This 6-second lead II strip is irregular, so the 300 and 1500 methods do not work. Use the 6-second method to estimate the ventricular heart rate in beats per minute.
ECG rhythm strip interpretation worksheet with answers
1.Easy
lead II, 25 mm/s, 10 mm/mVstrip length 6 s (marks at 0, 3 and 6 s)
heart rate = bpm
2.Easy
Measure the PR interval on this lead II strip, in seconds.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
PR = s
3.Medium
Measure the QT interval and the R–R interval on this regular lead II strip, then calculate the rate-corrected QTc with Bazett’s formula QTc = QT/√RR (both in seconds). Give the answer in seconds to three decimals.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
QTc = s
4.Medium
This lead II strip shows atrial flutter with a fixed conduction ratio. Find the ventricular rate in beats per minute.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
ventricular rate = bpm
5.Hard
This lead II strip shows a conduction problem between the atria and the ventricles. Which rhythm is it? Click your answer.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
- 1first-degree AV block
- 2second-degree AV block type II
- 3second-degree AV block type I (Wenckebach)
- 4atrial fibrillation
6.Hard
Identify the rhythm on this lead II strip. Click your answer.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
- 1ventricular tachycardia
- 2atrial fibrillation
- 3ventricular fibrillation
- 4atrial flutter, 2:1 conduction
Answer key
- 1. heart rate = 70 bpm
- 2. PR = 0.36 s
- 3. QTc = 0.495 s
- 4. ventricular rate = 100 bpm
- 5. second-degree AV block type I (Wenckebach)
- 6. ventricular tachycardia
Full step-by-step solutions: practise this topic on examproblems.com.