S6 Biology Practice Paper: Nervous And Endocrine Coordination (12 Questions, 100 Marks, 75 Minutes)

S6 Biology Practice Paper: Nervous and Endocrine Coordination (12 questions, 100 marks, 75 minutes)

2026-11-09

S6 Biology: Nervous and Endocrine Coordination

S6 Biology practice paper: action potentials, synapses, reflex arcs, muscle contraction, insulin, thyroxine and ADH feedback. 12 questions, 100 marks, 75 minutes.

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Section A: The Nervous System and Coordination (28 marks)

Answer all four questions in Section A and state membrane potentials with their signs.

#	Question	Answer and marking notes
1	1. The graph shows the membrane potential of a motor neurone: the resting potential is -70 mV; a stimulus raises the membrane potential to the threshold of -55 mV, which triggers an action potential peaking at +30 mV. (a) Calculate the change in potential from rest to peak, and the difference between the resting potential and the threshold. (b) Explain how the resting potential is maintained and how ions move during depolarisation. (c) Explain why an action potential does not travel backwards along the axon.	(a) Change from rest to peak = 30 – (-70) = 100 mV; the difference between the resting potential and the threshold = -55 – (-70) = 15 mV. (2 marks) (b) The sodium-potassium pump actively moves sodium out and potassium in, and the membrane is more permeable to potassium, so the outside is positive relative to the inside and the resting potential of -70 mV is maintained. On depolarisation, voltage-gated sodium channels open and sodium ions flow in, raising the inside to +30 mV. (3 marks) (c) After an action potential that region of membrane is refractory: the sodium channels are still inactivated and cannot be excited again, so the impulse travels only from the cell body towards the axon terminal. (2 marks) Marking note: potentials must carry their signs; writing only that sodium enters, without the pump or the refractory period, gains at most half marks.
2	2. A neural pathway contains 3 synapses, each with a transmission delay of 0.4 ms. (a) Calculate the total synaptic delay of the pathway. (b) Conduction velocity is about 100 m/s in a myelinated axon and 1 m/s in an unmyelinated axon. Calculate the ratio and explain the difference. (c) Describe the full process from neurotransmitter release at the presynaptic neurone to the postsynaptic potential.	(a) Total synaptic delay = 3 x 0.4 = 1.2 ms. (2 marks) (b) Velocity ratio = 100 / 1 = 100 times. The myelin sheath insulates the axon and the action potential jumps between nodes of Ranvier, so conduction is far faster. (3 marks) (c) The action potential reaches the presynaptic terminal, calcium ions enter, synaptic vesicles fuse with the membrane and release neurotransmitter by exocytosis; the transmitter diffuses across the cleft, binds to receptors on the postsynaptic membrane and causes sodium ions to enter, producing a postsynaptic potential. (2 marks) Marking note: calcium entry and exocytosis 1 mark each, receptor binding and postsynaptic potential 1 mark each.
3	3. A withdrawal reflex takes about 0.08 s from stimulus to the arm moving, whereas the same movement performed voluntarily takes about 0.25 s. (a) Calculate the difference in time and the percentage by which the reflex is faster. (b) List the five components of the withdrawal reflex arc and state where the reflex centre lies. (c) Explain why a reflex protects the body without conscious control, and state the role of the cerebrum in the reflex.	(a) Time difference = 0.25 – 0.08 = 0.17 s; the reflex is faster by 0.17 / 0.25 x 100% = 68.0%. (2 marks) (b) Receptor (pain receptors in the skin), sensory neurone, reflex centre in the spinal cord (via an interneurone), motor neurone, effector (arm muscles). The reflex centre is in the grey matter of the spinal cord. (3 marks) (c) The pathway is short and does not pass through the cerebral cortex, so the response is fast enough to withdraw the hand before injury. The cerebrum also receives the signal, so pain is felt and the movement can be modified afterwards, but a completed reflex cannot be cancelled. (2 marks) Marking note: 0.5 mark for each of the five components and 0.5 for the centre.
4	4. Functional areas of the brain include the cerebral cortex, the cerebellum and the medulla. (a) State the main role of each in coordinating body activity. (b) A patient has damage to the cerebellum. State two observable symptoms and explain their cause. (c) Explain why damage to the medulla is more often fatal than localised damage to the cortex.	(a) Cerebral cortex: consciousness, thought, memory, language and voluntary movement; cerebellum: coordination of muscle activity, balance and posture; medulla: involuntary control of heartbeat, breathing and blood pressure. (3 marks) (b) Two symptoms: uncoordinated movement such as an unsteady gait, and loss of balance so that walking in a straight line is difficult. The cerebellum cannot integrate position information from muscles and the inner ear, so the force and direction of movement cannot be corrected. (2 marks) (c) The medulla controls heartbeat and breathing, functions needed for life, so damage can be fatal within minutes; localised cortical damage usually affects specific functions such as language or limb movement while other functions continue. (2 marks) Marking note: the answer needs the contrast between vital functions and localised function.

Section B: Effectors and Hormonal Regulation (32 marks)

Use the data provided in Section B and give percentages to one decimal place.

#	Question	Answer and marking notes
5	5. A sarcomere was measured before and after contraction: the A band stays at 1.6 micrometres, the I band shortens from 0.9 to 0.5 micrometres and the H zone shortens from 1.0 to 0.6 micrometres. (a) Calculate the sarcomere length before and after contraction, and the percentage shortening. (b) Using the sliding filament theory, explain why the A band stays the same length while the I band and H zone both shorten. (c) Explain the roles of ATP and calcium ions in contraction.	(a) Length before = 1.6 + 0.9 = 2.5 micrometres; after = 1.6 + 0.5 = 2.1 micrometres; percentage shortening = (2.5 – 2.1) / 2.5 x 100% = 16.0%. (3 marks) (b) The thick (myosin) and thin (actin) filaments do not change length. The A band is the full length of the thick filaments, so it stays the same. During contraction the thin filaments slide towards the centre of the sarcomere, so the region of overlap increases and both the I band and the H zone shorten, each by 0.4 micrometres, which matches the sliding filament theory. (3 marks) (c) Calcium ions bind to troponin, moving tropomyosin and exposing the binding sites; ATP powers the power stroke and detachment of the myosin heads and is also used by the calcium pump to return calcium to the sarcoplasmic reticulum. (2 marks) Marking note: writing that the filaments themselves shorten contradicts the theory and scores nothing for part (b).
6	6. After a starchy meal, a healthy adult blood glucose rose from 5.4 mmol/L fasting to a peak of 8.8 mmol/L and fell to 5.6 mmol/L after two hours; blood insulin rose from 12 to 68 mU/L. (a) Calculate the rise in blood glucose, its percentage rise, and the fold increase in insulin. (b) Explain how the rise in glucose triggers insulin secretion by negative feedback, and state two target organs of insulin with their actions. (c) State which hormone rises as glucose returns to the fasting level and what it does.	(a) Rise = 8.8 – 5.4 = 3.4 mmol/L; percentage rise = 3.4 / 5.4 x 100% = 63.0%; insulin increased 68 / 12 = 5.67 times. (3 marks) (b) The rise in glucose is detected by the beta cells of the islets of Langerhans, which secrete insulin; insulin promotes uptake of glucose by cells and conversion of glucose to glycogen in the liver and muscles, lowering blood glucose, and the fall then reduces insulin secretion, which is negative feedback. (3 marks) (c) As glucose falls, the alpha cells secrete glucagon, which promotes the breakdown of liver glycogen to glucose released into the blood, raising glucose back to the fasting level. (2 marks) Marking note: the two target organs are worth 1 mark each.
7	7. Thyroxine and thyroid stimulating hormone (TSH) are regulated by negative feedback. Before and after treatment a patient showed thyroxine rising from 55 to 145 nmol/L and TSH falling from 4.8 to 1.2 mU/L. (a) Calculate the percentage rise in thyroxine and the percentage fall in TSH. (b) Explain why TSH fell and name the gland that secretes it. (c) If the patient takes too much thyroxine for a long time, predict the effects on body weight, body temperature and heart rate, with explanations.	(a) Thyroxine rise = 90 / 55 x 100% = 163.6%; TSH fall = 3.6 / 4.8 x 100% = 75.0%. (3 marks) (b) The higher blood thyroxine inhibits the hypothalamus and the anterior pituitary by negative feedback, so less TSH is secreted. TSH is secreted by the anterior pituitary. (2 marks) (c) Body weight falls because metabolic rate and energy use rise; body temperature rises because more heat is produced; heart rate rises because sympathetic effects are enhanced. Each needs its mechanism. (3 marks) Marking note: part (b) must mention both negative feedback and the anterior pituitary.
8	8. Oral glucose tolerance results for three adults (mmol/L): normal reference is 5.5 fasting and 6.5 at two hours; person A is 8.2 fasting and 14.5 at two hours; person B is 6.9 fasting and 9.8 at two hours. (a) Calculate how far A and B exceed the reference values when fasting and at two hours. (b) Decide which type of diabetes each most likely has and give one piece of supporting evidence. (c) Explain why a type 1 diabetic must inject insulin rather than take it by mouth.	(a) A: fasting exceeds the reference by 2.7 mmol/L and the two-hour value by 8.0 mmol/L; B: fasting exceeds it by 1.4 mmol/L and the two-hour value by 3.3 mmol/L. (4 marks) (b) A most likely has type 1 diabetes: the two-hour value is far above the reference and fasting is already clearly raised, consistent with an absolute lack of insulin. B most likely has type 2 diabetes: fasting is only slightly raised and the two-hour rise is smaller, consistent with insulin resistance. (3 marks) (c) Insulin is a protein hormone and would be digested by pepsin and stomach acid if swallowed, so it must be injected to keep its structure intact and reach the blood. (1 mark) Marking note: in part (b) the type and the evidence must be paired.

Section C: Comparison, Application and Experimental Design (40 marks)

Section C requires a clear conclusion, a mechanism and a measurable prediction.

#	Question	Answer and marking notes
9	9. Compare nervous coordination with hormonal coordination. (a) Compare them in a table under four headings: transmission route, speed, duration of effect and scope of action. (b) Using a faster heart rate when startled as an example, explain how the two systems act in sequence. (c) Explain why the nervous system, although it uses electrical signals, still depends on hormones for long-term regulation.	(a) Route: nervous through neurones and synapses, hormonal through the blood; speed: nervous fast (up to 100 m/s), hormonal slow (minutes); duration: nervous brief, hormonal long lasting; scope: nervous only the effector it supplies, hormonal the whole body but only cells with receptors respond. (4 marks) (b) When startled, sympathetic nerves speed the heart within seconds; at the same time the adrenal medulla secretes adrenaline into the blood, which prolongs and strengthens the faster heart rate and raises blood glucose, so the two systems act in sequence and reinforce each other. (3 marks) (c) Nervous signals are brief and local and cannot sustain long-term change; growth, metabolic rate and reproduction need regulation lasting days to years, which requires hormones carried in the blood. The nervous system also controls hormone secretion, for example the hypothalamus controls the pituitary, so the two depend on each other. (3 marks) Marking note: part (a) needs comparison, not just four labels.
10	10. An organophosphate pesticide inhibits acetylcholinesterase. Sprayed insects first show muscle spasms and later become paralysed and die. (a) Explain the normal function of acetylcholinesterase. (b) Explain how the acetylcholine concentration in the synaptic cleft changes when the enzyme is inhibited, and why spasms come before paralysis. (c) Suggest two measures that reduce the risk to humans handling such pesticides, each with a reason.	(a) Acetylcholinesterase hydrolyses acetylcholine in the synaptic cleft into acetate and choline, so the transmitter is removed quickly, the postsynaptic membrane repolarises and continuous stimulation is prevented. (2 marks) (b) With the enzyme inhibited, acetylcholine accumulates and the postsynaptic membrane stays depolarised, so the muscle is stimulated continuously and goes into spasm; prolonged depolarisation desensitises the receptors, the signal can no longer be transmitted and the muscle becomes paralysed. (5 marks) (c) Two measures: wear protective clothing and gloves to reduce skin contact and absorption; wash exposed skin thoroughly and change clothing after spraying to remove residues and cut absorption through the skin. (3 marks) Marking note: part (b) needs both accumulation and desensitisation; each measure in (c) needs a reason.
11	11. Two healthy adults joined a water balance test: A drank 1.0 litre and B drank 3.0 litres per day. Results: A produced 0.5 litre of urine with an osmolarity of 900 mOsm/L and blood ADH of 6.5 pmol/L; B produced 2.2 litres with an osmolarity of 300 mOsm/L and blood ADH of 1.4 pmol/L. (a) Calculate the urine volume ratio, the urine osmolarity ratio and the ADH ratio. (b) Explain why B has the lower ADH, and state the positions of the receptor and the effector. (c) Predict the urine volume and thirst of a person who secretes too little ADH, with explanations.	(a) Urine volume ratio = 2.2 / 0.5 = 4.4 times; osmolarity ratio = 900 / 300 = 3 times; ADH ratio = 6.5 / 1.4 = 4.64 times. (3 marks) (b) B drank more water, so blood osmolarity fell; osmoreceptors in the hypothalamus detect the change and less ADH is secreted. The effector is the wall of the collecting duct and distal tubule in the kidney; permeability to water falls, less water is reabsorbed, so a large volume of dilute urine is produced. (4 marks) (c) Urine volume rises greatly, up to several litres a day, and thirst and fluid intake increase because much water is lost in urine and blood osmolarity rises, stimulating the thirst centre. (3 marks) Marking note: the receptor and the effector are worth 1 mark each.
12	12. A student investigated the effect of caffeine on resting heart rate. Preliminary data: mean heart rate was 68 beats per minute before caffeine and 79 beats per minute after a drink containing 200 mg of caffeine. (a) Calculate the rise in heart rate and its percentage rise. (b) Write a testable hypothesis and design a controlled experiment, stating controlled variables, sample size and the measuring method. (c) State two confounding factors and one improvement for each.	(a) Rise = 79 – 68 = 11 beats per minute; percentage rise = 11 / 68 x 100% = 16.2%. (2 marks) (b) Hypothesis: after 200 mg of caffeine the resting heart rate will be higher than in a control group. Design: randomly assign participants to an experimental and a control group of at least 15 each; the experimental group drinks a beverage containing 200 mg of caffeine and the control group the same volume without caffeine; control the drink temperature, volume, time of testing, age and exercise habits; measure heart rate three times before and after and compare mean rises between groups. (6 marks) (c) Confounding factors: participants know their group, causing a psychological effect, so use a double-blind design; exercise or emotion before testing alters heart rate, so rest for 10 minutes and use the same environment. (2 marks) Marking note: part (b) hypothesis 1 mark, grouping 1, controlled variables 2, measuring method 2.

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