A Level H2 Biology Human Physiology Quiz

A-Level Biology H2 Quiz – Human Physiology ANSWERS

Section A: Short-Answer and Structured Questions

1.

• Support/protection of internal organs (e.g., rib cage protecting heart and lungs) [1]
• Production of blood cells (haematopoiesis) in bone marrow [1]
  Accept any two valid functions, such as mineral storage (calcium, phosphate) or endocrine function (osteocalcin).

2.
(a) Action potential / nerve impulse [1]

(b)

• Voltage‑gated sodium ion channels open [1]
• Sodium ions (Na⁺) rapidly diffuse into the axon / neurone [1]
  Accept: influx of Na⁺ makes the inside of the membrane more positive (depolarisation).

3.
(a)

• The SAN cells have a leaky membrane to Na⁺ (or slow inward Na⁺ current) that causes a gradual depolarisation / pacemaker potential [1]
• When the membrane potential reaches threshold, voltage‑gated calcium ion channels open; influx of Ca²⁺ generates the action potential [1]

(b)

• Delay allows the atria to finish contracting and emptying blood into the ventricles before ventricular contraction begins [1]
  Accept: ensures atria and ventricles contract in sequence, not simultaneously.

4.
(a) Resting minute ventilation = tidal volume × breathing rate = 0.5 × 12 = 6.0 dm³ min⁻¹ [1]

(b)

• During exercise, the respiratory muscles (diaphragm and external intercostal muscles) contract more forcefully, increasing the volume change in the thoracic cavity [1]
• This creates a larger pressure gradient between the atmosphere and the lungs, drawing in a greater volume of air / deeper inhalation [1]
  Also accept: increased recruitment of accessory muscles, greater lung compliance.

5.

• In the lungs (high pO₂ ~12 kPa), haemoglobin is almost fully saturated (97%); the curve is near‑flat, so even if lung pO₂ falls a little, loading is still high. This ensures efficient oxygen uptake. [1]
• In resting tissues (pO₂ ~4 kPa), saturation drops to 75%, meaning a significant amount of oxygen is released (22% of the maximum). [1]
• In active muscle (low pO₂ ~2 kPa), saturation falls steeply to only 30%, so much more oxygen is unloaded (a further 45%) where it is most needed. The steep part of the curve ensures a large difference in O₂ release for a small drop in pO₂. [1]
  Award marks for linking the shape of the curve (sigmoid) to cooperative binding and the Bohr effect if mentioned.

6.
(a)

• Increase in blood osmotic pressure / decrease in water potential detected by osmoreceptors in the hypothalamus [1]
  Accept: decrease in blood volume / pressure, or thirst.

(b)

• ADH binds to receptors on the cell membrane of collecting duct cells, stimulating the insertion of aquaporin‑2 water channels into the membrane [1]
• This increases the permeability of the collecting duct to water, allowing more water to be reabsorbed into the medulla, resulting in a smaller volume of more concentrated urine [1]

7.
(a)

• Helper T cells have CD4 receptors and recognise antigens presented on antigen‑presenting cells (e.g., macrophages) via MHC class II molecules [1]
• They secrete cytokines (e.g., interleukins) that activate B lymphocytes (to produce antibodies) and cytotoxic T cells (to kill infected cells) [1]
  Accept: stimulation of clonal expansion and differentiation of lymphocytes.

(b)

• HIV destroys helper T cells, so the immune system cannot effectively activate B cells and cytotoxic T cells [1]
• Without a full immune response, the body cannot fight off pathogens that a healthy individual would normally resist, leading to increased susceptibility to opportunistic infections [1]

8.
(a)

• In the healthy individual, blood glucose rises after the meal but returns to normal within 2.5 hours; in the diabetic individual, blood glucose rises higher and remains elevated even after 4 hours [1]
• The healthy individual shows tight homeostatic regulation (negative feedback), while the diabetic shows impaired regulation / lack of effective insulin response [1]

(b)

• Rising blood glucose after a meal is detected by β‑cells in the islets of Langerhans, which secrete insulin [1]
• Insulin binds to receptors on target cells (liver, muscle), increasing glucose uptake and stimulating conversion of glucose to glycogen (glycogenesis), thereby lowering blood glucose back to normal [1]

9.

• Light is absorbed by photopigments (rhodopsin in rods) causing the retinal component to change from cis to trans form; this activates transducin.
• Transducin activates phosphodiesterase, which hydrolyses cGMP → cGMP levels fall → sodium ion channels in the outer segment close.
• The photoreceptor cell hyperpolarises, reducing the release of inhibitory neurotransmitter onto bipolar cells. The bipolar cells depolarise and release neurotransmitter onto ganglion cells, generating action potentials that travel along the optic nerve. [3]
  Mark points: photopigment change, hyperpolarisation, generation of action potential.

10.
Two marks for each change, one mark for description and one mark for explanation.

• Increased ventilation rate: The body breathes faster and deeper in response to low arterial pO₂ detected by peripheral chemoreceptors. This increases the alveolar pO₂, steepening the partial pressure gradient for oxygen to diffuse into the blood. [2]
• Increase in red blood cell production: The kidneys detect low tissue oxygen and secrete erythropoietin (EPO), which stimulates the bone marrow to produce more red blood cells. A higher haematocrit increases the oxygen‑carrying capacity of the blood. [2]
  Other acceptable changes: increased 2,3‑BPG in red blood cells (shifts oxygen‑dissociation curve to the right, promoting O₂ unloading), increased myoglobin in muscles, increased capillary density.

Section B: Data‑Based and Free‑Response Questions

11.
(a)

• There is a positive correlation: as blood thyroxine concentration increases, resting heart rate tends to increase [1]
• The correlation is weak/moderate; the points show considerable scatter, and one point is a clear outlier [1]

(b)

• The patient may be suffering from another condition that raises heart rate independently, e.g., anxiety, fever, recent exercise, or excess of another hormone such as adrenaline. [1]

12.
(a)

• Calcium ions cause synaptic vesicles containing acetylcholine to fuse with the presynaptic membrane, so that acetylcholine is released into the synaptic cleft by exocytosis. [1]

(b)

• If acetylcholine remained in the cleft, it would continue to bind to receptors on the postsynaptic membrane, causing prolonged depolarisation / continued firing of action potentials [1]
• Rapid breakdown ensures that the postsynaptic cell can respond to new signals, allowing precise control and preventing continuous stimulation [1]

(c)

• Neostigmine inhibits acetylcholinesterase, so acetylcholine is not broken down and accumulates in the synaptic cleft, leading to prolonged binding to postsynaptic receptors [1]
• This results in sustained depolarisation of the postsynaptic membrane, which may cause muscle spasms or, in severe cases, paralysis due to depolarisation block. [1]

13.
(a) Percentage increase = [(38.2 – 12.5) / 12.5] × 100 = (25.7 / 12.5) × 100 = 205.6% (accept 206%) [2]
Award 1 mark for correct difference, 1 mark for correct percentage calculation.

(b)

• Propranolol (a β‑blocker) almost completely prevents the effect of adrenaline: the contractile force with adrenaline + propranolol (14.0 mN) is similar to the force without adrenaline (12.5 mN) [1]
• This indicates that adrenaline exerts its positive inotropic effect by binding to β‑adrenergic receptors on the cardiac muscle cells; when the receptors are blocked, the response is lost [1]

(c)

• Non‑overlapping standard deviations suggest that the mean values are significantly different; the effect of adrenaline on contractile force is statistically significant (typically at p < 0.05). [1]

14.
(a) Breathing rate = number of breaths per minute. One cycle = 4.0 s, so 60 / 4.0 = 15 breaths per minute. [1]

(b) Vital capacity is the maximum volume of air that can be forcibly exhaled after a maximum inhalation. [1]

(c) From the trace: maximum inhalation = 3.8 dm³, forced exhalation minimum = 1.2 dm³. Vital capacity = 3.8 – 1.2 = 2.6 dm³. [1]

(d)

• Soda lime absorbs CO₂; if it were exhausted, CO₂ would accumulate in the spirometer circuit [1]
• The subject would be rebreathing exhaled CO₂, which stimulates the respiratory centre to increase ventilation rate; the trace would show an increase in breathing frequency/amplitude (the trace would progressively shrink downwards because the subject would be using more oxygen and producing CO₂ that is not absorbed, causing the total volume of gas in the spirometer to decrease). [1]

15.

• Urea is filtered freely at the glomerulus, so filtrate concentration is similar to plasma [1]
• As filtrate passes through the nephron, water is reabsorbed (especially in the proximal convoluted tubule and loop of Henle / collecting duct) but urea is not reabsorbed to the same extent (some urea diffusion does occur, but net reabsorption is much less than that of water) [1]
• As a result, the urea becomes increasingly concentrated as water is removed; additional urea is added to the tubular fluid in the collecting duct via active secretion. This combination results in urine urea concentration being many times higher than in the filtrate. [1]
  Accept explanation based on osmotic gradient and counter‑current multiplier effect, but focus on urea concentration.

16.
(a)

• Saltatory conduction is the mode of propagation of an action potential along a myelinated axon in which the impulse jumps from one node of Ranvier to the next [1]
• At the nodes, voltage‑gated ion channels are concentrated, and depolarisation occurs only at these points; the myelin sheath insulates the internodal regions, preventing ion leakage. [1]

(b)

• Saltatory conduction greatly increases the speed of impulse transmission because depolarisation occurs only at the nodes, reducing the need for continuous propagation along the whole axon membrane / it conserves energy as fewer ions need to be pumped to restore the gradient. [1]

17.
(a) Capillaries / arterial end of capillaries (specifically, plasma is filtered through capillary walls). [1]

(b)

• Loss of tissue fluid reduces blood plasma volume, decreasing total blood volume and therefore reducing venous return to the heart [1]
• According to Starling’s law, reduced venous return leads to decreased stroke volume and cardiac output, so arterial blood pressure falls. [1]

18.
(a) Motor neurone / efferent neurone. [1]

(b)

• The reflex arc involves only the spinal cord (or brain stem) and does not require processing by the cerebral cortex [1]
• The pathway typically contains only two or three synapses (receptor–sensory–relay–motor), whereas a voluntary response involves multiple synapses in the brain, so transmission is faster and the response time is shorter. [1]

19.
(a) Vasoconstriction of arterioles in the skin / peripheral vasoconstriction, piloerection (goosebumps) (accept one). [1]

(b)

• The hypothalamus acts as the body’s thermostat / thermoregulatory centre; it detects increased blood temperature (or input from peripheral thermoreceptors) and initiates heat‑generating / heat‑conserving responses [1]
• It stimulates sympathetic nerves to cause vasoconstriction, shivering, and increased metabolic rate, and reduces sweating; these responses raise core temperature back to the set point (37°C). [1]

20.
(a) CO₂ + H₂O ⇌ H₂CO₃ (carbonic acid) [1]
Accept reversible arrow; H₂CO₃ may dissociate further to H⁺ + HCO₃⁻, but the catalysed reaction is the hydration of CO₂.

(b)

• In respiring tissues, CO₂ diffuses into red blood cells, where carbonic anhydrase catalyses its conversion to carbonic acid, which dissociates into H⁺ and HCO₃⁻ [1]
• The HCO₃⁻ (hydrogen carbonate ion) diffuses out of the red blood cell into the plasma (in exchange for Cl⁻, the chloride shift) and is the main form in which CO₂ is transported to the lungs. In the lungs the reverse reaction occurs, releasing CO₂ to be exhaled. [1]
  Full explanation must link to transport and reverse reaction at lungs.

End of Answers