Level 5 — MasteryCirculatory System

Circulatory System

75 minutes60 marksprintable — key stays hidden on paper

Subject: Biology (with cross-domain Physics, Mathematics & Computation) Time Limit: 75 minutes Total Marks: 60

Instructions: Answer all three questions. Show reasoning, derivations, and code logic. Use ...... notation for equations. Calculators permitted.


Question 1 — Hemodynamics & Cardiac Output (20 marks)

The heart's mechanical performance can be modeled with physics. Consider a resting adult.

(a) Cardiac output is CO=HR×SVCO = HR \times SV, where HRHR is heart rate and SVSV is stroke volume. Given end-diastolic volume EDV=120 mLEDV = 120\text{ mL}, end-systolic volume ESV=50 mLESV = 50\text{ mL}, and HR=72 bpmHR = 72\text{ bpm}: (i) Compute stroke volume and ejection fraction (EF = SV/EDV). (3) (ii) Compute cardiac output in L/min. (3)

(b) Blood flow obeys a hydraulic analogue of Ohm's law: Q=ΔPRQ = \dfrac{\Delta P}{R}, where ΔP\Delta P is the mean pressure difference across the systemic circuit and RR is total peripheral resistance. Given mean arterial pressure MAP=93 mmHgMAP = 93\text{ mmHg}, central venous pressure CVP=3 mmHgCVP = 3\text{ mmHg}, and the COCO from part (a) as QQ, derive RR in mmHg·min/L. (4)

(c) Poiseuille's law states R=8ηLπr4R = \dfrac{8\eta L}{\pi r^4}. Relate the microscopic vessel geometry to part (b): if an arteriole's radius decreases by 20% (vasoconstriction) with all else constant, by what factor does its resistance change? Comment on why this makes arterioles the primary regulators of blood pressure. (4)

(d) Explain physiologically how the SA node, autonomic nervous system, and baroreceptors act to restore MAPMAP after a sudden drop, linking your answer to the variables HRHR, SVSV, and RR. (6)


Question 2 — Blood Groups: Probability & Genetics (20 marks)

(a) The ABO gene has three alleles (IAI^A, IBI^B, ii). State the genotype(s) for each of the four phenotypes and explain the dominance relationships involved. (4)

(b) In a population, allele frequencies are IA=0.30I^A = 0.30, IB=0.10I^B = 0.10, i=0.60i = 0.60. Assuming Hardy–Weinberg equilibrium: (i) Derive the expected frequency of phenotype O. (2) (ii) Derive the expected frequency of phenotype A. (3) (iii) Derive the expected frequency of phenotype AB. (2)

(c) A man with blood type AB (Rh⁺, genotype for Rh is heterozygous RrRr) marries a woman of type O (Rh⁻, rrrr). (i) List all possible ABO blood types of their children with probabilities. (3) (ii) Give the probability a child is Rh⁻. (2)

(d) Explain hemolytic disease of the newborn (HDN): why it typically affects the second Rh⁺ child of an Rh⁻ mother, and how sensitization occurs. Connect this to the antibody/antigen basis of the Rh system. (4)


Question 3 — Modeling Blood Flow & Clotting (Coding + Biology) (20 marks)

(a) Trace the complete path of a single oxygen molecule from a pulmonary alveolus to a systemic capillary in the big toe and back to the alveolus, naming every chamber, valve, and major vessel type in order. (6)

(b) Write pseudocode (or Python) for a function oxygen_carried(hb_conc, sat) that returns the oxygen bound to hemoglobin in mL O₂ per 100 mL blood, given the physiological constant 1.34 mL O21.34\text{ mL O}_2 carried per gram of fully saturated hemoglobin. Then compute the value for hb_conc=15 g/dLhb\_conc = 15\text{ g/dL} at sat=0.97sat = 0.97. (5)

(c) The clotting cascade can be modeled as a sequential activation chain. Given the simplified pathway Prothrombinthrombin activatorThrombinFibrinogenFibrin\text{Prothrombin} \xrightarrow{\text{thrombin activator}} \text{Thrombin} \xrightarrow{} \text{Fibrinogen} \to \text{Fibrin} identify which two enzymes/factors the pathway needs Ca2+Ca^{2+} and vitamin K for, and write a short logical rule (if/then) representing how a positive-feedback loop accelerates clot formation. (4)

(d) Compare open and closed circulatory systems across three axes: pressure achievable, efficiency of oxygen delivery, and control over flow distribution. Argue from a physics/engineering standpoint why a closed system supports higher metabolic rates. (5)


End of paper

Answer keyMark scheme & solutions

Question 1

(a)(i) SV=EDVESV=12050=70 mLSV = EDV - ESV = 120 - 50 = 70\text{ mL} (1). EF=70/120=0.583=58.3%EF = 70/120 = 0.583 = 58.3\% (2 — formula + value).

(a)(ii) CO=HR×SV=72×70=5040 mL/min=5.04 L/minCO = HR \times SV = 72 \times 70 = 5040\text{ mL/min} = 5.04\text{ L/min} (3 — conversion mL→L required).

(b) ΔP=MAPCVP=933=90 mmHg\Delta P = MAP - CVP = 93 - 3 = 90\text{ mmHg} (1). Using R=ΔP/Q=90/5.04=17.86 mmHg⋅min/LR = \Delta P / Q = 90 / 5.04 = 17.86\text{ mmHg·min/L} (3 — correct rearrangement + numeric). Why: the driving pressure across the whole systemic bed divided by flow gives total peripheral resistance, the vascular analogue of electrical resistance.

(c) R1/r4R \propto 1/r^4. New radius =0.8r= 0.8r, so factor =1/(0.8)4=1/0.4096=2.44= 1/(0.8)^4 = 1/0.4096 = 2.44 (3). Resistance rises ~2.44×. (1) Because Rr4R\propto r^{-4}, tiny radius changes produce large resistance changes; arterioles have thick smooth muscle allowing precise radius control → dominant site of pressure/flow regulation.

(d) (6 — 2 each)

  • Baroreceptors (carotid sinus, aortic arch) detect the pressure fall (reduced stretch) and reduce inhibitory firing to the medulla.
  • Autonomic response: sympathetic activation increases SA-node firing → ↑HRHR; increases contractility → ↑SVSV; causes vasoconstriction → ↑RR. Parasympathetic (vagal) tone withdraws.
  • Result: since MAPCO×R=(HR×SV)×RMAP \approx CO \times R = (HR \times SV) \times R, raising all three restores MAPMAP. SA node is the pacemaker whose intrinsic rate is modulated by these autonomic inputs.

Question 2

(a) (4) Type A: IAIAI^AI^A or IAiI^Ai; Type B: IBIBI^BI^B or IBiI^Bi; Type AB: IAIBI^AI^B; Type O: iiii. IAI^A and IBI^B are codominant (both expressed in AB); both are dominant over recessive ii.

(b) Let p=IA=0.30p=I^A=0.30, q=IB=0.10q=I^B=0.10, r=i=0.60r=i=0.60.

  • (i) O =r2=0.602=0.36= r^2 = 0.60^2 = 0.36 (2)
  • (ii) A =p2+2pr=0.09+2(0.30)(0.60)=0.09+0.36=0.45= p^2 + 2pr = 0.09 + 2(0.30)(0.60) = 0.09 + 0.36 = 0.45 (3)
  • (iii) AB =2pq=2(0.30)(0.10)=0.06= 2pq = 2(0.30)(0.10) = 0.06 (2)

(c)(i) AB (IAIBI^AI^B) × O (iiii): children get IAI^A or IBI^B from father, ii from mother → IAiI^Ai (type A) or IBiI^Bi (type B), each 50%. No O or AB children. (3)

(c)(ii) Rr × rr → 1/2 RrRr (Rh⁺), 1/2 rrrr (Rh⁻). P(Rh⁻) =0.5= 0.5 (2).

(d) (4) An Rh⁻ mother lacks the D antigen and makes anti-D antibodies only after exposure to Rh⁺ blood (sensitization), usually at delivery of a first Rh⁺ child when fetal blood mixes with maternal blood. The first child is normally unaffected because sensitization occurs at/after its birth. In a second Rh⁺ pregnancy, maternal IgG anti-D crosses the placenta and destroys fetal RBCs → hemolytic disease of the newborn. (Rh anti-D are IgG that cross the placenta, unlike large IgM ABO antibodies.)

Question 3

(a) (6 — deduct for order/omission) Alveolus → pulmonary capillary → pulmonary vein → left atriumbicuspid/mitral valveleft ventricleaortic (semilunar) valve → aorta → systemic arteries → arterioles → systemic capillary (big toe) → venules → veins → superior/inferior vena cava → right atriumtricuspid valveright ventriclepulmonary (semilunar) valve → pulmonary artery → pulmonary capillary → alveolus.

(b) (5)

def oxygen_carried(hb_conc, sat):
    return 1.34 * hb_conc * sat   # mL O2 per 100 mL blood
# oxygen_carried(15, 0.97) = 1.34*15*0.97 = 19.497 mL/dL

Value =1.34×15×0.97=19.49719.5 mL O2/100 mL= 1.34 \times 15 \times 0.97 = 19.497 \approx 19.5\text{ mL O}_2/100\text{ mL}. (3 code + 2 value)

(c) (4) Vitamin K is required for synthesis of prothrombin (factor II) (and factors VII, IX, X); Ca2+Ca^{2+} (factor IV) is required for thrombin activator/prothrombin activator complex formation. (2) Positive-feedback rule: (2)

IF thrombin_present:
    THEN thrombin activates more prothrombin_activator
    → produces more thrombin  (accelerating loop until fibrin mesh forms)

(d) (5)

  • Pressure: closed systems (blood in vessels) sustain high pressures; open systems (hemolymph in sinuses) are low-pressure.
  • Efficiency: closed → directed, rapid, high O₂ delivery; open → slow diffusive exchange in body cavity.
  • Control: closed systems regulate distribution via vasoconstriction/dilation of specific vessels; open systems cannot selectively route flow.
  • Physics argument: confining fluid in tubes lets the heart generate high ΔP\Delta P, driving high flow Q=ΔP/RQ=\Delta P/R to active tissues on demand — supporting the high, variable O₂ demand of endotherms/active animals. Open systems dissipate pressure in cavities, limiting flow rate and metabolic ceiling.
[
  {"claim":"Ejection fraction = 70/120 ≈ 0.5833","code":"EF=(120-50)/120; result=abs(float(EF)-0.58333)<1e-3"},
  {"claim":"Cardiac output = 72*70/1000 = 5.04 L/min","code":"CO=72*70/1000; result=abs(float(CO)-5.04)<1e-9"},
  {"claim":"Peripheral resistance = 90/5.04 ≈ 17.857","code":"R=(93-3)/(72*70/1000); result=abs(float(R)-17.857)<1e-2"},
  {"claim":"Resistance factor from 20% radius drop = 1/0.8**4 ≈ 2.4414","code":"f=1/Rational(8,10)**4; result=abs(float(f)-2.44140625)<1e-6"},
  {"claim":"Phenotype A freq = p^2+2pr = 0.45","code":"p,q,r=Rational(3,10),Rational(1,10),Rational(6,10); A=p*p+2*p*r; result=A==Rational(45,100)"},
  {"claim":"Phenotype O freq = r^2 = 0.36 and AB = 2pq = 0.06","code":"p,q,r=Rational(3,10),Rational(1,10),Rational(6,10); result=(r*r==Rational(36,100)) and (2*p*q==Rational(6,100))"},
  {"claim":"Oxygen carried = 1.34*15*0.97 = 19.497","code":"O2=1.34*15*0.97; result=abs(float(O2)-19.497)<1e-6"}
]