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Human Physiology - Muscle and motility (HL)

Grade 11IBBiology

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Antagonistic Muscle Pairs: Muscles can only exert force by contracting (shortening). Therefore, they work in pairs; while one muscle (agonist) contracts, the other (antagonist) relaxes to return the bone to its original position. Example: Biceps and Triceps.

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Structure of Skeletal Muscle: Muscles are composed of bundles of muscle fibers. Each fiber is a single multinucleated cell containing specialized organelles called myofibrils, which are made of repeating units called sarcomeres.

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Sarcomere Components: The sarcomere is the functional unit of contraction. It consists of thin actin filaments and thick myosin filaments. It is defined by ZZ-lines at each end, an AA-band (length of myosin), an II-band (actin only), and an HH-zone (myosin only).

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The Role of Ca2+Ca^{2+} and Regulatory Proteins: In a relaxed state, tropomyosin blocks myosin-binding sites on actin. When an action potential stimulates the sarcoplasmic reticulum, Ca2+Ca^{2+} ions are released. Ca2+Ca^{2+} binds to troponin, causing a conformational change that moves tropomyosin away from the binding sites.

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The Sliding Filament Theory: Muscle contraction occurs as actin filaments slide over myosin filaments. This is driven by the 'cross-bridge cycle' involving the hydrolysis of ATPATP.

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ATP Hydrolysis in Contraction: ATPATP binds to the myosin head, breaking the cross-bridge. ATPATP is then hydrolyzed to ADPADP and inorganic phosphate (PiP_i), 'cocking' the myosin head into a high-energy state for the next power stroke.

📐Formulae

ATP+H2O→ATPaseADP+Pi+EnergyATP + H_2O \xrightarrow{\text{ATPase}} ADP + P_i + \text{Energy}

Force∝Number of active cross-bridges\text{Force} \propto \text{Number of active cross-bridges}

Sarcomere length=Distance between Z-lines\text{Sarcomere length} = \text{Distance between } Z\text{-lines}

💡Examples

Problem 1:

An electron micrograph shows a sarcomere with a total length of 2.5 μm2.5\ \mu m. If the AA-band measures 1.6 μm1.6\ \mu m, calculate the length of the II-band visible within this single sarcomere.

Solution:

The II-band is split across the ZZ-lines. Within one sarcomere, the total length of the non-overlapping actin regions is calculated as: Total I-band length=Sarcomere length−A-band length\text{Total } I\text{-band length} = \text{Sarcomere length} - A\text{-band length} Total I-band=2.5 μm−1.6 μm=0.9 μm\text{Total } I\text{-band} = 2.5\ \mu m - 1.6\ \mu m = 0.9\ \mu m

Explanation:

The AA-band represents the length of the thick myosin filaments. Since the sarcomere is the distance between ZZ-lines, subtracting the AA-band length from the total length gives the total length of the two half II-bands found within that sarcomere.

Problem 2:

Explain the state of the sarcomere during a maximal contraction compared to a relaxed state.

Solution:

During contraction:

  1. The distance between ZZ-lines decreases.
  2. The HH-zone (myosin-only region) shortens or disappears.
  3. The II-band (actin-only region) shortens.
  4. The AA-band remains constant.

Explanation:

Because the filaments themselves do not shorten but rather slide past each other, the width of the filaments (the AA-band) does not change. However, the regions of non-overlap (HH and II bands) decrease as the degree of overlap increases.