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Locomotion and Movement - Skeletal muscle, contractile proteins and muscle contraction

Grade 11CBSEBiology

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

🔑Concepts

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Skeletal muscle is composed of muscle bundles (fascicles) held together by a collagenous connective tissue layer called fascia. Each bundle contains numerous muscle fibers lined by the sarcolemma and containing sarcoplasm with multiple nuclei (syncytium).

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The sarcoplasmic reticulum of the muscle fibers is the storehouse of calcium ions (Ca2+Ca^{2+}), which are essential for muscle contraction.

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A myofibril has alternate dark and light bands. The dark band is the 'A' or Anisotropic band (contains myosin), and the light band is the 'I' or Isotropic band (contains actin).

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The functional unit of contraction is the Sarcomere, which is the portion of a myofibril between two successive ZZ-lines.

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Thin filaments (Actin) are composed of two 'FF' (filamentous) actins, which are polymers of monomeric 'GG' (globular) actins. It also contains two filaments of tropomyosin and a complex protein called Troponin (Troponin TT, II, and CC).

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Thick filaments (Myosin) are polymers of meromyosin. Each meromyosin has two parts: a Heavy Meromyosin (HMM) or globular head with a short arm, and a Light Meromyosin (LMM) or tail.

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The myosin head functions as an ATPATPase enzyme and has binding sites for ATPATP and active sites for actin.

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The Sliding Filament Theory states that muscle contraction occurs by the sliding of thin filaments over the thick filaments, triggered by a neural signal at the neuromuscular junction (NMJ) releasing Acetylcholine.

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During contraction, Ca2+Ca^{2+} binds with the subunit Troponin CC on actin filaments, uncovering the active sites for myosin to form a cross-bridge.

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Changes during contraction: The II-band shortens, the HH-zone reduces or disappears, while the AA-band retains its length.

📐Formulae

Sarcomere=12(I-band)+A-band+12(I-band)\text{Sarcomere} = \frac{1}{2} (\text{I-band}) + \text{A-band} + \frac{1}{2} (\text{I-band})

ATP→Myosin ATPase, Ca2+,Mg2+ADP+Pi+EnergyATP \xrightarrow{\text{Myosin ATPase, } Ca^{2+}, Mg^{2+}} ADP + P_i + \text{Energy}

Total length of myofibril=n×Length of Sarcomere (where n is number of sarcomeres)\text{Total length of myofibril} = n \times \text{Length of Sarcomere} \text{ (where } n \text{ is number of sarcomeres)}

💡Examples

Problem 1:

During a muscle contraction, the distance between two ZZ-lines decreases from 2.5μm2.5 \mu m to 2.0μm2.0 \mu m. If the AA-band length is 1.5μm1.5 \mu m, calculate the change in the length of the II-band.

Solution:

The length of a sarcomere is given by S=A+IS = A + I. Initial: 2.5μm=1.5μm+Iinitial⇒Iinitial=1.0μm2.5 \mu m = 1.5 \mu m + I_{\text{initial}} \Rightarrow I_{\text{initial}} = 1.0 \mu m. Final: 2.0μm=1.5μm+Ifinal⇒Ifinal=0.5μm2.0 \mu m = 1.5 \mu m + I_{\text{final}} \Rightarrow I_{\text{final}} = 0.5 \mu m. Change in II-band length = 1.0μm−0.5μm=0.5μm1.0 \mu m - 0.5 \mu m = 0.5 \mu m.

Explanation:

According to the sliding filament theory, the length of the AA-band remains constant (1.5μm1.5 \mu m), while the II-band shortens as actin filaments slide into the HH-zone.

Problem 2:

What is the role of Ca2+Ca^{2+} ions in the initiation of the cross-bridge cycle?

Solution:

In a resting state, the binding sites for myosin on actin are masked by a troponin-tropomyosin complex. When an action potential reaches the sarcoplasmic reticulum, Ca2+Ca^{2+} is released. These ions bind to Troponin CC (TnCTnC), causing a conformational change that pulls Tropomyosin away from the myosin-binding sites on the FF-actin filament.

Explanation:

The binding of Ca2+Ca^{2+} is the 'on-switch' for contraction, allowing the myosin head to attach to actin and form a cross-bridge using energy from ATPATP hydrolysis.