OxfordAQA International AS and A-level Biology: Homeostasis, feedback and control of heart rate (9610)

This guide follows the OxfordAQA International AS and A-level Biology (9610) specification, Version 5.1, and covers sections 3.4.6.1, 3.4.6.2 and 3.4.8 of Unit 4: Control. Unit 4 is an International A-level only unit, assessed through the International A2 papers. Blood glucose (3.4.7) has its own pages and is not covered here.

Use the OxfordAQA A-level Biology hub and the printable checklist to track progress. When you finish, try the revision notes and the practice questions, then check yourself with the free diagnostics. Heart structure and the cardiac cycle are taught in blood vessels, the heart and coronary heart disease.

What this unit covers

Section	What you must be able to do
3.4.6.1	Define homeostasis as keeping the internal environment within restricted limits. Explain why a stable core temperature and blood pH matter for enzyme activity
3.4.6.2	Explain negative feedback, why separate mechanisms for each direction give greater control, and positive feedback. Interpret information on examples of both
3.4.8	Describe myogenic stimulation and the roles of the SAN, AVN and Purkinje tissue in the bundle of His. Describe the roles of chemoreceptors, pressure receptors, the autonomic nervous system and effectors in controlling heart rate

3.4.6.1 The principles of homeostasis

In mammals, homeostasis involves physiological control systems that maintain the internal environment within restricted limits. The internal environment is the blood and the tissue fluid that surrounds cells. It is not held at one fixed value. It fluctuates a little around a set point.

Core temperature and enzymes

Human core temperature is about 37 °C. Enzymes are the reason this matters.

• Too low: enzyme and substrate molecules move more slowly, so they collide less often and fewer enzyme-substrate complexes form each second. The rate of reaction falls. The enzyme is not damaged.
• Too high: the enzyme’s hydrogen bonds and other weak bonds break, the tertiary structure changes, and the active site no longer fits the substrate. The enzyme is denatured, and this is not reversible.

A small rise can speed reactions up. A large rise stops them. Both extremes threaten metabolism.

Blood pH and enzymes

Blood pH is held at about 7.4 (roughly 7.35 to 7.45). A change in pH alters the concentration of hydrogen ions, which changes the charges on R groups in the enzyme. Ionic and hydrogen bonds then break or form in the wrong places, the tertiary structure alters and the active site is distorted. Substrate binds less well, so enzyme activity falls. In severe cases the enzyme is denatured.

Worked example: pH and enzyme activity

An enzyme in cells has this relative activity (optimum = 100).

pH	6.8	7.0	7.2	7.4	7.6	7.8
Relative activity	22	55	88	100	91	60

1. Fall in activity at pH 7.0 compared with the optimum: 100 − 55 = 45 %.
2. Fall at pH 7.2: 100 − 88 = 12 %.
3. Conclusion: a fall of 0.4 pH units from normal removes almost half of the activity. This is why blood pH must stay within narrow limits. The curve is steeper on the acid side here, so a drop in pH is more damaging than an equal rise.

3.4.6.2 Feedback

Negative feedback

Negative feedback restores systems to their original level. A control loop has these parts:

1. A receptor detects a change (a departure from the set point).
2. A coordinator (such as the brain) processes the information.
3. An effector (a muscle or gland) produces a response that reverses the change.

When the level returns, the stimulus ends and the response switches off.

Separate mechanisms in each direction

The specification states that having separate mechanisms controlling departures in different directions gives a greater degree of control. Temperature is the standard example.

Direction of departure	Responses that restore it
Core temperature rises	Vasodilation of skin arterioles, more sweating, hairs lie flat
Core temperature falls	Vasoconstriction of skin arterioles, shivering, hairs raised, less sweating

Why this is better than one mechanism:

• With one mechanism that is simply switched on and off, the level can only be pushed back one way. If it overshoots, it can only fall slowly by waiting.
• With two opposing mechanisms, an overshoot is corrected at once in the other direction. The level is held in a narrower range and is restored faster.

The heart works the same way. Separate nerves speed it up and slow it down (see 3.4.8).

Worked example: interpreting a negative feedback record

A volunteer sat in a chamber held at 40 °C for an hour.

Time (min)	0	10	20	30	40	50	60
Core temperature (°C)	37.0	37.1	37.3	37.4	37.3	37.1	37.0
Sweat rate (g per hour)	30	80	210	360	420	330	250

1. Rate of rise from 0 to 30 min: 0.4 °C in 30 min = 0.4 ÷ 30 × 60 = 0.8 °C per hour.
2. Percentage rise in sweat rate from 0 to 30 min: (360 − 30) ÷ 30 × 100 = 1100 %.
3. Interpretation: warmth is detected, sweating increases, and evaporation cools the body. Temperature peaks at 30 min, then falls even though the chamber stays hot. Sweat rate peaks after temperature, because the response lags behind the stimulus. As temperature returns to 37.0 °C, sweating declines. This is negative feedback.

Positive feedback

Positive feedback results in greater departures from the original level. The change produces a response that increases the change. The specification notes that it is often associated with the breakdown of control systems, such as in the control of temperature.

Think of temperature running out of control. Enzyme-controlled reactions speed up as temperature rises, so more heat is released by metabolism, which raises temperature further. In the cold, the reverse happens: a falling temperature slows metabolic reactions, less heat is produced, and temperature falls further.

Positive feedback also occurs in normal body processes. When you are given an example you have not met, ask “does the response make the change bigger or smaller?” Bigger means positive.

Worked example: a runaway loop

Data supplied: at 37.0 °C a person’s metabolic heat output is 90 W. Assume it rises by 12 % of its current value for each 1.0 °C rise. The maximum rate at which the body can lose heat is 130 W.

Core temperature (°C)	37	38	39	40	41	42
Heat output (W)	90.0	100.8	112.9	126.4	141.6	158.6

1. Each value is the previous one × 1.12. At 40 °C: 90 × 1.12³ = 126.4 W.
2. At 40 °C, output (126.4 W) is still below the maximum loss (130 W), so negative feedback can still win.
3. At 41 °C, output (141.6 W) exceeds 130 W. Heat builds up faster than it can be lost, and each rise causes a bigger rise. Control has broken down.
4. Limit of the model: real enzymes denature at higher temperatures, which would eventually reduce heat production, but by then cells are damaged.

3.4.8 Control of heart rate

Myogenic stimulation

Cardiac muscle is myogenic. It contracts and relaxes rhythmically without needing a nerve impulse to start each beat. The stimulus arises within the muscle itself. Nerves only change the rate.

The conducting system

1. The sinoatrial node (SAN) is a small patch of tissue in the wall of the right atrium. It acts as the pacemaker and initiates a wave of electrical activity (depolarisation) that spreads across both atria. The atria contract (atrial systole).
2. Fibrous tissue that separates the atria from the ventricles cannot conduct the wave, so it cannot cross there.
3. The wave reaches the atrioventricular node (AVN), which holds the wave briefly before releasing it. This gives the atria time to finish emptying into the ventricles first.
4. The wave travels down the bundle of His, running down the septum between the ventricles. It divides into Purkinje tissue (fibres), which carries the wave to the apex of the heart and then up the ventricle walls.
5. The ventricles contract from the apex upwards (ventricular systole), driving blood towards the arteries.

Changing the rate: receptors, coordinator, effectors

The autonomic nervous system is the part of the nervous system that works involuntarily. It has two divisions, and they have opposite effects on the heart: a separate mechanism for each direction.

Part of loop	Structure	Role
Chemoreceptors	In the walls of the carotid arteries and aorta	Detect changes in blood pH, which follows carbon dioxide concentration
Pressure receptors	In the walls of the carotid arteries and aorta	Detect changes in blood pressure
Coordinator	Cardiac control centre in the medulla oblongata	Receives impulses and sends impulses to the heart
Effector	The SAN (and cardiac muscle)	Changes the rate at which the SAN generates waves

• To speed up the heart: impulses pass along sympathetic neurones to the SAN, which increases the rate.
• To slow the heart: impulses pass along parasympathetic neurones (the vagus nerve) to the SAN, which decreases the rate.

Worked example: exercise and recovery

A volunteer’s data:

Condition	Blood pH	Heart rate (beats per min)
Rest	7.40	68
After 5 min exercise	7.32	150
10 min into recovery	7.39	82

1. Percentage rise in heart rate during exercise: (150 − 68) ÷ 68 × 100 = 121 % (3 s.f.).
2. Time for one cardiac cycle: 60 ÷ heart rate. Rest: 60 ÷ 68 = 0.88 s. Exercise: 60 ÷ 150 = 0.40 s.
3. Explanation: respiring muscle releases more carbon dioxide, which lowers blood pH (by 0.08 here). Chemoreceptors detect this and send more impulses to the cardiac control centre. It sends more impulses along sympathetic neurones to the SAN, so the heart rate rises. The faster blood flow removes carbon dioxide faster. In recovery, pH rises towards 7.40, so chemoreceptor stimulation falls and the heart rate returns towards 68. This is negative feedback.

Worked example: pressure receptors

Data supplied from a resting volunteer:

Mean arterial pressure (mmHg)	60	100	140
Heart rate (beats per min)	78	70	60

• Change in heart rate from 60 to 140 mmHg: (60 − 78) ÷ 78 × 100 = −23.1 %.
• Pressure rises, pressure receptors send more impulses to the medulla, the medulla sends more impulses along the vagus (parasympathetic) nerve to the SAN, and the heart rate falls. That lowers pressure again. When pressure is low, fewer impulses arrive and sympathetic stimulation raises the heart rate.

Common errors

• Writing that enzymes are “killed” or that low temperature denatures them. Cold slows enzymes. Heat denatures them.
• Saying negative feedback “stops” the change rather than reversing it, or giving a response that does not return the level to the set point.
• Calling any increase positive feedback. Positive feedback means the response amplifies the change.
• Placing the SAN in the ventricle or the AVN in the atrium. The SAN is in the right atrium wall; the AVN is between atria and ventricles.
• Saying the nervous system “starts” the heartbeat. The heart is myogenic; nerves change the rate.
• Writing that Purkinje tissue delays the wave. The AVN delays it; Purkinje tissue conducts it quickly.
• Forgetting the full chain: stimulus, receptor, coordinator, effector, response.

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.