ECG Basic Principles Flashcard 2
Rationale
B. Rationale: Heart rate can be approximated by dividing 1,500 by the number of small squares between two R-R intervals. In this case, 1,500 divided by 15 equals 100. So the estimated heart rate in this patient is 100 bpm. This method is most accurate in patients with a regular rhythm.
Question
The R-R interval in a standard ECG spans 15 small squares. What is the heart rate of this patient?
a. 60 bpm
b. 100 bpm
c. 120 bpm
d. 150 bpm
Answer
b. 100 bpm
Rationale
B. Rationale: The Purkinje system is considered to be the terminal portion of the heart’s conduction system. It consists of a network of nerve fibers within the endocardium of both ventricles. The conduction system of the heart consists of the following in descending order of conduction: SA node, atria, AV node, bundle of His, right bundle branch/left bundle branch, and the Purkinje system. The ventricles are not part of this conduction system.
Question
The terminal portion of the heart’s conduction system is the:
a. Sinoatrial node
b. Purkinje system
c. Bundle of His
d. Ventricles
Answer
b. Purkinje system
Rationale
B. Rationale: If the electric current is traveling away from the positive terminal, then a downward deflection is observed. On the other hand, if the electric current is traveling towards the positive terminal of an ECG lead, an upward deflection is noted.
Question
When an impulse generated from the heart moves away from the ECG lead, what effect does this have on the ECG tracing?
a. The impulse generates a positive or upward deflection.
b. The impulse generates a negative or downward deflection.
c. The impulse generates a positive and negative or upward and downward deflection.
d. The impulse generates a deflection that is even with the isoelectric line.
Answer
b. The impulse generates a negative or downward deflection.
Rationale
C. Rationale: W en measuring the amplitude on a standard ECG tracing, one small square represents 0.1 mV. Hence, 10 small squares or (10 mm) represents 1.0 mV.
Question
When interpreting a standard ECG tracing, an amplitude of 10 small squares or (10 mm) is indicative of which approximate voltage?
a. 0.01 mV
b. 0.1 mV
c. 1.0 mV
d. 10 mV
Answer
c. 1.0 mV
Rationale
B. Rationale: One small square is equivalent to 1 millimeter. Five small squares or 5 millimeters are equal to 200 milliseconds, and 25 small squares or 25 millimeters are equal to 1 second. Therefore, the speed that the electrocardiogram records is 25 mm/second.
Question
When measuring the electrocardiogram, one small square represents 40 milliseconds. Given this information, the speed of the electrocardiogram when it records impulses is:
a. 20 mm/sec
b. 25 mm/sec
c. 35 mm/sec
d. 50 mm/sec
Answer
b. 25 mm/sec
Rationale
A. Rationale: When the sodium concentration is higher outside of the cardiac cell, the extracellular environment is more positive and the intracellular environment is negative. This is known as the polarized state. The polarized state makes the cardiac cells ready for action.
Question
Which electrolyte exits the cardiac cell leading to cellular polarization?
a. Sodium
b. Potassium
c. Calcium
d. Magnesium
Answer
a. Sodium
Rationale
A. Rationale: The QRS complex represents ventricular depolarization or activation.
Question
Which element of the ECG tracing represents ventricular depolarization?
a. QRS complex
b. ST segment
c. T wave
d. P wave
Answer
a. QRS complex
Rationale
D. Rationale: There are three areas in the heart that govern the cardiac rate: the SA node, AV node, and the Purkinje fibers. These three areas have their own inherent rate. The ventricles can conduct at a rate of 20–40 bpm with the Purkinje fibers. The SA node regulates rates between 60 and 100 bpm. The AV junction regulates rates between 40 and 60 beats per minute.
Question
Which landmark in the heart’s conduction system has the slowest inherent rate?
a. Sinoatrial node
b. Atrioventricular node
c. Accessory pathways
d. Purkinje fibers in the ventricles
Answer
d. Purkinje fibers in the ventricles
Rationale
B. Rationale: Count the number of small squares between two R waves (18 small squares in this example). The heart rate is 1,500 divided by the number of small squares between the two R waves.
Question
You see this rhythm on the monitor:
The heart rate for this rhythm is:
a. 98 bpm
b. 88 bpm
c. 76 bpm
d. 72 bpm
Answer
b. 88 bpm