Here are all the quiz questions from BF-3, formulated as MCQs with answers and explanations. This is based on the topics listed across all your messages:
BF-3 Written Quiz - Practice MCQs
HEMOGLOBIN & OXYGEN TRANSPORT
Q1. In sickle cell anemia, which amino acid substitution occurs in the beta-globin chain?
- A) Valine → Glutamate at position 6
- B) Glutamate → Valine at position 6
- C) Glutamate → Lysine at position 6
- D) Proline → Valine at position 6
✅ Answer: B - Glutamate (negatively charged, hydrophilic) is replaced by Valine (nonpolar, hydrophobic) at the 6th position of the beta chain. This causes HbS to polymerize under low O2 conditions, deforming the RBC into a sickle shape.
Q2. In sickle cell anemia, the RBC changes shape under deoxygenated conditions due to:
- A) Increased 2,3-BPG binding
- B) Polymerization of deoxygenated HbS into long fibers
- C) Oxidation of heme iron to Fe³⁺
- D) Loss of spectrin from the cytoskeleton
✅ Answer: B - Deoxygenated HbS molecules aggregate and form insoluble polymer fibers (tactoids) that distort the RBC membrane into the characteristic sickle shape.
Q3. Where does 2,3-BPG bind on the hemoglobin molecule?
- A) Alpha-alpha interface
- B) Central cavity between the two beta chains
- C) Heme iron of alpha subunit
- D) F-helix of all four subunits
✅ Answer: B - 2,3-BPG binds in the central cavity formed between the two beta chains in the T (tense/deoxy) state. It stabilizes the deoxy form and decreases O2 affinity (shifts curve right).
Q4. In a patient with a variant hemoglobin where the His residue in the beta chain is replaced by Methionine (affecting the 2,3-BPG binding site), what would you expect?
- A) Increased Bohr effect
- B) Decreased O2 affinity due to better BPG binding
- C) Loss of 2,3-BPG binding → increased O2 affinity → right shift of O2 dissociation curve
- D) Loss of 2,3-BPG binding → increased O2 affinity → left shift of O2 dissociation curve
✅ Answer: D - If the key His residues (His-1β and His-143β) that anchor 2,3-BPG are lost, BPG cannot bind. Without BPG stabilizing the T-state, hemoglobin stays in the R (relaxed/oxy) state with higher O2 affinity → left shift of the dissociation curve (harder to unload O2 to tissues).
Q5. Fetal hemoglobin (HbF) has a higher affinity for O2 than adult HbA because:
- A) HbF has alpha chains instead of beta chains
- B) HbF has gamma chains instead of beta chains; gamma chains bind 2,3-BPG less avidly
- C) HbF contains more heme groups
- D) HbF has a different iron oxidation state
✅ Answer: B - HbF has two gamma (γ) chains in place of beta chains. Gamma chains bind 2,3-BPG with much lower affinity, so HbF stays in the R-state more readily → higher O2 affinity → left-shifted curve. This allows the fetus to extract O2 from maternal HbA.
Q6. Which of the following has the LOWEST oxygen affinity?
- A) Oxygenated hemoglobin (R-state)
- B) Fetal hemoglobin (HbF)
- C) Tense state (T-state / deoxy-Hb)
- D) Relaxed state hemoglobin
✅ Answer: C - The T (Tense) state = deoxy-Hb = LOW O2 affinity. The R (Relaxed) state = oxy-Hb = HIGH O2 affinity. HbF is high affinity. So T-state has the lowest affinity.
Q7. The O2-hemoglobin dissociation curve shifts to the RIGHT (decreased affinity, easier O2 unloading) when:
- A) pH increases, PCO2 decreases, temperature decreases
- B) pH decreases, PCO2 increases, temperature increases, 2,3-BPG increases
- C) 2,3-BPG decreases, altitude increases
- D) HbF replaces HbA
✅ Answer: B - Right shift = CADET (CO2↑, Acid/H+↑, 2,3-DPG↑, Exercise/Temperature↑). This favors O2 release to active tissues.
Q8. Why does HbF have higher O2 affinity than HbA?
- A) HbF has δ chains instead of α chains
- B) Gamma chains of HbF have Ser at position 143 instead of His, reducing 2,3-BPG binding
- C) HbF has 3 heme groups instead of 4
- D) HbF is always in the R-state regardless of pH
✅ Answer: B - The key structural difference: β-143 His (which is critical for BPG binding in HbA) is replaced by Ser in the γ-chain of HbF. This weakens BPG binding → HbF remains high-affinity.
PO2 AND DISSOLVED OXYGEN
Q9. How are PO2 and dissolved oxygen in plasma related?
- A) PO2 is inversely proportional to dissolved O2
- B) Dissolved O2 in plasma is directly proportional to PO2 (Henry's Law)
- C) PO2 measures only oxygen bound to hemoglobin
- D) Dissolved O2 is independent of PO2
✅ Answer: B - By Henry's Law, the amount of gas dissolved in a liquid is proportional to its partial pressure. Dissolved O2 (mL/dL) = 0.003 × PO2. Most O2 is carried bound to Hb, but dissolved O2 is what creates the partial pressure gradient driving diffusion.
ANEMIA & RBC STRUCTURE
Q10. Microcytic, hypochromic anemia is most commonly caused by:
- A) Vitamin B12 deficiency
- B) Iron deficiency
- C) Folate deficiency
- D) Aplastic anemia
✅ Answer: B - Iron deficiency → insufficient heme synthesis → small (microcytic), pale (hypochromic) RBCs with low MCHC. The classic case: young woman with heavy menstrual bleeding, fatigue, low serum ferritin.
Q11. The biconcave disc shape of the RBC is important for gas exchange because:
- A) It reduces cell volume and prevents hemolysis
- B) It maximizes surface area-to-volume ratio, reducing diffusion distance for gases
- C) It allows RBCs to carry more hemoglobin
- D) It prevents RBC aggregation in capillaries
✅ Answer: B - The biconcave shape gives RBCs a high surface area-to-volume ratio (~140 µm² surface for ~90 fL volume). This minimizes diffusion distance to the center of the cell, allowing rapid O2 and CO2 exchange.
Q12. In beta-thalassemia, what is the primary defect?
- A) Structural abnormality of the beta-globin chain
- B) Reduced or absent synthesis of beta-globin chains
- C) Increased synthesis of alpha-globin chains
- D) Defective heme synthesis
✅ Answer: B - β-thalassemia results from mutations in the beta-globin gene causing reduced (β+) or absent (β0) synthesis of beta chains. This leads to excess alpha chains that precipitate, causing hemolysis and ineffective erythropoiesis.
Q13. Ankyrin and spectrin are important in RBCs because they:
- A) Carry oxygen from lung to tissues
- B) Form the cytoskeletal meshwork that maintains the biconcave shape and deformability
- C) Catalyze the conversion of CO2 to bicarbonate
- D) Act as ion channels for Na+/K+ exchange
✅ Answer: B - Spectrin (α and β) forms a flexible meshwork on the inner surface of the RBC membrane. Ankyrin anchors spectrin to Band 3 protein (anion exchanger). Together they maintain RBC shape, deformability, and membrane integrity. Defects cause hereditary spherocytosis.
Q14. Which type of malaria provides partial protection due to the presence of HbS (sickle cell trait)?
- A) Plasmodium vivax
- B) Plasmodium falciparum
- C) Plasmodium malariae
- D) Plasmodium ovale
✅ Answer: B - HbS (sickle cell trait, HbAS) provides protection against severe P. falciparum malaria. Inside RBCs with HbS, the parasite cannot thrive well under low O2 conditions, and infected cells are removed by the spleen more efficiently.
Q15. What is the residual volume (RV)?
- A) Air remaining after a normal expiration
- B) Air remaining in the lungs after a maximum (forced) expiration
- C) Total air inspired in one normal breath
- D) Air that can be forcibly expired after normal expiration
✅ Answer: B - Residual Volume (RV) ≈ 1200 mL. It is the air remaining in the lungs after maximum expiration. It cannot be measured by spirometry (because you can't expire it). It keeps alveoli open and prevents collapse.
CARDIAC PHYSIOLOGY
Q16. Stroke Volume (SV) is calculated as:
- A) Heart Rate × Cardiac Output
- B) End-Diastolic Volume (EDV) minus End-Systolic Volume (ESV)
- C) End-Systolic Volume minus End-Diastolic Volume
- D) Cardiac Output divided by Heart Rate
✅ Answer: B - SV = EDV − ESV. Normal values: EDV ≈ 120 mL, ESV ≈ 50 mL, SV ≈ 70 mL. This is the volume ejected per beat.
Q17. The average stroke volume in an adult at rest is approximately:
- A) 30 mL
- B) 50 mL
- C) 70 mL
- D) 120 mL
✅ Answer: C - Normal resting SV ≈ 70 mL. CO = SV × HR = 70 × 72 ≈ 5 L/min.
Q18. During inspiration, stroke volume:
- A) Decreases because intrathoracic pressure increases
- B) Increases because venous return to the right heart increases (due to negative intrathoracic pressure)
- C) Remains unchanged
- D) Decreases due to increased afterload
✅ Answer: B - During inspiration, intrathoracic pressure drops → great veins expand → venous return to right heart increases → by Frank-Starling mechanism, right ventricular SV increases. (There is slight lag to left side, but overall SV increases with inspiration.)
Q19. To increase End-Systolic Volume (ESV), which of the following would work?
- A) Increase contractility (positive inotrope)
- B) Decrease heart rate
- C) Increase afterload (e.g., increase mean arterial pressure)
- D) Increase preload
✅ Answer: C - ESV is what's LEFT after systole. If afterload (arterial pressure) increases, the ventricle cannot eject as much → more blood remains → ESV increases.
Q20. The Incisura (dicrotic notch) on the arterial pressure waveform represents:
- A) Opening of the mitral valve
- B) Closure of the aortic valve at the end of systole
- C) Peak of ventricular systole
- D) Opening of the aortic valve
✅ Answer: B - The incisura/dicrotic notch is a brief pressure dip followed by a small rise on the aortic pressure tracing. It occurs when the aortic valve closes at the end of systole, causing a brief backflow that closes the valve and creates this characteristic notch.
Q21. When the ventricles relax (after systole), this phase is called:
- A) Systole
- B) Isovolumetric contraction
- C) Diastole
- D) Ejection phase
✅ Answer: C - Diastole is the relaxation phase of the cardiac cycle when ventricles fill with blood. It includes isovolumetric relaxation and then ventricular filling.
Q22. The inotropic effect of digitalis:
- A) Decreases heart rate by blocking beta receptors
- B) Increases force of cardiac muscle contraction by inhibiting Na+/K+-ATPase → raising intracellular Ca²⁺
- C) Dilates coronary arteries
- D) Increases conductivity at the AV node
✅ Answer: B - Digitalis inhibits the Na+/K+-ATPase pump → intracellular Na+ accumulates → the Na+/Ca²⁺ exchanger cannot remove Ca²⁺ → intracellular Ca²⁺ rises → stronger contraction (positive inotropy).
Q23. Compliance is highest in which vessel type?
- A) Arterioles
- B) Capillaries
- C) Veins
- D) Arteries
✅ Answer: C - Veins have the highest compliance (most distensible per unit pressure change). They act as capacitance vessels, holding ~65-70% of total blood volume. Arteries are stiffer (less compliant) to withstand high pressures.
Q24. Capillaries have a large cross-sectional area because:
- A) They need to pump blood at high pressure
- B) The total cross-sectional area of all capillaries combined is huge, slowing blood flow velocity for efficient exchange
- C) They contain smooth muscle for vasodilation
- D) They have the highest compliance
✅ Answer: B - By the continuity equation, velocity = flow/area. The combined cross-sectional area of all capillaries is ~2500-4000 cm². This massively reduces velocity to ~0.3 mm/sec, allowing time for gas, nutrient, and waste exchange.
Q25. Which is NOT used in a pulmonary function test?
- A) Spirometer
- B) Plethysmograph
- C) Sphygmomanometer
- D) Peak flow meter
✅ Answer: C - A sphygmomanometer measures blood pressure, not lung function. Spirometers, body plethysmographs, and peak flow meters are all used in pulmonary function testing.
Q26. Which statement about intrapulmonary (alveolar) pressure during inspiration is NOT TRUE?
- A) It falls below atmospheric pressure
- B) It becomes negative (sub-atmospheric) to allow air to flow in
- C) It increases above atmospheric pressure
- D) It equalizes with atmospheric pressure at end-inspiration
✅ Answer: C - During inspiration, the diaphragm contracts, chest volume increases, alveolar pressure drops BELOW atmospheric (to about -1 to -3 mmHg), creating a pressure gradient that drives air inward. It does NOT increase during normal inspiration.
Q27. Alveolar type II pneumocytes produce:
- A) Mucus
- B) Pulmonary surfactant (dipalmitoylphosphatidylcholine)
- C) Immunoglobulins
- D) Collagen fibers
✅ Answer: B - Type II pneumocytes produce surfactant, which reduces alveolar surface tension, prevents alveolar collapse (atelectasis), and increases lung compliance. Deficiency causes Respiratory Distress Syndrome (RDS) in premature infants.
Q28. If the arterial pressure drops below 60 mmHg, what happens?
- A) Baroreceptors are activated and inhibit the vasomotor center
- B) Cerebral perfusion is severely compromised; below this level, autoregulation fails
- C) The kidneys increase urine output
- D) Pulmonary pressure increases proportionally
✅ Answer: B - Mean arterial pressure (MAP) below ~60 mmHg means cerebral autoregulation fails → inadequate cerebral perfusion → loss of consciousness and organ damage. This is the critical threshold below which vital organ perfusion cannot be maintained.
Q29. The pneumotaxic center (in the pons) functions to:
- A) Stimulate inspiration by activating the diaphragm
- B) Inhibit the apneustic center and switch off inspiration, limiting inspiratory time
- C) Detect CO2 levels in the blood
- D) Control expiratory muscle activity
✅ Answer: B - The pneumotaxic center (pontine respiratory group) sends inhibitory signals to the inspiratory center, terminating each inspiration and regulating respiratory rate and tidal volume.
Q30. Stroke volume can be measured by which device?
- A) Spirometer
- B) Echocardiography (or cardiac catheterization)
- C) Sphygmomanometer
- D) Pulse oximeter
✅ Answer: B - Echocardiography measures EDV and ESV, from which SV = EDV - ESV. Cardiac catheterization and thermodilution (Swan-Ganz) also measure SV clinically.
PROTEINS: COLLAGEN, ELASTIN, PLASMA PROTEINS
Q31. In scurvy (Vitamin C deficiency), which step in collagen synthesis is NOT affected?
- A) Transcription of the pro-alpha chain genes
- B) Hydroxylation of proline and lysine residues
- C) Assembly of the triple helix
- D) Secretion from the cell
✅ Answer: A - Vitamin C (ascorbate) is a cofactor for prolyl hydroxylase and lysyl hydroxylase. In scurvy, hydroxylation of Pro and Lys is impaired → unstable triple helix → poor collagen cross-linking. Transcription (gene expression) is NOT dependent on Vitamin C and proceeds normally.
Q32. Which is the basic secreted unit of collagen that goes extracellular and polymerizes?
- A) Procollagen
- B) Tropocollagen
- C) Alpha chain
- D) Collagen fibril
✅ Answer: B - Tropocollagen (after propeptide cleavage from procollagen) is the basic extracellular unit (~300 nm long, ~1.5 nm wide). Tropocollagen molecules self-assemble in a staggered fashion and are cross-linked to form collagen fibrils.
Q33. The key molecular difference between elastin and collagen is:
- A) Elastin contains hydroxyproline; collagen does not
- B) Collagen has a triple-helix structure; elastin is a random coil network cross-linked by desmosine, giving it elastic recoil
- C) Elastin has a quaternary structure; collagen does not
- D) Collagen is globular; elastin is fibrous
✅ Answer: B - Collagen: triple helix, Gly-X-Y repeat, rigid, tensile strength. Elastin: random coil, cross-linked by desmosine/isodesmosine (unique to elastin), highly elastic, returns to original shape after stretch. Collagen resists stretch; elastin enables stretch-and-recoil.
Q34. Elastin is rich in which amino acids?
- A) Glycine and Proline (like collagen)
- B) Glycine, Alanine, Valine, and Proline (non-polar, hydrophobic)
- C) Lysine and Hydroxylysine
- D) Arginine and Histidine
✅ Answer: B - Elastin is rich in non-polar amino acids: Glycine (~33%), Alanine, Valine, and Proline. It also contains Lysine (for desmosine cross-link formation). Note: Unlike collagen, elastin does NOT have the Gly-X-Y repeat and has very little hydroxyproline.
Q35. Globular proteins differ from fibrous proteins in that globular proteins:
- A) Are water insoluble
- B) Are folded into compact, roughly spherical shapes and are generally water-soluble
- C) Consist only of alpha-helices
- D) Cannot have enzymatic function
✅ Answer: B - Globular proteins (hemoglobin, albumin, enzymes, antibodies) fold into compact 3D structures, are usually water-soluble, and have diverse functions. Fibrous proteins (collagen, elastin, keratin, actin) are elongated, often insoluble, and serve structural roles.
Q36. What is C-Reactive Protein (CRP)?
- A) A clotting factor produced in platelets
- B) An acute-phase protein produced by the liver in response to inflammation (IL-6 stimulus), used as a marker of infection/inflammation
- C) A complement protein
- D) An immunoglobulin produced by B cells
✅ Answer: B - CRP is a pentraxin acute-phase reactant synthesized by the liver, triggered mainly by IL-6 (and IL-1, TNF). It binds phosphocholine on damaged cells and microbes, activating complement. Elevated in infection, inflammation, and tissue damage. Used clinically as an inflammatory marker.
Q37. Which cytokines stimulate the liver to produce acute-phase proteins (like CRP, fibrinogen)?
- A) IL-2 and IL-3
- B) IL-1, IL-6, and TNF-α
- C) IL-4 and IL-5
- D) IFN-γ only
✅ Answer: B - IL-1, IL-6, and TNF-α are the primary stimulators of the acute-phase response. IL-6 is the most potent inducer of CRP and other acute-phase proteins from hepatocytes.
EDEMA & FLUID BALANCE
Q38. In protein malnutrition (kwashiorkor), edema develops because:
- A) Increased ADH causes water retention
- B) Low plasma albumin → decreased plasma oncotic (colloid osmotic) pressure → fluid moves into interstitial space
- C) Increased lymphatic obstruction
- D) Increased capillary hydrostatic pressure
✅ Answer: B - Albumin is the main protein maintaining plasma oncotic pressure (~25 mmHg). In protein malnutrition, albumin synthesis falls → low oncotic pressure → fluid is not reabsorbed at the venous end of capillaries → pitting edema.
Q39. A patient with liver cirrhosis develops ascites (fluid in abdomen). The primary mechanism is:
- A) Increased lymphatic drainage
- B) Portal hypertension (increased hydrostatic pressure in portal/hepatic capillaries) + low albumin (low oncotic pressure) → fluid accumulation in peritoneal cavity
- C) Decreased aldosterone
- D) Increased ANP secretion
✅ Answer: B - In cirrhosis: (1) Portal hypertension → increased hydrostatic pressure → fluid pushed out of splanchnic capillaries. (2) Impaired albumin synthesis → low oncotic pressure → fluid not retained. (3) Secondary hyperaldosteronism also contributes Na+/water retention.
Q40. Analbuminemia (congenital absence of albumin) typically causes:
- A) Severe fatal edema at birth
- B) Mild to moderate edema, because other compensatory oncotic proteins (globulins) partially compensate
- C) No edema because the lymphatics compensate fully
- D) Ascites only, no peripheral edema
✅ Answer: B - Interestingly, patients with analbuminemia have only mild edema. The liver upregulates other plasma proteins (globulins, α2-macroglobulin), and the lymphatic system compensates. This shows the body has multiple compensatory mechanisms.
HIGH ALTITUDE & MLCK
Q41. At high altitude, which adaptation helps increase O2 delivery to tissues?
- A) Decreased 2,3-BPG production
- B) Increased 2,3-BPG → right-shifted O2 curve → better O2 unloading; also increased erythropoiesis via EPO
- C) Increased HbF production
- D) Decreased respiratory rate
✅ Answer: B - At altitude: PO2 falls → hypoxia → EPO release → more RBCs. Also, 2,3-BPG increases to shift the O2 curve right, aiding O2 unloading to tissues. Chronic adaptations include increased capillary density and mitochondrial efficiency.
Q42. MLCK (Myosin Light Chain Kinase) pathway in smooth muscle contraction:
- A) Ca²⁺ binds troponin C → conformational change → contraction
- B) Ca²⁺ binds calmodulin → Ca²⁺-calmodulin activates MLCK → MLCK phosphorylates myosin light chain → cross-bridge cycling → contraction
- C) Ca²⁺ directly activates myosin ATPase
- D) Ca²⁺ activates phospholipase C to generate IP3
✅ Answer: B - In smooth muscle (unlike skeletal muscle, which uses troponin): Ca²⁺ enters → binds calmodulin → Ca²⁺-calmodulin complex activates MLCK → MLCK phosphorylates the 20-kDa myosin light chain → activates myosin ATPase → actin-myosin cross-bridge cycling → contraction.
Q43. Total Lung Capacity (TLC) minus Vital Capacity (VC) equals:
- A) Tidal Volume
- B) Inspiratory Reserve Volume
- C) Residual Volume (RV)
- D) Functional Residual Capacity
✅ Answer: C - TLC = VC + RV. Therefore TLC - VC = RV. RV ≈ 1200 mL (the air you cannot expire even with maximum effort).
Q44. Adult hemoglobin (HbA) structure is:
- A) α₂β₂ with 2 heme groups
- B) α₂β₂ with 4 heme groups (one per subunit)
- C) α₂γ₂ with 4 heme groups
- D) α₄ with 4 heme groups
✅ Answer: B - HbA = α₂β₂: two alpha chains + two beta chains. Each subunit contains ONE heme group (iron-porphyrin). Total = 4 heme groups per Hb molecule → can carry 4 O2 molecules. HbF = α₂γ₂; HbA2 = α₂δ₂.
Q45. Each heme group (one subunit of hemoglobin) contains:
- A) Two iron atoms and a globin chain
- B) One iron atom (Fe²⁺) at the center of a protoporphyrin IX ring
- C) One iron atom (Fe³⁺) at the center of a protoporphyrin ring
- D) A magnesium atom at the center of a porphyrin ring
✅ Answer: B - Heme = protoporphyrin IX + Fe²⁺ (ferrous). Each of the 4 subunits has one heme. Fe²⁺ is what binds O2. If Fe²⁺ is oxidized to Fe³⁺, you get methemoglobin, which cannot bind O2.
SUMMARY TABLE
| Topic | Key Fact |
|---|
| Sickle cell | Glu→Val at β6; HbS polymerizes when deoxygenated |
| β-Thalassemia | Reduced/absent beta chain synthesis |
| 2,3-BPG | Binds central cavity of beta chains; stabilizes T-state; right shifts curve |
| HbF vs HbA | HbF has γ chains; less BPG binding; higher O2 affinity; left shift |
| Lowest O2 affinity | T-state (deoxy-Hb) |
| Microcytic hypochromic | Iron deficiency anemia |
| Biconcave RBC | Max surface area:volume ratio → fast gas exchange |
| Spectrin/Ankyrin | RBC cytoskeleton; defect → hereditary spherocytosis |
| Collagen synthesis | Vit C needed for Pro/Lys hydroxylation; transcription unaffected |
| Elastin | Gly, Ala, Val, Pro; desmosine cross-links; elastic recoil |
| Globular protein | Compact, water-soluble, folded |
| CRP | Acute phase protein; liver; IL-1/IL-6/TNF-α stimulus |
| Edema in malnutrition | Low albumin → low oncotic pressure |
| Cirrhosis ascites | Portal HTN + low albumin |
| SV | EDV - ESV ≈ 70 mL |
| Diastole | Ventricular relaxation phase |
| Incisura notch | Aortic valve closure |
| Compliance highest | Veins (capacitance vessels) |
| Surfactant | Type II pneumocytes; prevents alveolar collapse |
| Residual volume | Air after max expiration; not measurable by spirometry |
| MLCK | Ca²⁺ → calmodulin → MLCK → phosphorylates myosin LC → smooth muscle contraction |
| Digitalis | Inhibits Na/K-ATPase → ↑Ca²⁺ → positive inotropy |
| HbS and malaria | Protection against P. falciparum |
| High altitude | ↑2,3-BPG, ↑EPO, ↑RBCs, right-shifted curve |
Good luck on your exam! The trickiest ones to watch for are: the 2,3-BPG/His→Met substitution question (left shift, not right), the analbuminemia question (mild not severe edema), and the collagen question (transcription is NOT affected in scurvy - only hydroxylation is).