Myocardial ischemia occurs when coronary blood flow is inadequate to meet the requirements of the heart. This commonly results from atherosclerotic obstruction of one of the coronary arteries. When a region of the heart becomes ischemic, it quickly loses its ability to maintain transmembrane ionic pumps and the cells become depolarized. This movement of positive charge into the cells causes the ischemic region of the heart to become negative with respect to the rest of the heart. Thus, a dipole is created during diastole and it is termed the current of injury. The amplifier in the electrocardiograph is not DC coupled, which means that there is always an ambiguity as to where zero voltage actually is on the recording. As a result, it is impossible to tell whether a current of injury is present by simply examining whether the baseline is at zero volts. A little logic, however, can overcome this problem. Where the QRS complex ends is called the J point. At that time, all of the ventricular cells are depolarized and no voltage differences exist on the heart. For that reason, the voltage in the ECG returns to zero for a moment before the onset of the T wave. The J point is more commonly called the ST segment (the flat region between the end of the QRS and the start of the T wave). In a normal ECG, the ST segment should always be at the same level as the baseline, zero volts.
Now consider the case in which there is an ischemic region on the heart. Because the injured region is always depolarized, it generates a current of injury between beats when the rest of the heart is polarized. At the J point, however, all of the heart is depolarized, including the injured region. Therefore, the voltage from the heart at the J point is truly zero. If the ST segment is above or below the baseline in any lead, then the above described injury is present. The actual location of the injured region on the heart and thus the coronary branch involved can be determined by plotting the ST segment shifts vectorially.
Consider the example in Fig. 8. The greatest ST segment shift is seen in aVF and in that lead the baseline
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