Here is your complete MCQ set built directly from your quiz notes, with every answer explained clearly.
π§ HIGH-YIELD PHYSIOLOGY MCQ SET
Based on Your Actual Quiz Questions
π« RESPIRATORY PHYSIOLOGY
Q1. Which of the following is NOT true about intrapulmonary pressure during inspiration?
A) It decreases below atmospheric pressure
B) It drives air into the lungs
C) It increases above atmospheric pressure β
D) It returns to atmospheric at end of inspiration
Answer: C
During inspiration, the diaphragm contracts and lung volume increases β intrapulmonary (alveolar) pressure DECREASES below atmospheric (~β1 to β3 mmHg) β air flows in along the pressure gradient. It never increases during normal inspiration.
Q2. During inspiration, intrapulmonary pressure is:
A) Equal to atmospheric
B) Greater than atmospheric
C) Less than atmospheric β
D) Equal to intrapleural pressure
Answer: C
Boyle's Law: β lung volume β β pressure. Alveolar pressure drops below atmospheric, creating the gradient for airflow inward.
Q3. The intrapleural space does NOT normally contain:
A) Serous fluid
B) Air β
C) Negative pressure
D) Lymphatic drainage
Answer: B
The intrapleural space is a potential space containing only a thin film of serous fluid. Air in this space = pneumothorax, which collapses the lung.
Q4. Air remaining in the lungs after maximum expiration is called:
A) Tidal Volume
B) Expiratory Reserve Volume
C) Residual Volume β
D) Functional Residual Capacity
Answer: C
RV (~1200 mL) cannot be expelled even with maximal forced expiration. It keeps alveoli from collapsing completely.
Q5. The difference between Total Lung Capacity (TLC) and Vital Capacity (VC) is:
A) Tidal Volume
B) Inspiratory Reserve Volume
C) Functional Residual Capacity
D) Residual Volume β
Answer: D
TLC = VC + RV β therefore TLC β VC = RV. This is the classic formula relationship.
Q6. Which instrument is NOT used in pulmonary function testing?
A) Spirometer
B) Body plethysmograph
C) Peak flow meter
D) Sphygmomanometer β
Answer: D
A sphygmomanometer measures blood pressure, not lung function. Spirometry, body plethysmograph, and peak flow meters all assess respiratory parameters.
Q7. Which respiratory center is responsible for LIMITING inspiration?
A) Dorsal respiratory group
B) Ventral respiratory group
C) Apneustic center
D) Pneumotaxic center β
Answer: D
The pneumotaxic center (parabrachial nucleus, upper pons) sends inhibitory signals to the inspiratory center β turns off inspiration β limits depth and duration of each breath.
Q8. Type II alveolar cells (pneumocytes) are responsible for producing:
A) Mucus
B) Surfactant β
C) Fibronectin
D) Immunoglobulins
Answer: B
Type II pneumocytes secrete surfactant (dipalmitoylphosphatidylcholine, DPPC) which reduces alveolar surface tension and prevents atelectasis. Type I cells cover most of the alveolar surface for gas exchange.
Q9. In obstructive pulmonary disease, which values are typically ELEVATED?
A) FVC and FEVβ
B) FEVβ/FVC ratio
C) Residual Volume and TLC β
D) Inspiratory Reserve Volume
Answer: C
In obstruction (COPD, asthma), air trapping occurs β RV β and TLC β. FEVβ/FVC ratio is REDUCED (<70%). Barrel chest = chronically elevated TLC.
Q10. What is NOT true about the relationship between POβ and dissolved oxygen?
A) Dissolved Oβ is proportional to POβ (Henry's Law)
B) Dissolved Oβ is the minor form of oxygen transport in blood
C) Most oxygen is dissolved in plasma β
D) Dissolved Oβ contributes to partial pressure measurement
Answer: C
Only ~1.5% of oxygen is dissolved in plasma. The vast majority (~98.5%) is bound to hemoglobin. Dissolved Oβ obeys Henry's Law (proportional to POβ) but is physiologically negligible for transport.
β€οΈ CARDIOVASCULAR PHYSIOLOGY
Q11. Stroke Volume is calculated as:
A) HR Γ EDV
B) EDV + ESV
C) EDV β ESV β
D) CO / HR
Answer: C
SV = EDV β ESV. Normal: 120 mL β 50 mL = 70 mL. Ejection Fraction = SV/EDV = 70/120 β 58%.
Q12. An increase in preload causes:
A) Decreased cardiac output
B) Increased cardiac output β
C) Decreased stroke volume
D) No change in cardiac output
Answer: B
Frank-Starling Law: β preload (β EDV) β β fiber stretch β β force of contraction β β SV β β CO. This is the fundamental mechanism matching CO to venous return.
Q13. If End-Systolic Volume (ESV) decreases, what happens to Stroke Volume?
A) Decreases
B) Stays the same
C) Increases β
D) First increases then decreases
Answer: C
SV = EDV β ESV. If ESV β (heart ejects more completely), SV β. This happens with β sympathetic stimulation (β contractility) or β afterload.
Q14. Compliance is HIGHEST in which blood vessels?
A) Arteries
B) Arterioles
C) Capillaries
D) Veins β
Answer: D
Veins are ~20Γ more compliant than arteries. They act as capacitance (reservoir) vessels and hold ~64% of total blood volume. Arteries are stiff pressure reservoirs (Windkessel).
Q15. Ventricular diastole begins when:
A) Mitral valve opens
B) AV node fires
C) Semilunar valves close β
D) Ventricular pressure exceeds aortic pressure
Answer: C
When the ventricle relaxes and ventricular pressure falls below aortic pressure, the aortic (semilunar) valve closes β this marks the START of ventricular diastole (isovolumetric relaxation begins).
Q16. The incisura (dicrotic notch) on the aortic pressure waveform represents:
A) Opening of the mitral valve
B) Peak systolic pressure
C) Closure of the aortic valve β
D) Ventricular filling
Answer: C
The dicrotic notch is caused by the brief backflow of blood that snaps the aortic valve shut at the end of systole. It marks the transition from systole to diastole on the arterial waveform.
Q17. Which structure has the SLOWEST conduction velocity in the heart?
A) SA node
B) Bundle of His
C) Purkinje fibers
D) AV node β
Answer: D
AV node conduction velocity = ~0.05 m/s (slowest). This creates the critical PR interval delay, allowing atria to finish contracting before ventricular filling. Purkinje fibers are the fastest (~4 m/s).
Q18. The inotropic effect of digitalis:
A) Decreases heart rate
B) Increases the force of cardiac contraction β
C) Acts via Ξ²β-adrenergic receptors
D) Reduces preload
Answer: B
Digitalis (digoxin) inhibits NaβΊ/KβΊ-ATPase β β intracellular NaβΊ β β NaβΊ/CaΒ²βΊ exchanger activity β β intracellular CaΒ²βΊ β β contractility (positive inotropy). It is NOT sympathetic - it works independently.
Q19. Why do capillaries have the slowest blood flow velocity?
A) They have the smallest diameter
B) They have the thinnest walls
C) They have the largest total cross-sectional area β
D) They have precapillary sphincters
Answer: C
Flow velocity = Flow rate / Cross-sectional area. The billions of capillaries in parallel give an enormous total cross-section (~2500 cmΒ²) β velocity slows dramatically. This allows time for gas/nutrient exchange.
Q20. When MAP falls below 60 mmHg, what happens to the kidney?
A) GFR increases due to vasodilation
B) Autoregulation maintains GFR
C) Autoregulation fails β GFR decreases β
D) Renin secretion decreases
Answer: C
Renal autoregulation works between MAP 60β180 mmHg. Below 60 mmHg, the myogenic and tubuloglomerular mechanisms fail β afferent arteriole cannot dilate enough β perfusion pressure drops β GFR falls β acute kidney injury risk.
Q21. During inspiration in a standing person, what happens to stroke volume in the LEFT ventricle?
A) Increases
B) Stays the same
C) Decreases slightly β
D) Doubles
Answer: C
Inspiration β β venous return to RIGHT heart β RV output β β blood pools in pulmonary vasculature (which expands with lung inflation) β LESS blood immediately returns to LEFT atrium β LV preload β slightly β LV SV β transiently. This creates the normal inspiratory drop in systolic BP (exaggerated in cardiac tamponade = pulsus paradoxus).
Q22. Average stroke volume in a healthy adult is:
A) 50 mL/beat
B) 70 mL/beat β
C) 100 mL/beat
D) 120 mL/beat
Answer: B
Normal SV β 70 mL. Normal HR β 72 bpm. CO = SV Γ HR = 70 Γ 72 β 5 L/min.
Q23. Stroke volume is best measured clinically by:
A) Sphygmomanometer
B) ECG
C) Echocardiography β
D) Spirometry
Answer: C
Echocardiography (ultrasound) measures EDV and ESV directly, allowing SV and ejection fraction calculation. It is the gold standard for cardiac function assessment.
Q24. The pacemaker of the heart is:
A) AV node
B) Bundle of His
C) SA node β
D) Purkinje fibers
Answer: C
The SA node (sinoatrial node) fires at 60β100 bpm intrinsically and is the primary pacemaker. AV node fires at 40β60 bpm (escape rhythm). Purkinje/ventricular cells at 20β40 bpm.
Q25. Average venous pressure is approximately:
A) 120 mmHg
B) 80 mmHg
C) 15 mmHg
D) 2 mmHg β
Answer: D
Central venous pressure (right atrial pressure) β 0β5 mmHg, average ~2 mmHg. This low pressure is essential for maintaining the pressure gradient that drives venous return to the heart.
Q26. Cerebral blood flow regulation is primarily controlled by:
A) Neural (sympathetic) mechanisms only
B) Hormonal regulation
C) Myogenic and metabolic mechanisms β
D) Baroreceptors
Answer: C
Cerebral autoregulation uses: (1) Myogenic mechanism - vessels constrict/dilate based on wall tension; (2) Metabolic mechanism - COβ and HβΊ are the most potent vasodilators. Sympathetic innervation has minimal effect on cerebral vessels under normal conditions.
Q27. In metabolic acidosis, ventilation:
A) Decreases to retain COβ
B) Stays the same
C) Increases (Kussmaul breathing) β
D) Becomes irregular
Answer: C
Metabolic acidosis β β pH β peripheral chemoreceptors stimulated β β ventilation (Kussmaul breathing) β blow off COβ β β carbonic acid β compensatory β pH. This is respiratory compensation for metabolic acidosis.
Q28. What is the potent regulator related to renin-angiotensin in BP control?
A) Vasopressin
B) Endothelin
C) Aldosterone β
(via RAAS) / ANP (as counter-regulator) β
D) Bradykinin
Answer: The RAAS cascade produces Aldosterone (via Angiotensin II β adrenal cortex), which retains NaβΊ/water β β blood volume β β BP. ANP (Atrial Natriuretic Peptide) is the potent COUNTER-regulator - it promotes natriuresis and vasodilation β β BP.
π©Έ HEMATOLOGY & HEMOGLOBIN
Q29. The biconcave disc shape of RBCs is important because:
A) It reduces friction during flow
B) It increases flexibility in capillaries
C) It maximizes surface area for gas exchange β
D) It prevents RBC aggregation
Answer: C
The biconcave shape gives RBCs a surface area of ~140 Β΅mΒ² (much more than a sphere of the same volume ~98 Β΅mΒ²). This maximizes the area available for Oβ and COβ diffusion.
Q30. Microcytic hypochromic anemia is caused by:
A) Vitamin B12 deficiency
B) Folate deficiency
C) Iron deficiency β
D) Hemolysis
Answer: C
Iron deficiency β β heme synthesis β β Hb per cell β small (microcytic), pale (hypochromic) RBCs. Also caused by thalassemia (microcytic) but thalassemia typically has normal/low MCV without the same hypochromia pattern.
Q31. The mutation in sickle cell anemia is:
A) Glutamine β Valine in Ξ±-globin
B) Glutamate β Valine in Ξ²-globin β
C) Glutamate β Lysine in Ξ²-globin
D) Valine β Glutamate in Ξ²-globin
Answer: B
Position 6 of the Ξ²-globin chain: Glutamate (hydrophilic, charged) β Valine (hydrophobic). This single amino acid change causes HbS to polymerize when deoxygenated, distorting RBCs into sickle shape.
Q32. Sickle cells polymerize when hemoglobin is:
A) Oxygenated (R-state)
B) Bound to CO
C) Deoxygenated (T-state) β
D) Bound to 2,3-BPG only
Answer: C
In the deoxy (T/tense) state, the hydrophobic valine is exposed β HbS molecules aggregate into long polymer fibers β sickle shape β vaso-occlusion, hemolysis.
Q33. HbF has increased Oβ affinity compared to HbA because:
A) It has more iron atoms
B) It binds 2,3-BPG more weakly β
C) It has Ξ± chains instead of Ξ² chains
D) It is smaller in molecular weight
Answer: B
HbF = Ξ±βΞ³β. The Ξ³-chains bind 2,3-BPG weakly (compared to Ξ²-chains). Since 2,3-BPG stabilizes the deoxy (T) state and reduces Oβ affinity, weak binding β HbF stays in the oxy (R) state longer β higher Oβ affinity. This allows the fetus to extract Oβ from maternal blood.
Q34. The binding site of 2,3-BPG on hemoglobin is:
A) Ξ±-globin chains
B) The heme group
C) The central cavity between Ξ²-globin chains β
D) The N-terminal of Ξ±-chains
Answer: C
2,3-BPG fits precisely into the central cavity formed between the two Ξ²-globin chains in the deoxy conformation. It stabilizes the T (tense/deoxy) state β reduces Oβ affinity β right shift of Oβ dissociation curve.
Q35. The difference between HbF and HbA is:
A) HbF has Ξ΄ chains instead of Ξ² chains
B) HbF = Ξ±βΞ³β; HbA = Ξ±βΞ²β; HbF has higher Oβ affinity β
C) HbF has more heme groups
D) HbA has higher Oβ affinity than HbF
Answer: B
HbA = Ξ±βΞ²β (adult). HbF = Ξ±βΞ³β (fetal). The Ξ³-chains weakly bind 2,3-BPG β HbF has higher Oβ affinity (left-shifted curve) β essential for fetal Oβ extraction from maternal blood.
Q36. FeΒ³βΊ hemoglobin that CANNOT bind oxygen is called:
A) Carboxyhemoglobin
B) Deoxyhemoglobin
C) Methemoglobin β
D) Sulfhemoglobin
Answer: C
Methemoglobin contains FeΒ³βΊ (oxidized iron). Only FeΒ²βΊ can bind Oβ. Causes: nitrites, dapsone, benzocaine (local anesthetics), primaquine. Treatment: Methylene blue (reduces FeΒ³βΊ back to FeΒ²βΊ).
Q37. A dental patient received benzocaine and developed cyanosis. The cause is:
A) Allergic reaction causing bronchospasm
B) CO poisoning
C) Conversion of FeΒ²βΊ to FeΒ³βΊ β methemoglobin β
D) Carboxyhemoglobin formation
Answer: C
Benzocaine (local anesthetic) can oxidize FeΒ²βΊ β FeΒ³βΊ in hemoglobin β methemoglobin β cannot carry Oβ β chocolate-brown blood β cyanosis unresponsive to Oβ. Treat with IV methylene blue.
Q38. Which hemoglobin provides protection against Plasmodium falciparum malaria?
A) HbA
B) HbF
C) HbS (sickle cell trait) β
D) HbC
Answer: C
Sickle cell trait (HbAS - one normal + one sickle gene) protects against P. falciparum malaria. Infected RBCs sickle and are cleared by the spleen before the parasite completes its cycle. High prevalence of HbS in sub-Saharan Africa due to this selective advantage.
Q39. One subunit of hemoglobin contains:
A) 2 heme groups and 2 polypeptide chains
B) 1 heme group, 1 polypeptide chain, 1 FeΒ²βΊ ion β
C) 1 heme group with FeΒ³βΊ
D) 4 heme groups
Answer: B
Each hemoglobin subunit = 1 globin polypeptide chain + 1 heme group + 1 FeΒ²βΊ ion. Hemoglobin is a tetramer of 4 subunits = 4 heme groups total = can carry 4 Oβ molecules.
Q40. Ξ²-thalassemia is characterized by:
A) Absent Ξ²-globin synthesis only
B) Structurally abnormal Ξ²-globin
C) Decreased (or absent) synthesis of Ξ²-globin chains β
D) Mutation at position 6 of Ξ²-globin
Answer: C
Ξ²-thalassemia = quantitative defect in Ξ²-globin production (Ξ²βΊ = reduced; Ξ²β° = absent). Excess Ξ±-chains precipitate β hemolysis. Different from sickle cell = qualitative (structural) defect.
Q41. In Ξ²-thalassemia, what happens when the Ξ²-chain is replaced or deficient?
A) Oβ affinity decreases
B) HbS forms
C) Increased Oβ affinity due to less 2,3-BPG binding β
(if Ξ³-chains compensate = HbF)
D) Methemoglobin forms
Answer: C
When Ξ²-chains are absent, Ξ³-chains persist (HbF production β as compensation). HbF binds 2,3-BPG weakly β β Oβ affinity (left shift). Also: without Ξ²-chains, 2,3-BPG has nowhere to bind β even with Ξ±-chain tetramers (Hb Barts = Ξ³β), Oβ affinity is abnormally high.
Q42. The hemoglobin state with LOWEST Oβ affinity is:
A) R (relaxed) state
B) HbF state
C) T (tense/deoxy) state β
D) Carboxyhemoglobin
Answer: C
T (tense) state = deoxy conformation = low Oβ affinity. Stabilized by: 2,3-BPG, HβΊ (Bohr effect), COβ, high temperature.
R (relaxed) state = oxy conformation = HIGH Oβ affinity.
Q43. On the Oβ-hemoglobin dissociation curve:
A) Left shift = decreased Oβ affinity
B) Right shift = increased Oβ affinity
C) Left shift = increased Oβ affinity; Right shift = decreased Oβ affinity β
D) 2,3-BPG causes a left shift
Answer: C
Left shift (β Oβ affinity, Hb holds Oβ): β 2,3-BPG, β COβ, β pH (alkalosis), β temperature, HbF, CO poisoning.
Right shift (β Oβ affinity, Hb releases Oβ): β 2,3-BPG, β COβ, β pH (Bohr effect), β temperature, exercise.
Q44. Spectrin is:
A) A plasma clotting protein
B) A type of hemoglobin
C) The major RBC membrane cytoskeletal protein maintaining biconcave shape β
D) A protein involved in iron transport
Answer: C
Spectrin forms the submembrane cytoskeleton of RBCs, giving them their biconcave shape and the flexibility to squeeze through capillaries (2.8 Β΅m wide). Defective spectrin β hereditary spherocytosis.
Q45. General causes of INCREASED hemoglobin Oβ affinity include:
A) β 2,3-BPG, β pH, β COβ
B) β 2,3-BPG, β pH, β COβ, β temperature β
C) Fever, exercise, high altitude
D) Acidosis and hypercapnia
Answer: B
High Oβ affinity (LEFT shift) = Hb picks up Oβ easily but releases it poorly:
- β 2,3-BPG (less T-state stabilization)
- β pH / β HβΊ (alkalosis)
- β COβ (less Bohr effect)
- β Temperature
- HbF (fetal)
- CO binding (carboxyhemoglobin)
π§± PROTEINS: COLLAGEN & ELASTIN
Q46. In scurvy, which step in collagen synthesis is NOT affected?
A) Hydroxylation of proline and lysine
B) Triple helix formation
C) Glycosylation (carbohydrate addition) β
D) Cross-linking of collagen fibrils
Answer: C
Scurvy = Vitamin C deficiency β hydroxylation of proline and lysine FAILS β unstable triple helix β collagen degraded. Glycosylation occurs in the RER BEFORE hydroxylation, and does not require Vitamin C, so it proceeds normally.
Q47. Why is tryptophan NOT found in collagen?
A) Tryptophan is too large for the triple helix
B) Collagen requires glycine at every third position (Gly-X-Y) β
C) Tryptophan destabilizes disulfide bonds
D) Collagen has no aromatic amino acids
Answer: B
Collagen's triple helix = repeating (Gly-X-Y)β sequence. Every third residue MUST be glycine (smallest AA, fits in the center). Tryptophan (bulky, aromatic) cannot fit. X = usually proline; Y = usually hydroxyproline.
Q48. Which enzyme deficiency affects BOTH collagen and elastin cross-linking?
A) Prolyl hydroxylase
B) Lysyl hydroxylase
C) Lysyl oxidase β
D) Collagenase
Answer: C
Lysyl oxidase (copper-dependent) oxidatively deaminates lysine residues β forms aldehyde groups β spontaneous cross-links (desmosine in elastin, pyridinoline in collagen). Copper deficiency or Menkes disease β weak connective tissue (aortic aneurysm, skin laxity).
Q49. The molecular difference between collagen and elastin is:
A) Collagen has random coil; elastin has triple helix
B) Collagen = triple helix (tensile strength); Elastin = random coil with desmosine cross-links (elasticity) β
C) Elastin is triple-stranded; collagen is single-stranded
D) Both have the same structure but different amino acids
Answer: B
- Collagen: triple helix of 3 Ξ±-chains β resistant to stretching β tensile strength (tendons, bones, skin)
- Elastin: random coil network cross-linked by desmosine (unique to elastin) β stretches and recoils β elasticity (lungs, arteries, skin)
Q50. Elastin is rich in which amino acids?
A) Glutamate and aspartate
B) Glycine, proline, and alanine (also lysine for cross-linking) β
C) Cysteine and methionine
D) Tryptophan and phenylalanine
Answer: B
Elastin is rich in: Glycine, Proline, Alanine, Valine (nonpolar/hydrophobic) β gives the random coil flexibility. Lysine is needed for lysyl oxidase cross-linking into desmosine.
Q51. Which is the basic secreted unit of collagen that is released extracellularly?
A) Collagen fibril
B) Collagen fiber
C) Tropocollagen β
D) Procollagen
Answer: C
Collagen synthesis pathway:
Pre-procollagen β Procollagen (in RER/Golgi, has N and C propeptides) β secreted β Procollagen peptidases cleave N and C terminals β Tropocollagen β spontaneous self-assembly + lysyl oxidase cross-linking β Collagen fibrils β fibers.
Q52. If N and C terminal propeptides of procollagen are NOT cleaved, what fails to form?
A) Tropocollagen
B) Collagen fibrils β
C) Triple helix
D) Procollagen
Answer: B
The propeptides must be cleaved extracellularly by procollagen peptidases to produce tropocollagen. Without cleavage, tropocollagen cannot self-assemble β fibrils do NOT form. This occurs in Dermatosparaxis (type VIIC Ehlers-Danlos).
Q53. If secretion from the cell is blocked, which molecule fails to form?
A) Pre-procollagen
B) Procollagen β
β therefore Tropocollagen also fails
C) Triple helix
D) Desmosine cross-links
Answer: B
Tropocollagen forms EXTRACELLULARLY after procollagen is secreted and its propeptides cleaved. If secretion is blocked β procollagen stays intracellular β tropocollagen cannot form β no fibrils, no fibers.
Q54. The simplest structural unit of the collagen triple helix is:
A) Procollagen
B) Alpha chain
C) Triple helix (of 3 Ξ±-chains) β
D) Collagen fibril
Answer: C
The triple helix (3 left-handed polyproline II helices wound into a right-handed superhelix) is the fundamental structural unit. It requires Gly-X-Y repeats and hydroxyproline for stability (H-bonding).
π§ EDEMA & PLASMA PROTEINS
Q55. What causes edema in protein malnutrition (kwashiorkor)?
A) Increased capillary hydrostatic pressure
B) Lymphatic obstruction
C) Decreased plasma oncotic pressure (hypoalbuminemia) β
D) Increased capillary permeability
Answer: C
Without adequate protein intake β β albumin synthesis β β plasma oncotic pressure β capillary hydrostatic pressure exceeds oncotic pressure β fluid leaks into interstitium β edema (especially abdomen in kwashiorkor).
Q56. What causes edema in proteinuria?
A) Increased capillary pressure
B) Loss of plasma proteins into urine β β oncotic pressure β edema β
C) Kidney produces too much fluid
D) Lymphatic overload
Answer: B
Nephrotic syndrome β massive proteinuria (>3.5 g/day) β β plasma albumin β β oncotic pressure β fluid escapes capillaries β edema (periorbital, pitting edema of legs, ascites).
Q57. Analbuminemia (absent albumin) causes surprisingly:
A) Severe, life-threatening edema
B) Only mild to moderate edema β
C) No edema at all
D) Pulmonary edema
Answer: B
Despite having NO albumin, patients with analbuminemia develop only mild to moderate edema. This is because compensatory increases in other plasma proteins (globulins, lipoproteins) partially maintain oncotic pressure.
Q58. In liver cirrhosis with ascites, the primary cause of fluid accumulation is:
A) Increased lymphatic flow
B) Portal hypertension only
C) Decreased plasma oncotic pressure due to reduced albumin production β
(combined with portal hypertension)
D) Kidney failure
Answer: C
Cirrhosis β β liver function β β albumin synthesis β β plasma oncotic pressure + portal hypertension (β hydrostatic pressure in portal/splanchnic vessels) β ascites (fluid in peritoneal cavity). Both factors together are needed for full picture.
Q59. C-reactive protein (CRP) is:
A) A complement protein
B) An acute-phase protein produced by the liver in response to inflammation β
C) An immunoglobulin
D) A clotting factor
Answer: B
CRP is synthesized by the liver in response to IL-6 (and IL-1, TNF-Ξ±). It rises within hours of inflammation/infection. Used clinically as a marker of acute inflammation and cardiovascular risk.
Q60. Which cytokine primarily stimulates acute-phase protein production?
A) IL-2
B) IL-4
C) IL-6 β
D) IL-10
Answer: C
IL-6 is the major inducer of acute-phase proteins (CRP, fibrinogen, serum amyloid A, haptoglobin, Ξ±β-antitrypsin, complement). IL-1 and TNF-Ξ± also contribute but IL-6 is the primary driver of hepatic acute-phase response.
π§ SMOOTH MUSCLE
Q61. In smooth muscle, calcium binds to:
A) Troponin C
B) Calmodulin β
C) Troponin I
D) Actin directly
Answer: B
Smooth muscle lacks troponin. Instead: CaΒ²βΊ enters β binds calmodulin (4 CaΒ²βΊ per calmodulin) β CaΒ²βΊ-calmodulin complex activates myosin light-chain kinase (MLCK) β phosphorylates myosin light chains β cross-bridge cycling β contraction.
β‘ RENIN-ANGIOTENSIN-ALDOSTERONE & ANP
Q62. The renin-angiotensin system ultimately produces which hormone for BP control?
A) Vasopressin
B) Cortisol
C) Aldosterone β
D) Epinephrine
Answer: C
RAAS cascade: β BP / β NaβΊ β Kidney secretes Renin β cleaves Angiotensinogen β Angiotensin I β ACE (lung) β Angiotensin II β adrenal cortex β Aldosterone β β NaβΊ/water retention β β blood volume β β BP.
Q63. Which is the potent counter-regulator of the RAAS?
A) Vasopressin
B) Endothelin
C) Atrial Natriuretic Peptide (ANP) β
D) Renin
Answer: C
ANP is released by atrial cardiomyocytes when they are stretched (β blood volume/pressure). ANP causes: natriuresis, diuresis, vasodilation, inhibits renin and aldosterone β β BP and β blood volume. It directly opposes the RAAS.
π SUMMARY REFERENCE CARD
| Concept | Key Answer |
|---|
| Intrapulmonary pressure during inspiration | DECREASES (below atmospheric) |
| RV measurement method | Helium dilution OR body plethysmograph (NOT spirometry) |
| TLC β VC = | RV |
| Frank-Starling: β preload β | β CO |
| SV = | EDV β ESV |
| Slowest cardiac conduction | AV node |
| Capillary slowest flow because | Largest total cross-sectional area |
| Sickle cell mutation | Glu β Val at Ξ²-6 |
| HbF higher affinity because | Weak 2,3-BPG binding (Ξ³ chains) |
| 2,3-BPG binds to | Central cavity of Ξ²-chains |
| FeΒ³βΊ Hb = | Methemoglobin (cannot carry Oβ) |
| Benzocaine β cyanosis | FeΒ²βΊ β FeΒ³βΊ (methemoglobin) |
| Malaria protection | HbS (sickle trait) |
| Ξ²-thalassemia | β Ξ²-globin synthesis |
| Scurvy: unaffected step | Glycosylation |
| Lysyl oxidase deficiency | Affects BOTH collagen + elastin crosslinking |
| Smooth muscle CaΒ²βΊ binds | Calmodulin (not troponin) |
| Edema in malnutrition | β Plasma oncotic pressure |
| Analbuminemia edema | Mild to moderate only |
| IL-6 β | Acute-phase proteins (CRP) |
| Pneumotaxic center | Limits/stops inspiration |
| Type II pneumocytes | Produce surfactant |
| RAAS end product | Aldosterone |
| ANP function | Natriuresis, opposes RAAS, β BP |
63 MCQs total, fully explained. Every answer maps directly to your quiz content. Good luck! π―