Cell & Nerve Physiology
7 questionsIn a resting neuron, the membrane potential is -70 mV, which is close to but less negative than the equilibrium potential for potassium (-90 mV) and far from the equilibrium potential for sodium (+60 mV). Which of the following best explains why the resting potential lies closer to the potassium equilibrium potential than to the sodium equilibrium potential?
- AThe resting membrane has much higher conductance for potassium than for sodium
- BThe Na+/K+ ATPase pumps three sodium ions out for every two potassium ions pumped in
- CThe extracellular concentration of sodium is higher than that of potassium
- DThe intracellular concentration of potassium is higher than that of sodium
Show answer
The resting membrane potential is determined mainly by the relative permeability (conductance) of the membrane to different ions, as described by the Goldman-Hodgkin-Katz equation. At rest, the membrane has many open potassium leak channels but very few open sodium channels, so potassium conductance dominates and pulls the membrane potential close to the potassium equilibrium potential. The Na+/K+ ATPase maintains the concentration gradients but contributes only a small direct electrogenic component to the resting potential. Concentration gradients alone do not determine membrane potential; permeability does.
- B. The pump's electrogenic contribution is small and does not primarily set the resting potential; it maintains the concentration gradients.
- C. Extracellular ion concentrations alone, without considering membrane permeability, cannot explain the resting potential.
- D. Concentration gradients establish the equilibrium potentials for each ion but do not by themselves determine which ion dominates the resting potential.
A physiology student stimulates an isolated nerve fiber with two successive supramaximal electrical shocks separated by a short interval and records the compound action potential. When the interval between the two shocks is reduced below a certain value, the second shock fails to produce any action potential no matter how strong it is made. This finding is best explained by the second shock falling within which phase of the first action potential?
- ASubnormal period, when potassium conductance is still elevated
- BSupernormal period, when excitability is increased
- CRelative refractory period, when some sodium channels have recovered
- DAbsolute refractory period, when sodium channels are inactivated
Show answer
During the absolute refractory period, which corresponds to the period of sodium channel inactivation following the upstroke, no stimulus of any strength can elicit a new action potential because insufficient sodium channels are available to open. This period limits the maximum frequency at which a nerve fiber can conduct impulses. The relative refractory period follows, during which a stronger-than-normal stimulus can trigger a response because some sodium channels have recovered from inactivation. The supernormal and subnormal periods relate to smaller changes in threshold occurring later in the cycle and would not produce complete failure to respond.
- A. The subnormal period shows reduced excitability but a strong enough stimulus can still generate a response, unlike the absolute refractory period.
- B. The supernormal period is a phase of increased, not absent, excitability.
- C. In the relative refractory period a sufficiently strong stimulus can still elicit a response, so failure regardless of stimulus strength would not occur here.
Which type of nerve fiber conducts fast pain and temperature sensation and has the slowest conduction velocity among myelinated fibers?
- AA-alpha fibers
- BA-delta fibers
- CA-beta fibers
- DC fibers
Show answer
A-delta fibers are thinly myelinated, small-diameter fibers that conduct impulses at 5-30 m/s and carry fast, sharp, well-localized pain and temperature sensation, making them the slowest among the myelinated A fibers. A-alpha fibers are the largest, heavily myelinated fibers with the fastest conduction velocity, carrying proprioceptive and motor information. C fibers are unmyelinated and conduct the slowest of all, carrying dull, poorly localized ('slow') pain. A-beta fibers are intermediate-sized myelinated fibers that carry touch and pressure sensation.
- A. A-alpha fibers are the largest and fastest-conducting myelinated fibers, carrying motor and proprioceptive signals, not pain.
- C. A-beta fibers carry touch and pressure, conduct faster than A-delta fibers, and are not primarily associated with pain.
- D. C fibers are unmyelinated and conduct even slower than A-delta fibers, carrying dull aching pain rather than fast sharp pain.
A researcher studies three agents at the neuromuscular junction. Agent X binds nicotinic acetylcholine receptors on the motor end plate without activating them, competitively blocking acetylcholine. Agent Y binds the same receptors but produces sustained depolarization, leading to initial fasciculations followed by flaccid paralysis as the end plate becomes unresponsive. Agent Z acts presynaptically to prevent calcium-dependent fusion of acetylcholine-containing vesicles with the terminal membrane. Agent Y most closely corresponds to which drug?
- ASuccinylcholine
- BTubocurarine
- CNeostigmine
- DBotulinum toxin
Show answer
Succinylcholine is a depolarizing neuromuscular blocker that binds nicotinic receptors at the motor end plate and causes sustained depolarization, producing initial transient fasciculations followed by a depolarization block as the end plate becomes refractory to further stimulation. Tubocurarine is a non-depolarizing, competitive antagonist at the same receptors, corresponding to agent X in the vignette. Botulinum toxin acts presynaptically by cleaving SNARE proteins required for calcium-dependent vesicle fusion, preventing acetylcholine release, corresponding to agent Z. Neostigmine is an acetylcholinesterase inhibitor that increases acetylcholine availability at the synapse and is not described by any of the three mechanisms given.
- B. Tubocurarine is a competitive antagonist that blocks the receptor without activating it, matching agent X, not agent Y.
- C. Neostigmine inhibits acetylcholinesterase and does not match any of the three described mechanisms.
- D. Botulinum toxin acts presynaptically on vesicle fusion, matching agent Z, not the postsynaptic depolarizing action of agent Y.
A skeletal muscle is set up in an experimental apparatus and stimulated under two different conditions. In condition one, the muscle is allowed to shorten against a constant light load, lifting a weight. In condition two, both ends of the muscle are fixed so that it cannot shorten, and it is stimulated to contract against an immovable load. Which statement correctly distinguishes these two types of contraction?
- AIn condition one, both muscle length and tension remain constant; in condition two, only tension changes
- BIn both conditions, muscle length and tension change simultaneously by the same proportion
- CIn condition one, muscle tension remains constant while muscle length changes; in condition two, muscle length remains constant while tension changes
- DIn condition one, muscle length remains constant while tension changes; in condition two, tension remains constant while length changes
Show answer
Condition one describes isotonic contraction, in which the muscle shortens and moves a load while developing tension that stays constant and equal to the load being lifted, so length changes while tension remains fixed. Condition two describes isometric contraction, in which the muscle generates tension against a fixed, immovable load without any change in overall muscle length, so tension changes while length remains fixed. This distinction is fundamental to the length-tension and force-velocity relationships studied in muscle physiology. Isotonic contractions do mechanical work (force times distance), whereas isometric contractions do no external mechanical work despite consuming energy.
- A. This reverses the relationship and incorrectly states that condition one shows no change in either variable.
- B. Neither contraction type involves simultaneous proportional change in both length and tension by definition.
- D. This reverses the correct pairing of length and tension changes between the two conditions.
A skeletal muscle fiber is depolarized by an action potential that spreads along the sarcolemma and into the T-tubules. This triggers a conformational change in the dihydropyridine receptor, which mechanically opens the ryanodine receptor channel on the adjacent sarcoplasmic reticulum, releasing calcium into the cytoplasm. Binding of this released calcium to which protein directly initiates the cycle of cross-bridge formation?
- AMyosin light chain kinase
- BCalmodulin
- CTroponin C
- DTropomyosin
Show answer
Calcium released from the sarcoplasmic reticulum binds to troponin C, a subunit of the troponin complex on the thin filament, causing a conformational change that displaces tropomyosin away from the myosin-binding sites on actin. This exposes the actin binding sites, allowing myosin cross-bridges to attach and cycle, producing contraction. Tropomyosin itself does not bind calcium directly; it moves only as a consequence of the calcium-induced change in troponin. Calmodulin and myosin light chain kinase mediate calcium-dependent regulation of contraction in smooth muscle rather than in the troponin-based mechanism of skeletal muscle.
- A. Myosin light chain kinase is activated by calcium-calmodulin in smooth muscle, not in the troponin-based skeletal muscle mechanism.
- B. Calmodulin mediates calcium signaling in smooth muscle contraction, not in skeletal muscle excitation-contraction coupling.
- D. Tropomyosin is moved as a result of the conformational change in troponin but does not itself bind calcium.
A motor neuron in the spinal cord has a resting membrane potential of -70 mV and a firing threshold of -55 mV. A single excitatory presynaptic terminal, when stimulated alone, produces an excitatory postsynaptic potential (EPSP) of 5 mV that decays within a few milliseconds. When the same single terminal is stimulated repeatedly in rapid succession within this short time window, the EPSPs add together to bring the membrane potential to threshold and trigger an action potential. This phenomenon is best described as which type of summation?
- APost-tetanic potentiation
- BFacilitation
- CTemporal summation
- DSpatial summation
Show answer
Temporal summation occurs when successive EPSPs from a single presynaptic terminal arrive close enough in time that their individual depolarizations overlap and add together before the first one has fully decayed, raising the membrane potential to threshold. Spatial summation instead involves simultaneous input from multiple different presynaptic terminals converging on the same postsynaptic neuron. Facilitation refers to an increase in the amount of neurotransmitter released by a presynaptic terminal with repeated stimulation over a short interval, which is a related but distinct presynaptic mechanism. Post-tetanic potentiation describes enhanced synaptic transmission following a period of high-frequency tetanic stimulation, persisting over a longer timescale.
- A. Post-tetanic potentiation refers to enhanced transmission after a preceding tetanic stimulus, a longer-lasting phenomenon distinct from simple summation of successive EPSPs.
- B. Facilitation describes a presynaptic increase in neurotransmitter release, not the postsynaptic adding of successive potentials.
- D. Spatial summation requires input from multiple different synapses acting simultaneously on the postsynaptic neuron, not repeated firing of one synapse.
Cardiovascular Physiology
8 questionsA researcher records transmembrane action potentials from two isolated cardiac cell types. Cell A shows a slow spontaneous depolarization of the resting membrane potential leading to threshold, with the upstroke mediated mainly by an inward current through T-type and L-type channels, and no true stable resting potential. Cell B maintains a stable resting potential near -90 mV, a rapid upstroke, and a prolonged plateau phase. Which ionic current is primarily responsible for the plateau phase seen in Cell B?
- AFast inward sodium current through voltage-gated sodium channels
- BTransient outward potassium current responsible for phase 1 notch
- CFunny current (If) through hyperpolarization-activated cyclic nucleotide-gated channels
- DSlow inward calcium current through L-type calcium channels
Show answer
In the ventricular myocyte (Cell B), the plateau (phase 2) is sustained by a slow inward Ca2+ current through L-type calcium channels, which balances outward K+ currents and keeps the membrane depolarized for 200-300 ms, allowing adequate time for calcium-induced calcium release and contraction. This is distinct from the SA node (Cell A), where the funny current and T/L-type calcium channels drive pacemaker depolarization without a true plateau. The long plateau also explains the long refractory period of ventricular muscle, preventing tetanic contraction.
- A. The fast Na+ current causes the rapid upstroke (phase 0) in ventricular cells, not the plateau.
- B. The transient outward K+ current produces the brief phase 1 notch, not the sustained plateau.
- C. The funny current drives pacemaker (phase 4) depolarization in nodal tissue, not the ventricular plateau.
In a cardiac physiology experiment on an isolated heart preparation, increasing the venous return progressively increases left ventricular end-diastolic volume from 120 mL to 160 mL, and stroke volume rises correspondingly from 70 mL to 100 mL without any change in contractility. When a positive inotropic agent is then added at the same end-diastolic volume of 160 mL, stroke volume increases further to 120 mL. Which statement best explains this additional increase in stroke volume with the inotropic agent?
- AHeart rate increased, allowing more time for ventricular filling
- BAfterload decreased, allowing greater ejection fraction at the same preload
- CVenous return increased further, raising end-diastolic volume beyond 160 mL
- DThe entire Frank-Starling curve shifted upward and to the left due to increased calcium sensitivity of myofilaments
Show answer
A positive inotropic agent increases myocardial contractility by enhancing intracellular calcium availability or myofilament calcium sensitivity, producing a greater stroke volume at the same end-diastolic volume (preload). Graphically, this shifts the entire Frank-Starling curve upward and to the left, distinguishing it from simple preload-mediated increases in stroke volume, which move along the same curve. Afterload and heart rate are not responsible since the scenario specifies the same end-diastolic volume and no mention of changed ejection resistance or rate. This distinction between a shift along the curve (preload effect) versus a shift of the curve itself (contractility effect) is fundamental to understanding cardiac performance.
- A. Heart rate change is not mentioned and filling time is a preload-related mechanism, not a contractility effect.
- B. Afterload change is not described in the scenario and would not explain a curve shift.
- C. The stem states end-diastolic volume was kept the same at 160 mL when the inotrope was added.
During simultaneous recording of left ventricular pressure, aortic pressure, and phonocardiogram in a healthy adult, a heart sound is recorded at the exact moment the left ventricular pressure curve crosses below the aortic pressure curve, just before the ventricular pressure falls below left atrial pressure. Which event corresponds to this heart sound?
- AOpening of the aortic valve producing an ejection click
- BClosure of the mitral valve producing the first heart sound
- CClosure of the aortic valve producing the second heart sound
- DOpening of the mitral valve producing an early diastolic sound
Show answer
The second heart sound (S2) occurs when ventricular pressure falls below aortic pressure, causing the aortic valve to close; this marks the end of ventricular systole and the start of isovolumetric relaxation, after which ventricular pressure continues to fall until it drops below atrial pressure to permit mitral valve opening. S1 occurs earlier, at mitral valve closure when ventricular pressure first exceeds atrial pressure. Mitral valve opening is silent in the normal heart and does not generate a heart sound. An ejection click, if present, would occur at aortic valve opening, which happens when ventricular pressure exceeds aortic pressure at the start of systole, not at this described crossover point.
- A. Aortic valve opening occurs earlier in systole, when ventricular pressure first exceeds aortic pressure.
- B. Mitral closure (S1) occurs at the start of systole when ventricular pressure exceeds atrial pressure, not at this point.
- D. Mitral valve opening is normally silent and occurs later, once ventricular pressure falls below atrial pressure.
A jugular venous pressure tracing is recorded along with a simultaneous ECG in a healthy 28-year-old volunteer. A positive deflection is noted to occur immediately after the QRS complex, coinciding with the period of isovolumetric ventricular contraction, and is attributed to bulging of the closed tricuspid valve into the right atrium. Which component of the JVP waveform does this represent?
- Av wave
- Bc wave
- Ca wave
- Dy descent
Show answer
The c wave in the jugular venous pulse occurs during isovolumetric ventricular contraction, just after the QRS complex, and results from the closed tricuspid valve bulging backward into the right atrium as ventricular pressure rises sharply. The a wave precedes it and is produced by right atrial contraction, occurring just before the QRS complex. The v wave occurs later in systole, reflecting atrial filling against a closed tricuspid valve, while the y descent represents the fall in atrial pressure after tricuspid valve opening and rapid atrial emptying into the ventricle.
- A. The v wave occurs later in ventricular systole, reflecting passive atrial filling, not isovolumetric contraction.
- C. The a wave occurs before the QRS, due to atrial contraction, not after it.
- D. The y descent is a downward deflection following tricuspid valve opening, not a positive wave after the QRS.
A 34-year-old man sustains a stab injury and loses approximately 1200 mL of blood over 20 minutes. His heart rate rises from 76/min to 128/min, and peripheral vascular resistance increases, while mean arterial pressure is partially maintained at 78/52 mmHg. Which mechanism most directly initiates this compensatory response?
- ADecreased stretch of carotid sinus and aortic arch baroreceptors reducing afferent firing to the nucleus tractus solitarius
- BIncreased stretch of atrial volume receptors suppressing antidiuretic hormone release
- CIncreased renal perfusion pressure activating the macula densa tubuloglomerular feedback
- DDecreased arterial PCO2 stimulating peripheral chemoreceptors directly
Show answer
A fall in blood volume reduces arterial pressure, decreasing stretch of baroreceptors in the carotid sinus and aortic arch; this lowers afferent firing to the nucleus tractus solitarius in the medulla, reducing vagal (parasympathetic) tone and increasing sympathetic outflow. The net effect is tachycardia, increased myocardial contractility, and generalized arteriolar vasoconstriction to defend arterial pressure. Atrial volume receptor stretch would decrease (not increase) with hemorrhage, promoting ADH release rather than suppressing it. Chemoreceptor stimulation by hypoxia or hypercapnia is a secondary, less immediate mechanism compared to the baroreceptor reflex in moderate hemorrhage.
- B. Atrial receptor stretch decreases, not increases, with hypovolemia, and this would promote ADH release, not suppress it.
- C. Tubuloglomerular feedback regulates renal autoregulation and does not directly drive the immediate systemic tachycardia and vasoconstriction seen here.
- D. Hemorrhage does not primarily lower arterial PCO2, and chemoreceptors are not the main early trigger in moderate blood loss.
During moderate exercise, myocardial oxygen consumption increases nearly fourfold. Coronary blood flow increases proportionally to match this demand, even though aortic diastolic pressure (the main driving pressure for coronary perfusion) rises only slightly. Which locally produced substance is considered the most important mediator of this increase in coronary flow?
- AThromboxane A2 released from activated platelets in coronary vessels
- BEndothelin-1 released from coronary endothelial cells
- CAdenosine released from myocardial cells due to ATP breakdown
- DAngiotensin II generated locally in coronary vascular smooth muscle
Show answer
Adenosine, formed from the breakdown of ATP when myocardial oxygen demand exceeds supply, is the principal metabolic mediator of coronary vasodilation; it diffuses from myocytes to adjacent vascular smooth muscle, activates A2 receptors, and causes vasodilation through increased potassium conductance and smooth muscle relaxation. This metabolic (local) autoregulation allows coronary flow to track myocardial oxygen consumption closely, independent of changes in perfusion pressure within the autoregulatory range. Angiotensin II and endothelin-1 are vasoconstrictors, not dilators, and thromboxane A2 similarly promotes vasoconstriction and platelet aggregation rather than the coronary vasodilation seen during increased myocardial work.
- A. Thromboxane A2 causes vasoconstriction and platelet aggregation, the opposite of the vasodilation required here.
- B. Endothelin-1 is one of the most potent vasoconstrictors and opposes coronary vasodilation.
- D. Angiotensin II is a potent vasoconstrictor and would reduce, not increase, coronary flow.
A 24-year-old asymptomatic medical student undergoes a routine ECG during a physiology practical. The interval from the onset of the P wave to the onset of the QRS complex measures 220 milliseconds, with every P wave followed by a QRS complex and a constant interval throughout the recording. Which physiological process is primarily reflected in this measured interval?
- ATotal duration of ventricular repolarization
- BTime taken for atrial depolarization alone
- CConduction time through the atrioventricular node and His-Purkinje system
- DTime taken for ventricular depolarization to be completed
Show answer
The PR interval represents the time from the onset of atrial depolarization to the onset of ventricular depolarization, and its major component is the conduction delay through the atrioventricular node, along with conduction through the His bundle and Purkinje system; the normal range is 120-200 ms. A PR interval of 220 ms indicates a delay in AV nodal conduction, consistent with first-degree AV block. Atrial depolarization alone is reflected only in the P wave duration, while ventricular depolarization corresponds to the QRS complex duration, and ventricular repolarization is reflected in the QT interval and T wave, not the PR interval.
- A. Ventricular repolarization is reflected in the QT interval and T wave, which are separate from the PR interval.
- B. Atrial depolarization alone is represented by the P wave duration, not the PR interval, which also includes AV nodal delay.
- D. Ventricular depolarization is represented by the QRS complex duration, not the PR interval.
In a patient with hypoalbuminemia secondary to nephrotic syndrome, plasma albumin falls to 1.8 g/dL. Generalized pitting edema develops over the lower limbs and periorbital region. Which alteration in capillary Starling forces best explains the increased net filtration of fluid into the interstitial space in this patient?
- AIncreased capillary hydrostatic pressure due to arteriolar vasodilation
- BDecreased plasma colloid osmotic pressure reducing the force drawing fluid back into capillaries
- CIncreased interstitial colloid osmotic pressure drawing fluid out of capillaries
- DDecreased interstitial hydrostatic pressure increasing outward filtration gradient
Show answer
Plasma albumin is the major contributor to plasma colloid osmotic (oncotic) pressure, which normally opposes capillary hydrostatic pressure and favors fluid reabsorption into capillaries. In nephrotic syndrome, heavy proteinuria lowers plasma albumin, reducing plasma oncotic pressure; this diminishes the inward force according to the Starling equation, so net filtration of fluid out of capillaries into the interstitium increases, producing edema. Capillary hydrostatic pressure and interstitial forces are not the primary abnormality in this condition; the edema in nephrotic syndrome is classically oncotic-pressure-driven rather than hydrostatic-pressure-driven.
- A. Hydrostatic pressure changes are not the primary mechanism in hypoalbuminemic edema of nephrotic syndrome.
- C. Interstitial oncotic pressure rising is not the main pathophysiology here; the key change is the fall in plasma oncotic pressure.
- D. Interstitial hydrostatic pressure does not typically fall in this condition and is not the primary described defect.
Respiratory Physiology
6 questionsAn arterial blood sample is analyzed under two conditions. In sample A, pH is 7.20, PaCO2 is 60 mmHg, and temperature is 39°C. In sample B, pH is 7.50, PaCO2 is 28 mmHg, and temperature is 36°C. At the same PaO2 of 40 mmHg, which statement correctly describes the oxygen saturation in these two samples?
- ASample B will have lower oxygen saturation than sample A due to increased 2,3-BPG production
- BSample A will have higher oxygen saturation than sample B due to a leftward shift of the dissociation curve
- CSample A will have lower oxygen saturation than sample B due to a rightward shift of the dissociation curve
- DBoth samples will have identical oxygen saturation because PaO2 is the only determinant of saturation
Show answer
Sample A has acidosis, hypercapnia, and fever, all of which shift the oxyhemoglobin dissociation curve to the right, lowering hemoglobin's affinity for oxygen and decreasing saturation at a given PaO2 (Bohr effect). Sample B has alkalosis, hypocapnia, and lower temperature, which shift the curve to the left, increasing affinity and saturation at the same PaO2. This rightward shift facilitates oxygen unloading at the tissues in conditions like exercising or acidotic tissue, while a leftward shift favors oxygen loading, as seen in fetal hemoglobin. 2,3-BPG also shifts the curve but is not the variable changed here.
- A. 2,3-BPG levels were not given as different between the two samples in this scenario.
- B. This reverses the actual relationship; sample A is shifted right, not left.
- D. PaO2 alone does not determine saturation; pH, PaCO2, and temperature also alter hemoglobin affinity.
A 58-year-old man with a 30-pack-year smoking history undergoes spirometry. His FEV1 is 1.4 L (55% predicted), FVC is 2.8 L (78% predicted), and FEV1/FVC ratio is 50%. Total lung capacity measured by body plethysmography is 130% predicted. Which pattern best describes these findings?
- AObstructive pattern with air trapping
- BRestrictive pattern with reduced lung volumes
- CMixed obstructive-restrictive pattern
- DNormal spirometry with age-related decline
Show answer
A reduced FEV1/FVC ratio below 70% is the hallmark of an obstructive ventilatory defect, as seen in chronic obstructive pulmonary disease from long-standing smoking. The disproportionately reduced FEV1 compared to FVC, combined with an increased total lung capacity from hyperinflation and air trapping due to loss of elastic recoil and small airway collapse, confirms obstruction rather than restriction. In restrictive disease, both FEV1 and FVC are reduced proportionately with a preserved or increased FEV1/FVC ratio, and total lung capacity is reduced. This patient's elevated TLC excludes a restrictive process.
- B. Restrictive disease shows reduced TLC with a normal or increased FEV1/FVC ratio, the opposite of this picture.
- C. A mixed pattern would show reduced TLC alongside a reduced FEV1/FVC ratio, but TLC here is increased, not reduced.
- D. The FEV1/FVC ratio of 50% is well below the normal cutoff of 70%, indicating clear obstruction, not a normal study.
In an upright healthy adult, both ventilation and perfusion are greater at the base of the lung than at the apex, but the ventilation-perfusion ratio is higher at the apex than at the base. Which mechanism best explains this regional difference in the V/Q ratio?
- AGravity causes a proportionally greater reduction in blood flow than in ventilation toward the apex
- BAlveoli at the apex have greater surfactant concentration, increasing their ventilation disproportionately
- CAirway resistance is lower at the base, causing ventilation to exceed perfusion at the apex
- DHypoxic pulmonary vasoconstriction is more pronounced at the base, reducing perfusion there
Show answer
Gravity causes a vertical gradient in both ventilation and perfusion in the upright lung, but pulmonary blood flow is affected far more than ventilation because pulmonary vascular pressures are low and easily influenced by hydrostatic forces. At the apex, perfusion falls almost to zero in some zones while ventilation falls only modestly, raising the V/Q ratio above 1 (West zone 1 physiology). At the base, both are higher but perfusion increases more steeply than ventilation, lowering the V/Q ratio below 1. This explains relatively higher alveolar PO2 and lower PCO2 at the apex compared to the base.
- B. Surfactant distribution does not differ regionally in this manner and is not the basis for the V/Q gradient.
- C. Airway resistance differences are not the primary determinant of the normal gravitational V/Q gradient.
- D. Hypoxic pulmonary vasoconstriction redirects blood from poorly ventilated alveoli, it does not create the baseline gravitational gradient.
At the pulmonary capillaries, hemoglobin becomes increasingly saturated with oxygen as it passes through the alveoli. Which direct consequence of this oxygenation facilitates the release of carbon dioxide from blood into the alveolar air?
- AOxygenation lowers the partial pressure of CO2 dissolved in plasma independent of hemoglobin binding
- BOxygenation increases red blood cell carbonic anhydrase activity, accelerating CO2 hydration
- COxygenated hemoglobin directly displaces bicarbonate from plasma into red blood cells
- DOxygenated hemoglobin has a reduced affinity for carbon dioxide and hydrogen ions, promoting CO2 release
Show answer
The Haldane effect describes how the binding of oxygen to hemoglobin reduces hemoglobin's affinity for both carbon dioxide (carried as carbaminohemoglobin) and hydrogen ions. As blood passes through the pulmonary capillaries and hemoglobin becomes oxygenated, it releases bound CO2 and H+, which shifts the carbonic acid-bicarbonate equilibrium to regenerate CO2 from bicarbonate, allowing it to diffuse into the alveoli and be exhaled. This effect works in conjunction with the Bohr effect, which operates in the opposite direction at the tissues, together optimizing both oxygen loading and CO2 unloading. The Haldane effect is responsible for a larger proportion of CO2 transport variation than simple physical solubility changes.
- A. The partial pressure change in plasma is a result of the shifted equilibrium caused by hemoglobin's altered affinity, not an independent event.
- B. Carbonic anhydrase activity itself is not directly altered by the oxygenation state of hemoglobin.
- C. Bicarbonate movement across the red cell membrane via the chloride shift is a downstream consequence, not the direct mechanism of the Haldane effect.
A male infant is born at 29 weeks of gestation and develops progressive respiratory distress with grunting, nasal flaring, and subcostal retractions within hours of birth. Chest radiograph shows a diffuse ground-glass appearance with air bronchograms. Which underlying physiological abnormality best explains the markedly reduced lung compliance in this infant?
- APremature closure of the ductus arteriosus increasing pulmonary vascular resistance
- BDeficiency of surfactant produced by type II pneumocytes, increasing alveolar surface tension
- CIncreased elastin content in the immature lung parenchyma increasing elastic recoil
- DExcess pulmonary surfactant causing alveolar flooding and impaired gas exchange
Show answer
Type II pneumocytes begin producing surfactant in adequate quantities only after about 34-35 weeks of gestation, so infants born prematurely, as at 29 weeks, often have insufficient surfactant. Surfactant, composed mainly of dipalmitoylphosphatidylcholine, lowers alveolar surface tension according to the Laplace relationship, preventing alveolar collapse at low lung volumes and maintaining compliance. Without adequate surfactant, surface tension rises, smaller alveoli tend to collapse, and the lung becomes stiff with reduced compliance, producing the diffuse atelectasis and ground-glass appearance seen in neonatal respiratory distress syndrome. Antenatal corticosteroids given to the mother accelerate fetal surfactant production and are used to reduce this risk when preterm delivery is anticipated.
- A. Ductus arteriosus closure affects pulmonary blood flow and cardiac physiology but does not directly explain the reduced lung compliance of RDS.
- C. Elastin content being increased would raise elastic recoil but is not the primary defect in surfactant-deficient neonatal lung disease.
- D. Surfactant deficiency, not excess, is the pathology in RDS; excess surfactant is not a recognized cause of reduced compliance.
A 45-year-old man with suspected pulmonary embolism undergoes respiratory function testing. His arterial PCO2 is 45 mmHg, mixed expired PCO2 is 27 mm Hg, and tidal volume is 500 mL. Using Bohr's method, what is his physiological dead space volume?
- A100 mL
- B250 mL
- C200 mL
- D150 mL
Show answer
Bohr's equation for physiological dead space is VD = VT × (PaCO2 − PECO2)/PaCO2. Substituting the given values: VD = 500 × (45 − 27)/45 = 500 × 18/45 = 200 mL. This method uses arterial PCO2 as a surrogate for ideal alveolar PCO2, so any ventilation-perfusion mismatch (as in pulmonary embolism) that raises alveolar dead space will increase the calculated physiological dead space compared to the anatomical dead space measured by Fowler's method.
- A. Underestimates the dead space; does not match the calculation from the given values.
- B. Overestimates the dead space; arithmetic error in applying the Bohr equation.
- D. Underestimates the dead space; arithmetic error in applying the Bohr equation.
Renal Physiology
6 questionsA physiologist wants to measure glomerular filtration rate in a research subject with the highest possible accuracy, independent of tubular handling. Which substance is the gold standard marker used for this purpose?
- AInulin
- BPara-aminohippuric acid
- CCreatinine
- DUrea
Show answer
Inulin is a fructose polymer that is freely filtered at the glomerulus and is neither reabsorbed nor secreted nor metabolized by the renal tubules, making its clearance exactly equal to GFR. This makes it the gold standard exogenous marker for GFR measurement, though its use requires continuous intravenous infusion and is impractical for routine clinical use. Creatinine clearance is used clinically as a convenient endogenous estimate of GFR but slightly overestimates true GFR because of a small amount of tubular secretion. PAH clearance measures effective renal plasma flow, not GFR, because it is both filtered and almost completely secreted by the proximal tubule.
- B. Filtered and extensively secreted; used to measure renal plasma flow, not GFR.
- C. Endogenous marker used clinically but overestimates true GFR due to tubular secretion.
- D. Filtered but partly reabsorbed by the tubules, so its clearance underestimates GFR.
In the renal medulla, the loop of Henle generates a hyperosmotic interstitium that allows concentration of urine. Which property of the thick ascending limb is primarily responsible for establishing this osmotic gradient?
- AActive secretion of hydrogen ions into the tubular lumen
- BPassive reabsorption of urea driven by a high medullary urea concentration
- CActive transport of sodium, potassium, and chloride out of the tubular lumen with low water permeability
- DHigh permeability to water throughout its length
Show answer
The thick ascending limb actively transports Na+, K+, and Cl− out of the lumen via the Na-K-2Cl cotransporter (NKCC2) while remaining impermeable to water. This single effect of diluting the tubular fluid while concentrating the surrounding interstitium is repeated and multiplied along the length of the loop because of the countercurrent arrangement between the descending and ascending limbs, generating the medullary osmotic gradient that increases from cortex to papilla. Urea recycling from the inner medullary collecting duct contributes to the gradient but is a separate, additive mechanism rather than the primary multiplier. The descending limb is water permeable but does not actively transport solutes.
- A. Hydrogen ion secretion occurs in the proximal tubule and collecting duct for acid-base regulation, not for the countercurrent multiplier.
- B. Urea recycling adds to medullary hypertonicity but is not the primary multiplier mechanism of the loop.
- D. This describes the descending limb, which is water permeable but does not actively pump ions.
A 60-year-old man with heart failure and reduced renal perfusion develops activation of the renin-angiotensin-aldosterone system. Angiotensin-converting enzyme, which converts angiotensin I to angiotensin II, acts predominantly in which location?
- AProximal convoluted tubule brush border
- BZona glomerulosa of the adrenal cortex
- CPulmonary vascular endothelium
- DJuxtaglomerular cells of the afferent arteriole
Show answer
Angiotensin-converting enzyme (ACE) is bound to the luminal surface of vascular endothelial cells and is particularly abundant in the pulmonary circulation, which has an extensive capillary bed that converts most of the angiotensin I generated systemically into angiotensin II during a single pass through the lungs. Juxtaglomerular cells secrete renin in response to decreased renal perfusion pressure, decreased sodium delivery to the macula densa, or sympathetic stimulation, but they do not contain ACE. Angiotensin II subsequently acts on the zona glomerulosa to stimulate aldosterone secretion, which is a downstream effect, not the site of ACE activity.
- A. This site reabsorbs sodium and water under aldosterone's influence but is not where angiotensin I is converted to angiotensin II.
- B. This is the target tissue where angiotensin II stimulates aldosterone secretion, downstream of ACE action.
- D. This is the site of renin secretion in response to reduced perfusion pressure, not the site of ACE activity.
A 55-year-old woman with small cell lung cancer is found to have serum sodium of 118 mEq/L, serum osmolality of 242 mOsm/kg, urine osmolality of 450 mOsm/kg, and urine sodium of 60 mEq/L. She is euvolemic on examination with normal renal, adrenal, and thyroid function. Which of the following best explains her condition?
- ASyndrome of inappropriate antidiuretic hormone secretion
- BNephrogenic diabetes insipidus
- CCentral diabetes insipidus
- DPrimary polydipsia
Show answer
The combination of hyponatremia with low serum osmolality, inappropriately concentrated urine (urine osmolality greater than serum osmolality), elevated urine sodium, and clinical euvolemia in a patient with a tumor known to secrete ectopic ADH is classic for SIADH. Small cell lung cancer is a well-recognized cause of ectopic ADH secretion, which causes water retention and dilutional hyponatremia without edema because the excess water is distributed intracellularly and the kidney continues to excrete sodium normally. Both forms of diabetes insipidus produce dilute urine despite high serum osmolality and hypernatremia, the opposite pattern seen here. Primary polydipsia causes dilute urine and hyponatremia but with suppressed, not elevated, urine osmolality.
- B. This condition also produces dilute urine despite high serum osmolality, not concentrated urine with hyponatremia.
- C. This condition produces dilute urine (low urine osmolality) and hypernatremia from water loss, the opposite of this patient's findings.
- D. This would show maximally dilute urine, not concentrated urine, since ADH secretion is appropriately suppressed.
A 30-year-old man presents with persistent vomiting for 5 days due to gastric outlet obstruction. Arterial blood gas shows pH 7.52, PaCO2 48 mm Hg, and HCO3− 38 mEq/L. Which acid-base disturbance best explains this picture?
- APrimary respiratory alkalosis with appropriate renal compensation
- BMixed metabolic alkalosis and respiratory acidosis
- CPrimary metabolic alkalosis with appropriate respiratory compensation
- DPrimary respiratory acidosis with appropriate renal compensation
Show answer
The elevated pH of 7.52 confirms alkalemia, and the markedly elevated HCO3− of 38 mEq/L identifies metabolic alkalosis as the primary disturbance, consistent with loss of hydrochloric acid from prolonged vomiting. Appropriate respiratory compensation for metabolic alkalosis is hypoventilation that raises PaCO2 by about 0.7 mm Hg for every 1 mEq/L rise in HCO3− above 24, predicting a PaCO2 of roughly 24 + 0.7×14 ≈ 34 + 14 ≈ 48 mm Hg, which matches the observed value closely. If this were respiratory acidosis, the pH would be low (acidemic), not high. A mixed disorder is not needed here because the degree of respiratory compensation is appropriate and does not require invoking a second primary process.
- A. Respiratory alkalosis would show a low PaCO2 from hyperventilation, not an elevated PaCO2 as seen here.
- B. The compensation is appropriate for the degree of metabolic alkalosis, so a second primary disorder need not be invoked.
- D. Respiratory acidosis would cause a low pH (acidemia) with primary elevation of PaCO2, not the alkalemia seen here.
A 58-year-old man with type 2 diabetes mellitus and chronic kidney disease is started on an ACE inhibitor for proteinuria. Two weeks later his serum potassium rises from 4.2 to 6.1 mEq/L. Which mechanism best explains this change?
- ADecreased aldosterone-mediated potassium secretion in the distal nephron
- BIncreased potassium reabsorption in the thick ascending limb
- CDecreased filtered load of potassium due to reduced GFR
- DShift of potassium from the extracellular to intracellular compartment
Show answer
ACE inhibitors block the formation of angiotensin II, which reduces aldosterone secretion from the zona glomerulosa. Aldosterone normally acts on the principal cells of the distal tubule and collecting duct to upregulate the epithelial sodium channel (ENaC) and the renal outer medullary potassium channel (ROMK), driving sodium reabsorption coupled to potassium secretion into the lumen. Reduced aldosterone activity therefore decreases potassium secretion, causing retention and hyperkalemia, a well-known risk of ACE inhibitors especially in patients with underlying CKD and reduced nephron mass. The thick ascending limb reabsorbs potassium via NKCC2 for recycling, not as the primary site of regulated potassium excretion, and this is not typically altered by ACE inhibitors.
- B. The thick ascending limb's NKCC2-mediated potassium handling is not the primary regulated site of potassium excretion affected by aldosterone or ACE inhibitors.
- C. While reduced GFR in CKD contributes to hyperkalemia risk, the acute rise after starting an ACE inhibitor is primarily due to reduced aldosterone-driven distal potassium secretion.
- D. ACE inhibitors do not directly cause a transcellular potassium shift; the mechanism is renal potassium retention.
Gastrointestinal Physiology
1 questionsA 45-year-old man feels the urge to defecate after rectal distension by stool. Reflex relaxation of the internal anal sphincter occurs through the rectosphincteric (recto-anal inhibitory) reflex, but defecation is voluntarily delayed until a socially convenient time. Which of the following best describes the neural control that allows this voluntary delay?
- ASustained contraction of the internal anal sphincter maintained by sympathetic fibers from the inferior mesenteric ganglion
- BInhibition of the myenteric plexus by parasympathetic fibers from the pelvic splanchnic nerves
- CTonic contraction of the puborectalis muscle mediated by the hypogastric nerve
- DVoluntary contraction of the external anal sphincter via the pudendal nerve, under control of the sacral spinal cord and cerebral cortex
Show answer
Rectal distension triggers the intrinsic rectosphincteric reflex, which relaxes the internal anal sphincter (smooth muscle, normally tonically contracted) via myenteric plexus activity, independent of extrinsic innervation. However, the external anal sphincter is skeletal muscle supplied by the pudendal nerve (S2-S4) and is under voluntary cortical control, allowing a person to consciously maintain its contraction and delay defecation despite internal sphincter relaxation. When it is appropriate, voluntary relaxation of the external sphincter together with increased intra-abdominal pressure (Valsalva) permits expulsion of feces. This dual sphincter arrangement is the physiological basis of continence and voluntary control of defecation.
- A. Sympathetic tone helps maintain internal sphincter contraction at rest, but it is not the mechanism of voluntary delay once the rectosphincteric reflex has relaxed it.
- B. Parasympathetic pelvic splanchnic activity promotes rectal contraction and defecation rather than inhibiting it.
- C. The puborectalis maintains the anorectal angle for continence but is not supplied by the hypogastric nerve and is not the basis of voluntary delay described here.
Endocrine Physiology
7 questionsWhich of the following hormone pairs correctly matches a hypothalamic releasing hormone with the anterior pituitary tropic hormone whose secretion it primarily stimulates?
- AGrowth hormone-releasing hormone - thyroid-stimulating hormone
- BGonadotropin-releasing hormone - prolactin
- CThyrotropin-releasing hormone - follicle-stimulating hormone
- DCorticotropin-releasing hormone - adrenocorticotropic hormone
Show answer
Corticotropin-releasing hormone (CRH) is secreted by the parvocellular neurons of the hypothalamic paraventricular nucleus and travels via the hypophyseal portal system to stimulate corticotroph cells of the anterior pituitary to release adrenocorticotropic hormone (ACTH). ACTH then acts on the adrenal cortex (zona fasciculata) to stimulate cortisol synthesis and release, which in turn exerts negative feedback on both the hypothalamus and pituitary. This is the classic hypothalamic-pituitary-adrenal axis tested extensively in physiology.
- A. Growth hormone-releasing hormone stimulates growth hormone release, not TSH.
- B. Prolactin secretion is mainly under tonic dopamine inhibition; GnRH stimulates LH and FSH, not prolactin.
- C. Thyrotropin-releasing hormone stimulates TSH (and prolactin), not FSH.
A researcher studies glucose-stimulated insulin secretion in isolated pancreatic beta cells. Raising extracellular glucose increases its entry via GLUT2 transporters and subsequent glycolysis, which raises the intracellular ATP/ADP ratio. Which of the following is the immediate downstream consequence of this rise in ATP/ADP ratio that leads to insulin release?
- AOpening of ATP-sensitive potassium channels, causing membrane hyperpolarization and closure of calcium channels
- BClosure of ATP-sensitive potassium channels, causing membrane depolarization and opening of voltage-gated calcium channels
- CIncreased cyclic AMP production, which opens voltage-gated sodium channels
- DDirect activation of protein kinase A, which phosphorylates insulin granule membrane proteins
Show answer
In pancreatic beta cells, a rise in the ATP/ADP ratio closes ATP-sensitive potassium (KATP) channels on the plasma membrane. This reduces potassium efflux and depolarizes the cell membrane, which opens voltage-gated L-type calcium channels, allowing calcium influx. The resulting rise in intracellular calcium triggers exocytosis of insulin-containing secretory granules. This KATP channel mechanism is also the target of sulfonylurea drugs, which close the channel independent of glucose to stimulate insulin release.
- A. Opening of KATP channels would hyperpolarize the membrane and suppress, not stimulate, insulin secretion; this is the opposite of what occurs.
- C. cAMP-dependent pathways modulate insulin secretion but act through calcium and PKA effects, not by opening sodium channels.
- D. Protein kinase A activation via cAMP potentiates insulin secretion (as with GLP-1) but is not the primary glucose-sensing pathway described here.
A 32-year-old woman with Graves' disease is started on an antithyroid drug that inhibits the enzyme responsible for coupling and oxidation of iodotyrosine residues within thyroglobulin. Inhibition of this enzyme most directly blocks which step of thyroid hormone synthesis?
- AActive transport of iodide into the thyroid follicular cell against its electrochemical gradient
- BOxidation of iodide and iodination of tyrosine residues on thyroglobulin, and coupling of iodotyrosines to form T3 and T4
- CProteolysis of thyroglobulin within the follicular cell to release free T3 and T4
- DPeripheral deiodination of T4 to the more active T3 in target tissues
Show answer
Thyroid peroxidase (TPO) catalyzes oxidation of iodide to its active form, iodination of tyrosine residues on thyroglobulin (organification), and the coupling of monoiodotyrosine and diiodotyrosine to form T3 and T4. Thioamide drugs such as propylthiouracil and methimazole inhibit TPO, blocking both organification and coupling, thereby reducing new hormone synthesis. Propylthiouracil additionally inhibits peripheral deiodination of T4 to T3, a property methimazole lacks. This explains why thioamides take weeks to produce clinical effect, since they do not block release of preformed hormone.
- A. Iodide uptake occurs via the sodium-iodide symporter (NIS), which is inhibited by perchlorate and thiocyanate, not thioamides.
- C. Proteolysis and release of stored hormone is blocked by iodine (Wolff-Chaikoff effect) and lithium, not by TPO inhibitors.
- D. Peripheral deiodination is catalyzed by deiodinase enzymes; only propylthiouracil (not methimazole) has a modest inhibitory effect on this step, and it is not the enzyme described in the stem.
A healthy 30-year-old volunteer has plasma cortisol measured at 8:00 AM and again at 11:00 PM as part of a physiology study. The 8:00 AM value is 18 micrograms/dL and the 11:00 PM value is 3 micrograms/dL. Which of the following best explains this pattern along with the expected metabolic action of cortisol at its peak?
- ACortisol secretion follows a circadian rhythm peaking at midnight, promoting glycogen synthesis and enhanced peripheral glucose uptake
- BCortisol secretion is secondary to feeding patterns and rises sharply within minutes after breakfast
- CThe high morning value indicates a pathological cortisol excess requiring immediate evaluation for Cushing syndrome
- DCortisol secretion follows a circadian rhythm driven by hypothalamic CRH pulses, peaking in early morning and promoting gluconeogenesis and insulin resistance
Show answer
Cortisol secretion shows a well-defined circadian rhythm governed by the suprachiasmatic nucleus acting through hypothalamic CRH and pituitary ACTH, with pulsatile secretion peaking around 6-8 AM and reaching a nadir around midnight. At its morning peak, cortisol promotes hepatic gluconeogenesis, inhibits peripheral glucose uptake causing relative insulin resistance, stimulates lipolysis, and has catabolic effects on muscle protein, helping mobilize substrates for the demands of the waking and active period. The values given (high AM, low PM) represent the normal physiological pattern, not disease.
- A. The rhythm is reversed from this description; cortisol is lowest, not highest, around midnight, and it opposes rather than promotes glycogen synthesis and glucose uptake.
- B. Cortisol's circadian rhythm is intrinsic to the HPA axis and persists even with fasting; it is not primarily a feeding-driven response.
- C. These values represent the normal diurnal variation, not pathological excess; Cushing syndrome is suggested by loss of diurnal rhythm and elevated 24-hour cortisol, not a normal morning peak.
A 50-year-old woman has a serum calcium of 7.2 mg/dL (normal 8.5-10.5 mg/dL). In response, parathyroid hormone secretion increases. Which of the following correctly describes a combined action of parathyroid hormone that helps restore serum calcium toward normal?
- ADirect stimulation of calcitonin release from parafollicular cells, which increases osteoclastic bone resorption
- BIncreased renal 1-alpha-hydroxylase activity raising calcitriol, combined with increased renal phosphate reabsorption and decreased calcium reabsorption
- CDecreased osteoblastic RANKL expression, reducing osteoclast activity and bone calcium release
- DIncreased renal 1-alpha-hydroxylase activity raising calcitriol, which enhances intestinal calcium absorption, combined with increased renal calcium reabsorption and decreased phosphate reabsorption
Show answer
Parathyroid hormone (PTH) raises serum calcium through three coordinated actions: it stimulates renal 1-alpha-hydroxylase, increasing conversion of 25-hydroxyvitamin D to active calcitriol (1,25-dihydroxyvitamin D), which enhances intestinal calcium absorption; it increases calcium reabsorption in the distal renal tubule; and it decreases phosphate reabsorption in the proximal tubule, causing phosphaturia. PTH also increases osteoblastic RANKL expression, which promotes osteoclastic bone resorption and releases calcium from bone. These combined effects raise serum calcium while lowering serum phosphate.
- A. PTH does not stimulate calcitonin release; calcitonin is released independently by parafollicular C cells in response to hypercalcemia and lowers serum calcium.
- B. PTH decreases, not increases, renal phosphate reabsorption, and it increases rather than decreases calcium reabsorption.
- C. PTH increases, not decreases, osteoblastic RANKL expression, which promotes rather than reduces osteoclastic bone resorption.
A 10-year-old boy undergoes sleep studies as part of a growth evaluation. Growth hormone levels are found to be highest during the early part of slow-wave (deep) sleep. Most of the growth-promoting effects of GH on long bone epiphyseal growth plates are mediated through which of the following mechanisms?
- ADirect binding of growth hormone to chondrocyte GH receptors, bypassing any need for insulin-like growth factor-1
- BStimulation of thyroid-stimulating hormone release, which indirectly promotes chondrocyte growth
- CStimulation of hepatic and local chondrocyte production of insulin-like growth factor-1, which promotes chondrocyte proliferation
- DDirect stimulation of osteoclast differentiation at the epiphyseal growth plate
Show answer
Growth hormone, released in pulses from somatotrophs with the largest surge during the first few hours of slow-wave sleep, acts mainly by stimulating hepatic synthesis of insulin-like growth factor-1 (IGF-1), which circulates and acts endocrinely, while GH also stimulates local IGF-1 production directly within the epiphyseal growth plate for paracrine action. IGF-1 promotes chondrocyte proliferation in the growth plate, driving linear bone growth. Other physiological stimuli for GH release include deep sleep, exercise, stress, and hypoglycemia, acting through hypothalamic growth hormone-releasing hormone and suppression of somatostatin.
- A. While GH does have some direct receptor-mediated effects on chondrocyte differentiation, the major proliferative drive on growth plates is mediated through IGF-1, not by bypassing it.
- B. GH does not act by stimulating TSH release; this is an unrelated axis.
- D. GH and IGF-1 promote chondrocyte and osteoblast activity for bone formation, not primarily osteoclast-driven resorption.
A 38-year-old woman presents with resistant hypertension and spontaneous hypokalemia (serum potassium 2.9 mEq/L). Plasma aldosterone concentration is markedly elevated at 35 ng/dL, and plasma renin activity is suppressed at 0.2 ng/mL/hr, giving an aldosterone-to-renin ratio above 30. Which of the following best explains the suppressed renin activity in this patient?
- AJuxtaglomerular cell hyperplasia secondary to diuretic use
- BAutonomous aldosterone secretion causing sodium retention and volume expansion
- CChronic sodium restriction stimulating the renin-angiotensin system
- DRenal artery stenosis causing reduced renal perfusion
Show answer
In primary hyperaldosteronism (such as an aldosterone-producing adenoma), aldosterone secretion is autonomous and independent of the renin-angiotensin system. The resulting sodium and water retention expands plasma volume and raises blood pressure, which suppresses renin release from the juxtaglomerular cells through negative feedback via increased renal perfusion pressure and sodium delivery to the macula densa. This gives the classic biochemical signature of high aldosterone with low renin, distinguishing primary hyperaldosteronism from secondary causes. Renal artery stenosis and diuretic use are secondary causes that raise both renin and aldosterone, the opposite pattern.
- A. Diuretic use causes volume depletion, which stimulates rather than suppresses juxtaglomerular renin release.
- C. Sodium restriction activates the renin-angiotensin system, raising renin rather than suppressing it.
- D. Renal artery stenosis reduces renal perfusion and stimulates renin release, producing high renin and high aldosterone (secondary hyperaldosteronism), not suppressed renin.
Reproductive & Fetal Physiology
5 questionsA 32-year-old man with infertility is found to have a normal serum testosterone level but an elevated serum FSH and a low serum inhibin B. Testicular biopsy shows markedly reduced germ cell numbers with preserved Leydig cells. Which of the following best explains the elevated FSH in this patient?
- AIncreased GnRH pulse frequency from hypothalamic dysfunction
- BDecreased androgen-binding protein secretion causing low intratesticular testosterone
- CLoss of negative feedback normally provided by testosterone from damaged Leydig cells
- DLoss of negative feedback normally provided by inhibin B from damaged Sertoli cells
Show answer
Sertoli cells, stimulated by FSH, support germ cell development and secrete inhibin B, which selectively suppresses FSH secretion from the anterior pituitary through negative feedback. When germ cell and Sertoli cell function is impaired (as in primary testicular failure affecting spermatogenesis), inhibin B production falls, removing this feedback and causing a selective rise in FSH while LH and testosterone remain normal because Leydig cell function and testosterone-mediated feedback on LH are preserved. This pattern localizes the defect to the seminiferous tubules rather than the Leydig cells. Androgen-binding protein, also secreted by Sertoli cells, concentrates testosterone within the tubule for spermatogenesis but does not directly regulate pituitary FSH feedback.
- A. Nothing in the history suggests a primary hypothalamic abnormality; the biopsy findings point to a testicular (Sertoli cell/germ cell) problem with intact Leydig cells.
- B. Reduced androgen-binding protein would impair intratesticular testosterone concentration and spermatogenesis but does not directly raise FSH via pituitary feedback.
- C. Testosterone is normal and Leydig cells are preserved, so loss of testosterone feedback does not explain the isolated FSH rise; testosterone feedback mainly suppresses LH.
A 26-year-old woman misses her menstrual period and a urine pregnancy test performed at home turns positive. The test detects a glycoprotein hormone in her urine that is first measurable about 8 days after fertilization. Which of the following best describes the primary physiological role of this hormone in early pregnancy?
- AIt maintains corpus luteum progesterone secretion until the placenta takes over steroidogenesis
- BIt suppresses maternal prolactin to prevent premature lactation
- CIt stimulates maternal thyroid-stimulating hormone receptors to increase basal metabolic rate
- DIt stimulates endometrial decidualization directly without involving the ovary
Show answer
Human chorionic gonadotropin (hCG) is secreted by the syncytiotrophoblast of the developing placenta shortly after implantation and is the hormone detected in urine pregnancy tests. Its principal physiological role in early pregnancy is to rescue and maintain the corpus luteum, sustaining its secretion of progesterone (and some estrogen) until roughly 8-10 weeks of gestation, when the placenta itself becomes the dominant source of progesterone (the luteal-placental shift). Progesterone from the corpus luteum is essential to maintain the decidualized endometrium and prevent menstrual shedding of the implanted conceptus. Although hCG shares an alpha subunit with TSH and has weak thyrotropic activity, this is a minor effect, not its primary role.
- B. hCG does not have a primary role in suppressing prolactin; lactation is actually suppressed in pregnancy mainly by high estrogen and progesterone levels inhibiting prolactin action on the breast.
- C. hCG has only weak cross-reactivity with the TSH receptor, which can mildly increase thyroid hormone levels in some pregnancies, but this is not its main physiological role.
- D. Decidualization is driven by progesterone and estrogen acting on the endometrial stroma, not directly by hCG.
Immediately after a full-term vaginal delivery, a healthy neonate takes its first breaths and begins crying vigorously. Over the next several minutes, pulmonary vascular resistance falls sharply and systemic vascular resistance rises. Which of the following changes is the most important direct trigger for functional closure of the foramen ovale at this time?
- AA reversal of the pressure gradient between the right and left atria
- BConstriction of the ductus venosus following umbilical cord clamping
- CA sudden rise in circulating maternal estrogen levels
- DA fall in arterial carbon dioxide tension following the first breath
Show answer
At birth, lung expansion and the first breaths dramatically reduce pulmonary vascular resistance, increasing pulmonary blood flow and venous return to the left atrium, which raises left atrial pressure above right atrial pressure. Simultaneously, clamping of the umbilical cord removes the low-resistance placental circuit, raising systemic vascular resistance and right atrial venous return decreases. This reversal of the atrial pressure gradient pushes the flexible septum primum against the septum secundum, producing functional closure of the foramen ovale, with anatomical fusion occurring over the following months. The ductus arteriosus closes separately in response to rising arterial oxygen tension and falling prostaglandin E2 levels, a distinct mechanism from the atrial septal closure.
- B. Ductus venosus closure is a separate event related to loss of umbilical venous flow and does not directly cause foramen ovale closure.
- C. Maternal estrogen levels are not the trigger for foramen ovale closure; this is a hemodynamic event driven by pressure changes.
- D. A fall in arterial carbon dioxide tension is not the direct mechanism for foramen ovale closure; it is the atrial pressure reversal that mechanically closes the flap valve.
A 24-year-old woman is breastfeeding her newborn. As the infant suckles at the breast, sensory impulses travel from nipple mechanoreceptors to the hypothalamus, ultimately leading to milk ejection from the alveoli into the ducts within seconds, even before significant new milk is synthesized. Which of the following best describes the mechanism responsible for this rapid milk ejection?
- ASuckling suppresses dopamine release from the hypothalamus, directly contracting the alveolar myoepithelium
- BSuckling directly stimulates local mechanical contraction of the lactiferous ducts without neural involvement
- CSuckling stimulates anterior pituitary release of prolactin, which increases milk protein synthesis in alveolar cells
- DSuckling stimulates posterior pituitary release of oxytocin, which contracts myoepithelial cells around the alveoli
Show answer
The milk ejection (let-down) reflex is mediated by afferent sensory signals from nipple stimulation reaching the hypothalamus, which triggers oxytocin release from the posterior pituitary (secreted by magnocellular neurons of the paraventricular and supraoptic nuclei). Oxytocin acts on myoepithelial cells surrounding the mammary alveoli, causing them to contract and eject already-synthesized milk into the ductal system within seconds, which is why this reflex is fast compared to milk synthesis. Prolactin, released from the anterior pituitary in response to suckling via reduced dopamine inhibition, promotes milk synthesis for the next feed but acts over a slower timescale and does not cause immediate ejection. The two reflex pathways, prolactin for synthesis and oxytocin for ejection, work together but have distinct mechanisms.
- A. Suppressed dopamine release permits increased prolactin secretion from the anterior pituitary, but it does not directly act on the myoepithelium; the direct contractile trigger is oxytocin.
- B. The ducts and alveoli do not contract on purely local mechanical stimulation; myoepithelial contraction requires oxytocin acting through the neuroendocrine reflex.
- C. Prolactin promotes milk synthesis in alveolar epithelial cells over a slower timescale and does not cause immediate contraction-driven ejection.
At the placental interface, maternal blood with a partial pressure of oxygen of about 40 mmHg delivers oxygen to fetal blood, which then becomes highly saturated with oxygen despite this relatively low oxygen tension. This efficient oxygen transfer is best explained by which property of fetal hemoglobin compared with adult hemoglobin?
- AFetal hemoglobin has a lower overall oxygen-carrying capacity that paradoxically increases oxygen extraction
- BFetal hemoglobin binds 2,3-bisphosphoglycerate less avidly, shifting its oxygen dissociation curve to the left
- CFetal hemoglobin has a higher concentration of beta-globin chains that increase cooperative oxygen binding
- DFetal hemoglobin binds carbon dioxide more avidly, enhancing the Bohr effect at the placenta
Show answer
Fetal hemoglobin (HbF) contains gamma chains in place of the beta chains found in adult hemoglobin, and these gamma chains bind 2,3-bisphosphoglycerate (2,3-BPG) less avidly. Since 2,3-BPG normally stabilizes the low-affinity deoxygenated state of hemoglobin, reduced binding in HbF shifts its oxygen dissociation curve to the left, giving it a higher oxygen affinity than maternal adult hemoglobin at the same oxygen tension. This allows fetal blood to extract oxygen efficiently from maternal blood across the placenta even at the relatively low oxygen tensions present there. This affinity difference reverses after birth as HbF is replaced by adult hemoglobin over the following months.
- A. Fetal hemoglobin concentration in fetal blood is actually higher, not lower, and oxygen-carrying capacity is not the explanation for the leftward shift in affinity.
- C. Fetal hemoglobin does not contain beta chains at all; it has gamma chains replacing beta chains, and cooperativity itself is not the main distinguishing mechanism here.
- D. The Bohr effect (increased carbon dioxide and hydrogen ion lowering hemoglobin affinity) applies similarly to both hemoglobins and is not the basis for fetal hemoglobin's higher oxygen affinity.
Neurophysiology (CNS Special Senses)
7 questionsA 30-year-old man undergoes testing of the knee jerk reflex. The examiner taps the patellar tendon with a reflex hammer, and the quadriceps contracts briefly. Which of the following receptors initiates this reflex response?
- APacinian corpuscle
- BGolgi tendon organ
- CMuscle spindle annulospiral ending
- DJoint capsule mechanoreceptor
Show answer
The stretch (myotatic) reflex begins when sudden stretch of the muscle activates the annulospiral (primary) endings wrapped around the nuclear bag and chain fibers of the muscle spindle. This generates a volley in the Ia afferent fiber, which synapses directly (monosynaptically) on the alpha motor neuron in the spinal cord, causing reflex contraction of the same muscle. This arc underlies deep tendon reflexes used clinically to assess the integrity of specific spinal segments.
- A. Pacinian corpuscles detect vibration and deep pressure in skin and deep tissue, not muscle length changes.
- B. Golgi tendon organs sense tension and mediate the inverse myotatic (autogenic inhibition) reflex, not the stretch reflex.
- D. Joint capsule receptors signal joint position and movement, not muscle stretch.
A 45-year-old man presents with weakness of the right lower limb for 3 weeks. Examination shows increased tone, exaggerated knee and ankle jerks, an extensor plantar response, and no muscle wasting or fasciculations. Which of the following best explains this pattern of findings?
- ALoss of descending inhibitory control over spinal reflex arcs
- BReduced acetylcholine release at the neuromuscular junction
- CDenervation of muscle fibers with loss of trophic influence
- DDegeneration of anterior horn cells
Show answer
Upper motor neuron lesions interrupt descending corticospinal and other suprasegmental pathways that normally inhibit spinal stretch reflexes and the extensor plantar response. Loss of this inhibition produces hyperreflexia, spasticity (increased tone), and a positive Babinski sign, while the lower motor neuron and muscle remain intact, so there is no wasting or fasciculation. In contrast, lower motor neuron lesions cause flaccid weakness, hyporeflexia, wasting, and fasciculations due to denervation.
- B. Reduced acetylcholine release at the neuromuscular junction causes fatigable weakness as in myasthenic syndromes, not hyperreflexia.
- C. Denervation with trophic loss explains lower motor neuron signs such as wasting and fasciculations, not seen here.
- D. Anterior horn cell degeneration is a lower motor neuron process producing flaccid weakness and wasting, not spasticity.
Which sensory pathway carries fine touch and vibration sense, and at what level does it decussate?
- ASpinothalamic tract, at the medulla
- BDorsal column, at the level of entry into the spinal cord
- CDorsal column, at the medulla
- DSpinothalamic tract, at the level of entry into the spinal cord
Show answer
The dorsal column (posterior column-medial lemniscus) pathway carries fine touch, vibration, and proprioception. First-order neurons ascend ipsilaterally in the gracile and cuneate fasciculi without crossing, synapse in the gracile and cuneate nuclei of the medulla, and second-order fibers decussate there as the internal arcuate fibers before ascending as the medial lemniscus to the thalamus. This contrasts with the spinothalamic tract, which carries pain and temperature and decussates within one or two segments of entering the spinal cord.
- A. The spinothalamic tract carries pain and temperature, and decussates near the level of spinal cord entry, not at the medulla.
- B. The dorsal column decussates in the medulla, not at the level of spinal entry.
- D. The spinothalamic tract carries pain and temperature, not fine touch and vibration.
A 50-year-old man with chronic alcohol use presents with a wide-based, staggering gait and inability to perform the finger-to-nose test smoothly, with overshoot and oscillation as his finger approaches the target. There is no limb weakness. Which of the following best explains these findings?
- ALoss of cerebellar feedback control over the timing and coordination of agonist-antagonist muscle activity
- BDysfunction of the vestibular labyrinth
- CDegeneration of the corticospinal tract bilaterally
- DLoss of proprioceptive input via the dorsal columns
Show answer
The cerebellum compares intended movement (from motor cortex) with actual movement (from proprioceptive feedback) and fine-tunes the timing, force, and coordination of agonist and antagonist muscle contractions. Damage to the cerebellum, as in chronic alcohol-related cerebellar degeneration affecting the anterior lobe, disrupts this feedback loop, producing dysmetria (overshoot), intention tremor, and a wide-based ataxic gait, without primary weakness. Corticospinal, dorsal column, or vestibular lesions produce distinct patterns: spasticity with weakness, loss of position/vibration sense with positive Romberg, or vertigo and nystagmus respectively.
- B. Vestibular dysfunction causes vertigo, nystagmus, and imbalance but not intention tremor or dysmetria on finger-to-nose testing.
- C. Corticospinal tract degeneration produces spastic weakness, not the coordination deficit described here.
- D. Dorsal column loss causes sensory ataxia with positive Romberg sign, typically with impaired vibration and position sense, not intention tremor.
Cerebrospinal fluid is primarily formed by active secretion from which structure, and is predominantly reabsorbed into the venous circulation at which site?
- APia-arachnoid membrane; ependymal lining of ventricles
- BArachnoid villi; choroid plexus
- CEpendymal lining of ventricles; choroid plexus
- DChoroid plexus; arachnoid villi
Show answer
CSF is mainly formed by the choroid plexus of the lateral, third, and fourth ventricles through active secretion involving the sodium-potassium ATPase pump and carbonic anhydrase, producing fluid with a composition distinct from plasma. CSF circulates from the ventricles through the foramina of Monro and the aqueduct of Sylvius into the fourth ventricle, then into the subarachnoid space. It is reabsorbed into the venous sinuses, chiefly the superior sagittal sinus, via arachnoid villi, which act as one-way valves allowing bulk flow when CSF pressure exceeds venous sinus pressure.
- A. The pia-arachnoid membrane and ependymal lining are not the primary sites of CSF formation or absorption.
- B. Arachnoid villi absorb CSF; they do not form it, and the choroid plexus forms it rather than absorbing it.
- C. The ependymal lining contributes minimally to CSF formation; the choroid plexus is the main source, and absorption occurs at arachnoid villi, not back at the choroid plexus.
In the resting (dark) state of a retinal rod photoreceptor, cyclic GMP keeps sodium channels open, producing a steady inward current. Which of the following events occurs upon exposure of the rod to light that leads to hyperpolarization of the photoreceptor?
- AIncreased glutamate release from the rod synaptic terminal
- BDirect depolarization of the rod by photon-induced calcium influx
- CIncreased synthesis of cGMP by guanylate cyclase, opening more sodium channels
- DActivation of transducin and phosphodiesterase, leading to breakdown of cGMP and closure of sodium channels
Show answer
Light absorbed by rhodopsin isomerizes 11-cis-retinal to all-trans-retinal, activating the G-protein transducin, which in turn activates phosphodiesterase. Phosphodiesterase hydrolyzes cGMP, lowering its intracellular concentration, causing closure of cGMP-gated sodium channels in the outer segment membrane. This stops the depolarizing dark current, hyperpolarizes the rod, and reduces glutamate release onto bipolar cells, the opposite of the resting dark state. Dark adaptation reflects the slow regeneration of rhodopsin from retinal and opsin, which gradually restores rod sensitivity after prolonged light exposure.
- A. Light exposure decreases, not increases, glutamate release from rod terminals because the cell hyperpolarizes.
- B. Rods hyperpolarize in response to light via closure of sodium channels, they do not depolarize through calcium influx.
- C. Light causes breakdown, not increased synthesis, of cGMP; guanylate cyclase activity rises only during recovery in the dark.
A 40-year-old man complains of hearing loss in his left ear following chronic otitis media. On tuning fork testing with a 512 Hz fork, air conduction is less than bone conduction in the left ear, and the Weber test shows lateralization to the left ear. Which of the following best describes this pattern?
- ASensorineural hearing loss in the left ear
- BSensorineural hearing loss in the right ear
- CConductive hearing loss in the left ear
- DMixed hearing loss with greater sensorineural component on the left
Show answer
In conductive hearing loss, the external or middle ear transmission of sound is impaired, so bone conduction (which bypasses this pathway) exceeds air conduction, giving a negative Rinne test in the affected ear. The Weber test, which assesses bone conduction centrally, lateralizes to the ear with conductive loss because that ear's background masking by ambient air-conducted sound is reduced, making bone-conducted sound relatively louder there. This combination of Rinne negative and Weber lateralizing to the same (affected) ear is characteristic of conductive hearing loss, as seen here in chronic otitis media affecting the middle ear.
- A. In sensorineural hearing loss, Rinne is positive (air conduction greater than bone conduction) in the affected ear, and Weber lateralizes to the normal ear, opposite to the findings here.
- B. Weber lateralizing to the right ear would suggest sensorineural loss on the left, not what is observed in this patient.
- D. Mixed loss would show Rinne results that vary with the frequency of forks tested and does not match this clean conductive pattern.
[Blood & Immunity Physiology]
3 questionsA 28-year-old woman presents with easy bruising and prolonged bleeding after dental extraction. Laboratory evaluation shows: Prothrombin time (PT) 13 seconds (control 12 seconds), activated partial thromboplastin time (aPTT) 58 seconds (control 30 seconds), and platelet count 250,000/µL. Which of the following is the most likely deficient factor?
- AFactor VII
- BFactor II (prothrombin)
- CFactor X
- DFactor VIII
Show answer
The aPTT tests the intrinsic pathway (factors XII, XI, IX, VIII) plus the common pathway, while the PT tests the extrinsic pathway (factor VII) plus the common pathway. An isolated prolonged aPTT with a normal PT localizes the defect to the intrinsic pathway, specifically factors VIII, IX, XI, or XII, since the common pathway factors (X, V, II, fibrinogen) would prolong both tests. Factor VIII deficiency (hemophilia A) is the most common cause of this pattern with a normal platelet count. Factor VII deficiency would prolong the PT, and factor X or II deficiency would prolong both PT and aPTT since they are common pathway factors.
- A. Factor VII is part of the extrinsic pathway; its deficiency would prolong the PT, which is normal here.
- B. Factor II (prothrombin) is part of the common pathway; its deficiency would prolong both PT and aPTT.
- C. Factor X is part of the common pathway; its deficiency would prolong both PT and aPTT.
A 34-year-old man with blood group A is mistakenly transfused with one unit of group B packed red cells. Within minutes he develops fever, chills, flank pain, and dark urine. Which of the following best explains the immediate mechanism of this reaction?
- APreformed anti-B IgM antibodies in the recipient's plasma bind donor B antigens and activate complement, causing intravascular hemolysis
- BRecipient T lymphocytes recognize donor red cell antigens and directly lyse the transfused cells
- CDonor plasma contains anti-A antibodies that coat the recipient's own red cells, leading to extravascular splenic destruction
- DDonor red cells trigger recipient B lymphocytes to produce new antibodies over several days, causing delayed hemolysis
Show answer
Individuals naturally develop antibodies against the ABO antigens they lack, through exposure to similar epitopes on gut bacteria and environmental antigens early in life. A group A individual has preformed IgM anti-B antibodies circulating in plasma. Transfusion of incompatible B red cells causes these antibodies to bind donor antigens and activate the classical complement pathway, forming the membrane attack complex and causing immediate intravascular hemolysis. This leads to hemoglobinemia, hemoglobinuria, fever, and flank pain from an acute hemolytic transfusion reaction, which can progress to disseminated intravascular coagulation and acute kidney injury.
- B. ABO antigens are carbohydrate structures recognized by antibodies, not by T-cell receptors; this is not a T-cell mediated process.
- C. This describes a mechanism more typical of minor ABO mismatch in plasma products, not the dominant mechanism in red cell transfusion incompatibility; the clinically relevant process is destruction of the transfused donor cells by recipient antibodies.
- D. This describes a delayed hemolytic transfusion reaction from an anamnestic response to a non-ABO antigen, not the immediate reaction seen with preformed ABO antibodies.
A small laceration occurs on a finger during a laboratory demonstration of hemostasis. Within seconds, platelets begin to adhere to the exposed subendothelial collagen at the injury site despite the high shear stress of flowing blood. Which of the following best explains the molecular basis of this initial platelet adhesion?
- AFibrinogen bridges adjacent platelets through glycoprotein Ib receptors, forming the initial platelet monolayer
- BVon Willebrand factor bound to exposed collagen binds platelet glycoprotein Ib receptors, anchoring platelets under high shear flow
- CPlatelet glycoprotein IIb/IIIa receptors bind directly to exposed collagen fibers, mediating firm platelet adhesion
- DThromboxane A2 released from endothelial cells binds platelet receptors, causing direct adhesion to the vessel wall
Show answer
When the vascular endothelium is breached, subendothelial collagen is exposed to flowing blood. Von Willebrand factor, synthesized by endothelial cells and platelets, binds exposed collagen and undergoes a conformational change that allows it to bind platelet glycoprotein Ib (part of the GPIb-IX-V complex). This interaction is essential for platelet adhesion under high shear stress conditions found in arterioles and capillaries, since direct platelet-collagen binding is too weak to withstand such flow. Subsequent platelet activation leads to degranulation, release of ADP and thromboxane A2, and conformational activation of glycoprotein IIb/IIIa, which binds fibrinogen to link platelets together and form the platelet plug.
- A. Glycoprotein Ib binds von Willebrand factor, not fibrinogen directly; fibrinogen bridging occurs via activated glycoprotein IIb/IIIa during aggregation, a later step.
- C. Glycoprotein IIb/IIIa binds fibrinogen (and vWF) to mediate platelet aggregation, not direct adhesion to collagen, and requires prior platelet activation.
- D. Thromboxane A2 is produced by activated platelets, not endothelial cells, and acts as a local mediator promoting platelet aggregation and vasoconstriction, not direct adhesion.
Frequently asked questions
Are these real NEET PG questions?
No. They are original practice questions written in the style of NEET PG for its content outline. Real exam items are confidential. Every question here was solved blind by a second, independent model and removed if the key was disputed.
How should I use these physiology questions?
Do the quiz first without looking at the answers, then read the explanation for every question you missed or guessed. The educational objective under each answer is the fact to remember.
Are the questions free?
Yes. All questions, answers and explanations on this page are free and need no account.
More NEET PG topics
Other exams
NEET PG is administered by National Board of Examinations in Medical Sciences (NBEMS). WhiteCoat Pro is not affiliated with or endorsed by them. All questions are original practice items written by the WhiteCoat Pro editorial team with AI assistance and checked by a second, independent AI model that solves every question blind; items with a disputed key are removed. They are not real exam questions. Educational use only, not medical advice.