How Does A Motor Unit Illicit A Muscular Contraction

Step 1: ACh is released a the Neuromuscular Junction Nervous impulse triggers the release of ACh via exocytosis The ACh diffuses to the space between the motor neuron and the motor end plate (synaptic cleft) The motor end plate is the area of the muscle that receives the ACh on the sarcolemma ACh diffuses across the synaptic cleft between the motor neuron and motor end plate

Step 2: ACh receptors become active Binding of just 2 ACh molecules at the motor end plate is all that is needed Once bound, ACh will cause ion channels to open up on the motor end plate These open channels allow Na+ ions to rush across the membrane

Step 3: Production of the Action Potential Sodium ions make the inside of the muscle fiber more positive than the outside cellular fluid This triggers the initiation of the action potential One action potential leads to one contractile response from the muscle cells Action potential propagates along the sarcolemma into the T-tubule system. Causes SR to release Calcium ions into the sarcolemma Leads to a contraction

Step 4: Termination ACh is broken down by Acetylcholinesterase (enzyme) AChE comes from collagen fibers located in the synaptic cleft When this is taking place, the fibers are inactive Once this passes, the muscle can respond to another cycle of ACh stimulation

Summary Of the Events All have to take place for a muscle unit to contract If another impulse comes by, steps 2 and 3 will repeat When action potentials stop, ACh release stops, AChE is no longer needed, and the SR stops releasing Ca +2 ions