Level 3 — ProductionCirculatory System

Circulatory System

45 minutes60 marksprintable — key stays hidden on paper

Level 3 Paper: Production (From-Scratch Explanations & Derivations)

Time limit: 45 minutes Total marks: 60

Instructions: Answer all questions. This paper asks you to produce explanations, trace pathways, and reason from first principles — not just recall definitions. Diagrams may be described in words where helpful. Show all reasoning steps.


Question 1 — Trace the full double circulation (12 marks)

Starting from deoxygenated blood entering the right atrium, trace the complete pathway of a red blood cell through both the pulmonary and systemic circuits until it returns to the same starting point.

(a) List every chamber, valve, and major vessel in correct sequence. (8 marks) (b) At which two points does gas exchange occur, and state the direction of O2O_2 and CO2CO_2 movement at each. (4 marks)


Question 2 — Derive the cardiac cycle from the conduction system (12 marks)

(a) Explain, from scratch, how a single heartbeat is initiated and coordinated. Begin at the SA node and end with ventricular relaxation, naming each structure and the delay that occurs. (7 marks) (b) A patient has a heart rate of 75 beats per minute. Calculate the duration of one cardiac cycle in seconds, then, using the standard ratio (atrial systole : ventricular systole : diastole = 0.1 s : 0.3 s : 0.4 s at 75 bpm), state what fraction of each cycle the heart spends in diastole. (5 marks)


Question 3 — Explain-out-loud: Hemoglobin & oxygen transport (10 marks)

Imagine explaining to a peer who has never heard of hemoglobin.

(a) Describe the structure of hemoglobin and how it binds oxygen. (4 marks) (b) Explain cooperative binding and why the oxygen dissociation curve is S-shaped (sigmoidal) rather than a straight line. (4 marks) (c) State one factor that shifts the curve to the right and explain its physiological benefit. (2 marks)


Question 4 — Blood clotting cascade from memory (10 marks)

(a) Produce the sequence of events in blood clotting from the moment a vessel is damaged to the formation of a stable clot. Name the key components: platelets, prothrombin, thrombin, fibrinogen, fibrin, and clotting factors. (7 marks) (b) Explain why Ca2+Ca^{2+} and vitamin K are essential to this process. (3 marks)


Question 5 — ABO/Rh blood group reasoning (10 marks)

(a) Construct a table showing, for each ABO blood group (A, B, AB, O), the antigens on the RBC surface and the antibodies present in the plasma. (4 marks) (b) Explain from first principles why group O is the "universal donor" and group AB is the "universal recipient." (3 marks) (c) An Rh⁻ mother carries an Rh⁺ fetus. Explain what may happen during a second Rh⁺ pregnancy and why. (3 marks)


Question 6 — Compare and justify (6 marks)

Compare open and closed circulatory systems on: (i) presence of blood-tissue contact, (ii) efficiency of oxygen delivery, and (iii) one example organism each. Then justify which system is better suited to a large, active animal and why. (6 marks)


End of paper

Answer keyMark scheme & solutions

Question 1 (12 marks)

(a) Pathway (8 marks — award 0.5 per correct structure in correct order, max 8):

Right atrium → tricuspid valve → right ventricle → pulmonary (semilunar) valve → pulmonary artery → lungs (pulmonary capillaries) → pulmonary veins → left atrium → bicuspid/mitral valve → left ventricle → aortic (semilunar) valve → aorta → systemic arteries → body tissues (systemic capillaries) → venae cavae (superior & inferior) → right atrium. Why: deoxygenated blood must be sent to lungs (pulmonary circuit) before oxygenated blood is pumped to the body (systemic circuit) — hence "double" circulation with two separate pumps in series.

(b) Gas exchange (4 marks):

  • At lung/pulmonary capillaries: O2O_2 diffuses into blood, CO2CO_2 diffuses out (2 marks).
  • At systemic/tissue capillaries: O2O_2 diffuses out to tissues, CO2CO_2 diffuses in to blood (2 marks). Why: diffusion follows partial-pressure gradients — high pO2pO_2 in alveoli, low in tissues.

Question 2 (12 marks)

(a) (7 marks):

  1. SA node (pacemaker, right atrium) fires spontaneously (1).
  2. Impulse spreads across both atria → atrial systole (contraction) (2).
  3. Impulse reaches AV node, which delays ~0.1 s, allowing atria to empty into ventricles (2).
  4. Signal passes down Bundle of Hisleft/right bundle branchesPurkinje fibres (1).
  5. Ventricles contract from apex upward (ventricular systole); then ventricles relax (diastole) (1). Why: the AV delay ensures atria finish emptying before ventricles contract — prevents backflow and maximises filling.

(b) (5 marks):

  • Duration of one cycle =6075=0.8 s= \dfrac{60}{75} = 0.8\ \text{s} (2 marks).
  • Check: 0.1+0.3+0.4=0.8 s0.1 + 0.3 + 0.4 = 0.8\ \text{s} ✓ (1 mark).
  • Fraction in diastole =0.40.8=0.5=50%= \dfrac{0.4}{0.8} = 0.5 = 50\% (2 marks). Why: HR and cycle duration are reciprocals: T=60/HRT = 60/\text{HR}.

Question 3 (10 marks)

(a) (4 marks): Hemoglobin = protein with 4 polypeptide (globin) chains (2 α, 2 β), each holding a haem group containing an Fe2+Fe^{2+} ion. Each iron binds one O2O_2 molecule → 4 O2O_2 per hemoglobin. Binding is reversible (oxyhaemoglobin).

(b) (4 marks): Cooperative binding — binding of the first O2O_2 changes the protein's shape, increasing affinity of the remaining sites (2). This makes uptake easy at high pO2pO_2 (lungs) and release easy at low pO2pO_2 (tissues), producing the sigmoidal (S-shaped) curve rather than linear (2).

(c) (2 marks): Right shift caused by e.g. increased CO2CO_2 / lower pH (Bohr effect) or increased temperature (1). Benefit: hemoglobin releases more O2O_2 to actively respiring tissues that need it most (1).


Question 4 (10 marks)

(a) (7 marks):

  1. Vessel damage exposes collagen; platelets adhere and form a plug (1).
  2. Damaged tissue + platelets release clotting factors (1).
  3. These activate conversion of prothrombin → thrombin (requires factors + Ca2+Ca^{2+}) (2).
  4. Thrombin (enzyme) converts soluble fibrinogen → insoluble fibrin (2).
  5. Fibrin threads form a mesh trapping RBCs/platelets → stable clot (1).

(b) (3 marks): Ca2+Ca^{2+} is a required cofactor for several steps in the cascade (prothrombin activation) (1.5). Vitamin K is needed by the liver to synthesise prothrombin and other clotting factors (1.5).


Question 5 (10 marks)

(a) Table (4 marks — 1 per row):

Group Antigen on RBC Antibody in plasma
A A anti-B
B B anti-A
AB A and B none
O none anti-A and anti-B

(b) (3 marks): Group O has no A or B antigens, so donated RBCs are not attacked by any recipient's antibodies → universal donor (1.5). Group AB has no anti-A or anti-B antibodies, so it can receive any ABO RBCs without agglutination → universal recipient (1.5).

(c) (3 marks): In the first pregnancy, fetal Rh⁺ blood mixing (usually at birth) sensitises the mother, producing anti-Rh (anti-D) antibodies (1.5). In a second Rh⁺ pregnancy these antibodies cross the placenta and attack fetal RBCs → haemolytic disease of the newborn (erythroblastosis fetalis) (1.5).


Question 6 (6 marks)

  • (i) Open: blood (haemolymph) bathes tissues directly in sinuses; Closed: blood stays within vessels, exchange via capillaries (2).
  • (ii) Open = lower efficiency, low pressure, slow delivery; Closed = high efficiency, higher pressure, faster targeted delivery (1).
  • (iii) Open example: insect/arthropod (e.g. grasshopper); Closed example: vertebrate/annelid (e.g. human, earthworm) (1).
  • Justification (2): A large active animal needs high metabolic O2O_2 delivery; the closed system provides high pressure, rapid, directed flow and regulated distribution — better suited.

[
  {"claim":"Cardiac cycle duration at 75 bpm is 0.8 s","code":"HR=75; T=Rational(60,HR); result = (T == Rational(4,5))"},
  {"claim":"Phase durations sum to one cycle","code":"result = (Rational(1,10)+Rational(3,10)+Rational(4,10) == Rational(8,10))"},
  {"claim":"Fraction of cycle in diastole is 50%","code":"result = (Rational(4,10)/Rational(8,10) == Rational(1,2))"},
  {"claim":"Hemoglobin binds 4 O2 molecules (one per haem iron)","code":"chains=4; o2_per_haem=1; result = (chains*o2_per_haem == 4)"}
]