Breathing and Exchange of Gases MCQs with Answers – Part 2 (Class 11 Biology)

Explanation: Quiet expiration follows relaxation of the diaphragm and external intercostal muscles after inspiration. During inspiration, the diaphragm and external intercostal muscles contract to enlarge the thoracic chamber. Their relaxation reverses these movements: the diaphragm becomes more dome-shaped, and the ribs and sternum return toward their resting positions. Thoracic volume decreases, and pulmonary volume follows the same direction through mechanical coupling. The reduced lung volume compresses pulmonary air and raises intra-pulmonary pressure slightly above atmospheric pressure. Air then moves outward along the pressure gradient. The Reason supplies the complete mechanical explanation for the Assertion, linking muscle relaxation with the volume and pressure changes that produce expiration. An open airway is necessary but not sufficient: without an appropriate volume change, no sustained pressure gradient develops.

Explanation: The graph supplies two observations during segment X: pulmonary volume decreases, and intra-pulmonary pressure is above the atmospheric-pressure reference. A decrease in lung volume compresses the air within the lungs during expiration. Air then has a higher pressure inside the lungs than outside, so it moves through the conducting passage from the lungs toward the atmosphere. The pressure trace is decisive for direction; a falling volume by itself describes the mechanical phase but does not replace the pressure comparison. As long as the pulmonary trace remains above the atmospheric line and the airway is open, the expected net flow is outward. At the end of expiration the pressure difference diminishes and net airflow stops when the pressures become equal. Segment X lies before that endpoint, so the graph describes the outward-flow part of expiration rather than an inspiratory interval or movement through the pleural space.

Explanation: Relaxation of the inspiratory muscles initiates quiet expiration. The thoracic boundaries return toward their resting positions, so thoracic volume decreases and pulmonary volume follows. Since the airway is temporarily closed, the compressed pulmonary air cannot yet escape; intra-pulmonary pressure rises above atmospheric pressure while airflow remains absent. Reopening the airway then provides a path from the higher-pressure lungs to the lower-pressure atmosphere. Air moves outward until the pressure difference is reduced and the two pressures again become equal. The temporary closure separates pressure generation from airflow and shows their dependency clearly: muscle relaxation and volume reduction can create the expiratory pressure gradient, but actual outward movement also requires an open conducting pathway. This changed condition differs from ordinary quiet expiration only in the delayed availability of the airflow route. Pressure equality at the final stage marks the end of net expiratory flow.

Explanation: Record P gives a fully consistent chain for quiet expiration. Relaxation of the inspiratory muscles reduces thoracic volume, and pulmonary volume decreases through thoracic-pulmonary coupling. Compression of pulmonary air raises intra-pulmonary pressure above atmospheric pressure, so air moves outward down the pressure gradient. The remaining records reverse at least one necessary relation: pulmonary expansion belongs to inspiration, pressure below atmospheric pressure favours inward flow, and inspiratory contraction does not initiate quiet expiration. All columns must describe the same phase, because one correct entry cannot rescue a row containing an incompatible volume, pressure or airflow direction. Record P alone preserves the causal sequence from muscle relaxation through volume reduction and pressure rise to outward airflow. The table therefore tests internal consistency across a mechanical pathway, not recognition of one isolated feature. The pressure relation must arise from the listed volume change and must predict the listed airflow.

Explanation: Quiet expiration normally begins as the diaphragm and external intercostal muscles relax and thoracic volume decreases. With normal mechanical coupling, the lungs also become smaller, compressing their air. Intra-pulmonary pressure then rises slightly above atmospheric pressure, producing outward airflow through the conducting passages. The device in the stem blocks the pulmonary-volume decrease while permitting thoracic volume to fall. It therefore interrupts the sequence before the usual compression of lung air can occur. The pressure rise will be reduced, and the driving force for outward airflow will likewise be reduced. Muscle relaxation identifies the attempted expiratory phase, but it is not itself the force that pushes air through the airway; the pressure difference generated by pulmonary-volume change is required. This prediction concerns the immediate response to the restraint, without assuming that alveolar walls, blood transport or atmospheric pressure have changed.

Explanation: Expiration requires a pressure gradient directed from the lungs toward the atmosphere. A decrease in pulmonary volume compresses the air inside the lungs and raises intra-pulmonary pressure above atmospheric pressure. Air then moves outward from the higher-pressure pulmonary space to the lower-pressure external environment. Equal pressures provide no net driving force, while pulmonary pressure below atmospheric pressure favours inward flow during inspiration. The completed relation includes both the pressure condition and the preceding volume change. It does not merely name a pressure inequality; it connects that inequality to the mechanical event that produces it during normal expiration. Thus, normal expiration depends on a coordinated volume-driven pressure reversal rather than an active pumping action by the alveoli.

Explanation: Quiet inspiration begins with contraction of the diaphragm and external intercostal muscles, giving P-2. Quiet expiration follows relaxation of these inspiratory muscles, so Q maps to 4. When greater strength of inspiration or expiration is required, additional abdominal muscles may contribute, producing R-1. Outward airflow requires intra-pulmonary pressure to exceed atmospheric pressure, giving S-3. The mappings distinguish ordinary muscle events from the extra recruitment available during stronger breathing. They also separate muscular action from the immediate pressure condition that determines airflow direction. Stronger breathing can alter the magnitude of respiratory movements, but air still follows the same higher-to-lower pressure rule. Pressure equality at completion of expiration stops airflow and establishes the starting state for the next inspiratory cycle.

Explanation: The diaphragm and intercostal muscles remain central to respiratory movements, but stronger inspiration and expiration can receive assistance from additional muscles in the abdominal region. Their recruitment increases the force of the mechanical changes involved in breathing. This does not convert pulmonary ventilation into cellular respiration; ventilation remains the movement of air, while cellular respiration remains an intracellular metabolic process. Pressure gradients also continue to determine the direction of airflow. Greater muscular effort changes how strongly thoracic and pulmonary volumes can be altered, which can produce larger pressure differences and more forceful air movement. The case identifies an increased-demand adjustment within the normal mechanical system rather than a new route of gas production or exchange.

Explanation: Additional abdominal muscles may be recruited when breathing must become more forceful, so statement I is valid. Their activity can strengthen respiratory movements beyond those of quiet breathing, and the established mechanism permits their contribution to both inspiration and expiration. Statement III is valid. Statement IV also remains true, since stronger muscular action ultimately produces airflow by generating larger volume changes and pressure gradients. Statement II is incorrect. Quiet inspiration is initiated mainly through contraction of the diaphragm and external intercostal muscles; abdominal muscles are not the sole participants in every normal breath. Statements I, III and IV distinguish additional muscular recruitment during increased effort from the basic mechanism of quiet ventilation.

Explanation: A healthy human at rest commonly breathes approximately \(12\text{–}16\,\mathrm{breaths\,min^{-1}}\). This value represents respiratory frequency, meaning the number of breathing cycles completed in one minute. A spirometer is used to estimate volumes of air involved in breathing movements and contributes to clinical assessment of pulmonary function. The two relations must agree: the numerical range identifies ordinary resting frequency, while the instrument identifies respiratory-volume measurement. Respiratory rate can be counted directly, but a spirometer supplies a volume record rather than measuring blood pressure, blood-cell number or heat production. The values are approximate physiological references and should not be treated as identical in every healthy individual under all conditions.

Explanation: The instrument described is functioning as a spirometer. Its tracing follows changes in respiratory air volume as the participant inspires and expires. Such a record can be used to estimate pulmonary volumes involved in breathing and can assist in assessment of respiratory function. The observation does not directly measure intracellular energy release, tissue-gas partial pressures or haemoglobin-molecule number. Those quantities would require different evidence and methods. The repeated rise and fall in the trace corresponds to air moving into and out of the respiratory system, making volume the directly recorded variable. A valid inference must remain within what the instrument measures rather than extending the tracing to unrelated cellular or blood properties.

Explanation: Respiratory frequency gives the number of complete breathing cycles in one minute. The person completes \(15\) cycles per minute for \(4\,\mathrm{min}\), so the total is: \[ 15\,\mathrm{cycles\,min^{-1}}\times4\,\mathrm{min}=60\,\mathrm{cycles} \] The minute units cancel, leaving a count of cycles. To judge the resting frequency, compare the original \(15\,\mathrm{cycles\,min^{-1}}\) with the common healthy range of \(12\text{–}16\,\mathrm{cycles\,min^{-1}}\); it lies inside that interval. The accumulated count of \(60\) is not a frequency and should not be compared directly with a per-minute range. A longer observation at the same breathing rate would increase the total count without changing whether the rate lies in the usual resting interval. Thus, the time-dependent total and the classification of the underlying respiratory frequency are two separate decisions made from the same record. The calculation can be checked by grouping the four minutes separately: each contributes fifteen full cycles, so four equal groups give sixty. No tidal-volume measurement is needed to count breaths..

Explanation: The stated reference interval includes values from \(12\) through \(16\,\mathrm{breaths\,min^{-1}}\). Individual P lies at the lower boundary and individual Q lies at the upper boundary, so both fall within the interval. Individual R is below it, while individual S is above it. The interval includes its endpoints, so values equal to \(12\) or \(16\) must be counted rather than excluded. This comparison does not by itself diagnose disease or establish that every person outside the range has a disorder. Respiratory frequency is influenced by physiological conditions, and the listed range is an approximate common resting value. From the supplied criterion alone, P and Q form the supported pair. The comparison concerns breaths per minute; without a tidal-volume value, the total volume of air moved cannot be inferred from this table.

Explanation: Initially, minute respiratory volume is \(M=TV_1\times f_1\). Under the changed condition, frequency doubles, so \(f_2=2f_1\), while minute volume remains unchanged: \[ TV_1f_1=TV_2\left(2f_1\right) \] Cancelling \(f_1\) from both sides gives: \[ TV_1=2TV_2 \] Hence: \[ TV_2=\frac{1}{2}TV_1 \] The relation shows an inverse adjustment under the stated constant-minute-volume condition. Doubling the number of breaths per minute must be balanced by halving the amount of air moved in each normal breath. This compensation is not assumed to occur in every physiological situation; it follows mathematically because \(M\) is explicitly held constant. Changing tidal volume with frequency fixed produces the same proportional change in total air moved per minute. A range of breathing frequencies produces a range of minute or hourly volumes rather than one exact value. Approximate physiological values describe a healthy resting pattern and are not identical for every individual.

Explanation: Tidal volume describes the air moved in one normal breath; respiratory frequency gives the number of breaths per minute. Multiply them to obtain the volume of air moved each minute: \[ 500\,\mathrm{mL\,breath^{-1}}\times12\,\mathrm{breaths\,min^{-1}} =6000\,\mathrm{mL\,min^{-1}} \] The stem says that both values remain constant for \(5\,\mathrm{min}\). Multiplying the minute volume by that time gives: \[ 6000\,\mathrm{mL\,min^{-1}}\times5\,\mathrm{min} =30000\,\mathrm{mL}=30\,\mathrm{L} \] The time unit cancels, so the final answer is a volume, not a volume-per-minute rate. This is cumulative tidal air passing through the respiratory passage across repeated breaths; it is not the amount of oxygen extracted from that air or the volume retained in the lungs at one instant. Keeping the per-breath, per-minute and five-minute stages distinct prevents a correct rate from being mistaken for the requested total. Equivalently, the five-minute interval contains \(12\times5=60\) breaths; \(60\times500\,\mathrm{mL}=30000\,\mathrm{mL}\), confirming the same total without changing the meaning of the answer.

Explanation: Minute respiratory volume is the product of tidal volume and respiratory frequency. At the initial frequency: \[ 500\,\mathrm{mL}\times12\,\mathrm{min^{-1}}=6000\,\mathrm{mL\,min^{-1}} \] At the higher frequency: \[ 500\,\mathrm{mL}\times16\,\mathrm{min^{-1}}=8000\,\mathrm{mL\,min^{-1}} \] The required increase is the difference: \[ 8000-6000=2000\,\mathrm{mL\,min^{-1}} \] Tidal volume is unchanged, so the increase results entirely from four additional breathing cycles each minute. The final value of \(8000\,\mathrm{mL\,min^{-1}}\) is not itself the increase; the original minute volume must first be subtracted. Thus, minute respiratory volume rises by \(2000\,\mathrm{mL\,min^{-1}}\), equivalent to \(2\,\mathrm{L\,min^{-1}}\). This quantity describes air moved by ventilation, not oxygen consumed directly by cells. Keeping the units as volume per minute confirms that the calculation measures a rate of air movement rather than a static pulmonary capacity.

Explanation: On a graph of cumulative air volume against time, slope represents the volume moved per unit time. Both patterns have the same tidal volume, so the slope difference must arise from respiratory frequency. Pattern P moves: \[ 500\times12=6000\,\mathrm{mL\,min^{-1}} \] Pattern Q moves: \[ 500\times16=8000\,\mathrm{mL\,min^{-1}} \] Q therefore accumulates air volume faster and produces the steeper line. The graph does not indicate a larger volume per breath because tidal volume is explicitly held constant. Instead, it shows that completing more equal-volume breaths each minute increases minute respiratory volume. The slope represents total ventilatory air movement and does not by itself specify how much of that air participates in gas exchange. Because time is on the horizontal axis, a steeper slope is the graphical equivalent of a larger calculated minute volume, not a larger retained lung volume.

Explanation: Tidal volume is the air inspired or expired during one normal breath, and its approximate reference value is \(500\,\mathrm{mL}\). Multiplying this per-breath volume by the number of breaths completed each minute gives minute respiratory volume. At a resting frequency of \(12\text{–}16\,\mathrm{breaths\,min^{-1}}\), the corresponding range is approximately: \[ 500\times12=6000\,\mathrm{mL\,min^{-1}} \] to \[ 500\times16=8000\,\mathrm{mL\,min^{-1}} \] Statements I, II and III therefore describe the definition, approximate value and derived relation correctly. Air remaining after forcible expiration is residual volume, so Statement IV confuses a retained lung volume with a time-dependent measure of air moved during breathing. The distinction is important because minute respiratory volume accumulates repeated tidal movements over time, whereas residual volume is present at a defined lung-volume state.

Explanation: Respiratory frequency is the number of breathing cycles completed per minute, giving P-3. A spirometer is used to estimate respiratory air volumes, so Q pairs with 4. Tidal volume is the amount inspired or expired during one normal breath, fixing R-1. Minute respiratory volume is the product of tidal volume and respiratory frequency, giving S-2. The mapping distinguishes a count, an instrument, a per-breath volume and a derived per-minute quantity. These terms are connected during respiratory assessment but are not interchangeable. A spirometer provides volume information, while respiratory frequency states how often that volume is moved. Combining the appropriate per-breath volume with the number of breaths gives total air movement per minute.

Explanation: Tidal volume is the air moved in one ordinary breath, and respiratory frequency tells us how many such breaths occur per minute. At the lower limit of the given resting range, calculate the minute volume: \[ 500\,\mathrm{mL\,breath^{-1}}\times12\,\mathrm{breaths\,min^{-1}}=6000\,\mathrm{mL\,min^{-1}}=6\,\mathrm{L\,min^{-1}} \] At the upper limit, the same calculation gives \(500\times16=8000\,\mathrm{mL\,min^{-1}}=8\,\mathrm{L\,min^{-1}}\). The question asks for air moved during a specific \(1\,\mathrm{h}\), so multiply each rate by the duration \(60\,\mathrm{min}\): \[ 6\,\mathrm{L\,min^{-1}}\times60\,\mathrm{min}=360\,\mathrm{L},\qquad 8\,\mathrm{L\,min^{-1}}\times60\,\mathrm{min}=480\,\mathrm{L} \] The units \(\mathrm{min^{-1}}\) and \(\mathrm{min}\) cancel, making \(360\text{–}480\,\mathrm{L}\) a cumulative volume. An independent check counts \(12\times60=720\) breaths at the lower limit and \(16\times60=960\) breaths at the upper limit; multiplying either count by \(500\,\mathrm{mL}\) gives the same endpoint volumes. The calculation assumes a constant tidal volume and a frequency held at one of the stated limits for the full hour. Keeping \(\mathrm{L\,h^{-1}}\) as the answer would describe a rate instead of the amount moved in the specified hour. This is total tidal air entering or leaving across many breaths; it is neither lung capacity at a single moment nor the oxygen actually consumed by tissues.