14. In smooth muscle calcium released by the smooth ER initiates contraction by
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14. In smooth muscle calcium released by the smooth ER initiates contraction by binding to what protein a. Actin b. Calmodulin c. Desmin d Myosin light chain kinase . e. Tropomyosin 15. Which feature typifies T-tubules? Evaginations of the sarcoplasmic reticulum a. b. Sequester calcium during muscle relaxation, releasing it during contraction c. Carry depolarization to the muscle fiber interior d. Overlie the A-I junction in cardiac muscle cells eRich supply of acetylcholine receptors 16. Which characteristic is unique to smooth musele? a. T-tubules lie across Z lines Each thick filament is surrounded by six thin filaments Thin filaments attach to dense bodies b. c. d. Cells are multinucleated e Cells have centrally located nuclei 17. In one type of muscle, numerous gap junctions, desmosomes, and adherens junctions are specifically localized in which structures? a. Myofilaments b. Dense bodies - c. Sarcomeres d. Neuromuscular spindles e. Intercalated discs 18. A 66-year-old man who lives alone has a severe myocardial infarction and dies during the night. The medical examiner's office is called the following morning and describes the man's body as being in rigor mortis. This state of rigor mortis is due to which one of the following? Inhibition of Ca2+ leakage from the extracellular fluid and sarcoplasmic reticulum a. b. Enhanced retrieval of Ca by the sarcoplasmic reticulum c. Failure to disengage tropomyosin and troponin from the myosin active sites d. Absence of ATP preventing detachment of the myosin heads from actin e. Increased lactic acid production 19. A 5-year-old boy sustains a small tear in his gastrocnemius muscle when he is involved in a bicycle accident. Regeneration of the muscle will occur through which of the following mechanisms? a. Dedifferentiation of muscle cells into myoblasts b. Differentiation of muscle satellite cell:s c. Fusion of damaged myofibers to form new myotubes d. Hyperplasia of existing muscle fibers e. Differentiation of fibroblasts to form myoblasts 20. A healthy 32-year-old man lifts weights regularly as part of his workout. In one of his biceps muscle fibers at rest, the length of the I band is 1.0 m and the A band is 1.5 m. Contraction of that muscle fiber results in a 10% shortening of the length of the sarcomere. What is the length of the A band after the shortening produced by muse contraction? a. 1.50 um b. 1.35 m c. 1.00 umm d. 1.90 ?mExplanation / Answer
14. Smooth muscle contraction is initiated by the release of Ca2+. Ca2+ first binds to calmodulin and activates it, which results in calmodulin binding to caldesmon, and the reaction proceeds further. Therefore, your answer is (b).
15. It has been found in research studies that t-tubules play a central role in the increase and subsequent decrease of Ca(2+) during the systolic Ca(2+) transient. Therefore, your answer is (b)
16. The thin filaments that form part of the contractile machinery are predominantly composed of ?- and ?-actin. and these actin filaments of contractile units are attached to dense bodies. Therefore, your answer is (c)
17. Three types of cell junctions that make up an intercalated disc are adherens junction, desmosomes and gap junctions. Therefore, your answer is (e)
18. As the person had died a night before, the concentration of ATP in his body must have been decline. Since ATP is required to cause separation of the actin-myosin cross-bridges during relaxation of muscle, the muscle can no longer relax now and rigor mortis sat in. Therefore, your answer is (d)
19. Myogenic regeneration mainly has two repair mechanisms. Mononucleated cells fuse together to give rise to newly formed multinucleated myotubes that further develop to striated myofibers or repair of injured muscle fibers by the possible fusion of mononucleated cells with their necrotic cut ends. Therefore, your answer is (c)
20. Since the contraction of the muscle causes 10% shortening, hence the band A will be shortened by 0.15microm (10% of 1.5 microm). Therefore, your answer is (1.5 - 0.15 = 1.35 microm), (b)
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