Read the pH: below 7.35 is acidosis, above 7.45 alkalosis.
Find the value that explains it: PaCO₂ moving opposite to the pH is respiratory, HCO₃⁻ moving with the pH is metabolic.
Look at the other value: normal means uncompensated, abnormal in the compensating direction means partial compensation, and a normal pH with both abnormal means full compensation.
In a metabolic acidosis, compare PaCO₂ with Winter’s formula, 1.5·HCO₃⁻ + 8 ± 2, and calculate the anion gap Na⁺ − (Cl⁻ + HCO₃⁻).
Check a high anion gap with the delta ratio and, when albumin is low, with the corrected gap.
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.
Evidence-based practice statistics
Sensitivity and specificity
Se = a/(a + c), Sp = d/(b + d)
Predictive values
PPV = a/(a + b), NPV = d/(c + d)
Accuracy
(a + d)/n
Likelihood ratios
LR+ = Se/(1 − Sp), LR− = (1 − Se)/Sp
PPV at prevalence p
Se·p / (Se·p + (1 − Sp)(1 − p))
Odds
odds = p/(1 − p); p = odds/(1 + odds); post-test odds = pre-test odds × LR
Risk measures
RR = EER/CER, ARR = CER − EER, RRR = ARR/CER
Number needed to treat
NNT = 1/ARR, rounded up; NNH = 1/(EER − CER)
Odds ratio
OR = ad/(bc)
Incidence rate
new cases / person-time
Method
Draw the 2 × 2 table and label a, b, c and d before you calculate.
Read sensitivity and specificity down the columns, predictive values along the rows.
For another prevalence, imagine 10 000 people and fill the table with Se, Sp and the prevalence.
For a likelihood ratio, convert the probability to odds, multiply, and convert back.
For a trial, find EER and CER first; then RR = EER/CER, ARR = CER − EER and NNT = 1/ARR rounded up.
2Practice problems
1.Human anatomyMedium
Which is the largest nerve in the human body?
AFemoral nerve
BMedian nerve
CSciatic nerve
DVagus nerve
2.Human anatomyHard
After a difficult delivery with traction on the neck, a baby's arm hangs by the side, internally rotated with the forearm pronated ("waiter's tip"). Which part of the brachial plexus is injured?
ALower trunk (C8–T1)
BUpper trunk (C5–C6)
CMedial cord
DPosterior cord
3.Human physiologyMedium
Why does glucose appear in the urine in uncontrolled diabetes mellitus?
AThe collecting duct actively secretes glucose
BInsulin is needed for glucose reabsorption in the tubule
CThe filtered load exceeds the tubular transport maximum
DThe glomerulus stops filtering glucose
4.Human physiologyHard
In long-standing aortic stenosis, how does the left ventricle adapt?
AThinning of the wall with dilation
BAtrophy of the muscle due to less work
CNo change because the valve is not part of the ventricle
DConcentric hypertrophy with a thicker wall
5.MicrobiologyMedium
Tetanus and diphtheria vaccines are toxoids. What does that mean?
AThey contain live bacteria that cannot make toxin
BThey contain ready-made antibodies against the toxin, giving passive immunity
CThey contain the bacterial DNA only
DThey contain an inactivated toxin that still triggers an immune response
6.MicrobiologyHard
Staphylococcus aureus food poisoning starts only 1–6 hours after eating. Why so fast?
AThe immune system reacts to the bacteria within minutes
BThe bacteria made a heat-stable toxin in the food before it was eaten
CThe bacteria multiply very fast inside the gut and invade the intestinal wall
DThe bacteria invade the bloodstream immediately
7.PharmacologyMedium
Why is IV promethazine given with great caution?
AIt causes immediate kidney failure
BIt is a strong opioid that causes addiction
CIt causes a severe rise in blood pressure when it is given intravenously
DIt can cause severe tissue injury if it leaks or enters an artery
8.PharmacologyHard
Why is intravenous insulin given with glucose in severe hyperkalaemia?
AGlucose binds potassium in the blood
BInsulin protects the heart muscle directly from potassium
CInsulin moves potassium from the blood into cells
DInsulin increases potassium excretion by the kidney
9.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.43, PaCO₂ 31 mmHg (4.1 kPa) and HCO₃⁻ 20 mmol/L. The primary disorder is respiratory alkalosis. How far is it compensated? Click your answer.
pH 7.43PaCO₂ 31 mmHg (4.1 kPa)HCO₃⁻ 20 mmol/L
1uncompensated
2partially compensated
3fully compensated
10.Arterial blood gas interpretationHard
A practice gas shows a metabolic acidosis: pH 7.31, PaCO₂ 29 mmHg (3.9 kPa), HCO₃⁻ 14 mmol/L. Compare the measured PaCO₂ with Winter’s formula. Click your interpretation.
pH 7.31PaCO₂ 29 mmHg (3.9 kPa)HCO₃⁻ 14 mmol/L
1appropriate compensation
2added respiratory acidosis
3added respiratory alkalosis
11.ECG rhythm strip interpretationMedium
Interpret this lead II rhythm strip. 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
1junctional rhythm
2normal sinus rhythm
3sinus bradycardia
4first-degree AV block
12.ECG rhythm strip interpretationHard
This patient has a pacemaker. Interpret the 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 paced rhythm
2idioventricular rhythm
3dual-chamber (AV) paced rhythm
4sinus rhythm with wide QRS
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a randomised trial, 650 patients received a new treatment and 890 received the control treatment. The table shows how many in each group had the outcome (an unwanted event such as a stroke). Find the relative risk RR of the event with treatment. Round to two decimal places.
treatment: 50/650control: 98/890
RR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A patient has a pre-test probability of 25% for a disease. A test with sensitivity 74% and specificity 71% comes back negative. Find the post-test probability p′ of the disease as a percentage. Round to one decimal place.
pre-test 25%Se = 74%Sp = 71%negative
p′ = %
3Solutions
1.Human anatomy
Answer: C · Sciatic nerve
Why this is the answer
The sciatic nerve (L4–S3) leaves the pelvis through the greater sciatic foramen and supplies the hamstrings and all the muscles below the knee. Skin on the inner side of the leg is supplied by the saphenous nerve instead.
Answer
C · Sciatic nerve
2.Human anatomy
Answer: B · Upper trunk (C5–C6)
Why this is the answer
Erb's palsy follows stretching of the upper trunk (C5–C6), paralysing deltoid, the external rotators and biceps. Lower trunk injury (Klumpke's) affects the small hand muscles instead.
Answer
B · Upper trunk (C5–C6)
3.Human physiology
Answer: C · The filtered load exceeds the tubular transport maximum
Why this is the answer
Proximal tubule SGLT carriers can reabsorb only so much glucose. When plasma glucose is high, the filtered load exceeds this maximum and the excess is excreted.
Answer
C · The filtered load exceeds the tubular transport maximum
4.Human physiology
Answer: D · Concentric hypertrophy with a thicker wall
Why this is the answer
Pressure overload makes the ventricular wall thicken (concentric hypertrophy) to keep wall stress normal. Volume overload, as in regurgitation, causes dilation instead.
Answer
D · Concentric hypertrophy with a thicker wall
5.Microbiology
Answer: D · They contain an inactivated toxin that still triggers an immune response
Why this is the answer
A toxoid is a toxin treated (for example with formaldehyde) so it is harmless but still antigenic. The immune system makes antibodies that neutralise the real toxin.
Answer
D · They contain an inactivated toxin that still triggers an immune response
6.Microbiology
Answer: B · The bacteria made a heat-stable toxin in the food before it was eaten
Why this is the answer
S. aureus produces enterotoxins in food left at warm temperatures. The toxin is already there when the food is eaten and survives reheating, so symptoms start quickly.
Answer
B · The bacteria made a heat-stable toxin in the food before it was eaten
7.Pharmacology
Answer: D · It can cause severe tissue injury if it leaks or enters an artery
Why this is the answer
Promethazine is a vesicant. Deep intramuscular injection is the preferred parenteral route, and if given IV it is diluted, given slowly into a well-running large vein, and stopped at once if the patient reports burning or pain.
Answer
D · It can cause severe tissue injury if it leaks or enters an artery
8.Pharmacology
Answer: C · Insulin moves potassium from the blood into cells
Why this is the answer
Insulin activates the Na⁺/K⁺-ATPase and drives potassium into cells, lowering plasma levels within minutes. Glucose is given to prevent hypoglycaemia. The effect is temporary; potassium must still be removed from the body.
Answer
C · Insulin moves potassium from the blood into cells
9.Arterial blood gas interpretation
Answer: fully compensated
Reading a blood gas (ROME): first the pH: below 7.35 is acidaemia, above 7.45 alkalaemia. Then find the value that explains it. PaCO₂ (normal 35–45 mmHg, 4.7–6.0 kPa) is the respiratory part and moves Opposite to the pH in a respiratory disorder; HCO₃⁻ (normal 22–26 mmol/L) is the metabolic part and moves in the same direction as the pH (Respiratory Opposite, Metabolic Equal). The other value shows the compensation: normal means uncompensated; abnormal in the direction that pulls the pH back while the pH is still abnormal means partially compensated; a normal pH with both values abnormal means fully compensated. Ranges differ slightly between laboratories; these are practice values.
ROME, value by value
pH 7.43: within 7.35–7.45
PaCO₂ 31 mmHg: low (35–45)
HCO₃⁻ 20 mmol/L: low (22–26)
Compensation
fully compensated
Answer
fully compensated
10.Arterial blood gas interpretation
Answer: appropriate compensation
Expected compensation: in metabolic acidosis Winter’s formula predicts PaCO₂ = 1.5·HCO₃⁻ + 8 ± 2 mmHg; a measured value above that range points to an additional respiratory acidosis, below it to an additional respiratory alkalosis. The anion gap AG = Na⁺ − (Cl⁻ + HCO₃⁻), normally about 8–12 mmol/L, separates high-gap from normal-gap metabolic acidosis; low albumin hides a high gap (add 2.5 per g/dL that albumin is below 4.0). The delta ratio (AG − 12)/(24 − HCO₃⁻) between about 0.8 and 2 fits a pure high-gap acidosis (some texts use 1 instead of 0.8), below that an added normal-gap acidosis, and above 2 an added metabolic alkalosis.
Expected PaCO₂
1.5·14 + 8 = 29 → 27–31 mmHg
Compare
measured: 29 mmHg → appropriate compensation
Answer
appropriate compensation
11.ECG rhythm strip interpretation
Answer: junctional rhythm
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.
Five steps
1. Rate: R–R = 33 small boxes, 1500 ÷ 33 ≈ 45 per minute.
2. Rhythm: regular, every R–R is 1.32 s.
3. P waves: none visible.
4. PR interval: not measurable.
5. QRS: 0.08 s, narrow (below 0.12 s).
Interpretation
junctional rhythm
Answer
junctional rhythm
12.ECG rhythm strip interpretation
Answer: ventricular paced rhythm
Paced rhythms: a pacemaker impulse shows as a narrow vertical pacing spike. After a ventricular spike the QRS is wide (0.12 s or more), because the impulse spreads from the lead through the muscle instead of the conduction system; with right-ventricular pacing it is often mostly negative in lead II. A spike before each P wave means atrial pacing, and a dual-chamber (AV) pacemaker shows an atrial spike and a P wave, then, after the programmed AV delay, a ventricular spike and a wide QRS. Capture means every spike is followed by a depolarisation. Failure to capture is a spike without a P wave or QRS after it; failure to sense (spikes that ignore the patient’s own beats) and failure to pace (no spike when one is due) are the other classic malfunctions.
Five steps
1. Rate: R–R = 22 small boxes, 1500 ÷ 22 ≈ 68 per minute.
2. Rhythm: regular, every R–R is 0.88 s.
3. P waves: none visible; a sharp pacing spike comes just before every QRS.
4. PR interval: not measurable.
5. QRS: 0.16 s, wide, and every spike is followed by a QRS (capture).
Interpretation
ventricular paced rhythm
Answer
ventricular paced rhythm
The paced QRS is wide by nature; do not call it a bundle branch block or a ventricular rhythm.
13.Evidence-based practice statistics
Answer: RR = 0.7
Likelihood ratios and risk measures: LR+ = sensitivity/(1 − specificity), LR− = (1 − sensitivity)/specificity. In a trial, EER and CER are the event rates with and without treatment: RR = EER/CER, ARR = CER − EER, RRR = ARR/CER and NNT = 1/ARR, rounded up. In a case-control study only the odds ratio OR = ad/(bc) can be used. An incidence rate = new cases/person-time.
Event rates
EER = 50/650 = 0.07692
CER = 98/890 = 0.1101
Relative risk
RR = 0.07692/0.1101 = 0.6986 ≈ 0.7
Answer
RR = 0.7
14.Evidence-based practice statistics
Answer: p′ = 10.9 %
Predictive values depend on prevalence: PPV = Se·p/(Se·p + (1 − Sp)(1 − p)) and NPV = Sp(1 − p)/(Sp(1 − p) + (1 − Se)p). With likelihood ratios: pre-test odds = p/(1 − p), post-test odds = pre-test odds × LR, probability = odds/(1 + odds). For a population of N, true positives = N·p·Se and false positives = N·(1 − p)·(1 − Sp). NNT = 1/(CER × RRR), rounded up.