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Identify the correct sequence of the following events. (a) Myosin generates a power stroke. (b) Ca+ binds to troponin. (c) ATP recharges the myosin head. (d) Troponin removes tropomyosin from G actin. (e) The sarcomere shortens. (f) Myosin binds to actin.

User Moe Sweet
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Answer:

This is the order: B - D - A - F - E - C

Step-by-step explanation:

We describe the differences between fylaments:

THIN FILAMENTS: They contain actin, are bound by Tropomyosin and are covered by some Troponin molecules.

THICK FILAMENTS: They contain myosin, which in their heads ADP molecules are attached

At the begining of contraction, it's necessary that the calcium ion be released from the sarcoplasmic reticulum (own organelle in muscle cells). When this happens, Ca2 + is released to the cytoplasm and binds with Troponin C

Remember that Troponin is a protein of 3 subunits:

Tt. Tropomyosin binding.

Ti. It binds with actin and covers the binding site with myosin.

Tc: It's free but it will bind to calcium to make union between actin and myosin, happens.

The binding of Ca2 + with Tc causes the troponin-actin union to weaken allowing tropomyosin to move and make visible the active site where actin will bind to myosin. Under resting conditions, the troponin-tropomyosin complex prevents such binding. On the other hand, each actin-G molecule contains another active site that keeps the troponin-tropomyosin binding stable.

Once the active site of myosin binding with actin is discovered, the myosin ADP is released so it contracts automatically by performing the power stroke. This causes the sarcomere be short because myosin reduces the length of actin. When the muscle relaxes, it needs energy, so the ADP that was lost from the myosin head is recovered at the expense of the sarcoplasma ATP. When it occupies its position, the myosin head is released from the actin and the sarcomere recovers its initial length

User Getaway
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