Respiratory System — MCQs

Respiratory System Indian Medical PG Practice Questions and MCQs

Question 11: According to Poiseuille's law, reducing the radius of an airway to one-third will increase its resistance how many fold?

A. 3
B. 3
C. 9
D. 81 (Correct Answer)

Explanation: ### Explanation **1. The Underlying Concept: Poiseuille’s Law** In respiratory physiology, the resistance to airflow ($R$) in the airways is governed by **Poiseuille’s Law**. The formula states that resistance is inversely proportional to the fourth power of the radius ($r$): $$R \propto \frac{1}{r^4}$$ When the radius of an airway is reduced to **one-third ($1/3$)** of its original size, the new resistance ($R_{new}$) is calculated as: $$R_{new} \propto \frac{1}{(1/3)^4} = \frac{1}{1/81} = 81$$ Thus, the resistance increases by **81-fold**. This demonstrates that even minor changes in airway caliber (due to bronchoconstriction or edema) lead to massive increases in the work of breathing. **2. Analysis of Incorrect Options** * **Option A & B (3):** This assumes a linear relationship ($R \propto 1/r$), which is incorrect for fluid/air flow dynamics. * **Option C (9):** This assumes resistance is inversely proportional to the square of the radius ($R \propto 1/r^2$), which describes the relationship with cross-sectional area, not resistance. **3. Clinical Pearls & High-Yield Facts for NEET-PG** * **Site of Maximum Resistance:** Although individual small airways have high resistance, the **medium-sized bronchi (generations 2-5)** are the site of *maximum* total airway resistance. * **Small Airway Paradox:** The terminal bronchioles (small airways) contribute very little to total resistance because they are arranged in **parallel**, greatly increasing the total cross-sectional area. * **Autonomic Control:** Parasympathetic stimulation (ACh) causes bronchoconstriction (decreasing radius, increasing resistance), while Sympathetic stimulation ($\beta_2$ receptors) causes bronchodilation. * **Density vs. Viscosity:** According to Poiseuille's law, resistance is also proportional to gas viscosity. However, in **turbulent flow** (high Reynolds number), density becomes more important than viscosity.

Question 12: Lungs do not collapse during expiration because of the presence of?

A. Hyaline membrane
B. Dipalmitoyl phosphatidyl choline (Correct Answer)
C. Macrophages
D. Interstitial fluid

Explanation: **Explanation:** The correct answer is **Dipalmitoyl phosphatidyl choline (DPPC)**, which is the primary phospholipid component of **Pulmonary Surfactant**. **1. Why DPPC is correct:** According to the **Law of Laplace ($P = 2T/r$)**, smaller alveoli have a higher collapsing pressure ($P$) due to surface tension ($T$). During expiration, as the radius ($r$) of the alveoli decreases, the risk of collapse (atelectasis) increases. Surfactant, secreted by **Type II Pneumocytes**, reduces surface tension. DPPC molecules are amphipathic; they crowd together as the alveolus shrinks, significantly lowering surface tension and preventing the lungs from collapsing at low lung volumes. **2. Why other options are incorrect:** * **Hyaline membrane:** This is a pathological finding (not a physiological one) seen in Neonatal Respiratory Distress Syndrome (NRDS) or ARDS, where protein-rich fluid leaks into alveoli, actually *hindering* gas exchange. * **Macrophages:** Alveolar macrophages (Dust cells) are part of the immune system and are responsible for clearing debris and pathogens; they do not influence surface tension. * **Interstitial fluid:** Excessive fluid in the interstitium leads to pulmonary edema, which decreases lung compliance and impairs respiration. **Clinical Pearls for NEET-PG:** * **L/S Ratio:** A Lecithin (DPPC) to Sphingomyelin ratio of **>2:1** in amniotic fluid indicates fetal lung maturity. * **Glucocorticoids:** These are administered to mothers in preterm labor to accelerate surfactant production by stimulating Type II pneumocytes. * **Surfactant Proteins:** SP-A and SP-D are involved in innate immunity, while SP-B and SP-C are essential for the film-forming properties of surfactant.

Question 13: Given an arterial blood oxygen content of 20 mL per 100 mL of blood and a venous blood oxygen content of 15 mL per 100 mL of blood, calculate the amount of oxygen transferred from the blood to the tissues if the blood flow is 200 mL/min?

A. 5 mL/min
B. 10 mL/min (Correct Answer)
C. 15 mL/min
D. 20 mL/min

Explanation: ### Explanation The amount of oxygen delivered to the tissues per minute is determined by the **Fick Principle**, which states that the uptake of a substance by an organ is the product of the blood flow to that organ and the difference in the concentration of that substance between the arterial and venous blood. **1. Why Option B is Correct:** To calculate the oxygen transfer (Oxygen Consumption or $\dot{V}O_2$), use the following formula: $$\text{Oxygen Transfer} = \text{Blood Flow} \times (\text{Arterial } O_2 \text{ content} - \text{Venous } O_2 \text{ content})$$ * **Step 1:** Calculate the Arteriovenous (A-V) Oxygen difference: $20\text{ mL}/100\text{ mL} - 15\text{ mL}/100\text{ mL} = 5\text{ mL of } O_2 \text{ per } 100\text{ mL of blood.}$ * **Step 2:** Apply the blood flow rate ($200\text{ mL/min}$): Since $5\text{ mL}$ is extracted per $100\text{ mL}$, for $200\text{ mL}$ of flow: $(5\text{ mL} / 100\text{ mL}) \times 200\text{ mL/min} = \mathbf{10\text{ mL/min}}.$ **2. Why Other Options are Incorrect:** * **Option A (5 mL/min):** This represents the A-V difference per 100 mL, but fails to account for the total blood flow of 200 mL/min. * **Options C & D (15 & 20 mL/min):** These values result from mathematical errors or incorrectly using only the venous or arterial content without calculating the difference. **3. Clinical Pearls for NEET-PG:** * **Normal Values:** In a healthy resting adult, the A-V $O_2$ difference is typically $5\text{ mL}/100\text{ mL}$ of blood. With a standard Cardiac Output of $5\text{ L/min}$, the total resting $O_2$ consumption is approximately $250\text{ mL/min}$. * **Utilization Coefficient:** The fraction of oxygen given up to the tissues (normally $\sim25\%$) is called the utilization coefficient. It increases significantly during strenuous exercise. * **Fick Principle Application:** This principle is also the gold standard for measuring Cardiac Output ($CO = \dot{V}O_2 / [Ca_{O2} - Cv_{O2}]$).

Question 14: What is the partial pressure of O2, which constitutes approximately 21% of inspired air?

A. 160 mmHg (Correct Answer)
B. 240 mmHg
C. 580 mmHg
D. 760 mmHg

Explanation: **Explanation:** The partial pressure of a gas in a mixture is determined by **Dalton’s Law**, which states that the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the individual gases. To calculate the partial pressure of Oxygen ($PO_2$) in dry inspired air at sea level: 1. **Total Atmospheric Pressure ($P_{atm}$):** 760 mmHg. 2. **Fraction of Oxygen ($FiO_2$):** 21% or 0.21. 3. **Calculation:** $PO_2 = P_{atm} \times FiO_2 = 760 \times 0.21 \approx \mathbf{159.6 \text{ mmHg}}$ (rounded to **160 mmHg**). **Analysis of Incorrect Options:** * **B (240 mmHg):** This value does not correspond to any physiological gas pressure at sea level under normal atmospheric conditions. * **C (580 mmHg):** This is the approximate partial pressure of **Nitrogen** ($PN_2$) in the atmosphere ($760 \times 0.79$). * **D (760 mmHg):** This represents the **total atmospheric pressure** at sea level, not the pressure of an individual gas component. **NEET-PG High-Yield Pearls:** * **Humidification Effect:** As air enters the upper airways, it is saturated with water vapor ($PH_2O = 47 \text{ mmHg}$). The $PO_2$ in **humidified tracheal air** drops to $\approx 149 \text{ mmHg}$ $[(760 - 47) \times 0.21]$. * **Alveolar Air:** In the alveoli, $PAO_2$ is further reduced to $\approx \mathbf{100 \text{ mmHg}}$ due to the continuous diffusion of $O_2$ into the blood and the addition of $CO_2$. * **Fractional Concentration:** Note that while partial pressure changes with altitude, the *percentage* of $O_2$ (21%) remains constant.

Question 15: What is the most important form of carbon dioxide transport in the blood?

A. Carboxyhemoglobin
B. Dissolved CO2
C. Bicarbonates (Correct Answer)
D. CO2 molecules attached to hemoglobin

Explanation: **Explanation:** Carbon dioxide (CO₂) is transported from the tissues to the lungs in three primary forms. Understanding the distribution of these forms is a high-yield concept for NEET-PG: 1. **Bicarbonate Ions (70%):** This is the **most important and predominant form**. CO₂ enters RBCs and reacts with water to form carbonic acid ($H_2CO_3$), catalyzed by the enzyme **Carbonic Anhydrase**. This acid dissociates into $H^+$ and $HCO_3^-$. The bicarbonate then diffuses into the plasma in exchange for chloride ions (the **Chloride Shift** or Hamburger phenomenon). 2. **Carbamino Compounds (23%):** CO₂ binds directly to the amino groups of hemoglobin (forming **carbaminohemoglobin**) and plasma proteins. 3. **Dissolved Form (7%):** A small fraction is carried physically dissolved in the plasma. **Analysis of Incorrect Options:** * **A. Carboxyhemoglobin:** This is a trap. Carboxyhemoglobin refers to **Carbon Monoxide (CO)** bound to hemoglobin, which is a toxic state, not a physiological CO₂ transport mechanism. * **B. Dissolved CO₂:** While CO₂ is 20 times more soluble than oxygen, this form only accounts for ~7% of total transport. * **D. CO₂ molecules attached to hemoglobin:** This refers to carbaminohemoglobin. While significant (23%), it is not the "most important" or majority form. **High-Yield Clinical Pearls:** * **Haldane Effect:** Deoxygenation of blood increases its ability to carry CO₂. In the lungs, when $O_2$ binds to Hb, it promotes the release of CO₂. * **Carbonic Anhydrase:** It is one of the fastest enzymes in the body; its absence would make CO₂ transport insufficient to sustain life. * **Chloride Shift:** Remember that $Cl^-$ moves **into** the RBC in systemic tissues and **out** of the RBC in pulmonary capillaries.

Question 16: Which of the following causes a right shift of the oxygen-hemoglobin dissociation curve?

A. Hypothermia
B. Hypoxia (Correct Answer)
C. Alkalosis
D. Hemoglobin Fetal (HbF)

Explanation: The oxygen-hemoglobin (O₂-Hb) dissociation curve represents the relationship between the partial pressure of oxygen (PO₂) and the percentage saturation of hemoglobin. A **right shift** indicates a decreased affinity of hemoglobin for oxygen, facilitating oxygen unloading to the tissues. ### Why Hypoxia is Correct **Hypoxia** (low oxygen levels) leads to a right shift primarily through the production of **2,3-Bisphosphoglycerate (2,3-BPG)**. When tissues are hypoxic, RBCs increase anaerobic glycolysis, producing more 2,3-BPG. This molecule binds to the beta chains of deoxyhemoglobin, stabilizing the "T" (Tense) state and promoting the release of oxygen to oxygen-starved tissues. ### Why Other Options are Incorrect * **Hypothermia (A):** Low temperature stabilizes the "R" (Relaxed) state of hemoglobin, increasing its affinity for oxygen and causing a **left shift**. * **Alkalosis (C):** An increase in pH (decreased H⁺ concentration) causes a **left shift** (the Bohr effect). Conversely, acidosis causes a right shift. * **Hemoglobin Fetal (HbF) (D):** HbF has a higher affinity for oxygen than adult hemoglobin (HbA) because it binds 2,3-BPG poorly. This results in a **left shift**, allowing the fetus to extract oxygen from maternal blood. ### NEET-PG High-Yield Pearls: "CADET, face Right!" To remember the factors that cause a **Right Shift** (decreased affinity), use the mnemonic **CADET**: * **C** – CO₂ increase (Hypercapnia) * **A** – Acidosis (Low pH) * **D** – 2,3-DPG (or 2,3-BPG) increase * **E** – Exercise * **T** – Temperature increase (Fever) **Note:** While acute hypoxia causes a right shift via 2,3-BPG, chronic high-altitude exposure also utilizes this mechanism to improve peripheral oxygen delivery.

Question 17: What is a characteristic finding in restrictive lung disease?

A. Decreased FEV1
B. Increased TLC
C. Normal FEV1/VC (Correct Answer)
D. All of the above

Explanation: **Explanation:** In **Restrictive Lung Diseases** (e.g., Pulmonary Fibrosis, Kyphoscoliosis, Sarcoidosis), the primary pathology is reduced lung compliance or chest wall expansion, leading to a decrease in all lung volumes. **1. Why the Correct Answer (C) is Right:** In restrictive disease, both the **FEV1** (Forced Expiratory Volume in 1 second) and the **FVC** (Forced Vital Capacity) decrease proportionately. Because both the numerator and denominator decrease, the **FEV1/FVC ratio remains normal (typically >0.7) or may even be increased** due to increased radial traction on the airways, which keeps them open during expiration. **2. Why the Other Options are Wrong:** * **Option A (Decreased FEV1):** While FEV1 is indeed decreased in restrictive disease, it is not a *distinguishing* characteristic because FEV1 is also significantly decreased in obstructive diseases (like Asthma/COPD). The hallmark of restriction is the preserved ratio, not just the drop in FEV1. * **Option B (Increased TLC):** Total Lung Capacity (TLC) is **decreased** in restrictive disease. An increased TLC is characteristic of obstructive diseases due to air trapping and hyperinflation. **High-Yield Clinical Pearls for NEET-PG:** * **Gold Standard for Diagnosis:** A decrease in **TLC** (<80% of predicted) is the definitive diagnostic criterion for restrictive lung disease. * **Flow-Volume Loop:** Shows a characteristic **"Witch’s Hat"** appearance (narrower loop shifted to the right with preserved peak flow). * **DLCO:** Usually decreased in intrinsic restrictive diseases (interstitial lung disease) but normal in extrinsic/extrapulmonary causes (obesity, neuromuscular weakness).

Question 18: In response to ascent to high altitude, the blood pH normalizes. What is the mechanism behind this normalization?

A. Increased erythropoiesis leads to increased buffering by hemoglobin.
B. Increased excretion of HCO3- by the kidneys.
C. Increased levels of 2,3-DPG.
D. Retention of bicarbonate by the kidneys. (Correct Answer)

Explanation: ### Explanation **Correct Answer: B. Increased excretion of HCO3- by the kidneys.** *(Note: There appears to be a discrepancy in the provided key. At high altitude, the body develops respiratory alkalosis; to normalize pH, the kidneys must **excrete** bicarbonate, not retain it. The explanation below clarifies the physiological process.)* **Mechanism of pH Normalization:** Upon ascent to high altitude, low partial pressure of oxygen ($PO_2$) stimulates peripheral chemoreceptors, leading to **hyperventilation**. This causes excessive washout of $CO_2$ (hypocapnia), resulting in **Respiratory Alkalosis** (increased blood pH). To compensate and normalize the pH, the kidneys decrease the secretion of $H^+$ and **increase the excretion of bicarbonate ($HCO_3^-$)**. This renal compensation typically begins within 24–48 hours, effectively lowering the plasma pH back toward normal. **Analysis of Incorrect Options:** * **A. Increased erythropoiesis:** This is a long-term adaptation to improve oxygen-carrying capacity, not a primary mechanism for acute pH normalization. * **C. Increased 2,3-DPG:** This shifts the oxygen-dissociation curve to the right, facilitating oxygen unloading at tissues. It does not directly normalize blood pH. * **D. Retention of bicarbonate:** This is incorrect. Retaining bicarbonate would worsen the existing alkalosis. The kidneys must **waste** bicarbonate to counteract the high pH. **High-Yield Clinical Pearls for NEET-PG:** * **Acetazolamide:** A carbonic anhydrase inhibitor used for Acute Mountain Sickness (AMS). It works by forcing bicarbonate excretion (alkaline diuresis), mimicking/speeding up the natural compensatory mechanism. * **Oxygen Dissociation Curve:** Initially shifts **Left** (due to alkalosis) but later shifts **Right** (due to increased 2,3-DPG). * **Periodic Breathing:** Cheyne-Stokes respiration is common at high altitudes during sleep due to the conflict between hypoxic drive and hypocapnic inhibition of breathing.

Question 19: Which of the following statements is true concerning normal expiration during resting conditions?

A. Expiration is generated by the contraction of expiratory muscles.
B. Alveolar pressure is less than atmospheric pressure.
C. Intrapleural pressure gradually falls (becomes more negative) during expiration.
D. The flow velocity of gas in the large airways exceeds that in the terminal bronchioles. (Correct Answer)

Explanation: ### Explanation **1. Why Option D is Correct:** The flow velocity of gas is governed by the principle of continuity: **Velocity = Flow Rate / Total Cross-sectional Area**. While the diameter of an individual terminal bronchiole is small, the **total cross-sectional area** of all terminal bronchioles combined is massive (thousands of times greater than the trachea). Because the same volume of air must pass through both levels, the velocity significantly decreases as air moves toward the periphery. Thus, gas moves rapidly in the large airways (bulk flow) and slows down to a crawl in the terminal bronchioles, where diffusion becomes the primary mechanism of gas exchange. **2. Why the Other Options are Incorrect:** * **Option A:** Normal expiration at rest is a **passive process**. It is driven by the elastic recoil of the lungs and the relaxation of the diaphragm, not by active muscle contraction. Expiratory muscles (like internal intercostals and abdominals) are only used during forced expiration. * **Option B:** For air to flow out of the lungs, **alveolar pressure must be greater than atmospheric pressure** (positive relative to the atmosphere). If it were less, air would flow into the lungs (inspiration). * **Option C:** During expiration, as the chest wall recoils inward, intrapleural pressure becomes **less negative** (moves from approximately -7.5 cm H₂O back toward -5 cm H₂O). It becomes more negative during inspiration. **3. NEET-PG High-Yield Pearls:** * **Dead Space:** The conducting zone (trachea to terminal bronchioles) constitutes the anatomical dead space (~150 ml) where no gas exchange occurs. * **Transition Zone:** The first site of gas exchange is the **respiratory bronchiole**. * **Compliance:** Emphysema increases lung compliance (loss of elastic recoil), making expiration difficult, whereas Pulmonary Fibrosis decreases compliance.

Question 20: What is the pacemaker that regulates the rate of respiration, being a component of the respiratory control pattern generator responsible for automatic respiration?

A. Pneumotaxic centre
B. Dorsal group of nucleus
C. Apneustic centre
D. Pre-Botzinger Complex (Correct Answer)

Explanation: ### Explanation The regulation of respiration is controlled by the **Respiratory Control Center** in the brainstem. **1. Why Pre-Bötzinger Complex is Correct:** The **Pre-Bötzinger Complex (pre-BötC)**, located in the ventrolateral medulla (part of the Ventral Respiratory Group), is identified as the **rhythm generator** or the **pacemaker** of respiration. It contains specialized neurons that exhibit spontaneous pacemaker activity, similar to the SA node in the heart. These neurons discharge rhythmically to initiate the basic respiratory cycle, specifically driving the inspiratory phase of automatic breathing. **2. Why Other Options are Incorrect:** * **A. Pneumotaxic Centre:** Located in the upper pons (Nucleus Parabrachialis), its primary role is to act as an "off-switch" for inspiration. It limits the duration of inspiration, thereby increasing the respiratory rate. It modulates the rhythm but does not generate it. * **B. Dorsal Respiratory Group (DRG):** Located in the Nucleus Tractus Solitarius (NTS), the DRG is primarily responsible for **inspiration**. While it sends the primary rhythmic drive to the diaphragm via the phrenic nerve, it receives its underlying rhythm from the pre-BötC. * **C. Apneustic Centre:** Located in the lower pons, it promotes long, deep inspirations (apneustic breathing). It is normally inhibited by the pneumotaxic center and vagal afferents. **3. High-Yield Clinical Pearls for NEET-PG:** * **Location:** Pre-BötC is situated between the nucleus ambiguus and the lateral reticular nucleus. * **Opioid Sensitivity:** The Pre-BötC is highly sensitive to **opioids and barbiturates**, which is why respiratory depression is the hallmark of overdose. * **Hering-Breuer Reflex:** This reflex (via stretch receptors in the lungs) prevents over-inflation by inhibiting the DRG, similar to the function of the Pneumotaxic center. * **Ondine’s Curse:** A clinical condition (Congenital Central Hypoventilation Syndrome) where automatic respiration is lost, requiring conscious effort to breathe.

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