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/mV

This lead II strip shows an AV block, so the atria and the ventricles beat at different rates. Measure the P–P interval and find the atrial rate in beats per minute (1500 method, whole number).

lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
atrial rate = bpm
3.Human physiologyMedium

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

  1. APacinian corpuscle
  2. BGolgi tendon organ
  3. CMuscle spindle
  4. DNociceptor
4.Human physiologyHard

Which transporter in the thick ascending limb of the loop of Henle is key to building the medullary concentration gradient?

  1. AThe Na⁺–K⁺–2Cl⁻ cotransporter (NKCC2)
  2. BThe epithelial sodium channel (ENaC)
  3. CAquaporin-2 water channels
  4. DThe sodium–glucose cotransporter (SGLT2)

3Solutions

1.ECG rhythm strip interpretation

Answer: QT = 0.42 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

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

Convert

QT = 10.5 × 0.04 s = 0.42 s

Answer

QT = 0.42 s

A rough rule: at normal rates the QT should be less than half the R–R interval (here R–R = 1.04 s).

2.ECG rhythm strip interpretation

Answer: atrial rate = 75 bpm

AV blocks: march out the P waves and check which ones are followed by a QRS. First-degree: every P conducted, PR constant and above 0.20 s. Second-degree type I (Mobitz I, Wenckebach): the PR lengthens from beat to beat until one P wave is not conducted, then the cycle starts again (group beating). Second-degree type II (Mobitz II): the PR of the conducted beats stays constant, and P waves are suddenly not conducted; the QRS is often wide. Third-degree (complete) block: P waves and QRS complexes are each regular but independent, the PR varies at random and the atrial rate is faster than the escape rate (junctional 40–60/min with a narrow QRS, ventricular 20–40/min with a wide QRS). A 2:1 block cannot be called type I or type II from one strip.

Atrial rate

P–P = 20 small boxes = 0.80 s → 1500 ÷ 20 = 75 ≈ 75 per minute

Ventricular rate

R–R = 40 small boxes = 1.60 s → 1500 ÷ 40 = 37.5 ≈ 38 per minute

Answer

atrial rate = 75 bpm

Answer

atrial rate = 75 bpm

2:1 AV block: every second P wave is not conducted, so the atrial rate is twice the ventricular rate.

3.Human physiology

Answer: C · 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

C · Muscle spindle
4.Human physiology

Answer: A · The Na⁺–K⁺–2Cl⁻ cotransporter (NKCC2)

Why this is the answer

NKCC2 moves Na⁺, K⁺ and Cl⁻ out of the tubule into the interstitium without water, building the medullary gradient. Loop diuretics such as furosemide block it.

Answer

A · The Na⁺–K⁺–2Cl⁻ cotransporter (NKCC2)

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