301. A person has \(TV=500\,\mathrm{mL}\), \(IRV=2700\,\mathrm{mL}\), \(ERV=1000\,\mathrm{mL}\) and \(RV=1200\,\mathrm{mL}\). What is \(TLC-FRC\), and which capacity does it equal?
ⓐ. \(2200\,\mathrm{mL}\), equal to \(FRC\)
ⓑ. \(4200\,\mathrm{mL}\), equal to \(VC\)
ⓒ. \(1700\,\mathrm{mL}\), equal to \(EC\)
ⓓ. \(3200\,\mathrm{mL}\), equal to \(IC\)
Correct Answer: \(3200\,\mathrm{mL}\), equal to \(IC\)
Explanation: Total lung capacity is:
\[
TLC=IRV+TV+ERV+RV
\]
\[
TLC=2700+500+1000+1200=5400\,\mathrm{mL}
\]
Functional residual capacity is:
\[
FRC=ERV+RV=1000+1200=2200\,\mathrm{mL}
\]
The difference is:
\[
TLC-FRC=5400-2200=3200\,\mathrm{mL}
\]
This is equal to inspiratory capacity:
\[
IC=TV+IRV=500+2700=3200\,\mathrm{mL}
\]
Subtracting the air remaining after normal expiration from the total volume at maximum inspiration leaves the maximum volume that can be inspired from the normal-expiratory level. The result is \(3200\,\mathrm{mL}\), equal to \(IC\). Algebraically, the shared \(ERV+RV\) portion is removed from \(TLC\), leaving only \(IRV+TV\). Physiologically, this is the volume that can be inspired after a normal expiration, so the numerical and state-based interpretations agree. The calculation uses all four primary volumes and then checks the result against an independently defined capacity. This double check reduces the chance of confusing \(FRC\), which is retained after normal expiration, with \(IC\), which is inspired above that level.
302. Which symbolic identity correctly isolates expiratory reserve volume by two independent capacity routes?
ⓐ. \(ERV=VC-EC=FRC-TV\)
ⓑ. \(ERV=VC-IC=FRC-RV\)
ⓒ. \(ERV=TLC-VC=IC-TV\)
ⓓ. \(ERV=EC-IRV=TLC-FRC\)
Correct Answer: \(ERV=VC-IC=FRC-RV\)
Explanation: Start with the component definitions of vital and inspiratory capacities:
\[
VC=IRV+TV+ERV,\qquad IC=IRV+TV
\]
Subtracting the common inspiratory reserve and tidal components leaves:
\[
VC-IC=ERV
\]
A second independent route uses the volume remaining after normal expiration:
\[
FRC=ERV+RV
\]
Residual volume is the part still present after maximal expiration. Removing it from the functional residual capacity again isolates the expiratory reserve component:
\[
FRC-RV=ERV
\]
Thus \(ERV=VC-IC=FRC-RV\). The first route compares movable air between maximum and normal expiration; the second compares air remaining at normal and maximal expiratory levels. Both differences represent the extra air that can be expelled after an ordinary expiration. Checking the shared endpoint makes the identity meaningful: subtracting quantities that refer to unrelated respiratory levels would not necessarily isolate one named primary volume. For example, adding the same \(RV\) to both ends of the second relation would obscure the particular extra expiratory movement being sought.
303. A person retains normal \(TV\), \(IRV\) and \(RV\), but \(ERV\) is reduced markedly. Which set of changes must occur from the capacity definitions?
ⓐ. \(IC\) and \(TLC\) increase, while \(FRC\) remains constant
ⓑ. \(IC\) decreases, while \(EC\) and \(FRC\) increase
ⓒ. \(EC\), \(FRC\), \(VC\) and \(TLC\) decrease, while \(IC\) remains unchanged
ⓓ. Only \(RV\) changes because reserve volumes do not contribute to capacities
Correct Answer: \(EC\), \(FRC\), \(VC\) and \(TLC\) decrease, while \(IC\) remains unchanged
Explanation: Expiratory reserve volume contributes to several capacities:
\[
EC=TV+ERV
\]
\[
FRC=ERV+RV
\]
\[
VC=IRV+TV+ERV
\]
\[
TLC=IRV+TV+ERV+RV
\]
A reduction in \(ERV\), with all other primary volumes unchanged, decreases each of these four sums. Inspiratory capacity is:
\[
IC=TV+IRV
\]
and contains no \(ERV\), so it remains unchanged. Residual volume is explicitly held constant. One altered component propagates through several capacities, so the relations must be followed together rather than memorised separately. The only supported pattern is simultaneous reduction of \(EC\), \(FRC\), \(VC\) and \(TLC\) with preservation of \(IC\). This changed-component analysis is useful; capacity names alone can hide shared terms. Tracking \(ERV\) through every formula reveals which measurements must change and prevents a false conclusion based only on the direction of breathing. For example, a lower \(ERV\) shortens the movable interval below normal expiration and also lowers every total that contains that interval. No assumption about a change in inspiratory reserve or residual air is needed.
304. Four individuals have the same \(TLC\) of \(5500\,\mathrm{mL}\) but different residual volumes.
| Individual | \(RV\) |
|---|
| P | \(1000\,\mathrm{mL}\) |
| Q | \(1200\,\mathrm{mL}\) |
| R | \(1500\,\mathrm{mL}\) |
| S | \(900\,\mathrm{mL}\) |
Who has the greatest vital capacity?
ⓐ. Individual S
ⓑ. Individual P
ⓒ. Individual Q
ⓓ. Individual R
Correct Answer: Individual S
Explanation: Total lung capacity is related to vital capacity and residual volume by:
\[
TLC=VC+RV
\]
Therefore:
\[
VC=TLC-RV
\]
Since all four individuals have the same \(TLC\), the person with the smallest residual volume must have the greatest vital capacity. Individual S has the lowest \(RV\) at \(900\,\mathrm{mL}\), giving:
\[
VC_S=5500-900=4600\,\mathrm{mL}
\]
The corresponding values for P, Q and R are \(4500\,\mathrm{mL}\), \(4300\,\mathrm{mL}\) and \(4000\,\mathrm{mL}\), respectively. The table requires comparing a derived capacity rather than choosing the largest listed volume. Individual S has the greatest \(VC\). Numerically, the common \(TLC\) fixes the upper lung-volume level, while \(RV\) fixes the lowest voluntary expiratory level. A smaller retained volume leaves a larger movable interval between those levels, which is the vital capacity. This inverse relation applies only because total lung capacity is held equal across the four individuals. Without that control, the smallest residual volume would not by itself guarantee the greatest vital capacity.
305. The relevant respiratory values are shown below.
| Compartment | \(\mathrm{pO_2}\) | \(\mathrm{pCO_2}\) |
|---|
| Alveoli | \(104\,\mathrm{mm\,Hg}\) | \(40\,\mathrm{mm\,Hg}\) |
| Deoxygenated pulmonary blood | \(40\,\mathrm{mm\,Hg}\) | \(45\,\mathrm{mm\,Hg}\) |
| Oxygenated systemic blood | \(95\,\mathrm{mm\,Hg}\) | \(40\,\mathrm{mm\,Hg}\) |
| Tissues | \(40\,\mathrm{mm\,Hg}\) | \(45\,\mathrm{mm\,Hg}\) |
Which pair of directions is supported?
ⓐ. \(\mathrm{O_2}\) moves from alveoli to pulmonary blood, and \(\mathrm{CO_2}\) moves from tissues to systemic blood
ⓑ. \(\mathrm{O_2}\) moves from pulmonary blood to alveoli, and \(\mathrm{CO_2}\) moves from systemic blood to tissues
ⓒ. Both gases move from alveoli into pulmonary blood
ⓓ. Both gases move from systemic blood into tissues
Correct Answer: \(\mathrm{O_2}\) moves from alveoli to pulmonary blood, and \(\mathrm{CO_2}\) moves from tissues to systemic blood
Explanation: At the pulmonary interface, alveolar \(\mathrm{pO_2}\) is \(104\,\mathrm{mm\,Hg}\), while incoming deoxygenated blood has \(40\,\mathrm{mm\,Hg}\). Oxygen moves from alveoli into blood. At the tissue interface, tissue \(\mathrm{pCO_2}\) is \(45\,\mathrm{mm\,Hg}\), while oxygenated systemic blood has \(40\,\mathrm{mm\,Hg}\). Carbon dioxide moves from tissues into blood. Each gas must be evaluated using the compartments that meet at the relevant interface. Atmospheric values are unnecessary for these two local decisions, and the gases do not move in the same anatomical direction. The supported pair follows the higher-to-lower partial-pressure rule separately for oxygen and carbon dioxide. The normal pulmonary carbon-dioxide direction and systemic oxygen direction are not asked, but they would follow the same rule at their respective interfaces. Each decision must use the partial pressures of compartments that are in direct exchange contact.
306. The pulmonary oxygen gradient is \(104-40=64\,\mathrm{mm\,Hg}\), while the pulmonary carbon-dioxide gradient is \(45-40=5\,\mathrm{mm\,Hg}\). Which interpretation is correct?
ⓐ. The oxygen gradient is \(5\) times larger, so carbon dioxide cannot diffuse
ⓑ. The carbon-dioxide gradient is \(12.8\) times larger because its value is lower
ⓒ. Both gradients are equal after accounting for membrane thickness
ⓓ. Oxygen has a \(12.8\)-fold larger gradient; carbon dioxide has higher solubility
Correct Answer: Oxygen has a \(12.8\)-fold larger gradient; carbon dioxide has higher solubility
Explanation: The ratio of the gradient magnitudes is:
\[
\frac{64}{5}=12.8
\]
Thus, the pulmonary oxygen partial-pressure difference is \(12.8\) times the carbon-dioxide difference. Carbon dioxide nevertheless diffuses efficiently because it is approximately \(20\text{–}25\) times more soluble than oxygen. Solubility increases transfer per unit pressure difference and helps compensate for the smaller carbon-dioxide gradient. The calculation does not imply that gradient size is irrelevant; both gradient and solubility contribute. Membrane thickness is not stated to differ between the gases in this comparison. The correct interpretation combines the numerical gradient ratio with the gas-specific solubility advantage of carbon dioxide. The result explains why comparing gradient magnitudes alone can be misleading. Oxygen and carbon dioxide cross the same thin membrane, but their different solubilities mean that a much smaller carbon-dioxide gradient can still support physiologically effective transfer. Partial-pressure gradients establish direction, haemoglobin and bicarbonate provide transport, and local tissue conditions promote unloading or loading.
307. Two haemoglobin conditions are compared at the same tissue \(\mathrm{pO_2}\). Condition P has low \(\mathrm{pCO_2}\), low \(\mathrm{H^+}\) and lower temperature. Condition Q has higher values of all three and shows lower haemoglobin saturation. Which conclusion follows?
ⓐ. Condition P releases more oxygen because its haemoglobin saturation is higher
ⓑ. Both conditions deliver equal oxygen because \(\mathrm{pO_2}\) is identical
ⓒ. Condition Q promotes greater oxygen unloading at the same \(\mathrm{pO_2}\)
ⓓ. Condition Q increases the maximum of four oxygen molecules per haemoglobin
Correct Answer: Condition Q promotes greater oxygen unloading at the same \(\mathrm{pO_2}\)
Explanation: The two conditions share the same oxygen partial pressure, so their difference in haemoglobin saturation must arise from the modifying variables. Higher carbon-dioxide pressure, greater hydrogen-ion concentration and higher temperature favour oxyhaemoglobin dissociation. Condition Q therefore leaves haemoglobin less saturated and releases more oxygen than condition P at the same \(\mathrm{pO_2}\). The maximum molecular capacity of haemoglobin remains four oxygen molecules and is not altered by tissue chemistry. Equal oxygen pressure also does not guarantee equal saturation when the other binding factors differ. The case integrates the oxygen-dissociation relation with the local chemical and thermal environment of active tissues. In graph terms, Condition Q would lie at a lower saturation for the same tissue oxygen pressure. The greater fall from lung loading to tissue saturation represents additional oxygen released where carbon-dioxide production, acidity and heat are higher.
308. Two oxygen-dissociation curves are described. At \(\mathrm{pO_2}=95\,\mathrm{mm\,Hg}\), both show \(97\%\) saturation. At \(\mathrm{pO_2}=40\,\mathrm{mm\,Hg}\), Curve P shows \(75\%\) saturation and Curve Q shows \(60\%\). Which curve releases the greater fraction between these two values?
ⓐ. Curve P, because \(97-75=22\) percentage points
ⓑ. Curve Q, because \(97-60=37\) percentage points
ⓒ. Both curves, because they begin at the same saturation
ⓓ. Neither curve, because saturation cannot change with \(\mathrm{pO_2}\)
Correct Answer: Curve Q, because \(97-60=37\) percentage points
Explanation: The fraction released over the stated change in oxygen partial pressure is represented by the fall in haemoglobin saturation. For Curve P:
\[
97\%-75\%=22\text{ percentage points}
\]
For Curve Q:
\[
97\%-60\%=37\text{ percentage points}
\]
Curve Q unloads the greater fraction between the lung-like and tissue-like oxygen values. The equal starting saturation does not imply equal release because the curves differ at the lower \(\mathrm{pO_2}\). A condition with higher \(\mathrm{pCO_2}\), higher \(\mathrm{H^+}\) or higher temperature could produce the lower tissue saturation represented by Q. The comparison links graph position with functional oxygen delivery rather than identifying the curve from shape alone. If both curves represent the same amount of haemoglobin, the larger percentage-point fall for Q corresponds to more oxygen released per available carrier. The calculation converts a graphical separation into a functional comparison of tissue delivery.
309. Arrange the events by which one bicarbonate-derived carbon-dioxide molecule is released at the lungs.
P. Carbon dioxide diffuses from pulmonary blood into alveolar air.
Q. Carbonic acid is formed from bicarbonate and hydrogen ions.
R. Carbonic anhydrase facilitates formation of carbon dioxide and water.
S. Carbon dioxide is removed during expiration.
ⓐ. R → Q → P → S
ⓑ. Q → P → R → S
ⓒ. Q → R → P → S
ⓓ. P → Q → R → S
Correct Answer: Q → R → P → S
Explanation: Near the lungs, bicarbonate and hydrogen ions first contribute to the formation of carbonic acid. Carbonic anhydrase then facilitates the reverse reaction that produces carbon dioxide and water. Once regenerated, carbon dioxide can diffuse from pulmonary blood into alveolar air; blood normally has the higher \(\mathrm{pCO_2}\). Ventilation finally removes the gas during expiration. The correct sequence separates chemical regeneration from membrane diffusion and whole-lung elimination. Carbon dioxide cannot cross into alveoli before it has been regenerated from the bicarbonate route, and expiration cannot remove that molecule before it enters alveolar air. The order traces the pulmonary completion of the dominant carbon-dioxide transport pathway. The same enzyme that favoured bicarbonate formation near tissues now facilitates the reverse pathway; local conditions differ. This reversibility allows carbon dioxide to be transported in a soluble ionic form and later recovered as a diffusible gas.
310. Match each carbon-dioxide transport component with its principal pulmonary event. A Column II entry is used once.
| Column I | Column II |
|---|
| P. Dissolved \(\mathrm{CO_2}\) | 1. Diffuses into alveoli when blood \(\mathrm{pCO_2}\) is higher |
| Q. Carbamino-haemoglobin | 2. Dissociates under low \(\mathrm{pCO_2}\) and high \(\mathrm{pO_2}\) |
| R. Bicarbonate | 3. Is converted back toward carbon dioxide |
| S. Carbonic anhydrase | 4. Accelerates the reversible bicarbonate reaction |
ⓐ. P-2, Q-1, R-4, S-3
ⓑ. P-1, Q-2, R-3, S-4
ⓒ. P-3, Q-4, R-1, S-2
ⓓ. P-4, Q-3, R-2, S-1
Correct Answer: P-1, Q-2, R-3, S-4
Explanation: Dissolved carbon dioxide can diffuse directly from pulmonary blood into alveolar air when its blood partial pressure is higher, giving P-1. Carbamino-haemoglobin dissociates under the low-\(\mathrm{pCO_2}\), high-\(\mathrm{pO_2}\) conditions of the lungs, producing Q-2. Bicarbonate is converted back toward carbon dioxide so that the gas can be released, giving R-3. Carbonic anhydrase accelerates the reversible conversion without being consumed, producing S-4. The mappings show how three transport forms converge on pulmonary carbon-dioxide elimination while the enzyme facilitates one chemical route. They are simultaneous components of one integrated process rather than interchangeable names for the same transported substance. The match also clarifies that carbonic anhydrase is not itself a transport form. It is a catalyst that supports rapid interconversion, while dissolved gas, carbamino association and bicarbonate represent actual ways carbon dioxide is carried.
311. Blood entering pulmonary capillaries has low \(\mathrm{pO_2}\) and high \(\mathrm{pCO_2}\). After passing an alveolus, it leaves with higher \(\mathrm{pO_2}\) and lower \(\mathrm{pCO_2}\), then reaches systemic tissues where the opposite changes occur. Which circuit description is correct?
ⓐ. Oxygen moves from pulmonary blood into alveoli and carbon dioxide moves from blood into tissues
ⓑ. Both gases move from alveoli to blood and remain there throughout systemic circulation
ⓒ. Both gases follow the same route from lungs to tissues
ⓓ. Oxygen moves from alveoli to blood to tissues; carbon dioxide moves in reverse
Correct Answer: Oxygen moves from alveoli to blood to tissues; carbon dioxide moves in reverse
Explanation: At the lungs, alveolar oxygen partial pressure exceeds that of incoming deoxygenated blood, so oxygen enters pulmonary blood. Blood carries oxygen to systemic tissues, where a lower tissue \(\mathrm{pO_2}\) favours unloading. Carbon dioxide is produced in tissues, diffuses into systemic blood, travels to the lungs and then enters alveolar air for expiration. The two gases therefore follow opposite overall routes even though both use diffusion at local interfaces and blood transport between them. The observed changes in blood values before and after the alveolus confirm oxygen loading and carbon-dioxide unloading. The correct description integrates pulmonary exchange, circulation and systemic tissue exchange into one respiratory circuit. This opposite routing maintains continuity between cellular metabolism and external gas exchange. Oxygen is acquired from the environment for tissues, whereas metabolically produced carbon dioxide is collected from tissues and returned to the lungs for elimination.
312. Tissue metabolism raises blood \(\mathrm{CO_2}\) and \(\mathrm{H^+}\). Peripheral and central chemosensitive regions detect the change, their signals reach the medulla, and respiratory muscles remain functional, yet ventilation does not increase. Which failed link is most consistent?
ⓐ. The medulla fails to increase respiratory drive
ⓑ. Chemosensitive regions fail to detect the change
ⓒ. Sensory signals fail to reach the medulla
ⓓ. Respiratory muscles fail to respond
Correct Answer: The medulla fails to increase respiratory drive
Explanation: The case confirms several upstream and downstream components. Carbon-dioxide-related chemical changes occur, chemosensitive regions detect them, and their signals reach the medulla. Respiratory muscles are also structurally functional, so the missing response lies between arrival of sensory information and increased motor drive to breathe. A failure of medullary integration or output best fits this location. Carbon dioxide must already have entered blood for the chemical change to be detected, and oxygen transport is not the failed step described. Alveoli do not contain cartilage that blocks expansion. The diagnosis follows the feedback chain from metabolism through chemical sensing, neural processing and ventilatory correction, identifying the first unsupported transition. Because detection and signal transmission are explicitly intact, blaming the receptors would contradict the evidence. Because the muscles are functional, the defect is also not mechanical execution; it lies in generating the corrective neural command after sensory input arrives.
313. Arrange the events that increase carbon-dioxide elimination when cellular metabolism rises.
P. Ventilation increases.
Q. Cellular production of \(\mathrm{CO_2}\) rises.
R. Chemosensitive regions detect increased \(\mathrm{CO_2}\) and \(\mathrm{H^+}\).
S. More carbon dioxide is removed through the alveoli.
T. Respiratory neural activity is adjusted.
ⓐ. Q → R → T → P → S
ⓑ. R → Q → P → T → S
ⓒ. Q → T → R → S → P
ⓓ. T → Q → R → P → S
Correct Answer: Q → R → T → P → S
Explanation: Increased cellular activity first produces more carbon dioxide. The associated rise in \(\mathrm{CO_2}\) and \(\mathrm{H^+}\) is detected by the central chemosensitive area and peripheral chemical receptors. These sensors provide information to the respiratory neural centres, which adjust respiratory activity. The altered neural output increases ventilation, allowing more carbon dioxide to diffuse into alveolar air and leave during expiration. The correct order therefore begins with metabolic production and ends with increased elimination. Detection cannot occur before the chemical change develops, and greater removal cannot precede the ventilatory response that maintains a favourable pulmonary gradient. This sequence represents negative feedback; the response opposes the initial rise in carbon-dioxide-related signals. It connects cellular metabolism, blood chemistry, neural regulation, breathing mechanics and alveolar removal without treating any one stage as an isolated event.
314. Two groups undergo the same increase in cellular metabolism. Group P has functional chemosensitive receptors, while chemical signalling from both central and peripheral receptors is blocked in Group Q. Respiratory muscles and airways remain normal. Which observation is most likely?
ⓐ. Q raises ventilation faster without chemical feedback
ⓑ. Both groups show identical adaptive ventilation because receptors are unnecessary
ⓒ. P raises ventilation; Q shows a weakened chemical-feedback response
ⓓ. P loses its basic respiratory rhythm, while Q maintains normal chemical regulation
Correct Answer: P raises ventilation; Q shows a weakened chemical-feedback response
Explanation: Increased metabolism raises carbon-dioxide production and commonly increases \(\mathrm{CO_2}\) and \(\mathrm{H^+}\) signals in blood. Functional central and peripheral chemosensitive regions detect these changes and communicate with respiratory neural centres, allowing ventilation to rise. Group P retains this feedback pathway and should show an appropriate adaptive response. In Group Q, the receptors cannot provide the relevant chemical information even though the respiratory muscles and airways remain capable of carrying out breathing movements. Its increase in ventilation is reduced or poorly matched to the metabolic disturbance. The experiment does not remove the medullary rhythm centre, so basic rhythm need not disappear. It isolates the contribution of chemical sensing to adaptive regulation by controlling metabolic demand, muscular function and airway patency while changing the receptor signal. Restoration of chemical balance depends on feedback that detects the disturbance and alters respiratory muscle activity through neural centres.
315. A graph plots blood \(\mathrm{CO_2}\)-related chemical signal on the horizontal axis and ventilation on the vertical axis. Curve P rises as the chemical signal increases, whereas Curve Q remains nearly horizontal after chemosensitive signalling is impaired. Which interpretation is best supported?
ⓐ. Curve Q shows stronger correction of increased carbon dioxide than Curve P
ⓑ. Curve P proves that oxygen is the only chemical regulator of breathing
ⓒ. Both curves show ventilation independent of blood chemistry
ⓓ. P shows chemical feedback; Q has lost normal responsiveness
Correct Answer: P shows chemical feedback; Q has lost normal responsiveness
Explanation: Curve P shows that ventilation rises as the carbon-dioxide-related chemical signal becomes stronger. This pattern is expected when central and peripheral chemosensitive mechanisms detect increases in \(\mathrm{CO_2}\) and \(\mathrm{H^+}\) and stimulate respiratory adjustment. Curve Q remains nearly flat, indicating that ventilation changes little despite the growing chemical disturbance. The stated impairment of chemosensitive signalling accounts for this loss of responsiveness. The graph does not establish oxygen as the only regulator and does not show independence from blood chemistry. It compares a functioning feedback relation with a disrupted one. The rising curve represents an adaptive response that supports additional carbon-dioxide elimination, whereas the flat curve indicates failure to convert a chemical change into an appropriately increased ventilatory output. Within the process described, restoration of chemical balance depends on feedback that detects the disturbance and alters respiratory muscle activity through neural centres.
316. Assertion: Increased ventilation can reduce an elevated carbon-dioxide-related chemical stimulus.
Reason: Greater ventilation promotes carbon-dioxide movement into alveoli and removal during expiration.
ⓐ. Both statements are true, and the Reason correctly explains the Assertion
ⓑ. Both statements are true, but the Reason does not explain the Assertion
ⓒ. The Assertion is true, whereas the Reason is false and cannot explain it
ⓓ. The Assertion is false, whereas the Reason is true and cannot explain it
Correct Answer: Both statements are true, and the Reason correctly explains the Assertion
Explanation: Ventilation is the major route by which carbon dioxide produced in tissues is ultimately removed from the body. When respiratory activity increases, alveolar air is renewed more effectively and carbon dioxide can continue moving from pulmonary blood into alveoli. Expiration then carries that gas out of the body. The Reason accurately describes this removal mechanism and directly explains how increased ventilation opposes an elevated carbon-dioxide-related stimulus. As carbon dioxide is eliminated, associated hydrogen-ion signals can also be reduced. This relation forms a negative-feedback loop linking cellular metabolism, chemosensitive detection and respiratory adjustment. The response does not eliminate cellular production of carbon dioxide; instead, it increases the rate at which the respiratory system transfers and expels the additional load.
317. Three respiratory disorders are compared.
| Disorder | Principal structural change | Immediate functional effect |
|---|
| P | Inflamed bronchi and bronchioles | Restricted airflow and wheezing |
| Q | Destroyed alveolar walls | Reduced respiratory surface area |
| R | Dust-related inflammation and fibrosis | Damaged or thickened respiratory tissue |
Which identification is correct?
ⓐ. P emphysema; Q asthma; R pleural disorder
ⓑ. P occupational fibrosis; Q asthma; R emphysema
ⓒ. P asthma; Q emphysema; R occupational fibrosis
ⓓ. P asthma; Q occupational fibrosis; R emphysema
Correct Answer: P asthma; Q emphysema; R occupational fibrosis
Explanation: Record P describes inflammation of bronchi and bronchioles with restricted airflow and wheezing, which identifies asthma. Record Q describes destruction of alveolar walls and loss of respiratory surface area, the defining structural pattern of emphysema. Record R links prolonged dust exposure with inflammation and fibrosis, identifying an occupational respiratory disorder. The comparison distinguishes disorders by the structure first affected and the immediate physiological consequence. Asthma primarily compromises the conducting pathway, emphysema removes exchange surface, and occupational fibrosis damages or thickens respiratory tissue after prolonged exposure. General breathlessness may occur in all three, so symptoms alone are insufficient. Correct classification requires combining the anatomical site, structural alteration and functional result rather than selecting a disorder from one shared clinical effect. Airway narrowing, alveolar-wall destruction and fibrotic thickening impair different stages even though all can reduce respiratory performance.
318. Patient P has episodic wheezing with inflamed bronchioles and intact alveolar walls. Patient Q has no marked wheezing but shows extensive alveolar-wall destruction after prolonged cigarette smoking. Which comparison is correct?
ⓐ. P has an asthma-like airway defect; Q has emphysema-like surface loss
ⓑ. P has emphysema-like surface loss; Q has an asthma-like airway defect
ⓒ. Both have asthma-like conducting-airway defects
ⓓ. Both have emphysema-like alveolar-surface defects
Correct Answer: P has an asthma-like airway defect; Q has emphysema-like surface loss
Explanation: Patient P has inflamed bronchioles, which belong to the conducting pathway. Narrowing at this site interferes with movement of air into and out of the lungs and therefore creates a primary ventilation problem. The preserved alveolar walls indicate that the principal diffusion surface is not initially destroyed. Patient Q has widespread alveolar-wall destruction associated with prolonged smoking, identifying emphysema. The major immediate consequence is a reduction in total respiratory surface area. Both conditions can eventually reduce oxygen delivery to tissues, but they begin at different stages of the respiratory sequence. The comparison must therefore separate air conduction from alveolar exchange rather than group the patients together because both may experience breathlessness. The supplied evidence supports an airway disorder in P and an exchange-surface disorder in Q.
319. Three lung samples receive equal air volumes and equal capillary blood flow. Sample P has normal tissue, Sample Q has destroyed alveolar partitions, and Sample R has fibrotic thickening of the exchange tissue. Both Q and R show reduced gas transfer. Which inference is strongest?
ⓐ. Q and R must have identical structural damage because their transfer is reduced
ⓑ. Loss of surface area and increased diffusion distance can independently reduce exchange
ⓒ. Equal ventilation prevents structural lung damage from affecting gas transfer
ⓓ. Fibrosis improves diffusion by increasing the amount of tissue between air and blood
Correct Answer: Loss of surface area and increased diffusion distance can independently reduce exchange
Explanation: Sample Q loses alveolar partitions, which reduces the total surface available for gas exchange. Sample R retains tissue but develops fibrotic thickening, increasing the distance gases must cross. Both structural changes can reduce transfer even though their mechanisms differ. Equal air delivery and capillary flow control two other major determinants, allowing the observed reductions to be linked to exchange-surface structure. The experiment does not support the claim that Q and R have identical damage. It instead shows that efficient diffusion requires both extensive area and a thin barrier. Emphysematous wall destruction and occupational fibrosis can impair exchange through different structural pathways. Equal ventilation cannot compensate fully when the membrane itself provides too little area or too great a diffusion distance. The affected anatomical site determines whether the primary limitation involves ventilation, exchange area or diffusion distance.
320. Match each change with its most direct respiratory consequence. A Column II entry is used once.
| Column I | Column II |
|---|
| P. Bronchiolar inflammation decreases | 1. Airflow through conducting branches improves |
| Q. Alveolar walls remain destroyed | 2. Reduced exchange surface persists despite open airways |
| R. Fibrotic tissue thickens around alveoli | 3. Diffusion distance between air and blood increases |
| S. A protective mask is used before dust exposure | 4. Inhalation of workplace particles is reduced |
ⓐ. P-2, Q-3, R-4, S-1
ⓑ. P-3, Q-4, R-1, S-2
ⓒ. P-4, Q-1, R-2, S-3
ⓓ. P-1, Q-2, R-3, S-4
Correct Answer: P-1, Q-2, R-3, S-4
Explanation: Reduced inflammation of bronchi and bronchioles widens the conducting pathway and improves airflow, giving P-1. Destroyed alveolar walls do not reappear merely because an airway is open, so the loss of respiratory surface persists and Q pairs with 2. Fibrosis adds tissue to the air-blood barrier and increases diffusion distance, fixing R with 3. A protective mask acts before injury by reducing the amount of industrial dust inhaled, giving S-4. The four mappings distinguish a reversible airway improvement, a persistent surface-area defect, a membrane-thickness defect and a preventive measure. These effects occur at different points in respiration and should not be treated as interchangeable responses to breathlessness. Structure, stage and timing determine whether an intervention improves ventilation, preserves exposure control or leaves established exchange damage unchanged. The correct set follows the nearest causal consequence of each change rather than a later, nonspecific symptom shared by several disorders.
