ECG Interpretation: cram pack

Formula sheets, the key method steps for every topic and 4 practice problems, with full solutions at the back.

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1Formulas and methods

ECG rhythm strip interpretation

Paper (25 mm/s, 10 mm/mV)
1 small box = 0.04 s, 1 large box = 0.20 s, 10 mm = 1 mV
Heart rate, regular rhythm
300 ÷ large boxes = 1500 ÷ small boxes = 60 ÷ RR (s)
Heart rate, irregular rhythm
QRS complexes in 6 s × 10
Normal PR interval
0.12–0.20 s
Normal QRS duration
< 0.12 s
QTc (Bazett)
QT/√RR
QTc (Fridericia)
QT/∛RR
QTc (Framingham)
QT + 154·(1 − RR)
Sinus rates
bradycardia < 60, normal 60–100, tachycardia > 100 /min
Escape rates
junctional 40–60, ventricular 20–40 /min
Fast rhythms
SVT 150–250 /min (narrow), VT > 100 /min (wide)

Method

  1. Rate: for a regular rhythm divide 300 by the large boxes or 1500 by the small boxes between two R waves; for an irregular rhythm count the QRS complexes in 6 seconds and multiply by 10.
  2. Rhythm: compare the R–R intervals. Equal is regular; a repeating pattern is regularly irregular; no pattern at all is irregularly irregular.
  3. P waves: is there an upright P wave in lead II before every QRS, all alike? Look for fibrillatory or sawtooth waves, inverted P waves and P waves without a QRS.
  4. PR interval: from the start of P to the start of the QRS, normally 0.12–0.20 s and constant. Watch whether it lengthens, stays fixed before a dropped beat, or varies at random.
  5. QRS duration: below 0.12 s is narrow; 0.12 s or more is wide. Then name the rhythm, and for the QT use QTc = QT/√RR (Bazett) with times in seconds.

2Practice problems

1.ECG rhythm strip interpretationMedium
II25 mm/s · 10 mm/mV

Measure the QT 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
QT = s
2.ECG rhythm strip interpretationHard
II25 mm/s · 10 mm/mVraterhythmP wavesPR intervalQRS duration

Interpret this lead II strip with the five steps in order: rate, rhythm, P waves, PR interval, QRS duration. At which step do you find the first abnormal finding? Click it.

lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
  1. 1rate
  2. 2rhythm
  3. 3P waves
  4. 4PR interval
  5. 5QRS duration
3.Human physiologyMedium

Which receptor is stimulated when the patellar tendon is tapped in the knee-jerk reflex?

  1. ANociceptor
  2. BMuscle spindle
  3. CPacinian corpuscle
  4. DGolgi tendon organ
4.Human physiologyHard

Constriction of the efferent arteriole of a glomerulus, with the afferent arteriole unchanged, tends to:

  1. AStop filtration completely
  2. BHave no effect on GFR at all
  3. CRaise glomerular capillary pressure and GFR
  4. DLower glomerular capillary pressure and GFR

3Solutions

1.ECG rhythm strip interpretation

Answer: QT = 0.4 s

QT and QTc: the QT interval runs from the start of the QRS complex to the end of the T wave. It shortens as the heart rate rises, so it is corrected for rate with the R–R interval in seconds: Bazett QTc = QT/√RR and Fridericia QTc = QT/∛RR. Bazett is the formula most textbooks use; it over-corrects at fast rates, where Fridericia is preferred. Limits differ between references: a QTc above about 0.45 s in men or 0.46 s in women is usually called prolonged, and above 0.50 s markedly prolonged.

Measure

10 small boxes from the start of the QRS to the end of the T wave

Convert

QT = 10 × 0.04 s = 0.4 s

Answer

QT = 0.4 s

A rough rule: at normal rates the QT should be less than half the R–R interval (here R–R = 0.64 s).

2.ECG rhythm strip interpretation

Answer: P waves

Five steps for every strip: 1. rate (300 or 1500 method, or the 6-second method if irregular); 2. rhythm: are the R–R intervals equal; 3. P waves: present, upright in lead II, one before every QRS and all alike; 4. PR interval: 0.12–0.20 s and constant; 5. QRS duration: below 0.12 s. Only after the five steps name the rhythm. Normal sinus rhythm passes all five: 60–100/min, regular, upright P before every QRS, PR 0.12–0.20 s, QRS below 0.12 s.

Five steps

1. Rate: R–R = 17 small boxes, 1500 ÷ 17 ≈ 88 per minute.
2. Rhythm: regular, every R–R is 0.68 s.
3. P waves: inverted, just after each QRS.
4. PR interval: not measurable.
5. QRS: 0.06 s, narrow (below 0.12 s).

First abnormal step

P waves → accelerated junctional rhythm

Answer

P waves
3.Human physiology

Answer: B · Muscle spindle

Why this is the answer

Tapping the tendon stretches quadriceps muscle spindles. Their afferents synapse directly on motor neurons in a monosynaptic reflex that contracts the muscle.

Answer

B · Muscle spindle
4.Human physiology

Answer: C · Raise glomerular capillary pressure and GFR

Why this is the answer

Efferent constriction holds blood in the glomerulus, raising its pressure and GFR. Angiotensin II acts this way, which is why ACE inhibitors can lower GFR.

Answer

C · Raise glomerular capillary pressure and GFR

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