S5 Biology Paper: Nerve Conduction, Glucose Control, Thermoregulation, Immunity, Cardiac Output and Transpiration (100 marks)

2026-11-13

S5 Biology – physiology and plant transport

S5 Biology paper: 6 questions, 24 parts, 100 marks, 60 minutes. Covers nerve conduction, insulin, thermoregulation, antibodies, cardiac output and transpiration.

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Paper details　S5 Biology practice paper with six questions on nerve conduction speed, insulin and blood glucose, thermoregulation, antibody response, cardiac output and plant transpiration; every value is calculated.

Section A: Nerve Conduction and Reflex Action (16 marks)

When a finger touches a hot pan, a sensory neurone carries the signal 0.85 m to the spinal cord at 85 m/s. A motor neurone then carries it 0.65 m to the arm muscle at 65 m/s. The reflex arc has 2 synapses, each with a delay of 0.5 ms.

#	Question	Answer and marking notes
1	Work out the conduction time in the sensory neurone in milliseconds.	Answer: 0.85 / 85 = 0.010 s = 10 ms. (Marking: correct substitution 2, unit conversion 1, answer 1)
2	Work out the conduction time in the motor neurone in milliseconds.	Answer: 0.65 / 65 = 0.010 s = 10 ms. (Marking: substitution 2, units 1, answer 1)
3	Work out the total reflex time, including synaptic delay.	Answer: 10 + 10 + 2 x 0.5 = 21 ms. (Marking: adding both lengths 2, synaptic delay 1, answer 1)
4	If myelin damage slows the sensory neurone to 42.5 m/s while the motor speed is unchanged, by what percentage does the reflex time increase?	Answer: 0.85 / 42.5 = 0.020 s = 20 ms; new total 20 + 10 + 1 = 31 ms; increase (31 – 21) / 21 x 100% = 47.6%. (Marking: new time 2, percentage working 1, answer 1)

Section B: Insulin and Blood Glucose Control (16 marks)

A patient has a fasting blood glucose of 7.8 mmol/L; the normal range is 4.4 to 6.1 mmol/L. Every 10 g of carbohydrate eaten raises glucose by 1.7 mmol/L. One unit of insulin lowers glucose by 2.4 mmol/L. For safety glucose must fall by no more than 3.6 mmol/L per hour.

#	Question	Answer and marking notes
5	After eating 60 g of carbohydrate, how much does blood glucose rise?	Answer: 60 / 10 x 1.7 = 10.2 mmol/L. (Marking: converting to 10 g units 2, multiplying 1, answer 1)
6	What is the peak blood glucose after the meal, and is it above the normal upper limit?	Answer: 7.8 + 10.2 = 18.0 mmol/L, well above the 6.1 mmol/L upper limit. (Marking: addition 2, answer 1, judgement 1)
7	How many units of insulin are needed to bring glucose back to 6.0 mmol/L?	Answer: 18.0 – 6.0 = 12.0 mmol/L; 12.0 / 2.4 = 5 units. (Marking: drop 1, dividing by 2.4 2, answer 1)
8	At the safe maximum fall of 3.6 mmol/L per hour, what is the minimum time to reach 6.0 mmol/L?	Answer: 12.0 / 3.6 = 3.3 hours, about 3 hours 20 minutes. (Marking: same drop 2, dividing 1, answer 1)

Section C: Thermoregulation and Water Balance (16 marks)

A 60 kg athlete produces 12 kJ of heat per minute while exercising. The specific heat capacity of the body is 3.5 kJ/(kg °C). Each litre of sweat that fully evaporates removes 2,400 kJ of heat.

#	Question	Answer and marking notes
9	If no heat is lost, how much does body temperature rise after 10 minutes of exercise?	Answer: 12 x 10 = 120 kJ; 120 / (60 x 3.5) = 0.57 °C. (Marking: total heat 1, heat capacity 2, answer 1)
10	How much sweat must evaporate to remove 120 kJ?	Answer: 120 / 2,400 = 0.05 L, that is 50 mL. (Marking: division 2, converting to mL 1, answer 1)
11	The athlete actually sweats 0.4 L. How much heat does this remove?	Answer: 0.4 x 2,400 = 960 kJ. (Marking: substitution 2, units 1, answer 1)
12	In humid air evaporation is only 60% efficient. What is the effective heat loss, and how many times the heat produced is it?	Answer: 960 x 0.60 = 576 kJ; 576 / 120 = 4.8 times, still enough; below 12.5% efficiency body temperature would rise. (Marking: effective loss 2, ratio 1, comment 1)

Section D: Immunity and Antibody Response (16 marks)

After vaccination the blood antibody level is 15 units at week 2, 60 units at week 4 and 240 units at week 8. After a first natural infection the level reaches 8 units only on day 5, but on a second infection it reaches 8 units by day 3.

#	Question	Answer and marking notes
13	By what factor does the antibody level rise from week 2 to week 4?	Answer: 60 / 15 = 4 times. (Marking: division 2, answer 1, stating it is a factor 1)
14	From week 4 to week 8 the level rises 4 times over 4 weeks. What is the average weekly multiplying factor?	Answer: the fourth root of 4 = 1.41, so about 41% per week. (Marking: root relation 2, answer 1, percentage 1)
15	If the level keeps rising by a factor of 1.41 per week, how many weeks after week 8 does it reach 500 units?	Answer: 500 / 240 = 2.08; ln 2.08 / ln 1.41 = 0.734 / 0.347 = 2.1 weeks. (Marking: ratio 1, logs 2, answer 1)
16	By what percentage earlier does a second infection reach 8 units, and why?	Answer: (5 – 3) / 5 x 100% = 40%; memory B cells divide quickly and form plasma cells. (Marking: calculation 2, answer 1, mechanism 1)

Section E: Cardiac Output and Oxygen Transport (16 marks)

At rest the heart rate is 72 beats per minute and stroke volume 70 mL. During hard exercise the heart rate is 150 beats per minute and stroke volume 110 mL. Each litre of blood carries 200 mL of oxygen.

#	Question	Answer and marking notes
17	Work out the resting cardiac output in litres per minute.	Answer: 72 x 70 = 5,040 mL = 5.04 L per minute. (Marking: multiplication 2, converting 1, answer 1)
18	Work out the cardiac output during exercise in litres per minute.	Answer: 150 x 110 = 16,500 mL = 16.5 L per minute. (Marking: multiplication 2, converting 1, answer 1)
19	How many times the resting value is the cardiac output during exercise?	Answer: 16.5 / 5.04 = 3.27 times. (Marking: same units 2, division 1, answer 1)
20	How many more litres of oxygen are delivered per minute during exercise than at rest?	Answer: exercise 16.5 x 0.2 = 3.3 L; rest 5.04 x 0.2 = 1.01 L; increase 3.3 – 1.01 = 2.29 L per minute. (Marking: each value 1, subtraction 1, answer 1)

Section F: Transpiration and Photosynthesis (20 marks)

A tree has a total leaf area of 40 square metres and transpires 1.5 L per square metre per day. It fixes 264 g of carbon dioxide per day. Photosynthesis is 6CO2 + 6H2O -> C6H12O6 + 6O2, with molar masses CO2 44 g, H2O 18 g and glucose 180 g.

#	Question	Answer and marking notes
21	Work out the total water lost by transpiration each day.	Answer: 40 x 1.5 = 60 L. (Marking: multiplication 2, units 1, answer 1, working 1)
22	How many moles of carbon dioxide are fixed per day?	Answer: 264 / 44 = 6 mol. (Marking: correct molar mass 2, division 1, answer 1, units 1)
23	How many moles and how many grams of glucose are made per day?	Answer: 6 / 6 = 1 mol; 1 x 180 = 180 g. (Marking: equation ratio 2, moles 1, mass 2)
24	How many litres of water are needed to make this glucose, how many times the transpiration loss is that, and why?	Answer: 1 mol glucose needs 6 mol water = 6 x 18 = 108 g = 0.108 L; 60 / 0.108 is about 556 times; transpiration mainly transports minerals and cools the leaf. (Marking: water 2, ratio 1, explanation 2)

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