Using the sliding flament theory, explain (or draw) the process of sarcomere shortening. Start from the point where calcium woul

Answer:

1. Calcium attaches to troponin C
2. Troponin T shifts tropomyosin to reveal the binding sites
3. Myosin heads connect to actin, forming cross-bridges
4. ATP is converted to ADP and inorganic phosphate and releases energy
5. This energy drives the sliding of myofilaments, resulting in a power stroke
6. ADP detaches and a fresh ATP binds to the myosin heads, breaking the bond with the actin filament
7. ATP is then split into ADP and phosphate, storing energy in the myosin heads, thus beginning another cycle
8. Z-bands are drawn together, which shortens the sarcomere and the I-band, leading to muscle contraction.

Explanation:  

At rest, tropomyosin blocks the attraction between actin and myosin filaments. Contraction starts when an action potential depolarizes the interior of the muscle fiber. Calcium channels in the T tubules open, leading to the release of calcium into the sarcolemma. At this moment, tropomyosin obstructs the myosin binding sites on actin. Upon binding of calcium to troponin C, troponin T modifies the position of tropomyosin, exposing the binding sites. Myosin heads attach to the exposed actin sites forming cross-bridges, while ATP is converted into ADP and inorganic phosphate, which is then released. The sliding of myofilaments is driven by the chemical energy stored in myosin heads, resulting in a power stroke. The power stroke starts as the myosin cross-bridge binds to actin. During the slide, ADP is released. A new ATP connects to myosin heads, terminating the bond with the actin filament.  Then ATP is split into ADP and phosphate, and the energy generated is stored in the myosin heads, which initiates a new cycle of binding to actin. In the end, Z-bands pull together, which shortens the sarcomere and the I-band, causing muscle fiber contracture.