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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
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.
Rhythm: compare the R–R intervals. Equal is regular; a repeating pattern is regularly irregular; no pattern at all is irregularly irregular.
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.
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.
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
This lead II strip shows atrial flutter. How many flutter waves per QRS complex are there (the n of n:1 conduction)?
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
flutter waves per QRS =
2.ECG rhythm strip interpretationHard
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
1rate
2rhythm
3P waves
4PR interval
5QRS duration
3.Human physiologyMedium
In which form is most carbon dioxide carried in the blood?
AAs bicarbonate ions in plasma
BBound to the haem iron
CAs carbonic acid in platelets
DDissolved as CO₂ gas
4.Human physiologyHard
According to the length–tension relationship, why does a skeletal muscle fibre produce less force when overstretched?
AThe muscle runs out of ATP when stretched
BActin and myosin overlap less, so fewer cross-bridges form
CThe neuromuscular junction stops working
DCalcium is no longer released from the sarcoplasmic reticulum when stretched
3Solutions
1.ECG rhythm strip interpretation
Answer: flutter waves per QRS = 4
Rhythm criteria (lead II): sinus rhythm has an upright P wave before every QRS, a constant PR of 0.12–0.20 s and a narrow QRS: normal sinus rhythm at 60–100/min, sinus bradycardia below 60, sinus tachycardia above 100. Atrial fibrillation: no P waves, a fibrillatory baseline and an irregularly irregular R–R. Atrial flutter: sawtooth flutter waves at about 250–350/min with a fixed conduction ratio such as 2:1 or 4:1. First-degree AV block: sinus rhythm with a constant PR above 0.20 s. Junctional rhythm: narrow QRS at 40–60/min with absent or inverted P waves (PR below 0.12 s, or P after the QRS). A PVC is an early, wide (0.12 s or more), bizarre QRS without a P wave, with the T wave in the opposite direction and usually a full compensatory pause; in bigeminy every second beat is a PVC.
Flutter waves
F–F = 5 small boxes = 0.20 s → 1500 ÷ 5 = 300 per minute
QRS complexes
R–R = 20 small boxes = 0.80 s → 1500 ÷ 20 = 75 per minute
conduction ratio = 300 ÷ 75 = 4:1
Answer
flutter waves per QRS = 4
2.ECG rhythm strip interpretation
Answer: rate
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 = 12 small boxes, 1500 ÷ 12 ≈ 125 per minute.
2. Rhythm: regular, every R–R is 0.48 s.
3. P waves: upright, one before every QRS, all alike.
4. PR interval: 0.14 s, constant, within 0.12–0.20 s.
5. QRS: 0.08 s, narrow (below 0.12 s).
First abnormal step
rate → sinus tachycardia
Answer
rate
3.Human physiology
Answer: A · As bicarbonate ions in plasma
Why this is the answer
About 70% of CO₂ is carried as bicarbonate. Carbonic anhydrase in red cells converts CO₂ and water to carbonic acid, which dissociates into H⁺ and HCO₃⁻.
Answer
A · As bicarbonate ions in plasma
4.Human physiology
Answer: B · Actin and myosin overlap less, so fewer cross-bridges form
Why this is the answer
Maximum force occurs at an optimal sarcomere length with best filament overlap. Overstretching reduces overlap; too short a length makes filaments collide.
Answer
B · Actin and myosin overlap less, so fewer cross-bridges form