Respiration, Muscles and the Internal Environment

Edexcel International A-Level Biology· 7.1–7.22· 45 min read

1. Aerobic & Anaerobic Respiration★★☆☆☆⏱ 10 min

Aerobic respiration is a stepped, enzyme-controlled process that releases energy from organic molecules to synthesise ATP. Its four key stages occur in distinct cellular locations: glycolysis in the cytoplasm, link reaction and Krebs cycle in the mitochondrial matrix, and oxidative phosphorylation on the inner mitochondrial membrane (cristae).

Glycolysis: Phosphorylates hexose sugars, produces 2 ATP (substrate-level), reduced NAD, and pyruvate. Under anaerobic conditions, pyruvate is converted to lactate in animal cells, oxidising reduced NAD to allow glycolysis to continue.
Link reaction: Decarboxylates pyruvate (releases CO₂), produces acetyl coenzyme A, and reduced NAD.
Krebs cycle: Decarboxylates acetyl CoA, produces 2 ATP (substrate-level), reduced NAD and FAD per glucose molecule.
Oxidative phosphorylation: Uses electrons from reduced coenzymes to power the electron transport chain (ETC), creating a proton gradient across the inner mitochondrial membrane. Protons diffuse down their gradient via ATP synthase, driving ATP synthesis (chemiosmosis).

📘 Definition

Respiratory Quotient (RQ)

Ratio of the volume of carbon dioxide produced to the volume of oxygen consumed during respiration, used to identify the primary respiratory substrate. Typical RQ values: carbohydrates = 1.0, lipids = 0.7, proteins = 0.9.

📐 Worked Example

Calculate the RQ of a germinating seed that consumes 20 cm³ of oxygen and produces 14 cm³ of carbon dioxide. Identify the most likely respiratory substrate being used.

1
Step 1: Use the RQ formula:
$RQ = \frac{\text{Volume of } CO_2 \text{ produced}}{\text{Volume of } O_2 \text{ consumed}}$
2
Step 2: Substitute the given values:
$RQ = \frac{14}{20} = 0.7$
3
Step 3: Match to standard RQ values: an RQ of 0.7 indicates the primary substrate is lipid.

Core practical 15 uses redox indicators (e.g. methylene blue) to measure the rate of respiration in yeast, as reduced coenzymes decolourise the indicator. Core practical 16 uses a respirometer to measure respiration rate and calculate RQ by measuring changes in gas volume inside a sealed vessel.

Exam tip:

Always state the exact location of each respiration stage in exam answers: glycolysis = cytoplasm, link/Krebs = mitochondrial matrix, oxidative phosphorylation = inner mitochondrial membrane. Mentioning 'mitochondria' alone is not enough for full marks.

2. Muscle Structure and Contraction★★★☆☆⏱ 10 min

Skeletal muscles work in antagonistic pairs with tendons, ligaments and the skeleton to produce movement: flexors bend a joint, while extensors straighten it. Skeletal muscle fibres are categorised into slow twitch (type I) and fast twitch (type II) fibres, adapted for different activity types.

Feature	Slow Twitch Fibres	Fast Twitch Fibres
Primary energy source	Aerobic respiration	Anaerobic respiration
Contraction speed	Slow, sustained	Fast, short bursts
Mitochondria density	High	Low
Myoglobin content	High (red colour)	Low (pale colour)
Use case	Endurance activities (e.g. long-distance running)	Power activities (e.g. sprinting, weightlifting)

📘 Definition

Sliding Filament Theory

Theory explaining muscle contraction, where actin and myosin filaments slide past each other to shorten the sarcomere, without changing the length of either filament type.

The key steps of sliding filament contraction are: 1) An action potential releases Ca²⁺ from the sarcoplasmic reticulum. 2) Ca²⁺ binds to troponin, moving tropomyosin to expose myosin binding sites on actin filaments. 3) Myosin heads bind to actin, forming cross-bridges, and pull actin filaments towards the centre of the sarcomere using energy from ATP hydrolysis (catalysed by myosin ATPase). 4) ATP binds to myosin heads to break cross-bridges, and the cycle repeats.

📐 Worked Example

Explain why the A-band of a sarcomere stays the same length during muscle contraction, while the I-band narrows.

1
Step 1: Define each sarcomere region: the A-band spans the full length of the thick myosin filaments. The I-band contains only thin actin filaments, with no overlap with myosin.
2
Step 2: Recall the sliding filament mechanism: actin and myosin filaments slide past each other, rather than shortening themselves.
3
Step 3: Explain the length change: as actin slides towards the centre of the sarcomere, the region where actin is not overlapping with myosin (the I-band) becomes smaller. The myosin filaments do not change length, so the A-band remains constant.

Exam tip:

If asked to label a sarcomere diagram, remember: A-band = myosin length (constant), I-band = actin only (narrows), H-zone = myosin only (narrows). Mixing these up is a common mark loss point.

3. Cardiac Function and Exercise Response★★★☆☆⏱ 10 min

Cardiac muscle is myogenic, meaning it contracts without nervous stimulation. The electrical conduction pathway of the heart controls coordinated contraction: 1) The sinoatrial node (SAN, the natural pacemaker) in the right atrium initiates a wave of electrical excitation, causing atrial contraction. 2) The atrioventricular node (AVN) delays the impulse by ~0.1s to allow atria to fully empty before ventricles contract. 3) The impulse travels down the bundle of His in the septum, then spreads through Purkyne fibres in the ventricle walls, causing ventricular contraction from the apex upwards.

📘 Definition

Electrocardiogram (ECG)

A trace that records the electrical activity of the heart over time, used to diagnose abnormal heart rhythms. Key features: P wave (atrial depolarisation), QRS complex (ventricular depolarisation), T wave (ventricular repolarisation).

During exercise, the medulla oblongata controls increases in cardiac output (stroke volume × heart rate) and ventilation rate (tidal volume × breathing rate) to deliver more oxygen to working muscles. Adrenaline release during the fight-or-flight response also increases heart rate and ventilation rate to prepare the body for activity. Core practical 17 uses a spirometer to measure the effect of exercise on tidal volume, breathing rate, minute ventilation and oxygen consumption.

📐 Worked Example

A student has a resting tidal volume of 0.5 dm³ and resting breathing rate of 12 breaths per minute. After 5 minutes of running, their tidal volume increases to 2.2 dm³ and breathing rate to 30 breaths per minute. Calculate the percentage increase in their minute ventilation.

1
Step 1: Calculate resting minute ventilation:
$0.5 \times 12 = 6 \text{ dm}^3 \text{ min}^{-1}$
2
Step 2: Calculate exercise minute ventilation:
$2.2 \times 30 = 66 \text{ dm}^3 \text{ min}^{-1}$
3
Step 3: Calculate percentage increase:
$\frac{(66 - 6)}{6} \times 100 = 1000\%$

Exam tip:

When describing the cardiac conduction pathway, always list the structures in the correct order: SAN → AVN → bundle of His → Purkyne fibres. Transposing AVN and SAN will lose marks.

4. Homeostasis and Hormonal Regulation★★★☆☆⏱ 8 min

Homeostasis is the maintenance of a stable internal environment via dynamic equilibrium, controlled by feedback mechanisms. Negative feedback reverses a change in a physiological variable to return it to its set point, while positive feedback amplifies a change (e.g. uterine contractions during labour).

The hypothalamus controls thermoregulation during exercise: it detects increases in core body temperature, and triggers responses including sweating (evaporative cooling) and vasodilation of skin arterioles to increase heat loss. If core temperature drops too low, shivering and vasoconstriction of skin arterioles are triggered to generate and conserve heat.

Hormones regulate gene expression via two mechanisms: peptide hormones (e.g. insulin, ADH) bind to extracellular receptors on cell surface membranes, triggering secondary messenger pathways that activate transcription factors inside the cell. Steroid hormones (e.g. testosterone, oestrogen) are lipid-soluble, so they diffuse across cell membranes and bind to intracellular receptors, forming a hormone-receptor complex that acts as a transcription factor to switch genes on or off.

📐 Worked Example

Explain why peptide hormones cannot bind to intracellular receptors, while steroid hormones can.

1
Step 1: Recall the chemical properties of each hormone type: peptide hormones are water-soluble (hydrophilic), while steroid hormones are lipid-soluble (lipophilic).
2
Step 2: Relate to cell membrane structure: the cell membrane has a hydrophobic phospholipid bilayer core, which prevents water-soluble molecules from diffusing across it.
3
Step 3: Explain the difference: peptide hormones cannot cross the phospholipid bilayer, so they must bind to extracellular receptors. Steroid hormones can diffuse freely across the membrane to bind to intracellular receptors in the cytoplasm or nucleus.

Exam tip:

When explaining hormone action, clearly distinguish between peptide (extracellular receptors, secondary messengers) and steroid (intracellular receptors, direct transcription factor action) mechanisms. Confusing these is a common error.

5. Kidney Structure and Osmoregulation★★★★☆⏱ 7 min

The kidneys excrete waste products (including urea, produced by the liver from excess amino acids) and regulate plasma water potential via osmoregulation. Key nephron functions include ultrafiltration in the Bowman’s capsule, selective reabsorption in the proximal convoluted tubule, and water regulation in the loop of Henle and collecting duct.

Ultrafiltration: High hydrostatic pressure in the glomerulus forces small molecules (water, glucose, urea, ions) into the Bowman’s capsule, while large molecules (proteins, blood cells) stay in the blood.
Selective reabsorption: All glucose, most water and ions are reabsorbed from the proximal convoluted tubule back into the blood via active transport and diffusion.
Loop of Henle: Uses a countercurrent multiplier mechanism to create a high solute concentration in the medulla of the kidney, allowing water to be reabsorbed from the collecting duct.

📘 Definition

Antidiuretic Hormone (ADH)

Hormone released by the posterior pituitary gland when osmoreceptors in the hypothalamus detect high plasma solute concentration (low water potential). ADH increases the permeability of the collecting duct to water, increasing water reabsorption and producing more concentrated urine.

📐 Worked Example

A person drinks 2 litres of water in a short period of time. Explain how ADH regulates their plasma water potential via negative feedback.

1
Step 1: Detect the change: Drinking large amounts of water lowers plasma solute concentration (high water potential), detected by osmoreceptors in the hypothalamus.
2
Step 2: Adjust ADH release: Less ADH is released from the posterior pituitary gland.
3
Step 3: Target organ response: The collecting duct becomes less permeable to water, so less water is reabsorbed into the blood, and large volumes of dilute urine are produced.
4
Step 4: Outcome: Plasma water potential returns to its normal set point, completing the negative feedback loop.

Exam tip:

You do not need to recall the ornithine cycle for urea production, or the names of specific transporters in the nephron for IAL exams. Stick to the key processes outlined in the specification for full marks.

6. Common Pitfalls

Wrong move:

Stating that glycolysis occurs in the mitochondria, or that the Krebs cycle occurs in the cytoplasm.

Why:

Exam marks are only awarded for exact location matches per the specification.

Correct move:

Always specify: glycolysis = cytoplasm, link reaction + Krebs cycle = mitochondrial matrix, oxidative phosphorylation = inner mitochondrial membrane.

Wrong move:

Claiming the A-band of the sarcomere shortens during muscle contraction.

Why:

The A-band spans the full length of myosin filaments, which do not shorten during contraction.

Correct move:

State that only the I-band (actin only) and H-zone (myosin only) narrow during contraction, while the A-band remains constant in length.

Wrong move:

Calculating RQ as oxygen consumed divided by carbon dioxide produced.

Why:

The formula for RQ is explicitly CO₂ produced / O₂ consumed per the specification.

Correct move:

Memorise the RQ formula and match results to standard values (carb = 1.0, lipid = 0.7, protein = 0.9) to identify substrates.

Wrong move:

Describing the cardiac conduction pathway as SAN → bundle of His → AVN → Purkyne fibres.

Why:

The AVN delays the impulse before it travels down the bundle of His to ensure coordinated atrial and ventricular contraction.

Correct move:

Always list the pathway in the correct order: SAN → AVN → bundle of His → Purkyne fibres.

Wrong move:

Stating that ADH decreases the permeability of the collecting duct to water.

Why:

ADH is released to conserve water when plasma water potential is low.

Correct move:

Remember ADH increases collecting duct water permeability, leading to more water reabsorption and concentrated urine.

Wrong move:

Confusing peptide and steroid hormone action, claiming peptide hormones bind to intracellular receptors.

Why:

Peptide hormones are hydrophilic and cannot cross the phospholipid bilayer of the cell membrane.

Correct move:

State peptide hormones bind to extracellular receptors (secondary messengers), while steroid hormones bind to intracellular receptors (act as transcription factors).

7. Quick Reference Cheatsheet

Concept	Key Facts
Respiration locations	Glycolysis: cytoplasm; Link/Krebs: mitochondrial matrix; Oxidative phosphorylation: inner mitochondrial membrane
RQ Values	Carbohydrate = 1.0; Lipid = 0.7; Protein = 0.9
Sarcomere contraction	A-band = constant length; I-band + H-zone = narrow
Cardiac conduction	SAN → AVN → bundle of His → Purkyne fibres
ADH function	Increases collecting duct water permeability to conserve water
Hormone action	Peptide = extracellular receptors; Steroid = intracellular transcription factors

8. Frequently Asked

What is the difference between substrate-level and oxidative phosphorylation?

Substrate-level phosphorylation directly transfers a phosphate group from a high-energy intermediate to ADP to make ATP, and occurs in the cytoplasm (glycolysis) and mitochondrial matrix (Krebs cycle). Oxidative phosphorylation uses energy from the electron transport chain to create a proton gradient, which drives ATP synthase to make ATP on the inner mitochondrial membrane.

Which parts of the sarcomere change length during muscle contraction?

During contraction, the I-band (thin actin filaments only) and H-zone (thick myosin filaments only) narrow, while the A-band (full length of myosin filaments) stays the same length, as actin and myosin slide past each other.

Where does each stage of aerobic respiration occur?

Glycolysis takes place in the cytoplasm, the link reaction and Krebs cycle occur in the mitochondrial matrix, and oxidative phosphorylation occurs on the inner mitochondrial membrane (cristae).

Going deeper

What's Next

Now that you have mastered Topic 7 content, move on to Edexcel IAL Biology Unit 5 Topic 8 (Coordination and Gene Technology) which covers nervous system function, synapses, and recombinant DNA technology, building on the hormonal regulation and gene expression content you learned here. Practice full Unit 5 past papers to consolidate your knowledge of both topics, and focus on 6-mark extended response questions for high-mark topics including sliding filament theory and negative feedback mechanisms. Revise core practical methods and data analysis to prepare for practical-based exam questions, which make up ~15% of Unit 5 marks.