Free Pearson Edexcel International A-level Biology · Unit 5

Skeletal muscles, fibre types and sliding filaments

What you’ll learn

  • Relate skeletal structures to antagonistic movement.
  • Explain the cross-bridge cycle and calcium control.
  • Compare slow- and fast-twitch fibres.

Understand the idea

Tendons transmit force from muscle to bone, while ligaments connect bone to bone and stabilise joints. Muscles exert pulling forces, so antagonistic pairs produce opposite movements: a flexor reduces a joint angle and an extensor increases it. A skeletal muscle fibre is a long, multinucleate cell containing myofibrils. Repeating sarcomeres contain thin actin and thick myosin filaments; shortening many sarcomeres together shortens the fibre.

Excitation releases calcium ions from the sarcoplasmic reticulum. Calcium binds troponin, moving tropomyosin away from myosin-binding sites on actin. A myosin head binds actin and performs a power stroke. ATP binding detaches the head; ATP hydrolysis provides energy to reset it for another cycle. Filaments slide past each other without becoming shorter. Active calcium uptake into the sarcoplasmic reticulum promotes relaxation and also requires energy.

Slow-twitch fibres are adapted to sustained aerobic activity, with many mitochondria, abundant myoglobin and a dense capillary supply; they resist fatigue. Fast-twitch fibres contract rapidly and generate high power, with greater reliance on rapid ATP supply including anaerobic pathways in many fast-fibre types, but generally fatigue sooner. The proportion recruited depends on the force and duration required; actual muscle contains a mixture rather than a single universal fibre type.

Key formulae

percentage shortening=Linitial−LfinalLinitial×100%\text{percentage shortening}=\frac{L_{\mathrm{initial}}-L_{\mathrm{final}}}{L_{\mathrm{initial}}}\times100\%

Shortening percentage depends on the initial length

Conditions. Geometric calculation for fixed initial length 2.5 µm: shortening = 100(2.5 − L)/2.5. The finite 1.5–2.5 µm range is an arithmetic example, not a force–length or safety model.

Final sarcomere length / µm; Shortening / %.

At final length 2.0 µm shortening is 20%; at 2.5 µm it is zero. The plot does not imply that actin and myosin filaments themselves shorten.

Open the interactive lesson to view plotted curves and coordinate tables.

Worked example

Explain why the width of the A band stays approximately constant while a sarcomere shortens.

  1. The A band corresponds to the length of the thick myosin filaments.
  2. Cross-bridge cycling pulls actin farther between myosin filaments without shortening either filament.
  3. The overlap increases while the I band and H zone narrow and Z lines move closer.

Answer: The A band remains approximately constant because myosin filament length stays the same; shortening comes from sliding.

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