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A 45-year-old man suffers acute myocardial ischemia while visiting his brother in the cardiac care unit of the hospital. Dr. Cox immediately begins intraventricular thrombolytics. A biopsy of the ischemic tissue would most likely show:

Option 1: Coagulative necrosis.
Option 2: Caseous necrosis.
Option 3: Liquefactive necrosis.
Option 4: Fat necrosis.

1 Answer

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Final answer:

A biopsy of the ischemic tissue in a 45-year-old man with acute myocardial ischemia would most likely show coagulative necrosis, which is typical of myocardial infarction caused by severe blockage of coronary arteries.

Step-by-step explanation:

A 45-year-old man suffering from acute myocardial ischemia, which is a severe blockage of the coronary arteries leading to death of the cardiac muscle tissue, would most likely display coagulative necrosis upon biopsy of the ischemic tissue. Coagulative necrosis is the typical type of cell death seen in myocardial infarction (MI), commonly known as a heart attack. This type of necrosis results from the denaturation of proteins which causes cell death while retaining the structural outlines of the cells for several days.

Myocardial infarction is the formal term for heart attack and it typically results from a lack of blood flow (ischemia) and oxygen (hypoxia) to a region of the heart. This lack of blood flow and oxygen is usually caused by a complete blockage of a coronary artery due to a blood clot or the rupture of an atherosclerotic plaque. MI can lead to the death of cardiac muscle cells, presenting clinically as chest pain, pressure, and other symptoms; however, some myocardial infarctions may occur without any symptoms.

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