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Midterm

ES 330 Midterm: Muscle, Aging and Exercise exam 3
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Department
Exercise Science
Course Code
ES 330
Professor
Lou Gosselin

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1 Exam 3 Muscle, Aging, & Exercise 1. What is a satellite cell? What is its role? How does it differ from a muscle cell? a. A satellite cell is a precursor to a skeletal muscle cell b. Its main role is to repair damaged muscle cells or contribute to formation of new muscle tissue b.i. Upon injury or loading, satellite cells are activated and proliferate (rapidly in number and multiply) b.ii. Some of the proliferated daughter cells from the original satellite cells leave their position and pass through the sarcolemma and migrate to the site of injury b.iii. They then fuse with the existing fiber and engage in fiber repair and/or contribute to fiber growth/hypertrophy b.iv. Chemokines/cytokines induces a cell to migrate (inflammatory/satellite cells) b.v. Growth factor (FGF) stimulates cells to divide/proliferate c. Distinct location- sandwiched between the basal lamina & the sarcolemma (cell membrane) d. Satellite cells differ from muscle cells by: d.i. Satellite cell is mono-nucleated d.ii. Satellite cell has little to no cytoplasm d.iii. Satellite cell is capable of mitosis and migration 2. What happens to satellite cells with age? a. Satellite cell proliferation and/or differentiation may be impaired, which contributes to a lower regenerative potential b. Satellite cells decrease with age in certain muscles 3. What role does IGF-1 and myostatin have on muscle growth? a. Myostatin inhibits growth (negative regulator) b. IGF-1 (&GH) promotes growth (positive regulator) 4. How does a muscle cell become multi-nucleated? a. A mesenchymal cell proliferates and differentiates into myoblasts (precursor cell) b. One myoblast fuses with another and each nuclei from each myoblast come together c. This then becomes one long, multinucleated cell-- which begins the formation of a z-line. Nuclei get sandwiched as you add more myoblasts d. During post-natal development, some myoblasts become encapsulated in the basal lamina and fail to fuse with the parent fiber. These cells then become satellite cells. d.i. Satellite cells can divide and the daughter cells either fuse with the parent fiber and inject its DNA into the parent fiber &  the # of nuclei OR it is reserved for future use (ex. another division) e. With resistance training and hypertrophy, new nuclei are added to the muscle cell by satellite cells 5. How does the myonuclear domain (MD) affect a muscle cell? a. MD: Each nucleus controls a certain area in its surrounding territory b. MD size varies with fiber type b.i. Type 1 (slow oxidative): more mitochondria, greater oxidative capacity— small MD --- more myonuclei taking up smaller amounts of space b.ii. Type 2: less mitochondria, less oxidative capacity— larger MD --- less myonuclei taking up larger amounts of space c. MD size is inversely proportional with fiber oxidative capacity c.i. Fast fibers (type 2) have fewer nuclei, leading to a larger MD 2 d. If you  MD size,  number of myonuclei (from satellite cells) 6. Describe the structure of skeletal muscle connective tissue. a. Epimysium: surrounds whole muscle b. Perimysium: surrounds bundle of fibers c. Endomysium: surrounds individual fibers d. Myofibrils e. With age, more collagen cross links muscle gets stiffer f. Thick filaments (A-band/myosin) DON’T shorten g. Thin filaments (I-band/actin) shorten w contraction, bringing the z-lines closer together h. Every thick filament is surrounded by 6 thin filaments in a hexagon shape. Every thin filament is surrounded by 3 thick filaments. This is highly preserved (looks the same on everyone) I. Sarcomere Length: determines how much force the fiber can develop. If you change the length of the fiber, you change the amount of overlap. overlap=force b/c more cross bridges interact (Length/tension curve) 7. What are the different characteristics of a healthy cell vs. a damaged cell?*****? a. A healthy cell has the mononuclei on the outside b. A damaged cell has the mononuclei at the center 8. How does a muscle cell contract (from nerve stimulation to contraction)? a. Action potential reaches the terminal end of an axon at the NMJ b. Ca is released by SR into the boutin, which triggers synaptic vesicles containing ACh to release its contents via exocytosis. c. ACh is released into the synaptic cleft and travel into the junctional folds of the sarcolemma at the motor end plate d. ACh binds onto ACh-receptors, located on the sarcolemma, causing them to open e. Na enters the cell, creating an action potential f. AP travels across sarcolemma, 2+ich propagates down the T-tubule g. This allows voltage-gated Ca channels to open simultaneously h. Ca leaves the terminal cisternae of the sarcolemma and binds to troponin i. Troponin, located on the thin filaments (actin), moves tropomyosin off of the binding site j. ATP is converted to ADP + P (iiorganic phosphate), which binds to the thick filaments (myosin) k. Myosin binds to the actin binding sites l. Myosin pulls the actin towards the M-line (the center of the sarcomere), while myosin moves towards the Z-line l.i. This is known as the sliding filament theory l.i.1. Each sarcomere shortens as the thin filament slides close together between the thick filament so that the Z-disks are pulled closer together m. Myosin detaches from actin when: m.i. Ca binds to troponin again m.ii. ATP hydrolysis enables the myosin head to detach from the actin filament 9. What is the role of the T-tubule and sarcoplasmic reticulum (SR)? a. T-tubule – invagination of cell membrane, deep into the m. fiber. Allows AP to travel down through it to allow for spontaneous release of Ca from terminal cisternae in the sarcoplasmic reticulum, allowing for simultaneous contraction 2+ b. SR – Stores and releases Ca  Contains terminal cisternae, which is what contains the calcium ions necessary for contraction. 3 10. Can you draw and explain the length-tension curve? a.  thin/thick filament overlap,  force. (On Pic): a.i. 1A. Sarcomere length too short. Too much overlapping, inhibits cross bridges from binding so don’t get a lot of force a.ii. 1B. expose more cross bridge sites to thin filamentsmore force a.iii. 2. Max amount of cross bridges interacting with actin, optimal overlap of thick/thin filaments= optimal force optimal length (L ) o a.iv. 3/4. Start to pull thick/thin filaments away from each other  less force b. Assess m. function: b.i. In vitro force: fixed length, measure max force ISO contraction. Need to make sure m. is same length every time otherwise will get different forces. b
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