology Endocrinology Gastrointestinal system Musculoskeletal Gastrointestinal system Endocrinology 1 USMLE WORLD STEP 1 PHYSIOLOGY Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: 3 0 3 1 3 2 3 3 3 4 3 5 3 6 3 7 3 8 3 9 4 0 4 1 4 2 Cardiology Gastrointestinal system Pulmonology Cardiology Pulmonology Renal Endocrinology Cardiology Renal Neurology Renal Pulmonology Musculoskeletal Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: 7 2 7 3 7 4 7 5 7 6 7 7 7 8 7 9 8 0 8 1 8 2 8 3 8 4 Cardiology Endocrinology Musculoskeletal Renal Musculoskeletal Blood vessels Pulmonology Pulmonology Hematology Endocrinology Pulmonology Hematology Cardiology 2 USMLE WORLD STEP 1 PHYSIOLOGY Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 Renal Blood vessels Endocrinology Endocrinology Cardiology Renal Neurology Endocrinology Pulmonology Endocrinology Reproductive system Endocrinology Cardiology Pulmonology Endocrinology Cardiology Renal Musculoskeletal Renal Renal Genitourinary Hematology Renal Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: Physiology Q No: 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 Renal Renal Pulmonology Gastrointestinal system Hematology Pulmonology Gastrointestinal system Pulmonology Pulmonology Cardiology Cardiology Renal Cardiology Endocrinology Cardiology Pulmonology Cardiology Musculoskeletal Musculoskeletal Endocrinology Pulmonology Renal Gastrointestinal system 3 thus increasing circulating volume. Aldosterone release is stimulated by angiotensin II. respectively. Aldosterone increases sodium and water reabsorption in the collecting ducts of the nephron. Explanation: Aldosterone is a mineralocorticoid hormone synthesized and released by the zona glomerulosa cells of the adrenal cortex. and (to a small degree) ACTH. binds a cytosolic protein receptor and subsequently moves into the nucleus to modify gene expression. Which of the following is most likely to be decreased by the activity of this new drug? Proximal collecting tubule brush border enzyme activity B. which is normally activated in times of low blood pressure. Its net effect is to increase the number of Na/K-ATPase proteins and sodium channels in the cell membranes of the kidney’s cortical collecting ducts. Aldosterone is a component of the renin-angiotensin-aldosterone system. high serum potassium ion concentrations. As a consequence. Urea reabsorption in the collecting tubules A. Under the influence of aldosterone. (Choice A) The brush border enzyme activity of the proximal collecting tubules is responsible for reabsorbing of 2/3 of the sodium and water filtered from the 4 . sodium and water are removed from the tubular fluid actively and passively.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 1: A new agent has been shown to block aldosterone’s interaction with its receptors. A steroid hormone aldosterone crosses the cell membrane directly. low visceral perfusion and low plasma volume. Thus. Hydrogen ion secretion from the intercalated cells of the collecting tubules E. Na-R-2C cotransport in the thick ascending limb D. aldosterone promotes potassium ion and hydrogen ion secretion from the intercalated cells of the renal collecting tubules. Intracellular proximal collecting tubule hydrogen ion generation C. potassium and hydrogen ions are lost into the tubular fluid. g. furosemide. phosphate and lactate by cotransport with sodium. (Choice B) Intracellular hydrogen ion generation in the proximal collecting tubule occurs by the action of carbonic anhydrase.USMLE WORLD STEP 1 PHYSIOLOGY glomerular capillaries into Bowman’s capsule. amino acids. 5 . (Choice C) Na-R-2C1 cotransport in the thick ascending limb is the target of the potent class of diuretics known as loop diuretics (e. These drugs are used as diuretics and in the treatment of mountain sickness. Carbonic anhydrase inhibitor drugs like acetazolamide inhibit this enzyme. ethacrynic acid). These enzymes are also responsible for reabsorbing filtered glucose. (Choice E) ADH (vasopressin) increases urea reabsorption in the collecting tubules by contributing to the formation of the cortico papillary renal interstitial osmotic gradient. The HCO is transported into the blood and the H is transported into the tubular fluid in exchange for sodium. Educational Objective: Aldosterone is a component of the renin-angiotensin-aldosterone system that acts on the intercalated cells of the renal collecting tubules to cause resorption of sodium and water and loss of potassium and hydrogen ions. These agents decrease the activity of the Na-K-2C1 cotransporter leading to potassium loss and diuresis. Carbonic anhydrase synthesizes HCO and H from CO2 and H20 within the cells of the early proximal tubule. Urea. respectively? A. Inulin. Thus. Bicarbonate. Line on the above graph indicates a substance with a rapidly increasing concentration in the tubular fluid. Substances that behave in this fashion are freely 6 . an upward (positive) slope indicates an increasing concentration of that substance as fluid moves toward distal parts of the tubule which is usually the result of secretion or non-reabsorption of that substance. Which of the following substances are most likely to produce curves 2 and 4. A downward slope indicates resorption of that substance in the proximal tubule. Inulin. Potassium chloride D. bicarbonate B. sodium Explanation: The tubular fluid plasma ultra filtrate value in the graph above is calculated by taking the tubular fluid concentration of a given substance in the proximal tubule and dividing that value by the initial concentration of that substance within Bowman’s capsule. sodium E. PAH C.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 2: Tubular fluid/plasma ultrafiltrate concentration ratios for several substances are shown on the slide below. USMLE WORLD STEP 1 PHYSIOLOGY filtered from the glomerular capillaries and are poorly reabsorbed from the proximal tubule. Most notably. and amino acids decrease. in the proximal tubule. Urea is ultimately secreted in veri high concentrations. Bicarbonate is reabsorbed in the proximal tubule due to the activity of carbonic anhydrase within proximal tubular cells. Line 5 indicates solutes that are avidly reabsorbed in the proximal tubule. glucose. lines 2 and 4 represent the behavior of urea and bicarbonate. respectively. but less so than PAH or inulin. Line 3 indicates no concentration change along the proximal tubule. Line 4 represents the behavior of bicarbonate in the proximal tubule. as well as the sodium or potassium concentration. as is it a waste product of metabolism. their concentration in the proximal tubule increases. Line 2 is indicative of the behavior of urea in the proximal tubular fluid. Chemicals that behave in this manner include creatinine. 7 . sodium and potassium are reabsorbed in concentrations equal with water in the proximal tubule. Urea is freely filtered from the glomerular capillaries and is poorly reabsorbed from the proximal tubule. these include glucose and amino acids. This reabsorbing force causes the concentration of bicarbonate to decrease as fluid runs along the proximal tubule. and para-aminohippuric acid (PAH). inulin. Thus. inulin. This line could represent the osmolality of the tubular fluid in this segment. Educational Objective: The concentrations of PAHI creatinine. while the concentrations of bicarbonate. Because water is reabsorbed faster than these substances. and urea increase as fluid runs along the proximal tubule. 8) In a healthy individual at sea level. Tissue 002 production D. This value is calculated using the alveolar gas equation: P4O2 = [FiO2 x (P .25mm Hg and the PaO2is given as 70mm Hg. It is a measured value determined with an arterial blood gas analysis. (Choice D) The diffusion capacity of the lungs is decreased in conditions where the alveolar walls are thickened. Left-to-right cardiac shunting F. Pa02> 92 mmHg.g.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 3: A 34-year-old male living on the coast suffers from mild dyspnea. His arterial blood is drawn and shows a Pa02 of 70 mmHg and a PaCO2 of 35 mmHg. A-a gradient = PAO2 . The arterial P02 is measured with an arterial blood gas analysis. the A-a gradient does not exceed 10-15 mmHg. (Choice A) Lung compliance is generally increased in patients with emphysema. (Choice F) The mixed venous blood oxygen content is increased in conditions where abnormal hemoglobin binds with greater avidity to oxygen preventing unloading of oxygen in the tissues or in conditions where oxidative metabolism is inhibited such as in cyanide toxicity. Normally. Pa02 is the partial pressure of oxygen in the arterial blood. Thus. PAO2-P8O2 difference C. (Choice E) Left-to-right shunting of blood occurs with a patent ductus arteriosus. during exercise. Lung diffusing capacity E. and the alveolar P02 is determined using the alveolar gas equation. This value is much higher than normal.21 at seal level) R = The ratio of carbon dioxide production to oxygen consumption (normal = 0. the A-a gradient is 36. Educational Objective: Determining the difference between the alveolar and arterial P02 (A-a gradient) can help determine the cause of hypoxemia. atrial septal defect or ventricular septal defect. e. (Choice C) Tissue CO2 production is increased when the rate of cellular metabolism is elevated. Explanation: The alveolar-arterial oxygen gradient (A-a gradient) is the difference between the partial pressure of oxygen in the alveoli and the partial pressure of oxygen in the arterial blood. as in pulmonary fibrosis and hyaline membrane disease. In a healthy individual.(PaCO2 10.. the PAO2 is normally about 100 mmHg. Mixed venous blood oxygen content A.25 mmHg. Which of the following is increased in this patient? Lung compliance B.Pa02 P402 is the partial pressure of oxygen in the alveolar air. Calculating this value helps to determine the cause of hypoxemia. Using the values provided in the question the PAO2forthis patient is 106. 8 .(PaCO2 / R) Fi02 = Fraction of inspired air that is oxygen (normal = 0.8) PB = Barometric pressure (normal = 760 mm Hg at sea level) P = Water vapor pressure (normal = 47 mm Hg at sea level) This equation can be simplified by inserting the normal values at sea level: PAO2 = 150 . Simple diffusion — molecules move through a membrane without the help of carrier proteins. They bind extracellular ions and molecules forming a carrier-substrate complex.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 4: Two graphs illustrating the transport of low molecular weight solutes across the plasma membrane are shown on the slide below. Protein carrier Explanation: The image above illustrates the difference between the rate of transport of solute across the cell membrane in simple diffusion (line 2) and carrier-mediated transport (line 1). Binding is followed by movement of the complex across the cell membrane to the intracellular space. Oil/water partition coefficient E. Concentration difference across the membrane B. There are two types of diffusion: 1. Once all of the carrier molecules are bound by substrate. (Choices A—D) The other factors mentioned are not significantly different between simple diffusion and facilitated diffusion. no further diffusion can occur until a carrier protein is vacated. The higher the concentration gradient the higher the rate of transport of the molecule or ion across the membrane. saturation is an important property of carrier-mediated diffusion (facilitated diffusion). Carrier proteins are typically transmembrane proteins that possess binding sites for the substrate they transport. Facilitated diffusion — requires carrier proteins. 9 . Membrane thickness D. This maximum rate of transport is referred to as the transport maximum (Tm) and is similar in principle to the in standard enzyme kinetics. Because there is a finite number of carrier proteins in the cell membrane. Diffusion area C. The difference in the shape of the graphs is explained by the presence of a carrier protein in the scenario represented by line 1. and these carrier proteins require a discrete amount of time to pass a single molecule or ion through the membrane. where the substrate is released into the cytoplasm. The difference in the concentration of the substrate across the membrane is called concentration gradient (P). 2. Which of the following best explains the difference in the shape of the curves? A. Transport mechanisms utilizing proteins are able to be saturated. If the thickness of the membrane CT) is higher (eg. This characteristic is widely used to describe the pharmacokinetic properties of general anesthetics. Movement of substrate across the cell membrane by these mechanisms depends on the presence of carrier proteins in the membrane. 10 . A high oil-water partition coefficient means that a solute is much more soluble in oil than in water. High molecular weight (MW) compounds diffuse slowly. lung fibrosis) then the rate of diffusion would be slower. Educational Objective: Carrier-mediated transport includes facilitated diffusion and active transport.USMLE WORLD STEP 1 PHYSIOLOGY The larger the surface area (BA) the greater the rate of diffusion. Threshold B. Temporal summation refers to sequential impulses from the same neuron overtime (Choice B) whereas spatial summation refers to impulses from several different neurons (Choice C). Spacial summation D. A decrease in the space constant reflects increased charge dissipation along a nerve axon and will result in impaired stimulus transmission along an axon. Demyelination does not change the other parameters listed in the question. not multiple sclerosis.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 5: A 23-year-old woman with multiple sclerosis experiences diplopia due to a small area of axon demyelination detectable on a brain MRI. Space constant E. Demyelination decreases the space constant and is responsible for symptoms in multiple sclerosis. One can think of myelin as insulation around a wire. Temporal summation C. Educational Objective: The space constant is a measure of the ability of an impulse to travel down an axon. 11 . the more effective its transmission of electrical charges. A decrease in which of the following is most likely responsible for this patient’s symptoms? A. The purpose of myelin is to increase the space constant thus demyelination will decrease the space constant and result in ineffective signal transduction. Chloride channel dysfunction Explanation: The space constant is a measure of how far along an axon an electrical impulse will travel. the better insulated a wire. Threshold refers to the membrane potential needed for an action potential to develop (Choice A). (Choice E) Chloride channel dysfunction is the hallmark of cystic fibrosis. A low space constant decreases the length an impulse will travel. and upper airway trapping are mechanisms of clearing the largest dust particles. (Choices A and B) Coughing. (Choice C) Mucociliary transport is the prima means of eliminating medium-sized particles. including platelet-derived growth factor (PDGF) and insulin-like growth factor (1SF). Upper airway trapping C. The pneumoconioses are diseases that result from the inhalation of fine dust particles. Educational Objective: Dust particles smaller than 2µm in size reach the alveoli. Some of these cytokines induce injury and inflammation of alveolar cells. which stimulate fibroblasts to proliferate and produce collagen. Particles 2. The finest pat des (diameter less than 2 µm) reach the terminal bronchioli and alveoli and are phagocytized by macrophages. Inflammation with subsequent fibrosis results.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 6: A 45-year-old male has been working as a coal miner for the past 15 years where he is exposed to industrial dust. 12 .5-10 pm in size enter the trachea and bronchi and are cleared by mucociliary transport. They are taken up by macrophages and stimulate connective tissue growth. sneezing. Alveolar macrophages that take up dust particles become activated and release a number of cytokines. Mucociliary transport D. Phagocytosis E. Growth factors. Which of the following is the prima pulmonary defense mechanism that clears the respirators tract of deposited particles less than 2 µm in size? A. Pneumoconiosis (interstitial lung fibrosis secondary to inhalation of inorganic dust) arises by this mechanism. Cough and sneezing B. The clearance mechanisms utilized by the lung vary depending on the size of the particles. Particles 10-15 pm in size are trapped in the upper respiratory tract. Immunologic memory Explanation: Dust particles are constantly being inhaled and cleared by the respiratory tract. are also released. (Choice E) Dust particles do not induce immunologic memory. High potassium conductance only Explanation: The resting potential refers to the difference in charge across the cell membrane at rest (when there is no signaling activity). 13 . then chloride would need to be the only membrane permeable ion with no contribution from other ions.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 7: The resting membrane potential for an isolated muscle cell is -70 mV. High chloride conductance and some sodium conductance C. and the question stem states that the membrane potential is-JO mV. Every charged ion that the cell membrane is permeable attempts to push the resting membrane potential of the cell toward its equilibrium potential. High potassium conductance and some chloride conductance E. The membrane prevents the free diffusion of ions. (Choice D) High potassium conductance does occur at rest and is the largest contributor to the resting membrane potential. High calcium conductance and some chloride conductance D. in reality. This separation of positive and negative charges across the cell membrane gives rise to an electrical potential difference that ranges from -60 to -80 mV. which pumps three Na ions out of the cell for every 2 K ions allowed into the cell. Equilibrium potentials for important ions are given below: ENa= +60mV EK= -80mV ECl= -70mV ECa= +l25mV Which of the following most likely forms the resting membrane potential of the cell? A. In the resting cell the permeability of the membrane for K is much higher than that for Na. the greater its contribution to the resting membrane potential will be. There are. High potassium conductance and some sodium conductance B. plays little role in the determination of the membrane potential. The resting membrane potential can therefore be viewed as a combination of the equilibrium potentials of the ions that can penetrate the cell membrane. however. (Choice B) If the chloride equilibrium potential is -70 mV and the resting membrane potential is -70 mV. which approaches the equilibrium potential for K. the typical resting membrane potential of a cell is -70 mV. The greater the permeability of a given ion. Out flow of K-ions through nongated channels maintains the negative charge inside the membrane. The cytoplasm of a typical cell is characterized by a low concentration of Na and high concentration of K. Chloride plays little to no role in the determination of the resting membrane potential. Each permeable ion does not contribute equally to the membrane potential: itis obvious that the resting membrane potential does not equal the average of the different equilibrium potentials. and for this reason the resting potential of the membrane is always less negative than the equilibrium potential for K (Choice A). a small number of channels that allow the flow of Na into the cell. and generally there is an excess of positive ions outside the cell and an excess of negative ions in the cytoplasm. This difference in Na and K concentration is maintained by the NaK-ATPase. Chloride. The extracellular fluid is inversely rich in Na and low in K. (Choice C) High calcium conductance would make the resting membrane potential positive. Ions with little permeability will contribute very little to the resting membrane potential. This small influx of Na ions decreases the membrane potential. a high potassium efflux and some sodium influx are responsible forthe value of the resting potential. which is typically about-JO mV.USMLE WORLD STEP 1 PHYSIOLOGY (Choice E) If a high potassium conductance were the only contributor to the resting membrane potential.state conditions. 14 . then the membrane potential would be the same as the potassium equilibrium potential. Educational Objective: The resting membrane potential is the difference in the electrical charges across the cell membrane under steady. In general. The ions that are most permeable to the cell membrane make the largest contribution to the resting membrane potential. (Choice E) In the absence of mitral stenosis. Educational Objective: The pulmonary capillary wedge pressure (PCV\IP) measures the left atrial end diastolic pressure (LAEDP). The PCWP and LVEDP would remain approximately equal. Mitral stenosis elevates the LAEDP and PCWP relative to the LVEDP.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 8: A 38-year-old female is being evaluated for progressive exertional dyspnea. tamponade would elevate the PCWP and LVEDP to the same degree. keeping them equal. (Choice D) Restrictive cardiomyopathy causes diastolic dysfunction and similar elevations of the PCVVP and LVEDP. cardiac catheterization reveals a LA end-diastolic pressure (LAEDP) that is significantly greater than the LVEDP. (Choice C) Dilated cardiomyopathy would reduce the left ventricular peak systolic pressure and elevate the PCWP and LVEDP. Under normal conditions. The PCWP and LVEDP would remain approximately equal. Cardiac tamponade Explanation: The pulmonary capillary wedge pressure (PCWP) is the pressure in a pulmonary artery distal to the point of its occlusion by an inflated intravascular balloon. i. (Choice B) Isolated aortic stenosis would cause elevation of the left ventricular peak systolic pressure. Aortic stenosis C. The following cardiac catheterization data are obtained: Left ventricular end-diastolic pressure --. Dilated cardiomyopathy D. Because there is no significant blood flow towards the left atrium (LA) beyond this point of occlusion. the LA pressure is nearly equal to the left ventricular (LV) pressure since the open mitral valve offers minimal resistance to flow between the two chambers.10 mmHg (normal 3-12 mmHg) Left ventricular peak systolic pressure --. mitral stenosis. 15 . In this patient. This abnormal pressure gradient implies increased resistance flow between the LA and LV.36 mmHg (normal < 12 mmHg) Which of the following is the most likely cause of this patient’s symptoms? A. the LAEDP is nearly equal to the LV end-diastolic pressure (LVEDP). Restrictive cardiomyopathy E. the pressure at the tip of the “wedged” pulmonary artery catheter becomes nearly equal to the LA pressure. Mitral stenosis B. It could also elevate both the PCVVP and the LVEDPI but these two values would remain approximately equal.110 mmHg (normal 100-140 mmHg) Pulmonary capillary wedge pressure --. During normal diastole.e. ejection. Blood volume loss Explanation: Ventricular pressure-volume loops depict the relationship between pressure and volume in the left ventricle during systole and diastole. Abdominal aorta clamping D. These graphs represent left ventricular contraction. The rightward shift of the “ventricular filling” portion of the graph indicates that a larger than normal volume is being placed into the ventricle during diastole. Normal saline infusion C. This can occur during any state of fluid overload: renal failure congestive heart failure. Dobutamine infusion B. or after infusion of intravenous fluids. 16 . Loss of contracting myocardial mass E. relaxation and refilling as follows: The dashed line above reflects increased ventricular preload.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 9: The solid line below depicts the pressure-volume loop for a 45-year-old hospitalized patient. The change indicated by the dashed line may reflect which of the following? A. The graph would be narrow because less blood would be ejected during ventricular contraction compared to the unclamped patient. causing narrowing and upward shift of the ventricular ejection phase of the pressure-volume loop (the result of higher than normal pressures at the time of aortic valve opening and closure). The result would be widening of the pressure-volume loop. An increase in the circulating volume would increase preload and cause a rightward widening of the pressure-volume loop. causing the ventricular filling phase to be shorter than in the normal control. This would affect both the ventricular ejection and isovolumetric relaxation phases of the pressure-volume loop. The isovolumetric relaxation segment would therefore be shifted to the right indicating a smaller ejection fraction (a greater volume of blood remaining in the heart after contraction is complete). Higher pressures would be reached during the ventricular ejection phase of the pressure-volume loop and a greater volume of blood would be ejected during contraction. 17 . Clamping of the abdominal aorta would drastically increase afterload.USMLE WORLD STEP 1 PHYSIOLOGY Dobutamine infusion would cause an increase in contractility. (Choice E) Blood volume loss would have an effect opposite that depicted. Loss of contracting myocardial mass would result in decreased contractility. The ventricular ejection segment would likely not reach as high a pressure as a normal control and would not last as long as a normal control. Educational Objective: Pressure-volume loops represent the relationship between pressure and volume in the left ventricle during systole and diastole. USMLE WORLD STEP 1 PHYSIOLOGY Q NO 10: The cardiac and venous return curves of a healthy person are shown below in solid lines. Which of the following is the most likely cause of the change depicted by the dashed lines? A. Chronic arteriovenous fistula B. Acute gastrointestinal bleed C. Phenylephrine infusion D. Myocardial infarction E. Anaphylaxis Explanation: The dashed lines depict an increased cardiac output (increase in the height of the cardiac function curve), a decreased TPR (increased slope of both the cardiac function curve and the venous return curve) and an increased mean systemic pressure (rightward shift of the venous return curve on the x-axis). A chronic arteriovenous fistula would cause these changes. Acutely, a fistula causes a decrease in TPR resulting in an increased cardiac output and an increased venous return. The venous return curve does not immediately shift on the x-axis, but over time, the sympathetic nervous system and kidneys will begin to compensate for a chronic fistula by increasing TPR, cardiac contractility, and the circulating blood volume. These changes further increase the cardiac function curve and increase the right atrial pressure, causing a rightward shift of the venous return curve on the xaxis. (Choice B) Acute GI bleeds cause a sharp decrease in the circulating blood volume and shifts the venous return graph to the left. (Choice C) 18 USMLE WORLD STEP 1 PHYSIOLOGY A phenylephrine infusion would increase sympathetic tone resulting in vasoconstriction and an increase in TPR, as shown above. The increased TPR causes more blood to remain on the arterial side of the circulation thus decreasing venous return. Since afterload is high, cardiac output decreases as well. (Choice D) Myocardial infarction causes an isolated decrease in cardiac output with no change in blood volumes or venous return. (Choice E) Anaphylaxis causes widespread venous and arteriolar dilatation along with increased capillary permeability and third-spacing of fluids. This results in a serious drop in venous return. Educational Objective: A chronic arteriovenous shunt would increase cardiac output because of increased sympathetic stimulation to the heart, increased venous return, and decreased TPR. It would also cause the venous return curve to shift to the right because the circulating blood volume is increased by renal retention of fluids and the venous return is increased by sympathetic tone. 19 USMLE WORLD STEP 1 PHYSIOLOGY Q NO 11: A researcher measures alveolar ventilation in a healthy volunteer. Which of the following values accounts for the difference between the minute ventilation and the alveolar ventilation? A. Tidal volume B. Breath rate C. Functional residual capacity D. Residual volume E. Dead space Explanation: The minute ventilation is the volume of air that enters or leaves the lungs and conducting airways in one minute. The minute ventilation accounts for all air, including air that does not participate in gas exchange, entering or leaving the airways per minute. The minute ventilation can be calculated using the following equation: Minute ventilation (L/min) z Tidal Volume (L) x breaths/minute The volume of air that does not participate in gas exchange is known as the physiologic dead space. The physiologic dead space can be calculated with the following formula: VD = VT X ((Pa CO2 — PECO2) / PaCO2) where VD is the dead space volume. VT is the tidal volume, P3C02 is the arterial partial pressure of C02, and PECO2 is the partial pressure of CO2 in the expired air. The alveolar ventilation differs from the minute ventilation in that the alveolar ventilation refers only to the volume of air participating in gas exchange per minute. Thus, the alveolar ventilation does not include the dead space volume while the minute ventilation does. The alveolar ventilation can be calculated with the following formula: Alveolar ventilation (L/min) = (Tidal Volume — Dead space volume) x breaths/minute (Choice A) The tidal volume is the volume of air that is inhaled or exhaled with each normal, unforced breath and is a factor in calculating both the minute ventilation and the alveolar ventilation. (Choice B) The breath rate is the number of breaths taken per minute and is a factor in calculating both the minute ventilation and the alveolar ventilation. (Choice C) The functional residual capacity is the volume of air remaining in the lungs after a normal exhalation. It is the sum of the residual volume and the expiratory reserve volume. (Choice D)The residual volume is the volume of air remaining in the lungs after forced exhalation. The residual volume cannot be measured by spirometry. Educational Objective: The minute ventilation is equal to the product of the tidal volume and the respiratory rate and includes dead space ventilation. The alveolar ventilation does not take into account the air in the physiologic dead space. It is the product of the respiratory rate and the difference between the tidal volume and the dead space volume. 20 The spheres will become equal in size B. What should be expected when the clamp is opened? A. The smaller sphere will increase in size C. as the radius of a sphere with constant surface tension decreases.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 12: Two spheres with elastic properties similar to alveoli are connected as shown below. Thus. The smaller sphere will collapse E The larger sphere will collapse Explanation: Laplace’s law tells us that the pressure inside a bubble is directly proportional to the surface tension (T) and inversely proportional to the radius (R). The surface tension within the two spheres is identical. An important role of surfactant is to decrease surface tension and thus prevent alveolar collapse. P α T/R Assuming constantly. the smaller sphere will tend to collapse and the larger sphere will inflate further. The larger sphere will decrease in size D. This situation models what would happen in the lungs if two alveoli of different sizes without surfactant were connected and then opened to the atmosphere. 21 . a bubble with a smaller radius will have a higher collapsing pressure. Educational Objective: According to Laplace’s law. Surfactant counteracts alveolar collapse by decreasing surface tension as the alveolar radius decreases. when the clamp above is opened. the collapsing pressure increases. Neither is lined with surfactant. The initial readings were most likely obtained from which of the following locations? A. we can see that the catheter was probably initially positioned in the right ventricle (RV). left-sided pressure are greater than right-sided pressures. Advancement of the catheter from this point would place the catheter in the pulmonary artery (PA). Right ventricle C. the maximum pressure is normally close to 25mm Hg.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 13: A 46-year-old patient is undergoing cardiac catheterization. Recall that in general. Using the above table. Pulmonary artery Explanation: This question tests our knowledge of normal pressure variations in the chambers of the heart and pulmonary artery during the cardiac cycle. The second set of pressure readings are consistent with normal PA pressures. The left ventricular and aortic maximum pressures are normally close to the systolic blood pressure. Left atrium E. the catheter records periodic pressure changes with a maximum of 27 mmHg and minimum of 2 mmHg. Left ventricle D. the maximum pressure is normally about 25mm Hg. The catheter is advanced further. Within the atria. the normal pressure maximum is close to 10mm Hg. The patient’s RV pressures are in the normal range. and within the pulmonary artery. Right atrium B. Representative (adult) values are provided below. Educational Objective: Normal (adult) pressures in the cardiac chambers and pulmonary artery are as follows: 22 . Initially. Within the right ventricle. and then shows periodic pressure changes with a maximum of 26 mmHg and a minimum of 10 mmHg. Gastrin D. It functions to increase pancreatic enzyme secretion (by acinar cells) and gallbladder contraction. gastrin. (Choice F) Somatostatin is a hormone made in numerous tissues (hypothalamus. Somatostatin inhibits the release of growth hormone (GH) and thyroid stimulating hormone (TSH). counteracts gastrin in the stomach.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 14: A 43-year-old obese woman presents to the emergency room with severe epigastric pain that started after a heavy meal.g. and vagal stimulation. In cholecystitis. Educational Objective: Cholecystokinin (OCR) is the hormone responsible for gallbladder contraction. and suppresses the release of myriad other molecules (e. 23 . intestine. Moreover. Cholecystokinin (CCK) is the hormone responsible for gallbladder contraction. (Choice D) Vasoactive intestinal peptide (VIP) is produced in the pancreas and stimulates intestinal water secretion. stomach. insulin. She has had several milder episodes before related to fatty food consumption. and to decrease gastric empting. glucagon). and pancreas) in response to low pH. Which of the following hormones most likely provoked the current attack in this patient? A. (Choice A) Secretin promotes bicarbonate secretion from pancreatic ductal epithelium. secretin. VIP. Itis produced by l cells of the duodenum and jejunum when fat-protein-rich chyme enters the duodenum. It has no effect on pancreatic enzyme secretion. Physical examination reveals marked tenderness in the right subcostal area. It is made in the duodenum and jejunum in response to fatty acids and amino acids. CCK. fatty foods increase CCK production and pain occurs when an inflamed and/or obstructed gallbladder contracts. she has three well-known risk factors for gallbladder disease (“forty” “fat” and “female”). It situates gastric acid secretion and motility. and promotes bicarbonate secretion for the pancreas. Vasoactive intestinal peptide (VIP) E. Its production is inhibited by vagal stimulation. Motilin F. Cholecystokinin C. (Choice E) Motilin is a hormone made in the small intestine that promotes intestinal motility. (Choice C) Gastrin is a hormone made in the stomach in response to stomach distention peptides. Secretin B. Somatostatin Explanation: This patient presents with classic signs and symptoms of biliary colic. exercise causes a uniform simultaneous increase in both ventilation and perfusion of individual alveoli. but does not significantly increase the physiologic dead space. The venous blood pH is decreased. respectively. Educational Objective: During aerobic exercise. In healthy individuals. The venous blood CO2 content is increased due to increased CO2 production. CO2 content in the arterial blood B. increased skeletal muscle CO2 production increases the PCO2 of mixed venous blood. pH of the arterial blood E. O2 content in the arterial blood C. These increases are balanced by increases of the cardiac output/skeletal muscle perfusion and ventilation. Homeostatic mechanisms maintain arterial 02 and CO2 contents and arterial pH near normal resting values but there are significant changes in the venous blood O2 and CO2 contents and pH. (Choice E) Physiologic dead space corresponds to regions of alveolar ventilation that do not engage in gas exchange due to relatively low alveolar perfusion and/or alveolar hyperaeration. 24 . Which of the following do you expect to increase at the peak of his exertion? A.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 15: A 23-year-old male runs five miles. CO2 content in the mixed venous blood D. The active skeletal muscles increase their rate of both oxygen consumption and carbon dioxide production. The hyperventilation that accompanies exercise may slightly increase the anatomic dead space by stretching the conducting airways with fuller inspirations. Physiologic dead space Explanation: Long distance running and other forms of aerobic exercise cause increased oxidative metabolism of glucose and fatty acids in skeletal muscle. Because exercising muscles extract additional O2 the venous blood 02 content is decreased. Homeostatic mechanisms maintain arterial blood gas levels and arterial pH near the resting values. (Choice D) The arterial pH is usually normal in moderate exercise but may decrease during strenuous exercise due to lactic acidosis. PG refers to the hydrostatic pressure in the glomerular capillaries. This will decrease RPEI but will increase glomerular capillary hydrostatic pressure as the fluid “backs up” in the glomerulus. 25 . The renal plasma flow (RPE) is the volume of plasma that is delivered to the kidney per unit time. Hyperproteinemia B. Bladder neck obstruction C. this increased glomerular hydrostatic pressure will cause an increase in the filtration fraction. Dilation of the efferent arteriole Explanation: Constriction of the efferent (outgoing) arteriole will impede blood flow through the kidney. Constriction of the afferent arteriole E. The GER is dependent on hydrostatic and oncotic pressures in the glomerular capillaries and Bowman’s space and can be calculated with the following equation: GER = Kf ((PG — PB) — (TTG — TTB)) Where Kf refers to the coefficient of filtration. Using the equation above. increasing hydrostatic pressure.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 16: A researcher is studying the effect of various manipulations on kidney blood flow and glomerular filtration. Clinically. As described below. The RPE is theoretically calculated by subtracting the amount of erythrocytes (hematocrit) from the total renal blood flow. Increases in the glomerular capillary hydrostatic pressure or the Bowman’s space oncotic pressure will increase GFR. The RPF is provide by the renal blood flow. EPE is linked to the glomerular filtration rate (GER) and the filtration fraction (FE) by the following equation: FE= GFR/RPF The filtration fraction refers to the proportion of the RPE that is filtered from the glomerular capillaries into Bowman’s space and is expressed as a percentage. however RPE is generally estimated by calculating the para-aminohippuric acid (PAH) clearance. one can observe that increases in GER or decreases in RPF will increase the FE. Which of the following is most likely to both decrease renal plasma flow and increase the filtration fraction? A. PB refers to the hydrostatic pressure in Bowman’s space. TTG refers to the oncotic pressure in the glomerular capillaries and TTB refers to the oncotic pressure in Bowman’s space. which delivers both erythrocytes and plasma to the kidney. Constriction of the efferent arteriole D. (Choice B) Bladder neck obstruction causes an increase in Bowman’s space hydrostatic pressure thereby decreasing GFR. and the FF is decreased due to the decrease in GFR and increase in RPE. The FE remains unchanged due to decreases in both GFR and RPE. (Choice A) Hyperproteinemia causes increased glomerular capillary oncotic pressure thereby decreasing GFR. but will cause a decreased FF due to the decreased GFR. The RPE is increased by this process. (Choice E) Dilation of the efferent arteriole causes decreased glomerular capillary hydrostatic pressure leading to a decrease in GFR. RPE is also decreased by this process. This will have no election the RPF but will cause a decreased FF due to the decreased GFR. The filtration fraction (FF) can be calculated by dividing the GFR by the renal plasma flow (RPF). 26 .USMLE WORLD STEP 1 PHYSIOLOGY while increases in capillary oncotic pressure or Bowman’s space hydrostatic pressure will decrease GFR. This too will have no effect on the RPF. Increases in GFR or decreases in RPF will increase the FF. (Choice D) Constriction of the afferent (incoming) arteriole causes a decrease in glomerular capillary hydrostatic pressure leading to a decreased GFR. Educational Objective: Increases in the capillary hydrostatic pressure or the Bowman’s space oncotic pressure will increase GFR while increases in capillary oncotic pressure or Bowman’s space hydrostatic pressure will decrease GFR. not insulin. with insulin inhibiting glucagon release. There is classically a fruity odor on their breath due to the presence of ketone bodies. severe abdominal pain. the body perceives hypoglycemia and a starved state despite high serum glucose levels.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 17: A 23-year-old male is brought to the emergency room with weakness and confusion. Glucagon. DKA can be the initial presentation of DM. especially in patients lacking regular health follow-up. The three cardinal signs of diabetes mellitus (DM) are polyuria. Insulin and glucagon normally act in opposition to one another. He has experienced polyuria. inhibits ketoacid formation and inhibits lipolysis. (Choice D) Insulin promotes glucose storage in the form of glycogen. Increases renal glucose production E. polydipsia and polyphagia. dry mucous membranes and lethargy. Patients with DM (especially type I DM) however. Increases the renal threshold for glucose reabsorption Explanation: This patient is exhibiting classic signs and symptoms of diabetic ketoacidosis (DRA). Glucagon stimulates ketoacid synthesis in adipose tissue because during starvation ketoacids can be used by cells for energy in place of glucose. 27 . lipolysis and ketone body production while insulin increases glucose. This patient is deficient in a substance that A. Patients with DRA clinically exhibit nausea and vomiting. (Choice A) Factors that increase glucagon secretion are hypoglycemia. tachycardia. Glucagon also increases glycogenolysis. gluconeogenesis. lipolysis. Facilitates the action of glucagon D. Increases glucagon secretion B. Decreases glucagon secretion C. and urea production. His breath has a fruity odor. DRA results from a deficiency of insulin and an excess of glucagon coupled with adrenergic activation and increased levels of cortisol and growth hormone. adrenergic stimulation cholecystokinin and acetylcholine. polydipsia and increased appetite recently. stimulates glucose production (gluconeogenesis). Because glucose is inadequately transported into cells in patients with DM and DRA. gluconeogenesis. amino acid and potassium uptake by cells. increased serum amino acid concentration. (Choice E) Normally all glucose filtered from the blood in the glomerular capillaries into Bowman’s capsule is reabsorbed in the proximal convoluted tubule. Educational Objective: Insulin opposes glucagon action. Undiagnosed DM can progress to DKA. are unable to synthesize sufficient insulin to prevent hyperglycemia and to inhibit glucagon’s effects. Adrenergic nervous system activation and increased glucagon production result. Glucagon stimulates glycogenolysis. tachypnea. Insulin does not affect this process. K efflux occurs via nongated K channels (leak channels). Hyperpolarizing afterpotential (Choice E): Occurs due to the fact that for a short moment following repolarization both the voltage-gated and the non-voltage-gated potassium channels are open. 2. Potassium efflux is responsible for returning the membrane potential back to the resting potential. Educational Objective: The action potential results from changes in the membrane conductance for K and Na ions. The membrane potential of an excitable cell (nerve. A B C D E Explanation: The graph above depicts the potential (voltage / conductance) changes across a cell membrane. D.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 18: A recording of the membrane potential of a giant axon is shown on the slide below. and V1 and V2 represent the corresponding flow velocities)? . This results in a sharp decrease in the conductance of sodium and a significant increase in potassium conductance to levels even higher than those seen in the resting membrane potential. Itis maintained by a high membrane permeability to Rand a low permeability to Na. Repolarization occurs due to closure of voltage-gated Na channels and opening of voltage-gaited K channels. Depolarization causes the membrane potential to become positive (approximately +35mV). 4. Which of the points indicated on the slide most likely corresponds to the highest membrane conduction for potassium A. E. 3. The large influx of Na ions that occurs after opening of the voltage-gated sodium channels leads to an increased negative charge outside the membrane known as depolarization (Choice B). Depolarization: Occurs due to opening of voltage-gated Na channels with rapid influx of sodium ions into the cell. The membrane potential becomes more negative than the normal resting potential and approaches the K equilibrium potential of-85 mV. Q NO 19: In order to maintain constant fluid flow through a tube with varying diameters which of the following would be true (where Al and A2 represent 28 cross-sectional areas. hyperpolarization and resting potential) are collectively known as the action potential. C. Potassium ion conduction is highest during the repolarization phase of the action potential. During the resting potential the inner side of the membrane is negatively charged with respect to the outer surface of the membrane. Depolarization results from massive influx of Na through voltage-gated channels. muscle) undergoes the following stages: 1. Resting potential (Choice A): Usually equal to-JO mV. Repolarization (Choice D): Results from closure of Na channels and simultaneous opening of K channels. The action potential occurs due to changes in the membrane permeability to Na and K ions. Overshoot (Choice C) refers to the maximal value of the action potential where the membrane potential is a positive value. these changes (depolarization repolarization. B. When the voltage-gated K-channels close the membrane potential returns to the resting value. V1=A1*A2/V2 E. 29 .g. This principle is applied in clinical practice when the cross sectional aortic valve area is calculated from Doppler echocardiographic measurements of the velocity of the jet of blood flow at the aortic valve. This is because numerous capillaries arranged in parallel receive flow from a given feeder vessel: thus the functional crosssectional area of those capillaries combined is actually much larger than that of the feeder vessel from which they receive their flow Educational Objective: The law of conservation of mass applied to the steady state flow of an incompressible fluid through a system of cylinders of varying cross sectional areas tells us that: Total Flow = Flow Velocity x Cross Sectional Area = Constant This principle may be applied to blood flow in the cardiovascular system. Vol in = A1*V1 = Vol out = A2*V2 (Choice C) is the only answer choice consistent with the above equation. the steady state velocity of fluid (e. the cardiovascular system).g. density may be considered constant. A2=A1*V1/V2 D. volume in (Vol in) = volume out (Vol out). Applying this law to the cylinders above: Q in = grams of fluid material in per unit time = volume in x density (in)/time Q out = grams of fluid material out per unit time = volume out x density (out) time For a relatively incompressible fluid (e. This equation basically states that in a system of connected cylinders of varying radii (e. In other words. Therefore in any steady state.USMLE WORLD STEP 1 PHYSIOLOGY A. blood). V1*A2=V2*A1 Explanation: The law of conservation of mass as applied to fluid dynamics tells us that the total flow of mass into a contained system must be equal to the total outflow of mass from the system in the steady state. blood) flow at a given point is inversely related to the cross sectional area at that point. V1=A1*V2 C.g. V1=V2 B. the flow velocity increases at points of narrowing. Although individual capillaries have a small cross-sectional area the flow velocity through these vessels is slow. (Choice E) Myosin light chain kinase phosphorylates the myosin light chain. This attachment causes immediate: Calcium binding to troponin C B. radiolabeled ATP is injected into skeletal muscle. ATP binding to myosin causes release of the myosin head from its binding site on the actin filament. (Choices A and B) Calcium binding to troponin C shifts tropomyosin away from the myosin binding site on actin. Myosin light chain phosphorylation by a specific enzyme A. 30 . Myosin head detachment from the actin filament D. These steps do not depend directly on ATP. Educational Objective: During the skeletal muscle contraction cycle. the role of ATP in skeletal and cardiac muscle contraction may be to release the myosin head from its actin binding site and then to energize a conformational change that resets the myosin head to “contract” again the next time it binds to actin. Cross-bridge formation E. the labeled ATP is observed to attach to the sarcomere.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 20: As part of an experiment. There is no myosin head phosphorylation by a specific kinase in skeletal muscle contraction. During muscle contraction. the cross-bridge between myosin and actin will persist (rigor mortis).. Explanation: According to the model posited by Rayment et al. Tropomyosin displacement from the groove on the actin molecule C. activating myosin to bind actin filaments in smooth muscle cells. (Choice D) If ATP is not available. as blood moves through the lungs. the majori of blood from the bronchial arteries is returned to the left heart in 31 . has a lower p02 (100 mmHg) than blood in the pulmonary capillaries. This decrease is due to the admixture of deoxygenated bronchial blood with blood in the pulmonary veins and is represented on the above graph by the downward deflection identified by the arrow. Perfusion limitation C. Blood is subsequently pumped from the right ventricle into the pulmonary arteries for oxygenation in the lungs. Which of the following most likely explains the tension change indicated by the arrow? A. which is generally about 104 mmHg in the pulmonary capillary beds. As blood enters the right heart from the venous system. Shift in the hemoglobin dissociation curve Explanation: This graph illustrates the p02 of blood as it moves from the venous system through the lungs and into the systemic circulation. it has a low p02 because much of its oxygen has been removed for use in the tissues. While bronchial veins do exist. Blood in the left atrium however. Normally. Wasted ventilation (dead space) D. Venous admixture E. This is represented by the curve’s first upward deflection. These vessels carry oxygenated blood to the bronchi and bronchioles and together with the pulmonary veins form the dual blood supply to the lungs. The left and right bronchial arteries arise from the descending thoracic aorta. Diffusion limitation B. blood returns to the left heart via the pulmonary veins. it becomes progressively oxygenated until it equilibrates with the alveolar p02 (p02).USMLE WORLD STEP 1 PHYSIOLOGY Q NO 21: The blood oxygen tension curve for an apparently healthy individual is shown below. After becoming fully oxygenated in the pulmonary capillaries. The bronchial veins return blood to the right heart via the azygous. there would be no alveolar blood p02 plateau. adult respiratory distress syndrome. emphysema or hyaline membrane disease of the infant. In diffusion-limited gas exchange. Increases in pCO2. In graph form. temperature and/or 23diphosphoglycerate concentration all cause shifting of the Hb dissociation curve to the right. (Choice A) Diffusion limitation is a cause of hypoxia that typically results from pulmonary fibrosis. The physiologic (anatomic) dead space refers to the volume of air that remains in the conducting parts of the airways. the blood p02 does not equilibrate with the alveolar p02 by the end of the alveolar capillary. The opposite conditions shift the curve to the left and favor oxygen loading in the lungs. accessory hemiazygous or intercostals veins. 32 . Educational Objective: The p02 in the left atrium is lower than that in the pulmonary veins because deoxygenated blood from the bronchial arteries mixes with oxygenated blood in the pulmonary veins. (Choice E)The hemoglobin dissociation curve shifts to the right when the tissue environment favors oxygen unloading. such as the bronchi and bronchioles. Alveolar dead space refers to the volume of alveolar air that cannot transfer oxygen to the blood due to inadequate blood flow. (Choice C) Wasted ventilation (dead space) refers to a volume of inspired air that is not available for gas exchange.USMLE WORLD STEP 1 PHYSIOLOGY deoxygenated form via the pulmonary veins. loop diuretics also stimulate prostaglandin release. It is an important mediator in the body’s inflammatory response. Prostaglandins B. loop diuretics also increase renal blood flow leading to increased GFR and enhanced drug delivery. loop diuretics also stimulate prostaglandin release. Aldosterone promotes Na reabsorption in the distal tubule and the collecting duct. (Choice B) Endothelin is a peptide found in smooth muscles that causes smooth muscle contraction. naproxen. Although loop diuretics do stimulate renin release leading to angiotensin II formation. Thus concurrent use of NSAIDs with loop diuretics can result in a decreased diuretic response. Angiotensin II D. Rather. (Choice D) Aldosterone is a hormone secreted by the adrenal cortex. loop diuretics also increase renal blood flow leading to increased GFR and enhanced drug delivery. vasodilation. Educational Objective: Furosemide is a loop diuretic that works by inhibiting Na-K-2C1 symporters in the loop of Henle effectively causing increased Na. Natriuretic peptides do not play a role in the effect of loop diuretics. They cause vasodilation. a week later the patient receives high-dose ibuprofen for joint pains and soon develops worsening abdominal distention. aldosterone antagonist diuretics like spironolactone work by inhibiting aldosterone. Natriuretic peptides F. Prostaglandins have vasodilatory effects. By stimulating renal prostaglandin release. and indomethacin inhibit prostaglandin synthesis. (Choice E) Natriuretic peptides like ANP and BNP are natural proteins that help regulate Na and fluid balance in the body. and fluid excretion. Additionally. Blunting of the diuretic response in this patient is due to interruption of which of the following substances? A. (Choice C) Angiotensin II is a peptide responsible for vasoconstriction and aldosterone release. Endothelin C. Both factors ultimately enhance diuretic response. He develops massive ascites and lower extremity edema that responds well to furosemide therapy. Cl and fluid excretion. Thus concurrent use of NSAID5 with loop diuretics can result in a decreased diuretic response. Thus by stimulating renal prostaglandin release. However. and diuresis in response to volume expansion. (Choice F) Bradykinin is a peptide that stimulates pain. Bradykinin Explanation: Furosemide is a loop diuretic that works by inhibiting Na-K-2C1 symporters in the ascending limb of the loop of Henle. 33 . It binds to symporters and effectively blocks Na and Cl transport resulting in increased Na. Nesiritide is a BNP analog that is used in the setting of acute decompensated heart failure for its vasodilatory and diuretic effects. Cl. angiotensin II does not play a role in the diuretic response of loop diuretics. Aldosterone E. Endothelin does not play a role in the effects of loop diuretics. natriuresis. Non-steroidal anti-inflammatory drugs like ibuprofen. Bradykinin does not play a role in the effects of loop diuretics. and increased vascular permeability.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 22: A 43-year-old male is diagnosed with hepatic cirrhosis due to prolonged heavy alcohol consumption. It is a very potent vasoconstrictor. Loop diuretics do not affect aldosterone. Additionally. Glucose is an example of a substance that is aggressively reabsorbed in the proximal tubule. the secretion of PAH by proximal tubular epithelial cells increases—but only up to a maximum value of approximately 80 mg/mm. Filtration fraction is decreased E. (Choice B) The maximal reabsorption rate does not apply to PAH. despite the fact that glucose is filtered into Bowman’s space. Once the plasma concentration of PAH is increased above this level the extraction ratio decreases progressively. which is a secreted substance. however is not enzyme or protein mediated so it does not show a maximal value. (Choice A) Excretion of a substance is defined as the filtration plus the secretion minus the reabsorption. As the blood concentration of PAH increases. Renal plasma flow is decreased Explanation: Para-aminohippuric acid (PAH) is an organic acid that is avidly filtered from the blood in the glomerulus and secreted by the cells lining the proximal tubule of the nephron. Once carrier-mediated secretion and reabsorption mechanisms are saturated these processes are at their maximal rate. Because filtration can not be saturated neither can excretion. 34 . Maximal reabsorption rate is reached C. At any given blood concentration. PAH is also taken up by the proximal tubular epithelial cells from the peritubular capillaries and secreted into the tubular fluid. Which of the following best explains the observed decrease in the PAH extraction ratio? A. Carrier transport is saturated D. a constant proportion of PAH will be filtered from the glomerular capillaries into the tubular fluid in Bowman’s space.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 23: It is estimated that for para-aminohippuric acid (PAH) the extraction ratio (arterial plasma PAH minus venous plasma PAH divided by arterial plasma PAH) is near 90% at arterial plasma concentrations lower than 20 mg/dL. Under normal conditions no glucose is lost in the urine. As the blood PAH concentration increases the tubular fluid PAH concentration in Bowman’s space will also increase on a linear basis. This maximum secretion rate corresponds with the transport maximum (TMAX) of the secretion enzymes and at this level secretion plateaus and any increases in the urine PAH concentration are due to increased filtration. the calculated clearance of this acid can be used to estimate the renal plasma flow (RPF). Maximal excretion rate is reached B. In uncontrolled diabetes however glucose is seen in the urine because the reabsorption carrier proteins become saturated at very high glucose concentrations. Filtration. Because PAH is both filtered and secreted by the glomerulus and renal tubules. This secretion is a carrier enzyme-mediated process and therefore is able to be saturated. In addition to being freely filtered. The filtration of PAH is not a process that can be saturated. USMLE WORLD STEP 1 PHYSIOLOGY (Choice D) The filtration fraction is decreased by decreases in GFR or increases in PF. The secretion of PAH can be saturated at high blood concentrations. Itis also secreted from the blood into the tubular fluid by the cells of the proximal tubule by a carrier Proteinmediated process. Educational Objective: Para-aminohippuric acid (RAH) is freely filtered from the blood in the glomerular capillaries to the tubular fluid in Bowman’s space. 35 . (Choice E)The renal plasma flow (RPF) is decreased by factors that constrict blood vessels in the kidney such as epinephrine norepinephrine and angiotensin II. Calcium D. Sodium and calcium Explanation: The graphs above illustrate the action potentials of two different cell types. Potassium C. Movement of which of the following ions or combinations of ions creates the deflection indicated by the arrow? A.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 24: Special electrodes are used to record the membrane potential changes of different cardiac muscle cells (see below). The arrow points to an area of the pacemaker cell action potential. Sodium B. Sodium and potassium F. Chloride E. 36 . Pacemaker cell action potentials do not have phases 1 or 2. 37 . which results from an outward potassium current. Phase 3 is the repolarization phase1 where potassium ions rush out of the cell and calcium conductance is decreased. Pacemaker cell depolarization is followed by phase 3 of the action potential. This differs from phase 0 of ventricular cells. which results from an inward sodium ion current. Depolarization is caused by a calcium ion influx in pacemaker cells. Phase 2 is the plateau phase of the action potential. in contrast to the Ca-mediated depolarization of pacemaker cells. In the ventricular myocyte. Phase 0 of the ventricular action potential is mediated by a rapid sodium ion influx. phase 4 corresponds to the resting membrane potential and is typically stable at approximately-90 mV (near the equilibrium potential of potassium). where inward calcium current counteracts the outward movement of potassium ions that started in phase 1. As the funny current brings the membrane potential closerto-40mV. sometimes referred to as the “funny current (L) . phase 0 depolarization occurs. phase 0 depolarization is mediated by an inward flux of calcium ions. Educational Objective: In pacemaker cells.USMLE WORLD STEP 1 PHYSIOLOGY Pacemaker cells are unique in that they exhibit automaticity (an inherent ability to depolarize without any external incense). The arrow above indicates phase 0 of the pacemaker action potential. Phase 1 is an early short repolarization that occurs immediately after the rapid depolarization. mediated by an increased outflow of potassium ions and a decreased sodium ion conductance. Automaticity is made possible by a slow inward sodium ion current that occurs during phase 4 of the pacemaker action potential. within the seminiferous tubules. as inhibin B cannot be produced without proper functioning of ESH and the ESH receptors (see above diagram). respectively. (Choices B and D) LH and testosterone levels are likely to be normal in this patient because ESH does not play a role in either LH or testosterone feedback (see above diagram). DHEAS and androstenedione are also not affected by ESH. as well as low inhibin B levels. inhibin B D. Testosterone E. because ESH is responsible for both spermatogenesis and inhibin B production. FSH B. Defective ESH receptors will cause loss of spermatogenesis. LH C. 38 . If the FSH receptor defect is truly isolated this patient will have normal testosterone and LH levels. ESH also stimulates the Sertoli cells to produce something called “androgen binding protein” locally. Educational Objective: LH stimulates the release of testosterone from the Leydig cells of the testes: ESH stimulates the release of inhibin B from the Sertoli cells in the seminiferous tubules. (Choice E) The circulating levels of other androgens such as DHEA. This androgen-binding protein is responsible for the high local testosterone concentration. DHEA Explanation: Pulsatile secretion of gonadotrophin releasing hormone from the hypothalamus stimulates the release of follicular stimulating hormone (ESH) and luteinizing hormone (LH) from the gonadotroph cells in the anterior pituitary. Which of the following substances is most likely decreased in this patient? A. (Choice A) ESH will be high in patients with ESH receptor defects because there will not be any negative feedback from inhibin B.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 25: A 25-year-old male is evaluated for infertility. Lab studies reveal a low sperm count. LH stimulates the release of testosterone from the Leydig cells of the testicles and FSH stimulates the release of inhibin B from the Sertoli cells of the seminiferous tubules of the testicles. respectively. A more thorough evaluation elucidates a rare genetic abnormally that leads to the inactivation of ESH receptors. High local levels of testosterone and ESH are necessary for spermatogenesis. Testosterone and inhibin B provide negative feedback of LH and ESH. Testosterone and inhibin B create negative feedback on LH and ESH. Increased sympathetic tone can cause some vasoconstriction of the larger pulmonary arteries. Hypoxic pulmonary vaso constriction might then decrease total lung perfusion (0). Acetylcholine also acts on the M3 receptors of mucous glands in the bronchial submucosa. Pulmonary arterial resistance E. (Both α. (Choice D)The autonomic nervous system exercises only weak control over the pulmonary circulation. Expiratory flow rates B. causing increased secretions.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 26: Bronchi are richly innervated by the autonomic nervous system.and β-adrenergic receptors are present on the pulmonary vasculature. These effects would increase airway resistance and the work of breathing. Educational Objective: Stimulation of the vagus nerve branches supplying the lung would cause bronchoconstriction and increased bronchial mucus secretion. There would not be a net increase in the total V/0 ratio. V/Q ratio Explanation: The main pulmonary effect of increased vagus nerve efferent activity is bronchial smooth muscle constriction.) (Choice E) Increased airway resistance might decrease total lung ventilation (V) due to increased work of breathing. (Choice B) Increasing vagus nerve activity would not significantly affect lung elastic recoil. This effect is mediated by acetylcholine from postganglionic parasympathetic neurons acting on muscarinic receptors. Bronchoconstriction and increased mucus secretion would increase airflow resistance and therefore increase the work of breathing. Which of the following parameters would increase most as a result of vagal stimulation? A. 39 . Elastic recoil of the lungs C. Work of breathing D. Increased parasympathetic activity may be weakly vasodilatory. Thus at the ERCI the airway pressure is zero and there is no tendency for air to flow either into or out of the lungs at this point. resulting in a positive transmural pressure across the chest wall at all times. 40 . +10mmHg B. 0 D. -10mmHg Explanation: The graph in the question stem is a pressure-volume curve that demonstrates the compliance of the lung and the chest wall (identified by labels) and the net combined compliance of these structures during inspiration (the curve marked by point A). the alveolar transmural pressure is always positive. Due to the elasticity of the lungs. Inversely the chest wall tends to transmit an expanding force at a baseline. this is why the curve marked “lung” is always a positive value. +5mmHg C. except during maximal inspiration. Which of the following is the best estimate of the intrapleural pressure at point A? A. the positive alveolar transmural pressure and the negative chest wall transmural pressure oppose one another equally at the functional residual capacity (FRC) resulting in an airway pressure of zero.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 27: The lung and chest wall compliance curves for a healthy individual are shown on the slide below. as illustrated on the above graph. These two forces. resulting in a perpetual collapsing force on the lungs. -5mmHg E. however. as may occur during subclavian central catheter placement. The presence of this negative pressure can be illustrated clinically during a traumatic injury that extends into the intrapleural space. it asks what the intrapleural pressure is most likely to be at the FRC. 41 . a pneumothorax will develop as the lung springs inward (collapses) and the chest wall springs outward. the negative intrapleural pressure equilibrates with atmospheric pressure favoring air entry into the intrapleural space. and it identifies the resting state where the airway pressure equals zero. i. does not ask about the airway pressure. At the ERCI the intrapleural pressure is slightly negative with a value of —5 mmHg. Educational Objective: The center of the airway pressure-volume curve is the functional residual capacity (FRC) of the lungs. the intrapleural pressure is negative with a value of approximately-5 mmHg.USMLE WORLD STEP 1 PHYSIOLOGY The question stem. such as a stab wound or by the accidental insertion of a needle.e. When puncture of the pleura allows communication with atmospheric pressure. At the FRC. More specifically. Secondly a reduction in the circulating blood volume (especially in excess of 10%) causes an intense increase in plasma ADH. Vasopressin is released due to blood osmolarity changes C. Vasopressin is released due to blood volume changes B.not inhibited -. Such drops in blood volume are detected by arterial baroreceptors (eg. an osmotic diuretic with rapid onset. (Choices A and B) Changes in blood volume or blood osmolality alone do not entirely explain the stimulus for vasopressin release. mannitol can quickly induce a hyper osmolal state. one of the most commonly used for this purpose is mannitol. the hypothalamic osmoreceptor cells shrink and send signals to the supraoptic and paraventricular nuclei. Vasopressin is released due to hyporeninemia E. metabolic suppression. In response to the mannitol-induced rise in blood osmolality and decrease in blood volume the posterior pituita releases vasopressin (antidiuretic hormone [ADH])to limit the amount of water excreted by the kidneys. (Choice D) Diminished renal blood flow stimulates the release of renin. Vasopressin secretion is inhibited due to blood osmolarity changes Explanation: Moderate or severe head injuries often result in increased intracranial pressure secondary to cerebral swelling. Diuretics are extremely effective at decreasing brain volume (and. 42 . Therefore hyperreninemia (not hyporeninemia) can contribute to an increase in vasopressin levels. intravenous mannitol is administered. (Choices E and F) Vasopressin is stimulated -.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 28: A 34-year-old construction worker loses consciousness after head trauma. and decompressive surgical procedures. These nuclei in turn send impulses to the posterior pituitary that prompt the release of ADH. Following an initial evaluation that included a head CT scan. when osmolality increases. Its diuretic effect is so powerful that electrolytes and serum osmolality must be closely monitored during usage to ensure that the patient remains euvolemic. carotid sinus) and through decreased stretch of the left atrium and the pulmonary veins. Vasopressin is released due to blood volume and osmolarity changes D. therefore at lowering intracranial pressure). which in turn increases angiotensin II production. Which of the following is most likely to happen shortly after the infusion starts? A.by changes in blood volume or blood osmolality. Educational Objective: The administration of mannitol induces hyperosmolarity and reduced blood volume both of which are strong stimuli for vasopressin release. A sugar alcohol that pulls the excess water from the brain into the intravascular compartment. Multiple treatments exist to reduce the total brain volume including diuretic administration perfusion augmentation. Angiotensin II acts on the brain to stimulate ADH release. Vasopressin secretion is inhibited due to blood volume changes F. The rate of blood flow through the pulmonary circulation must equal the rate of blood flow in the systemic circulation at all times. (Choice A) The arterial resistance in the systemic circulation is considerably higher than that in the pulmonary circulation in all scenarios. if the flow of blood were faster in the pulmonary circulation than in the systemic circulation the left ventricle would soon be overloaded. The normal systemic blood pressure is 120180 mmHg while the normal mean pulmonary arterial pressure is 14 mmHg. C and D) The mean arterial pressure diastolic arterial pressure and driving pressure for blood flow (systolic arterial pressure) are considerably different in the pulmonary and systemic circulations both at rest and during exercise. Arterial oxygen content Explanation: Under normal conditions in a healthy adult living at sea level. Arterial resistance B. If the flow of blood through the pulmonary circulation were slower than the flow of blood in the systemic circulation the left ventricle would soon empty completely. (Choices B. in patients with idiopathic pulmonary hypertension. Alternatively. 43 . The arterial pressures and oxygen contents of the pulmonary and systemic arterial systems are considerably different both at rest and during exercise. The low alter load pressures of the pulmonary circulation allow the thin right ventricle to keep pace with the more substantial left ventricle. The pulmonary arterial blood pressure increases by only a few mm Hg during maximal exercise despite the fact that the rate of blood flow increases 5-to 7-fold. Educational Objective: The pulmonary circulation is part of a continuous circuit with the systemic circulation. At the peak of the exercise. the pulmonary circulation is a low-pressure high-flow system. This is true for conditions of both exercise and rest. Mean arterial pressure C. Diastolic arterial pressure D. which of the following parameters is the same in the systemic and pulmonary circulations? A. Because the circulatory system is a continuous circuit the volume output of the left ventricle must equal the output of the right ventricle at all times. (Choice F)The arterial oxygen contents of the pulmonary and systemic circulations are dramatically different due to the fact that the blood in the pulmonary arterial circulation is deoxygenated and the blood in the systemic arterial circulation is oxygenated. Driving pressure for blood flow E. The resistance of the pulmonary vasculature maybe increased in individuals living at high altitudes. Blood flow per minute F. In order to provide continuous blood flow to the body the blood flow per minute (mL/min) in the pulmonary circulation must be identical to the blood flow in the systemic circulation.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 29: A 23-year-old military recruit begins special training that involves high-i intensity physical exercise. etc. 44 . Accessory C. He reports that this has happened several times in the past. His blood pressure was 70/40 mmHg and his heart rate was 45/mm during one of the episodes. Baroreceptors sense arterial wall stretch. Stimulation of which of the following nerves is most likely responsible? Hypoglossal B. Explanation: Arterial baroreceptors provide constant monitoring of the systemic blood pressure. When stimulated by stretch the nerves innervating baroreceptors produce an increased number of signals.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 30: A 65-year-old male loses consciousness while buttoning a tight shirt collar. which releases inhibition of the vasoconstrictor center and increases the blood pressure. The result is a blood pressure decrease through peripheral vasodilatation and decreased cardiac output (decreased contractility/stroke volume and decreased heart rate). Conversely hypotension leads to decreased baroreceptor signaling. an indicator of the systemic pressure in the arterial circulation. Vagal A. Glossopharyngeal E. Trigeminal D. Fibers from the aortic arch baroreceptors run within the vagus nerve (CNX) (Choice E). hypotension. which is a branch of the glossopharyngeal nerve (CN IX) (Choice D). 45 . Educational Objective: The carotid sinus is a dilatation of the internal carotid artery that lies at the bifurcation of the carotid artery. The hypoglossal nerve primarily controls the muscles of the tongue. and sometimes syncope. as described in the question stem. B and C) The hypoglossal (CN XII). External pressure on the carotid sinuses. causing bradycardia. the accessory nerve controls the sternocleidomastoid and the trapezius muscles and the trigeminal nerve primarily mediates facial sensation. Afferent fibers from both sets of baroreceptors terminate in the solitary nucleus of the medulla. a decrease in heart rate and contractility. (Choices A.USMLE WORLD STEP 1 PHYSIOLOGY The carotid sinus and aortic arch baroreceptors are especially important in blood pressure control. The carotid sinus is a dilatation of the internal carotid artery located just above the bifurcation of the common carotid artery. They increase their inhibitory discharges in response. Afferent fibers from carotid sinus stretch receptors form a small carotid sinus nerve. Hering’s nerve. causes the baroreceptors to react as if there was a systemic blood pressure increase. accessory (CN Xl) and trigeminal (CNV) nerves do not transmit signals from the carotid sinus baroreceptors. and a decrease in blood pressure. Blood pressure increases or external pressure on the carotid sinuses stimulate baroreceptors in the carotid sinus walls leading to vasodilatation. Such carotid sinus syncope tends to occur in patients with very sensitive carotid sinuses. 46 . Cephalic phase C. and is triggered by the thought. (Choice F) Postprandial alkaline tide is defined as an increase in plasma HC03 and decrease in plasma CF secondary to the surge of acid within the gastric lumen. Other factors that inhibit acid secretion include somatostatin and prostaglandins. It is not an important factor in the down-regulation of postprandial gastric secretion. sight. The gastric phase is mediated by the presence of gastrin (which stimulates histamine secretion and therefore indirectly. The intestinal phase is initiated when protein-containing food enters the duodenum but this phase plays only a minor role in stimulating gastric acid secretion. Which of the followings helps most to down-regulate gastric secretion after a meal? Basal secretion B. Educational Objective: The cephalic and gastric phases stimulate gastric acid secretion while intestinal influences tend to reduce gastric acid secretion. It is not an important factor in the down-regulation of postprandial gastric secretion.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 31: Experimental gastric function monitoring in healthy subjects before. smell. and intestinal. Postprandial alkaline tide A. (Choices B and C) The cephalic and gastric phases both stimulate gastric acid secretion not down-regulate it. and is triggered by the chemical stimulus of food and distension of the stomach. acid secretion). Intestinal influences E. and then decline in gastric acid production. the stimulation of acid secretion within the stomach is separated into three phases: cephalic. The ileum and colon release peptide W. intestinal influences are effective in down-regulating gastric acid secretion after a meal. Explanation: Classically. which binds to receptors on the endocrine. Basal gastric secretion is not an important factor in the downregulation of postprandial gastric secretion. The cephalic phase is mediated primarily by cholinergic and vagal mechanisms. Gastrin release D. histamine-containing cells described as enterochromaffin-like (ECLs). peak. (Choice E) Receptive relaxation is a reflex that allows the gastric fundus to dilate in anticipation of food passing through the pharynx and esophagus. and taste of food. Receptive relaxation F. the “baseline” gastric acid secretion). during and after a meal shows an initial rise. Such binding counteracts the cephalic and gastric phases of acid secretion by inhibiting gastrin-stimulated histamine release from ECLs. gastric. In fact. (Choice A) Basal gastric secretion has been defined as the gastric juices secreted in the absence of intentional or avoidable stimuli (essentially. Which of the following is the best drug to use as a provocation test in this patient? Scopolamine B. Phenoxybenzamine D. Methacholine is a muscarinic cholinergic agonist that acts by inducing bronchial smooth muscle contraction and increased bronchial mucous production. or cold air inhalation can be used to provoke asthma symptoms. Spirometry findings are normal in this patient. such as asthma. including physical. but his older brother suffers from atopic dermatitis. Explanation: The patient described in the question stem is experiencing respiratory difficulties that may be attributable to asthma. It is typically used in an inhaled form to treat bronchoconstriction in patients with obstructive lung diseases. (Choice D) Isoproterenol is a nonspecific beta-adrenergic agonist. He has no history of allergies. and allergenic irritants. but has normal spirometry values. Educational Objective: The methacholine challenge test can be used to induce bronchoconstriction in patients with asthma. agents such as methacholine. Methacholine C. chemical. The diagnosis of asthma is typically made by spirometry. (Choice A) Scopolamine is an antagonist of muscarinic cholinergic receptors.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 32: A 23-year-old male complains of occasional nighttime dyspnea and wheezing. 47 . (Choice E) Ipratropium bromide. histamine. is an antagonist of muscarinic cholinergic receptors. History can aid in raising the suspicion of asthma if a patient has a personal or family history of other diseases in the “allergic triad i. typically an inhaled betaadrenergic agonist. but can act as an antiemetic and a sedative. exercise.” allergic rhinitis or atopic dermatitis. a decrease in EFV by more than 20% after a methacholine challenge indicates the diagnosis of bronchial asthma. Patients with asthma will demonstrate a decreased FEV and peak expiratory flow rate on spirometry. Asthma is an obstructive air way disease that occurs due to hypersensitivity of the conducting airways to various stimuli. These changes are typically reversible with the use of a bronchodilator. like albuterol. Ipratropium bromide A. This medication has a bronchodilatory effect in asthmatic patients. Isoproterenol E. thereby allowing demonstration of the disease on spirometry. It is most commonly used to dilate the pupil for ophthalmologic exams. like scopolamine. When a patient presents with a history consistent with asthma. Methacholine is a muscarinic cholinergic against that causes bronchoconstriction and increased airway secretions.e. (Choice C) Phenoxybenzamine is an alpha-1-adrenergic antagonist used in the treatment of hypertension and benign prostatic hyperplasia. Anaphylaxis Explanation: The graph above combines cardiac and vascular function curves. Increases in blood volume shift this graph to the right and increases in the TPR decrease the slope of the curve (in part due to the increased afterload in the systemic circulation). Acute hemorrhage C. The Frank-Starling effect states that as cardiac muscle is increasingly stretched. such as a myocardial infarction. This is essentially a length-tension relationship. This relationship reaches its limit at the flat portion of the curve. Myocardial infarction E. 48 . This indicates either the action of a negative inotropic drug or injury to the myocardium inhibiting contraction. An isolated decrease in cardiac output indicates decreased contractility that is not the result of decreased preload (because the venous return line is unchanged). gives the mean systemic pressure where it intersects with the x axis and the total peripheral resistance (TPR) with its slope. Chronic anemia D. Excessive hydration B.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 33: The cardiac output and venous return curves of a healthy person are shown below with solid lines. labeled venous return. labeled as cardiac output (CO) illustrates the Frank-Starling effect. Which of the following is the most likely cause of the change depicted by the dashed lines? A. The dashed lines depict decreased cardiac output and unchanged venous return (unchanged blood volume and TPR). the cardiac output increases (up to a limit). The cardiac function curve. The vascular function curve. 49 . This causes an increase in the slope of the cardiac output graph. On a cardiac function curve. This results in a serious drop in venous return. (Choice B) Acute hemorrhage would cause a sharp decrease in the circulating blood volume and would shift the venous return curve to the left.USMLE WORLD STEP 1 PHYSIOLOGY (Choice A) Excessive hydration or a blood transfusion would increase the blood volume. (Choice C) Chronic anemia causes an increase in cardiac output in an effort to meet the metabolic demands of the tissues. the right atrial pressure is also elevated. Because there is more volume in the right atrium. Educational Objective: Myocardial infarction causes a sharp decrease in cardiac output due to loss of function of a zone of myocardium. (Choice E) Anaphylaxis causes widespread venous and arteriolar dilatation along with increased capillary permeability and third-spacing of fluids. myocardial infarction would decrease both the slope and the maximal height of the line. shift the venous return graph to the right and change the cardiac output from point A to point B. Medium and smallsized bronchi greater than 2mm in diameter (the first 10 generations of bronchi) provide the greatest summated frictional resistance to airflow in normal airways. larynx. In contrast airways smaller than 2 mm in diameter normally contribute less than 20% of the 50 . Which of the following curves best corresponds to the airway resistance pattern in normal lungs? Explanation: Beginning at the trachea there are 23 generations of bronchi. and trachea) accounting for the curve’s early parabolic shape. Airway generations 2-5 (including the segmental bronchi) normally have a higher regional resistance than the upper respiratory tract (nasal passages.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 34: A small volunteer study is conducted to assess airflow resistance along the bronchial tree. mouth phanx. (Choice C) The pattern shown here is essentially an inverse of the normal pattern. (Choice B) This pattern could apply to a patient with severe distal small airway obstructive disease. Airway resistance is minimal in bronchioles. including the segmental bronchi. This is because the smallest bronchi and bronchioles have a larger total cross sectional area as they continue to divide into further generations. This would be a good representation of airway conductance versus airway caliber. Educational Objective: The majority of total frictional airway resistance is localized to the medium and small-sized bronchi greater than 2mm in diameter in normal individuals.USMLE WORLD STEP 1 PHYSIOLOGY total airway frictional resistance. (Choice E)This pattern could apply to a patient with significant distal small airway obstructive disease. Regional airway resistance is maximal in the second to fifth generation airways. 51 . 0 mEq/L 5.0 mg/dl Glucose 0 80. 200 ml/min C.0)/0. The RBF is dependent upon the pressure difference between the renal artery and renal vein as well as the resistance in the renal vasculature. Thus the fraction of the blood volume occupied by plasma is equivalent to ( 1-hematocrit). 100 ml/min B. 500 ml/min D. The RBF can therefore be determined by dividing the RPF by the fraction of the blood volume occupied by plasma as follows: RBF = (PAH clearance) / (1 — hematocrit) For the purposes of the question above: RBF = [(100 x 1.5 = 1000 ml/mm Educational Objective: The renal blood flow (RBF) refers to the volume of blood that flows through the kidney per unit time and can be calculated by dividing the renal plasma flow by (1 — hematocrit).0 mg/dl Potassium 50. Whereas the RBF is the complete blood volume that flows through the kidney per unit time. Urine Serum Inulin 200. The RPF can be calculated using the clearance of para-aminohippuric acid (PAH) as follows: RPF = PAH clearance = (urine [PAH] x urine flow rate) / plasma [PAH] The hematocrit is the fraction of the blood volume occupied by erythrocytes. which of the following is the best estimate of this patient’s renal blood flow? A.0 mg/dl 3. A simpler means of calculating the RBF uses the renal plasma flow (RPF) and hematocrit values.0 mg/dl 2.21/ (1—0.0 mEq/L Uric acid 15. 52 . determining the values for the above equation would be invasive and difficult.5) = 500/0.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 35: The following measurements were taken from a healthy 33-year-old volunteer. The RBF is tightly regulated by mechanisms within the kidney that modulate renal vascular resistance to keep the RBF constant over a wide range of systemic blood pressures. The RBF can be calculated from pressure and resistance values as follows: RBF = (renal artery pressure — renal vein pressure) / renal vascular resistance In practice. the RPF is the plasma volume (plasma = blood— erythrocytes) that flows through the kidney per unit time. 1000 ml/min E.0 mg/dl PAH 100 mg/ml 0.2 mg/ml Assuming a hematocrit of 50% and a urine flow of ml/mm. 2000 ml/min Explanation: The renal blood flow (RBF) is the volume of blood that flows through the kidney per unit time. Metformin D. C peptide and insulin are secreted in equimolar concentrations. C peptide and mature insulin are subsequently packaged into secretory granules in the Golgi apparatus. Insulin is derived from single-chain proinsulin in pancreatic 13-cells. Thus. Acarbose B. however. His urine is negative for glucose and ketones. Treatment with which of the following agents would increase the C-peptide level in this patient’s blood? A. Enalapril F. C peptide is not significantly extracted on first pass through the liver. Essentially this question asks us which of the agents listed increases the rate of insulin secretion by residual pancreatic islet j3-cells in patients with type 2 diabetes.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 36: A 45-year-old male is diagnosed with diabetes mellitus during a routine check-up. He does not have glycosuria or ketosis because his islet cells are secreting some insulin. Low-calorie diet Explanation: This patient likely has type 2 diabetes with some residual pancreatic [3-cell function. Rosiglitazone E. but unlike insulin. its circulating levels in plasma can be used as a marker of the total rate of 13-cell endogenous insulin secretion under steady-state conditions. Glyburide C. 53 . Within the endoplasmic reticulum of these cells endopeptidases cleave proinsulin to form insulin and C peptide. acarbose decreases endogenous insulin secretion and C peptide levels in type 2 diabetics with residual pancreatic [3cell function. By reducing postprandial hyperglycemia. (Choices C and D) Metformin (a biguanide) and rosiglitazone (a thiazolidinedione) increase the sensitivity of target tissues to insulin. (Choice E)The ACE-inhibitor enalapril is a reno protective agent that decreases diabetic proteinuria. (Choice A) Acarbose inhibits a-glucosidase in the intestinal brush border. Sulfonylureas increase the rate of insulin secretion and C peptide levels in patients with type 2 diabetes.USMLE WORLD STEP 1 PHYSIOLOGY Glyburide is an oral hypoglycemic agent (a sulfonylurea) that stimulates insulin release in type 2 diabetics. Educational Objective: C peptide can be used as a marker of the total rate of endogenous 3-cell insulin secretion under steady-state conditions. 54 . They have little effect on insulin secretion. (Choice F) Fasting and decreased caloric intake would decrease the rate of endogenous insulin secretion and corresponding C peptide levels. which impairs the hydrolysis of sugars and postprandial absorption of ingested polymeric glucose. It does not have significant effects on blood glucose levels or endogenous insulin secretion. A B C D E 55 . E. Which of the following points corresponds to aortic valve opening? A.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 37: The volume and pressure tracings for the left ventricle of a 34-year-old male are shown below. C. B. D. At point D. blood is forcefully expelled from the atria into the ventricles causing the slight increase in ventricular pressure and volume seen on the far left of the graph.USMLE WORLD STEP 1 PHYSIOLOGY Explanation: The graph above depicts the left ventricular pressure and volume during the cardiac cycle. Educational Objective: Ventricular pressure and volume curves allow one to identify the phases of the cardiac cycle and to determine the exact time of opening and closure of the cardiac valves. Isovolumetric relaxation occurs between points C and D. Together these graphs can be used to plot the various phases of the cardiac cycle. When these two values are equal. The segment between points A and B corresponds to isovolumetric contraction. Now the ventricular volume begins to decrease sharply as blood is expelled. the aortic valve closes (point C). When the ventricular pressure exceeds the diastolic systemic blood pressure the aortic valve opens (point B). The ventricle then begins to contract. the mitral valve opens. In this phase. Next the mitral valve closes causing a slight depression of the ventricular pressure curve (marked as point A above). During this time the ventricular pressure increases but the ventricular volume remains the same because the aortic and mitral valves are closed. is atrial systole. 56 . First. initiating the phase of diastolic filling. on the left. The left ventricular pressure continues to increase between points B and C until the systolic maximum blood pressure is reached. The ventricle relaxes and both the aortic and mitral valves are closed. Thiazide diuretics. patients will reabsorb the majority of the filtered potassium in the proximal tubule and loop of Henle. Further reabsorption of potassium (about 20% of the filtered potassium load) occurs in the loop of Henle by the action of the Na/K/2C cotransporter. In a healthy person the potassium concentration within the glomerular capillaries will always be equal to the potassium concentration in the tubular fluid in Bowman’s space.intercalated cells reabsorb extra potassium via H*/K*-AT Passes. Patients with renal failure and a poor GER may develop elevated plasma potassium due to failure to filter potassium from the plasma to the urine. Even in hyperkalemic states. hence a shift of potassium into the tubular fluid from the tubular cells Educational Objective: The collecting duct is the primary site of regulation of potassium concentrations in the tubular fluid and urine. High dietary potassium consumption 2. 100% 150% 150% 150% 150% D. The degree of potassium filtration in the glomerulus depends upon the glomerular filtration rate (GER) and is relatively constant. Which of the following best describes the percentage of the Filtered potassium load remaining in the tubular fluid as it flows along the nephron in this patient? Bowman’s Proximal Loop of Distal Collecting capsule convoluted Henle convoluted duct tubules tubules A. In the hypokalemic state. The cortical collecting duct is the primary site of potassium regulation. Approximately 2/3 of the filtered potassium load is reabsorbed in the proximal convoluted tubules.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 38: A healthy 32-year-old male who has been consuming a high potassium diet for the past two weeks is found to have a urine potassium concentration that is higher than his plasma potassium concentration. These processes reabsorb potassium at a fixed rate. 100% 100% 100% 120% 150% Explanation: Each segment of the nephron plays a role in potassium homeostasis. 57 . 100% 30% 23% 15% 150% B. loop diuretics. Aldosterone: Stimulates the Na*/K*-pump which leads to sodium retention (and passive water retention) at the expense of K lost into the tubular fluid S. Alkalosis: In alkalosis potassium is lost in the urine in order to preserve protons (K and H freely exchange across the tubular cell membrane) 4. Conversely. 100% 30% 23% 150% 150% C. collecting duct alpha. and do not play a significant role in the regulation of potassium excretion in the urine. factors that increase potassium secretion into the tubular fluid include: 1. and any other cause of increased flow rate through the distal tubule: There is increased potassium loss in the setting of high flow through the collecting ducts because the fluid here is dilute. 100% 100% 100% 150% 150% E. is severe inspiratory airflow obstruction causing rapid diaphragmatic fatigue. Nasopharynx B. though less likely. The pattern illustrated above thus suggests that this patient’s diaphragmatic contractions are becoming progressively weaker with repetition since there is no apparent decrement in total phrenic nerve stimulation of the diaphragm during inspiration. Brainstem C.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 39: A patient is being evaluated for hypoventilation. or of the lung tissue and/or chest wall. decremental synaptic transmission) and/or abnormally rapid diaphragmatic muscle fatigue. This could be consistent with myasthenia gravis or restrictive lung or chest wall disease (would cause a normal diaphragm to fatigue rapidly). Peripheral nerves E. of the skeletal muscle itself.g. Neuromuscular junction Explanation: This physio graph shows decreasing amplitudes of cyclic intrapleural pressure changes as deep breathing continues. Diaphragmatic contraction is responsible for most of the intrapleural pressure fall during inspiration. Pericardial space D. and the following physiologic information is obtained: Which of the following locations is the most likely site of this patient’s disease? A. (Choice A) Nasopharyngeal lesions would be unlikely to cause severe enough inspiratory airflow obstruction to cause rapid diaphragmatic fatigue. He is asked to take several deep breaths in rapid succession. The weakening of inspiratory diaphragmatic contractions seen here is most likely due to some pathology of the neuromuscular junction. Also possible. (Choice D) Peripheral neuropathy affecting the phrenic nerve would cause uniformly weak inspirations during maximal voluntary ventilation. 58 . Educational Objective: Progressively weakening diaphragmatic contractions during maximal voluntary ventilation with constant phrenic nerve stimulation during inspiration indicate neuromuscular junction pathology (e. The filtration fraction using these numbers is 0. Compensatory activation of the renin-angiotensin mechanism in response to hypotension leads to constriction of the efferent (outgoing) arteriole to maintain GER. The patient described in this clinical vignette is severely hypovolemic. The filtration fraction which is equal to the GER divided by the RPEI increases in hypovolemia as the RPE drops proportionately more than does the GER due to the aforementioned compensatory mechanism.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 40: A 72-year-old male is brought to the ER with diarrhea and vomiting. which increases hydrostatic pressure in the glomerular capillaries and maintains the glomerular filtration rate.0.15. The RPE is the volume of plasma delivered to the kidney per unit time. Angiotensin II constricts the efferent glomerular arteriole. When the blood pressure drops below a certain point autoregulatory mechanisms become unable to maintain RPE. and the GER is approximately 125 mI/mm though these numbers vary by gender and from person to person. The symptoms started 24 hours ago and he has been unable to take much fluid during this period of time. Hypovolemia stimulates secretion of renin. which means that about 1/5 of the plasma that passes through the glomerular capillaries is filtered into Bowman’s capsule. Which of the following changes in renal plasma flow (RPE) glomerular filtration rate (GER) and filtration fraction (FE) do you expect most in this patient? Explanation: Filtration of fluid through the glomeruli depends on the renal plasma flow (RPE) and the glomerular filtration rate (GER). Renal plasma flow depends on the circulating blood volume and the resistance of the afferent and efferent arterioles. In a wide range of blood pressures (100-200 mm Hg) renal plasma flow is maintained at a constant level by local changes in vascular resistance. His mucous membranes are dry and his blood pressure is 90/60 mmHg. with a subsequent decrease in renal plasma flow. The filtration fraction is a ratio of glomerular filtration rate to renal plasma flow (EE=GER/RPE). Profuse diarrhea and vomiting led to a rapid decline in the circulating blood volume. Due to this compensatory mechanism the decrease in GFR in hypovolemia is less pronounced than the decrease in RPF resulting in an increased filtration fraction because FF=GFRl/RPFll. Normally the RPE is approximately 600-700 ml/mm. The GER is the volume of fluid filtered from the renal glomerular capillaries into Bowman’s capsule per unit time. 59 . which is followed by synthesis of angiotensin II.20. Educational Objective: Dehydration leads to a decrease in renal plasma flow and a decrease in the glomerular filtration rate. (choice A). and atrioventricular node. However. Panic attack Explanation: Increased heart rate and cardiac output with preservation of arterial partial pressures of 02 and CO2 are most consistent with the integrated cardiorespiratory response to exercise. (Choice E) A panic attack could raise heart rate and cardiac output via sympathetic stimulation of the heart. Cardiac output is thus decreased and compensatory tachycardia takes place. Changes in the partial pressures of 02 and CO2 during exercise occur predominantly in the venous blood. (Choice C) Pulmonary embolism (PE) causes low blood flow to the left side of the heart. Carotid sinus massage would not significantly affect the Pa02 or PaCO2. increased cardiac output. atrial myocytes.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 41: Increases in heart rate and cardiac output are observed in a 32-year-old woman. Simultaneous arterial blood gas studies show a normal Pa02 and PaCO2. Heart rate and cardiac output increase to meet increased tissue oxygen demands. (Choice A) At high altitude (condition of ambient hypoxia). heart rate and cardiac output can increase in order to improve oxygen delivery to tissues. while venous P02 is decreased and venous PCO2 is increased. The content of 02 and CO2 in arterial blood remains relatively close to resting values at most levels of exercise. (Choice D) Carotid sinus massage causes a reflex vagal discharge to the sinoatrial node. PE often causes hypoxemia (low Pa02) due to acute V/Q imbalances in the affected lung. the Pa02 and PaCO2 would be lower than normal due to the hypoxemia and resulting hyperventilation and respiratory alkalosis. Exercise C. and thus a fall in the PaCO2. These increases are coordinated so that arterial blood gases remain relatively constant. Pulmonary embolism D. and increased respiratory rate in order to balance the increased total tissue 02 consumption and CO2 production. Carotid sinus massage E. These changes are most likely part of an integrated response to which of the following? A. Educational Objective: The integrated cardiorespiratory response to exercise includes increased heart rate. High altitude B. and the respiratory rate increases to eliminate excess CO2 produced. 60 . However. there is usually associated hyperventilation and respiratory alkalosis. Hypoxemia and stimulation of lung vagal irritant receptors in ischemic lung parenchyma also tend to produce hyperventilation and a respiratory alkalosis with a lowered PaCO2. The heart rate and cardiac output are transiently reduced. Which of the following factors best explains the differences between the two curves? A. Altering environmental factors. Other factors. such as increasing physical activity and optimizing calcium and vitamin D intake. during childhood and early adulthood. Environmental factors do play a role. although polymorphism is likely. The inevitable loss of bone mineral density. Menopause onset Explanation: Bone mass is a major determinant of the risk for osteoporotic fractures. but cannot change their genetic makeup. (Choice D) A high-sodium diet increases calcium loss in the urine. bone mass progressively declines shortly after reaching its peak in the 20s or 30s. These life-style modifications should not be ignored. however. These modifications often slow bone loss and thus reduce fracture risk. On the graph below. Physical activity is not as significant a contributor to the bone density of an individual as genetics. (Choice B) Genetic factors are the most important determinator in peak bone mass. which can also contribute to low bone mineral density. puts these patients at an increased risk for osteoporosis later in life because bone is lost from a lower-than-average peak. Epidemiological and twin studies have shown that genetic factors are responsible for 15% of the variation in peak bone mass among individuals. Bone loss can be minimized by lifestyle changes. Vitamin supplementation E. Educational Objective: The differences in the bone density curves can be best explained by genetic differences. they do not establish an adequate bone density peak. Bone loss accelerates at menopause because estrogen levels decline significantly. daily weight-bearing physical activity. fracture-tree women. Genetics are more significant contributors to bone mineral density than environment. such as optimizing calcium and vitamin D intake. Physical activity is believed to contribute to bone mineral density during the teen and young adult years. Bone mass increases during childhood and adolescence. like calcium intake and daily physical activity. (Choice C) Physical activity also increases bone mass. as women at risk for osteoporosis can change their environments. play a smaller role. Statistically. The differences in bone density among the two women graphed above are most likely due to genetic factors. but environment is not as significant as genetics in the differences in bone density (graphed above). albeit a lesser one.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 42: Bone mass is an important determinant of fracture risk. Calcium intake C. Most subjects will lose a predictable amount of bone mineral density per year. (Choice E) As seen in both curves. Genetic factors B. The genes that contribute to this variation are largely unknown. then. Individuals with lower calcium and vitamin D intake have a lower peak bone mass because. and some scientists believe that exercise also diminishes the rate of bone mass loss experienced later in life. 61 . and restriction of sodium in the diet can augment peak bone mass. however. Physical activity level D. bone mass versus lifetime is shown for two healthy. Caucasians have lower bone densities than people of African decent. peaking between 20 and 30 years of age. On cardiac auscultation. Amyl nitrite inhalation Explanation: This patient presents with signs and symptoms of heart failure. The murmurs associated with mitral valve prolapse and hypertrophic cardiomyopathy become more audible and those associated with aortic stenosis become less audible. The 33 sound can be accentuated by having the patient lie in the left lateral decubitus position and exhale fully. Expiration C. The volume reduction would likely lessen the intensity of the 33 sound. Furosemide injection E. Standing D. Squatting and Valsalva release (maneuvers that increase venous return) have the opposite effect. It results from blood rushing into a partially filled ventricle or into a very stiff ventricle in patients with restrictive cardiomyopathy. (Choices A and C) The Valsalva maneuver (bearing down against a closed glottis) and the standing position can be used to differentiate between the various causes of a systolic murmur in the left heart.) Having the patient lie in the left lateral decubitus position makes it easier to hear 53. The 33 heart sound is a low frequency sound that occurs immediately following 32 during the phase of rapid passive ventricular filling. An 53. (Choice E) Amyl nitrite inhalation causes vasodilatation and decreased blood pressure. Which of the following is most likely to accentuate this physical examination finding? A. Heart failure can result from a variety of pathogenic processes. It would have the same effect as the Valsalva maneuver on the murmurs of mitral prolapse hypertrophic cardiomyopathy and aortic stenosis. In heart failure resulting from systolic dysfunction the ejection fraction is decreased and the heart is left with a considerable end-systolic volume within the ventricles. a diastolic sound is heard when the patient lies in the left lateral decubitus position. 62 . He has been sleeping in a recliner chair to relieve his shortness of breath. or ventricular gallop heart sound is commonly heard in patients with left ventricular failure (this can be a normal variant in healthy children and adolescents). Both maneuvers decrease venous return to the heart thereby reducing left ventricular volume and blood pressure. (The bell is well-suited to detect low frequency sounds while the diaphragm is best for hearing high-pitched sounds like S1 and S2. Valsalva maneuver B. Educational Objective: A 33 sound is a low frequency heart sound that can be physiologic in children but is typically pathologic in adults where it generally results from left ventricular failure or restrictive cardiomyopathy. (Choice D) Furosemide injection causes brisk diuresis. His past medical history is significant for a myocardial infarction two months ago. The 33 is best appreciated with the bell of the stethoscope pressed lightly against the skin in the region of the ventricular apex.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 43: A 53-year-old male presents to your office with difficulty breathing and increasing fatigue. Repeated bouts of myocardial ischemia are the most common cause. Having the patient exhale completely while in this position can make the sound even more audible by decreasing the volume of the lungs and bringing the heart closer to the chest wall. Examples of such processes include demyelinating diseases such as multiple sclerosis. Which of the following would you expect most on visual field examination? Homonymous hemianopsia B. and is the most common cause of blindness in people over 50 years old in the US. the fibers of the visual pathway linking the optic chiasm to the lateral geniculate body. most commonly by a pituitary adenoma. macular sparing is common in lesions of the occipital cortex. The term scotoma refers to any visual defect surrounded by a relatively unimpaired field of vision. (Choice A) Homonymous hemianopsia is a loss of vision in the same side of the visual field in both eyes. which. This may result from calcified carotid arteries. diabetic retinopathy and retinitis pigmentosa. Binasal hemianopsia C. Each macular cone synapses to a single bipolar cell. Explanation: The macula is a yellowish spot approximately 1. It is caused by pressure to the lateral areas of the optic chiasm. in turn. Central scotomas A. (Choice C) Bitemporal hemianopsia is a loss of vision in the temporal fields of both eyes. Lesions of the macula cause central scotomas. Due to this arrangement the visual acuity in the macula. It is caused by compression of the medial part of the optic chiasm where the nasal fibers decussate. 63 . synapses to a single ganglion cell.5 mm in diameter located near the center of the retina. and particularly the fovea. few overlying cells and no blood vessels. Scotomas occur due to pathologic processes that involve parts of the retina or optic nerve. Due to this peculiar cortical representation. Funduscopy reveals small yellow retinal lesions clustered in the macula. is greater than in any other area of the retina. Pathologic processes that involve the entire optic nerve lead to monocular blindness.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 44: A 65-year-old male complains of bilateral visual difficulty that has progressed over the last year. Arcuate scotomas E. The neural fibers that serve the macula transmit to an area of the occipital visual cortex that is separate from the area of representation of the peripheral fields (this area is also relatively large in size). Macular degeneration is characterized by progressive loss of central vision due to deposition of faulty tissue (drusen) behind the retina (dry MD) and neovascularization of the retina (wet MD). The patient described in this clinical vignette is likely to have macular degeneration (MD). Bitemporal hemianopsia D. This disorder is frequently age-related. The resulting visual field defect follows the arcuate shape of the nerve fiber pattern. Macular lesions impair central vision and result in central scotomata (Choice E). It is characterized histologically by the presence of densely packed cones. It occurs due to transection of the contralateral optic tract. (Choice B) Binasal hemianopsia is a loss of vision in the nasal fields of both eyes. Educational Objective: A scotoma is a visual field defect that occurs due to a pathologic process that involves parts of the retina or the optic nerve resulting in a discrete area of altered vision surrounded by zones of normal vision. (Choice D) Arcuate scotomas occur due to damage to a particular region of the optic nerve head. It manifests with muscle weakness and cramping during exercise resulting from failure of muscle glycogen breakdown due to a defect in myophosphorylase. (Choice A) Calcium released from the sarcoplasmic reticulum binds troponin C causing a movement of tropomyosin to reveal the myosin-binding sites on actin. No abnormalities are found on light microscopy. Poor force generation on repeated stimulation D. Impaired relaxation after a single contraction E. The uniform distribution of T-tubules in skeletal muscle fibers ensures that a depolarizing signal reaches each fiber at the same time. This autosomal-dominant disease occurs due to an abnormality of myotonic protein kinase resulting from trinucleotide repeat expansion. Electron microscopy of the specimen reveals no T-tubules in some muscle fibers. one T-tubule contacts two terminal cisterns forming a triad at the junction of the A-band and the I-band. Which of the following is the most likely consequence of the observed finding in the affected muscle cells? A. Because they are extensions of the cell membrane they consist of a phospholipid bilayer with a large number of voltage-gated calcium channels (dihydropyridine receptors) located in close proximity to ryanodine receptors on the terminal cisterns of the sarcoplasmic reticulum. The function of T-tubules is to transmit a depolarization from the sarcolemma to sarcoplasmic reticulum in a rapid and uniform manner. (Choice E) McArdle disease (glycogen storage disease type V) is an example of a disease with impaired energy production during muscle contraction. A lack of T-tubules in some myofibrils would lead to uncoordinated contraction of individual fibers (Choice B). The uniform distribution of T-tubules ensures coordinated contraction of all myofibrils. This leads to release of calcium from the sarcoplasmic reticulum and induction of muscle contraction. It occurs due to the presence of autoantibodies to acetylcholine receptors in the neuromuscular junction. They transmit depolarization signals to the sarcoplasmic reticulum and trigger the release of calcium. This coordination is necessary for muscle contraction. In skeletal muscle. (Choice D) Impaired relaxation after a single contraction occurs in myotonic dystrophy. Uncoordinated contraction of myofibrils C. Educational Objective: T-tubules are invaginations of the sarcolemma that extend into each muscle fiber. (Choice C) Poor contractile force on repeated muscle stimulation is a characteristic finding in myasthenia gravis. They run perpendicular to the long axis of the muscle fiber and lie in close proximity to extensions of the sarcoplasmic reticulum (terminal cisterns). T-tubules are therefore extensions of the extracellular space. No contraction in response to intracellular Ca2 B. 64 . An abnormality of troponin C or myosin would lead to absence of contraction in response to release of calcium.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 45: Skeletal muscle biopsy is performed on a 17-year-old male with proximal muscle weakness. Impaired energy production during contraction Explanation: T-tubules are invaginations of the muscle cell membrane (sarcolemma). Ryanodine receptors are calcium release channels that are opened under the influence of activated dihydropyridine receptors on the T-tubules. When the arterial blood pressure is maintained in the range of 60-140 mm Hg. Decreased arterial partial oxygen tension B. Decreased arterial pH E.) The p02 has much less of an influence on cerebral blood flow.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 46: A 32-year-old female with claustrophobia (fear of closed spaces) is stuck in a malfunctioning elevator. Carbon dioxide a potent cerebral vasodilator exerts the most sign 1H cant effects. (Choices A and B) Hyperventilation does not decrease the p02 or the arterial oxygen content. 65 . Other vasoactive substances in the systemic circulation do not affect cerebral blood flow because they do not cross the blood-brain barrier. or sepsis) and is not associated with panic attacks. weakness and blurred vision. cerebral perfusion remains constant. She experiences severe anxiety as well as dizziness. seizures. (Choice E) Increased arterial lactate content is a sign of tissue hypoxia (e. Decreased arterial partial CO2 tension D. Educational Objective: Panic attacks are associated with hyperventilation and decreased p002. Increased arterial lactate content Explanation: The patient in this vignette is having a panic attack. the arterial blood gases exert the most powerful influence on the brain circulation. If anything these parameters would increase. Lactic acidosis is also associated with metformin use. carbon monoxide poisoning.g. An increase in the pCO2 from 25 to 100 mm Hg causes a linear increase in cerebral blood flow and a decrease in the pCO2 causes a linear decrease in cerebral perfusion. marked by an increased arterial pH and a decreased p002. Which of the following is the most likely cause of this patient’s symptoms? A. Hypocapnia can cause decreased cerebral perfusion and thus neurological symptoms. Decreased arterial oxygen content C. For p02 in the range of 50-100 mm Hg. Panic attacks are typically accompanied by hyperventilation and decreased pCO2 (hypocapnia). (Choice D) Hyperventilation produces respiratory alkalosis. cyanide poisoning. Hypocapnia causes cerebral vasoconstriction and decreased cerebral blood flow. from ischemia. Severe hypoxia (p02 <50 mm Hg) leads to a rapid increase in cerebral blood flow and intracranial pressure. (Because hypocapnia can decrease cerebral blood flow patients with cerebral edema are often hyperventilated in order to decrease intracranial pressure. Renal blood flow C. Third. Educational Objective: Exercising muscles can receive up to 85% of the total cardiac output during periods of strenuous activity thanks to local release of vasodilatory factors. After swimming for 30 minutes his mean arterial pressure has risen only slightly to 115mm Hg. A decrease in which of the following during exercise most likely accounts for the observed finding? A. (Choice E) The right atrial pressure increases during prolonged exercise due to increased return of blood to the heart and decreased pooling of blood in the venous system. Changes in the renal blood flow do not contribute to blood pressure maintenance during exercise. effectively shunting blood toward the exercising muscle. Although sympathetic discharge during exercise causes an increase in cardiac output and increased contraction of blood vessels. the mean arterial pressure typically increases only 20-40 mmHg during full body exercise. (Choice C) The cardiac stroke volume is increased during exercise. neurohormonal influences directly enhance contractility. The mean arterial pressure remains fairly stable during exercise due to an adaptive decrease in the systemic vascular resistance (SVR). Despite these three sympathetic nervous system effects. potassium ions. Exercising muscle releases local vasodilatory factors including adenosine. Increased venous return means enhanced diastolic filling and more forceful systolic contractions through the Frank-Starling mechanism. (Choice A) Systolic blood pressure is increased slightly during exercise due to sympathetic nervous system effects. First. the increased metabolic demands of the exercising tissue require cardiac output (CO = SV x HR) to increase from a resting rate of 5 L/min to approximately 20 L/min during maximal exertion. sympathetic discharge causes contraction of the arterioles in all tissues except the actively working muscles. the kidneys receive a greater total blood flow due to the significant increase in cardiac output. ATPI CO2 and lactate. increasing venous return to the heart. (Choice B) Renal blood flow normally accounts for 25% of the total cardiac output. but a slightly smaller proportion of the CO because sympathetic stimulation causes vasoconstriction of the renal arterioles. There is also a small vasodilatory effect of sympathetic beta receptors. there is only a modest blood pressure increase because the vasodilatation within muscle so significantly decreases the total systemic vascular resistance. Second.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 47: A 23-year-old man has a mean arterial pressure of 95 mm Hg at rest. During exercise. Cardiac stroke volume D. Systolic blood pressure B. sympathetic stimulation causes contraction of the venous system. Additionally. Right atrial pressure Explanation: Stimulation of the sympathetic nervous system is essential during exercise for a number of reasons. 66 . Systemic vascular resistance E. in order to supply enough preload to the heart to meet this increased stroke volume. One of the patterns they obtain is shown on the slide below. Renal medulla E. The flow curve most likely reflects the flow pattern in which of the following tissues? A. The systolic reduction in coronary blood flow is greatest in the LV subendocardial myocardium. The great majority of blood flow through the vascular beds supplied by the coronary arteries to the left ventricle (LV) occurs during diastole. Renal cortex F. when the blood vessels are not compressed by myocardial contraction. Subcortical nuclei B. 67 . Subendocardial myocardium D. which corresponds to ventricular systole.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 48: A group of investigators studies patterns of blood flow in various tissues. The coronary circulation is unique in that only 10% of total perfusion through the myocardial capillaries of the left ventricle occurs during systole. making it the region most prone to ischemia and myocardial infarction. Adrenal medulla Explanation: The flow curve shows a cyclic reduction in blood flow during the OT interval. The systolic reduction in coronary blood flow is greatest in the subendocardial myocardium of the LV. while the majority of left ventricular blood flow occurs during diastole. Educational Objective: Only 10% of total perfusion through the myocardial capillaries of the LV occurs during systole. Brain cortex C. E. D. C. VT is represented by the space between points C and Don the graph above.USMLE WORLD STEP 1 PHYSIOLOGY Q NO 49: A 34-year-old male undergoes spirometry as a part of a pre-employment evaluation. The additional volume exhaled during maximal expiration is known as the expiratory reserve volume (ERV). represented by the space between points B and C on the graph. The remaining volume between points A and B is the residual volume. This additional inspired air is referred to as the inspiratory reserve volume (IRV). a constant volume of gas that remains in the lungs after maximal exhalation. represented by the space between points D and Eon the graph above. a person can draw more air into the lungs than occurs during normal tidal volume breathing. At which point on the graph below is the total pulmonary vascular resistance lowest? A. the intrathoracic pressure is atmospheric. A B C D E Explanation: The tidal volume (VT) is the volume of air moved into and out of the lungs during normal breathing. B. During maximal inspiration. and the alveoli do 68 . The pulmonary vascular resistance is lowest at the functional residual capacity (FRC) represented by point C on the graph because at this level of inflation the lung is maximally compliant. These volumes can be combined into lung capacities as follows: Total lung capacity (TLC) = IRV + VT + ERV + RV Vital capacity (VC) = IRV + VT + ERV Inspiratory capacity (IC) = IRV + VT Functional residual capacity = ERV + EN Hypoxia gravity and the level of lung expansion can affect the pulmonary blood flow. When air is forcefully exhaled positive pressure is placed on the lungs by the surrounding musculature resulting in compression of both airways and blood vessels. Educational Objective: Pulmonary vascular resistance (PVR) is lowest at the functional residual capacity. This also has the effect of increasing pulmonary vascular resistance. The alveoli expand as they fill with inspired air and the interstitial elastic tissue of the lungs is put under tension. the lung volume increases from the FRC. As air is drawn into the lungs during inhalation. Forced exhalation increases PVR due to the collapsing positive pressure placed on the lung parenchyma. These changes exert collapsing pressure and tension on the pulmonary vasculature leading to an increase in pulmonary vascular resistance. 69 . Inhalation increases PVR due to the pressure placed on pulmonary vessels by the expanding alveoli.USMLE WORLD STEP 1 PHYSIOLOGY not exert meaningful pressure on the pulmonary vasculature. USMLE WORLD STEP 1 PHYSIOLOGY Q NO 50: Invasive monitoring tools used on a 54-year-old study participant reveal the following: Arterial blood oxygen content Pulmonary artery diastolic pressure Left ventricular diastolic pressure Mean right atrial pressure Venous blood oxygen content 20 ml O2/100 ml blood 12 mmHg 10 mmHg 4 mmHg 15 ml O2/100 ml blood What additional information is necessary to calculate cardiac output in this patient? A. which calculates cardiac output by determining how much blood must flow through the pulmonary or systemic circulation to account for the difference in oxygen content between the arterial and venous circulations. the arterial O2 content. Arteriovenous difference in the oxygen content Explanation: The cardiac output. Educational Objective: The Fick principle can be used to calculate the cardiac output. and the level of tissue O2 consumption. Blood oxygen carrying capacity B. It is used to estimate the metabolic rate. is commonly calculated using the following formula: Cardiac output = Stroke volume x Heart rate Cardiac output can also be determined using the Pick principle. Respiratory quotient E. Oxygen consumption D. (Choice E) The arteriovenous difference in oxygen partial pressures can be calculated using the information provided in the question stem. Here. Blood hemoglobin concentration C. The normal value is typically 0. One must measure the venous O2 content (typically obtained from the pulmonary anew). (Choice B) The blood hemoglobin (Hb) concentration can be used in the following formula to estimate the cardiac output but this formula also requires that the patient’s body surface area (BSA) be known: CO = (135 x BSA) / [(13 x Hb) x (SaO2-SvO2)] (Choice D) The respiratory quotient is the ratio of CO2to 02 across the alveolar membrane. we know the arterial and venous O2 content. all that is needed is the oxygen consumption.8. expressed in liters per minute. It states that the cardiac output is equal to the oxygen consumption by the tissues divided by the arteriovenous oxygen difference. 70 . CO = O2 consumption / arteriovenous O2 difference Invasive methods are required to gather data for the Fick calculation of cardiac output.