Organisation -- Diagnostic Tests

Organisation — Diagnostic Tests

Unit Tests

UT-1: The Digestive System and Enzymes

Question: (a) Label the main organs of the digestive system in order from mouth to anus. (b) Explain where and how starch, proteins, and lipids are digested, naming the enzymes and the products. (c) Explain why the small intestine is adapted for absorption. (d) Calculate the surface area of the small intestine if it is 6 m long, has a diameter of 2.5 cm, and has villi that increase the surface area by a factor of 600.

Solution:

(a) Mouth $\to$ Oesophagus $\to$ Stomach $\to$ Small intestine (duodenum, then ileum) $\to$ Large intestine (colon) $\to$ Rectum $\to$ Anus. Accessory organs: liver, pancreas, gall bladder.

(b) Starch: Begins in the mouth (salivary amylase breaks down starch to maltose), continues in the small intestine (pancreatic amylase, then maltase on the intestinal wall breaks maltose to glucose).

Proteins: Stomach (pepsin breaks proteins to polypeptides, works best in acidic conditions pH 2), small intestine (trypsin breaks polypeptides to amino acids).

Lipids: Small intestine only — bile (from liver, stored in gall bladder) emulsifies lipids (increases surface area), then lipase (from pancreas) breaks lipids into glycerol and fatty acids.

(c) The small intestine has: (1) Villi — finger-like projections that increase the surface area for absorption. (2) Microvilli on the villi epithelial cells, further increasing surface area. (3) Thin walls (single layer of epithelial cells) for a short diffusion distance. (4) Dense capillary network to carry away absorbed glucose and amino acids quickly, maintaining a steep concentration gradient. (5) Lacteal (lymph vessel) in each villus to absorb fatty acids and glycerol.

(d) Cylinder surface area $= \pi d L = \pi \times 0.025 \times 6 = 0.471$ m $^2$ . With villi: $0.471 \times 600 = 282.7$ m $^2$ .

UT-2: The Circulatory System

Question: (a) Describe the structure and function of arteries, veins, and capillaries. (b) Explain the difference between double and single circulatory systems. (c) Calculate the cardiac output if the heart rate is 72 beats per minute and the stroke volume is 70 mL. (d) Explain why the left ventricle has a thicker muscular wall than the right ventricle.

Solution:

(a) Arteries: Carry blood away from the heart at high pressure. Thick walls (muscle and elastic tissue) to withstand and smooth out pressure. Narrow lumen. No valves (except semi-lunar valves at the heart).

Veins: Carry blood towards the heart at low pressure. Thinner walls (less muscle and elastic). Wide lumen. Valves prevent backflow. Blood is at low pressure after passing through capillaries.

Capillaries: Exchange vessels connecting arteries to veins. Walls are one cell thick (short diffusion distance). Very narrow lumen (slows blood flow for more exchange time). Permeable walls allow diffusion of substances.

(b) Single circulatory system (e.g., fish): Blood passes through the heart once per complete circuit. Heart $\to$ gills $\to$ body $\to$ heart. Blood pressure drops at the gills, so blood reaches the body at low pressure — limits activity.

Double circulatory system (e.g., mammals): Blood passes through the heart twice per complete circuit. Pulmonary circulation (heart $\to$ lungs $\to$ heart) and systemic circulation (heart $\to$ body $\to$ heart). The heart re-pressurises the blood for each circuit, maintaining high pressure for efficient delivery to the body.

(d) The left ventricle pumps blood to the entire body (systemic circulation) through the aorta, which requires high pressure to overcome the resistance of the long systemic circulation. The right ventricle pumps blood only to the lungs (pulmonary circulation), which is a much shorter circuit with less resistance. A thicker muscular wall in the left ventricle generates the higher pressure needed.

UT-3: The Respiratory System

Question: (a) Describe the pathway of air from the nose to the alveoli. (b) Explain how gas exchange occurs at the alveoli. (c) A person inhales 500 mL of air per breath, with 21% oxygen. They exhale 500 mL with 16% oxygen. Calculate the volume of oxygen absorbed per breath. (d) Explain two adaptations of alveoli for efficient gas exchange.

Solution:

(a) Nose $\to$ trachea $\to$ bronchi (left and right) $\to$ bronchioles $\to$ alveoli. The nose warms, moistens, and filters the air. The trachea and bronchi have C-shaped cartilage rings to keep them open. Bronchioles are smaller and have smooth muscle to control air flow.

(b) Oxygen diffuses from the alveolar air (high concentration) into the blood (low concentration) across the alveolar epithelium and capillary endothelium. Carbon dioxide diffuses from the blood (high concentration) into the alveolar air (low concentration). This maintains a concentration gradient because: (1) blood flow continuously removes oxygenated blood and brings deoxygenated blood, (2) breathing continuously refreshes alveolar air.

(c) Oxygen inhaled $= 500 \times 0.21 = 105$ mL. Oxygen exhaled $= 500 \times 0.16 = 80$ mL. Oxygen absorbed $= 105 - 80 = 25$ mL per breath.

(d) Two adaptations: (1) Large surface area: Millions of alveoli provide a huge total surface area (about 70 m $^2$ ) for gas exchange. (2) Thin walls: Alveolar epithelium and capillary endothelium are each one cell thick, giving a very short diffusion distance. (3) Dense capillary network: Maintains a steep concentration gradient. (4) Moist surface: Gases dissolve in the moisture before diffusing.

Integration Tests

IT-1: Enzyme Activity and Digestion (with Bioenergetics)

Question: An experiment investigates the effect of temperature on the activity of amylase. The time taken for starch to be completely digested is recorded:

Temperature ( $^\circ$ C)	10	20	30	40	50	60	70
Time (seconds)	300	180	90	45	60	240	No reaction

(a) Calculate the rate of reaction (1/time) at each temperature. (b) Plot a graph of rate against temperature and identify the optimum temperature. (c) Explain why the rate decreases above $40^\circ$ C. (d) Explain how the products of digestion (glucose, amino acids, fatty acids) are used in cellular respiration.

Solution:

(a) Rates (1/time in s $^{-1}$ $\times 1000$ ): 3.3, 5.6, 11.1, 22.2, 16.7, 4.2, 0.

(b) The graph peaks at $40^\circ$ C (rate 22.2), making this the optimum temperature.

(c) Above $40^\circ$ C, the enzyme’s active site begins to change shape due to the breaking of bonds maintaining the tertiary structure. This is denaturation — the substrate can no longer fit into the active site, so the reaction rate decreases. At $70^\circ$ C, the enzyme is completely denatured and cannot function.

(d) Glucose: Used in aerobic respiration (with oxygen) to produce ATP: glucose $+$ oxygen $\to$ carbon dioxide $+$ water $+$ ATP. Also used in anaerobic respiration: glucose $\to$ lactic acid $+$ ATP.

Amino acids: Absorbed into cells and reassembled into proteins (enzymes, structural proteins, antibodies, hormones) by ribosomes.

Fatty acids and glycerol: Used to build cell membranes (phospholipid bilayer), stored as fat for energy reserves, or used in hormone synthesis.

IT-2: Circulation and Gas Exchange Combined (with Cell Biology)

Question: Red blood cells contain haemoglobin and no nucleus. (a) Explain how the structure of red blood cells relates to their function in gas transport. (b) A red blood cell has a diameter of 7 $\mu$ M and must pass through a capillary of diameter 5 $\mu$ M. Explain how this is possible and why it is beneficial for gas exchange. (c) Haemoglobin binds oxygen in the lungs and releases it in the tissues. Explain how the different conditions (high O $_2$ in lungs, high CO $_2$ in tissues) drive this process. (d) Calculate how many red blood cells would fit along 1 cm of a capillary of diameter 5 $\mu$ M.

Solution:

(a) Biconcave disc shape: Maximises surface area for oxygen diffusion. No nucleus: Maximises space for haemoglobin (more oxygen carried per cell). Flexible membrane: Allows the cell to squeeze through narrow capillaries (diameter smaller than the cell). Contains haemoglobin: A pigment that binds oxygen reversibly (oxyhaemoglobin), allowing efficient transport.

(b) Red blood cells are flexible and can deform to pass through capillaries narrower than themselves. This is beneficial because: (1) It slows blood flow through capillaries, allowing more time for gas exchange. (2) The close contact with capillary walls reduces the diffusion distance for oxygen and carbon dioxide.

(c) In the lungs: High oxygen concentration drives oxygen to bind to haemoglobin, forming oxyhaemoglobin. High CO $_2$ concentration also promotes oxygen binding (Bohr effect: CO $_2$ causes haemoglobin to change shape, increasing its affinity for O $_2$ in the lungs).

In the tissues: Low oxygen concentration and high CO $_2$ concentration (from cellular respiration) cause oxyhaemoglobin to release oxygen. The lower pH (due to dissolved CO $_2$ forming carbonic acid) reduces haemoglobin’s oxygen affinity, promoting unloading.

(d) Capillary diameter $= 5 \mu$ M. Cell diameter $= 7 \mu$ M. Since cells deform, they pass through single file. Length of capillary $= 10,000 \mu$ M. If each cell occupies approximately 7 $\mu$ M when deformed: $10,000/7 \approx 1429$ cells.

IT-3: Nervous System and Homeostasis (with Organisation)

Question: (a) Describe the pathway of a nerve impulse in a reflex arc from touching a hot object to withdrawing the hand. (b) Explain the difference between sensory, relay, and motor neurons. (c) Calculate the speed of a nerve impulse that travels 1.5 m in 0.012 seconds. (d) Explain why reflex actions are faster than voluntary actions.

Solution:

(a) Reflex arc: (1) Receptors in the skin detect heat and generate an electrical impulse. (2) The sensory neuron transmits the impulse to the spinal cord. (3) The relay neuron in the spinal cord connects the sensory neuron to the motor neuron (impulse may also travel to the brain for conscious awareness). (4) The motor neuron carries the impulse from the spinal cord to the effector (bicep muscle in the arm). (5) The muscle contracts, withdrawing the hand from the hot object.

(b) Sensory neuron: Carries impulses from receptors to the CNS (brain and spinal cord). Cell body is in the middle of the neuron, with a long axon leading to the CNS.

Relay neuron: Found within the CNS. Connects sensory neurons to motor neurons. Short axons, many branches.

Motor neuron: Carries impulses from the CNS to effectors (muscles or glands). Cell body is at one end, with a long axon leading to the effector.

(d) Reflex actions are faster because: (1) The pathway is shorter — the impulse travels only to the spinal cord and back, not to the brain. (2) Fewer synapses (gaps between neurons) are involved, reducing transmission time. (3) The response is automatic and does not require conscious processing in the brain, which takes additional time. This speed is critical for survival — it allows the body to respond to danger (e.g., touching a hot object) before the brain has even registered the pain.

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Common Pitfalls

Confusing terminology or concepts that appear similar but have distinct meanings.
Overlooking key assumptions or boundary conditions that limit applicability.

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