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HKDSE Biology Practice Paper: Human Physiology, Coordination and Homeostasis (12 questions, 100 marks, 75 minutes)
HKDSE Biology: Human Physiology and Homeostasis
Original HKDSE Biology paper on human physiology: cardiac output, gas exchange, kidney filtration, heat transport, nerve conduction, blood glucose and thermoregulation.
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Paper details Level: S6 | Subject: Biology | Time: 75 minutes | Marks: 100 | Questions: 12
Section A (32 marks)
Circulation, breathing and excretion: show the calculation with units.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | An adult at rest has a heart rate of 75 beats per minute and a stroke volume of 70 mL. (a) Calculate the resting cardiac output in L/min. (b) During vigorous exercise the heart rate is 150 beats per minute and the stroke volume is 120 mL. Calculate the cardiac output and how many times the resting value it is. (c) State two benefits of the higher cardiac output for skeletal muscle. |
Answer: (a) Cardiac output = heart rate x stroke volume = 75 x 70 = 5,250 mL/min = 5.25 L/min. (b) 150 x 120 = 18,000 mL/min = 18.0 L/min; 18.0 / 5.25 is about 3.4 times. (c) More oxygen and glucose reach skeletal muscle each minute, supporting more aerobic respiration; carbon dioxide and heat are removed faster, so fatigue is delayed. Marking points: formula with units 2, resting value 1, exercise value 1, ratio 1, two benefits 3. |
| 2 | A person has a tidal volume of 500 mL, an anatomical dead space of 150 mL and a breathing rate of 15 per minute. (a) Calculate the total pulmonary ventilation per minute in L/min. (b) Calculate the alveolar ventilation in L/min. (c) If the tidal volume becomes 1000 mL at 8 breaths per minute, calculate the alveolar ventilation and compare. (d) Explain why deep slow breathing is more effective than shallow fast breathing. |
Answer: (a) 500 x 15 = 7,500 mL/min = 7.5 L/min. (b) (500 – 150) x 15 = 5,250 mL/min = 5.25 L/min. (c) (1000 – 150) x 8 = 6,800 mL/min = 6.8 L/min, which is 1.55 L/min higher. (d) Dead space volume is fixed for each breath, so deep slow breathing gives a smaller dead space fraction and a larger share of fresh air reaches the alveoli. Marking points: two formulas 3, two values 2, comparison 1, dead space explanation 3. |
| 3 | An adult has a glomerular filtration rate of 125 mL/min and produces 1.5 L of urine per day. (a) Calculate the volume of filtrate formed per day in litres. (b) Calculate the volume of water reabsorbed per day and the percentage reabsorbed. (c) Name two substances reabsorbed in the proximal convoluted tubule. (d) Explain why reabsorption is important for homeostasis. |
Answer: (a) 125 x 60 x 24 = 180,000 mL = 180 L. (b) Reabsorbed = 180 – 1.5 = 178.5 L; 178.5 / 180 x 100 is about 99.2 percent. (c) Glucose, amino acids, sodium ions, water (any two). (d) Useful substances stay in the blood instead of being lost in urine; water and electrolytes remain stable, so plasma osmotic pressure and pH are maintained and the internal environment stays constant. Marking points: 180 L 2, reabsorbed volume 1, percentage 1, two substances 2, homeostasis explanation 2. |
| 4 | Two cubes with sides of 1 cm and 3 cm model two animals. (a) Calculate the surface area to volume ratio for each cube. (b) Calculate how many times larger one ratio is than the other. (c) Explain why a small mammal needs a higher metabolic rate per unit body mass. (d) State two ways a mammal loses heat. |
Answer: (a) 1 cm cube: 6 / 1 = 6; 3 cm cube: 54 / 27 = 2. (b) 6 / 2 = 3 times. (c) A small animal has a larger surface area to volume ratio, so it loses heat faster per unit mass and needs a higher metabolic rate to replace the heat, hence greater oxygen uptake and food intake. (d) Sweating and evaporation; dilation of skin arterioles with more skin blood flow (also panting or behavioural change). Marking points: two ratios 3, multiple 1, metabolic explanation 3, two methods 2. |
Section B (36 marks)
Nerves, hormones, temperature and blood: explain the mechanism.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | Stimulating a person’s arm produces an action potential that reaches a recording electrode 60 mm away in 2.0 ms. An unmyelinated fibre in the same person conducts at 0.5 m/s. (a) Calculate the conduction speed of the fibre in m/s. (b) Calculate how many times faster it is than the unmyelinated fibre. (c) Explain how myelin speeds up conduction and name this type of conduction. |
Answer: (a) 60 mm / 0.002 s = 30,000 mm/s = 30 m/s. (b) 30 / 0.5 = 60 times. (c) Myelin is lipid and blocks ion movement across the axon membrane, so the action potential is regenerated only at the gaps between myelin (nodes of Ranvier). The impulse jumps from node to node, which is saltatory conduction, so speed increases greatly. Marking points: conversion with units 2, 30 m/s 2, ratio 1, myelin insulation 1, saltatory conduction and name 3. |
| 6 | A person has a fasting blood glucose level of 4.8 mmol/L which rises to 7.6 mmol/L one hour after a meal. (a) Calculate the rise in blood glucose in mmol/L. (b) Describe the changes in insulin and glucagon secretion after the meal. (c) Use negative feedback to explain how the glucose level falls again. (d) Name two target tissues and their responses. |
Answer: (a) 7.6 – 4.8 = 2.8 mmol/L. (b) Insulin secretion rises and glucagon secretion falls. (c) Insulin makes liver and muscle cells take up glucose and form glycogen and inhibits glycogen breakdown, so the level falls back to about 4.8 mmol/L; insulin secretion then decreases, which is negative feedback. (d) Liver (forms glycogen), skeletal muscle (takes up glucose and respires faster), adipose tissue (forms fat) (any two). Marking points: rise 1, two hormone changes 2, three step feedback 4, two targets and responses 2. |
| 7 | A person exercises for 30 minutes and sweats 0.8 L. Each litre of sweat removes 2,430 kJ of heat when it evaporates. (a) Calculate the total heat removed in kJ. (b) Calculate the average rate of heat loss in kW. (c) Explain why heat loss is less efficient in a hot humid environment. (d) State two responses of the body in that environment. |
Answer: (a) 2,430 x 0.8 = 1,944 kJ. (b) 1,944 kJ / 1,800 s = 1.08 kJ/s = 1.08 kW. (c) High humidity slows evaporation of sweat, so evaporative heat loss falls sharply and heat accumulates. (d) Dilation of skin arterioles with more skin blood flow, more sweat secretion, reduced activity (any two). Marking points: total heat 2, rate with units 3, evaporation explanation 2, two responses 2. |
| 8 | A person has a blood haemoglobin concentration of 150 g/L, a blood volume of 5.0 L, and each gram of haemoglobin binds 1.34 mL of oxygen. (a) Calculate the total volume of oxygen the blood can carry in mL. (b) State two ways the body increases oxygen supply to muscle during exercise. (c) Explain the mechanism of carbon monoxide poisoning. |
Answer: (a) 150 x 1.34 = 201 mL/L; 201 x 5.0 = 1,005 mL of oxygen. (b) Higher heart rate and cardiac output; deeper and faster breathing to raise alveolar ventilation; dilation of muscle capillaries to increase muscle blood flow (any two). (c) Carbon monoxide binds haemoglobin far more strongly than oxygen, occupies the binding sites and does not leave easily, so the oxygen carrying capacity of blood falls and tissues become short of oxygen. Marking points: per litre 1, total 2, two methods 3, competitive binding 3. |
Section C (32 marks)
Metabolism, comparison and experimental design: justify your answer.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | A person drinks too little water, so plasma osmotic pressure rises and daily urine output falls from 1.5 L to 0.5 L. (a) Name the organ that detects the change in plasma osmotic pressure and the hormone secreted. (b) Describe the action of this hormone in the kidney. (c) Compare the concentration of urine when water intake is sufficient and when it is not, and explain. (d) Explain the importance of reduced urine output for homeostasis. |
Answer: (a) The hypothalamus detects osmotic pressure; the posterior pituitary secretes antidiuretic hormone. (b) It increases the permeability of the collecting duct and distal tubule to water, so more water is reabsorbed into the blood. (c) With too little water the urine is small in volume and concentrated; with plenty of water the urine is large in volume and dilute because less antidiuretic hormone is secreted. (d) Water is retained so plasma osmotic pressure and blood volume stay stable, preventing cell dehydration; this is negative feedback. Marking points: two organs 2, hormone action 2, concentration comparison 2, homeostasis 2. |
| 10 | A person uses 5.0 L of oxygen and produces 4.0 L of carbon dioxide during exercise. (a) Calculate the respiratory quotient (RQ). (b) Identify the main energy source and explain. (c) Explain why RQ falls after prolonged exercise. (d) State what an RQ above 1.0 may indicate. |
Answer: (a) 4.0 / 5.0 = 0.8. (b) A mixture of carbohydrate and fat is being used, because RQ lies between 0.7 for fat and 1.0 for carbohydrate. (c) After long exercise liver and muscle glycogen are depleted, so fat becomes the main fuel; fat oxidation needs more oxygen per unit of carbon dioxide released, so RQ falls. (d) Heavy exercise with overbreathing that washes out carbon dioxide, or strong fat synthesis, can raise RQ above 1.0 for a time. Marking points: RQ 2, judgement with explanation 3, fat oxidation 2, RQ above 1 1. |
| 11 | (a) Compare nervous coordination and hormonal coordination in terms of transmission route, speed, duration of effect and range of action. (b) Give one example of each. (c) Explain how the two systems share the work of thermoregulation. (d) Explain why hormonal control suits long term regulation. |
Answer: (a) Nervous: through neurons and synapses; fast; short lived effect; local and precise. Hormonal: through the blood; slower; longer lasting; whole body. (b) Nervous example: reflex dilation of skin arterioles when body temperature is high; hormonal example: antidiuretic hormone regulating water balance. (c) The nervous system gives immediate responses such as sweating and vasoconstriction, while hormones such as thyroxine adjust metabolic rate and heat production over a longer period. (d) Hormones travel in the blood to the whole body and act for a long time, which suits keeping the internal environment stable over hours or days rather than second by second. Marking points: four comparisons 4, two examples 2, division of work 1, long term explanation 1. |
| 12 | A group of students studies the effect of exercise on heart rate. They measure heart rate three times at rest, after jogging for 5 minutes, and after resting for 2 minutes, then take the mean. (a) State the independent variable, the dependent variable and two controlled variables. (b) Explain why each reading is taken three times and averaged. (c) Suggest one safety precaution. (d) Suggest one way to improve the validity of the experiment. |
Answer: (a) Independent variable: state of activity (rest, after exercise, after rest). Dependent variable: heart rate. Controlled variables: same subject, same exercise intensity and duration, same posture, same room temperature (any two). (b) Repeated readings reduce random error and natural fluctuation, so the mean is more reliable. (c) Warm up before exercise, stop if unwell, avoid exercising in very hot conditions (any one). (d) Use more subjects and average the group, or use a heart rate monitor instead of manual counting to reduce human error. Marking points: three variables 3, averaging reason 2, safety 1, validity 2. |
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