OxfordAQA International AS and A-level Biology: Homeostasis, feedback, heart rate (9610) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover sections 3.4.6.1, 3.4.6.2 and 3.4.8 of the OxfordAQA International AS and A-level Biology (9610) specification, Version 5.1 (Unit 4: Control). Teaching is in the study guide and the revision notes. Course pages: hub and checklist.

Questions

1. A marathon runner’s core temperature rises from 37.0 °C to 41.5 °C in hot weather. Explain why this rise threatens the enzymes in her cells. [3]

2. A patient’s blood pH is 7.1. The normal value is about 7.4. Explain how this change could reduce the activity of enzymes. [4]

3.

(a) State what negative feedback is. [2] (b) Suggest why separate mechanisms for correcting a rise and a fall give better control than one mechanism. [2]

4. A volunteer sat in cold water. The table shows the results.

Time (min)	0	15	30	45	60
Core temperature (°C)	37.0	36.6	36.0	35.2	34.2
Metabolic heat production (W)	90	120	170	150	110

(a) Calculate the percentage increase in heat production between 0 and 30 minutes. [2] (b) Calculate the rate of fall in core temperature between 45 and 60 minutes in °C per hour. [2] (c) Explain what the data show about feedback in the first 30 minutes and after 45 minutes. [2]

5. During labour, stretching of the cervix sends impulses to the brain, which makes the pituitary gland release more oxytocin. Oxytocin increases contractions of the uterus, which stretch the cervix further.

(a) Identify the type of feedback and give a reason. [2] (b) Suggest how this loop is brought to an end. [2]

6.

(a) Cardiac muscle is described as myogenic. State what this means. [1] (b) Describe the role of the sinoatrial node (SAN). [2]

7. Describe how the wave of electrical activity from the SAN is passed on so that the ventricles contract after the atria and from the apex upwards. [6]

8. A runner’s heart rate rises from 72 to 156 beats per minute.

(a) Calculate the duration of one cardiac cycle at each rate, in seconds to 2 significant figures. [2] (b) Calculate the percentage reduction in the duration of the cycle. Round your answer to a whole number. [2]

9. The table shows data from a volunteer.

Condition	Lying down	10 s after standing	2 min after standing
Mean arterial pressure (mmHg)	93	78	90
Impulses per second from pressure receptors	30	19	27
Heart rate (beats per minute)	66	82	70

(a) Calculate the percentage increase in heart rate between lying down and 10 s after standing. Round your answer to a whole number. [2] (b) Use the data to explain the change in heart rate. [4]

10. Explain how heart rate is increased at the start of exercise and returned to its resting level afterwards. [7]

11. In an anaesthetised mammal, the nerves to the heart were cut or stimulated and the heart rate recorded.

Treatment	Heart rate (beats per minute)
Nerves intact	72
Vagus nerves cut	94
Sympathetic nerves to heart cut	71
Both cut	93
Vagus nerve stimulated	48
Sympathetic nerve stimulated	130

(a) Calculate the percentage increase in heart rate when the vagus nerves were cut. Express the answer to three significant figures. [2] (b) Explain what the results show about the nerves at rest. [3] (c) Explain how the results support the idea that separate mechanisms give greater control of heart rate. [3]

Answers

1. The high temperature breaks hydrogen bonds and other weak bonds in the enzyme [1]; its tertiary structure alters, so the active site changes shape [1]; substrate no longer fits, so fewer enzyme-substrate complexes form, the enzyme is denatured and reaction rate falls [1]. [3] Examiner insight: answers that say only “enzymes stop working” are too vague; name the bonds and the active site.

2. The change in pH alters the hydrogen ion concentration [1]; this changes the charges on R groups, so ionic and hydrogen bonds break or form in the wrong places [1]; the tertiary structure alters and the active site changes shape [1]; fewer enzyme-substrate complexes form, so activity falls (or the enzyme is denatured) [1]. [4] Examiner insight: pH answers tend to gain credit when they name hydrogen ions or R group charge, not just “the enzyme changes shape”.

3. (a) A receptor detects a change from the original level [1]; the response reverses the change and restores the original level [1]. [2] (b) Each direction is corrected actively and quickly, and an overshoot is reversed at once [1]; the level stays within narrower limits [1]. [2] Examiner insight: “greater control” needs a comparison with a single mechanism; saying “two mechanisms are better” without the reason gains nothing.

4. (a) (170 − 90) ÷ 90 × 100 [1] = 88.9 % [1]. (b) Fall = 35.2 − 34.2 = 1.0 °C in 15 min [1]; 1.0 ÷ 15 × 60 = 4.0 °C per hour [1]. (c) In the first 30 minutes heat production rises as temperature falls, which opposes the change: negative feedback [1]; after 45 minutes heat production falls as temperature falls, so the response adds to the change (positive feedback), because slower enzyme-controlled reactions release less heat [1]. [6] Examiner insight: in a calculation answer, show the substitution and the units; in interpretation, say whether the response makes the change smaller or bigger.

5. (a) Positive feedback [1]; the response (contractions) increases the stimulus (stretching), so the departure grows [1]. (b) Birth of the baby removes the stretching of the cervix [1]; less oxytocin is released, so contractions stop [1]. [4] Examiner insight: “positive” must be justified by what the response does to the stimulus; a bare label does not show understanding.

6. (a) It contracts and relaxes without needing a nerve impulse; the stimulus starts in the muscle itself [1]. (b) The SAN acts as the pacemaker and starts a wave of depolarisation [1] that spreads across the atria and makes them contract [1]. [3] Examiner insight: “the SAN controls the heart” is too general; say what it initiates and where the wave goes.

7. The wave spreads over the atria, which contract [1]; fibrous tissue between atria and ventricles blocks direct crossing [1]; the wave reaches the AVN [1]; the AVN holds it briefly, giving the atria time to empty before the ventricles contract [1]; the wave passes down the bundle of His in the septum [1]; Purkinje tissue carries it to the apex and up the walls, so the ventricles contract from the apex upwards [1]. [6] Examiner insight: the order matters in a “describe the route” answer; the delay belongs to the AVN, not Purkinje tissue.

8. (a) 60 ÷ 72 = 0.83 s [1]; 60 ÷ 156 = 0.38 s [1]. (b) (0.833 − 0.385) ÷ 0.833 × 100 [1] = 54 % [1]. [4] Examiner insight: rounding early can shift the final figure; carrying full values until the end is safer.

9. (a) (82 − 66) ÷ 66 × 100 [1] = 24 % [1]. (b) Standing lowers the arterial pressure (93 to 78 mmHg) [1]; pressure receptors send fewer impulses (30 to 19 per second) to the medulla [1]; the medulla sends more sympathetic (and fewer parasympathetic) impulses to the SAN, so heart rate rises to 82 [1]; this raises pressure again, and after 2 min the pressure (90), impulses (27) and heart rate (70) are near their lying values, which is negative feedback [1]. [6] Examiner insight: use numbers from the table at each link in the chain; an answer that uses only general statements loses the data-use credit.

10. Respiring muscle makes more carbon dioxide [1]; blood pH falls [1]; the fall is detected by chemoreceptors located in the carotid arteries and aorta [1]; they send more impulses to the cardiac control centre in the medulla [1]; impulses pass along sympathetic neurones to the SAN [1]; the SAN generates waves faster, so heart rate rises [1]; afterwards carbon dioxide falls and pH returns to normal, so impulses along the sympathetic neurones decrease and the vagus (parasympathetic) nerve slows the SAN back to the resting rate [1]. [7] Examiner insight: give both phases, the rise and the return; many answers stop after the rise.

11. (a) (94 − 72) ÷ 72 × 100 [1] = 30.6 % [1]. (b) Cutting the vagus raises the rate, so at rest the vagus is slowing the heart [1]; cutting the sympathetic nerves has almost no effect (71 against 72), so there is little sympathetic stimulation at rest [1]; the heart still beats at 93 with both cut, because the SAN generates the beat itself (myogenic) [1]. (c) Stimulating the vagus lowers the rate to 48 [1]; stimulating the sympathetic nerve raises it to 130 [1]; the two nerves act in opposite directions, so the rate can be corrected quickly whether it needs to go up or down [1]. [8] Examiner insight: in an evaluation of data, tie each conclusion to a quoted value and state what the nerve does, not just whether the number changed.

Where marks are usually lost

• Writing that cold denatures enzymes, instead of saying they slow.
• Describing enzyme changes without bonds, tertiary structure and the active site.
• Not saying that a negative feedback response reverses the change.
• Labelling positive feedback without explaining that the response increases the change.
• Percentage change calculated from the wrong starting value.
• Mixing up the SAN, AVN and Purkinje tissue.
• Forgetting the cardiac control centre in the medulla and writing “the brain” only.
• Forgetting the way back (recovery) in heart-rate explanations.
• Giving data answers without quoting figures.

Next steps

• Revision notes
• Study guide
• OxfordAQA A-level Biology hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

OxfordAQA International AS and A-level Biology (9610), Version 5.1, for International AS exams May/June 2017 onwards and International A-level exams May/June 2018 onwards, published by OxfordAQA. Sections 3.4.6.1, 3.4.6.2 and 3.4.8.