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 nerve is compressed in carpal tunnel syndrome?
AUlnar nerve
BMusculocutaneous nerve
CRadial nerve
DMedian nerve
2.Human anatomyHard
Where does the thoracic duct usually empty its lymph into the blood?
AInferior vena cava at the level of the kidneys
BJunction of the right subclavian and right internal jugular veins
CRight atrium directly
DJunction of the left subclavian and left internal jugular veins
3.Human physiologyMedium
How is the ejection fraction calculated?
AEnd-systolic volume ÷ stroke volume
BCardiac output ÷ heart rate
CEnd-diastolic volume ÷ end-systolic volume
DStroke volume ÷ end-diastolic volume
4.Human physiologyHard
Why does a deep sea diver risk decompression sickness when surfacing quickly?
AThe lungs collapse when the pressure falls
BNitrogen dissolved at high pressure forms bubbles as the pressure falls
COxygen becomes toxic at the surface
DCarbon dioxide cannot leave the lungs fast enough
5.MicrobiologyMedium
Which organism causes typhoid fever?
AVibrio cholerae
BShigella dysenteriae
CCampylobacter jejuni
DSalmonella Typhi
6.MicrobiologyHard
Corynebacterium diphtheriae causes diphtheria only if it produces diphtheria toxin. When does it make this toxin?
AWhenever it is exposed to antibiotics
BOnly when it is grown at high temperature
CWhen it carries a bacteriophage with the toxin gene
DOnly after it has conjugated with E. coli and received a plasmid
7.PharmacologyMedium
A nurse has a needlestick injury from a patient with HIV. When should post-exposure prophylaxis start?
AAfter one week, when the risk of infection is clearer
BOnly after the nurse's HIV test turns positive
COnly if the patient's viral load is above a set level
DAs soon as possible, and within 72 hours at the latest
8.PharmacologyHard
A patient who has taken lorazepam daily for many months stops it suddenly. Which serious complication may follow?
APinpoint pupils and coma
BWithdrawal seizures
CSevere bradycardia
DProfound respiratory depression
9.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.53, PaCO₂ 51 mmHg (6.8 kPa) and HCO₃⁻ 41 mmol/L. The primary disorder is metabolic alkalosis. How far is it compensated? Click your answer.
pH 7.53PaCO₂ 51 mmHg (6.8 kPa)HCO₃⁻ 41 mmol/L
1uncompensated
2partially compensated
3fully compensated
10.Arterial blood gas interpretationHard
At sea level, a practice gas taken on 30% oxygen shows PaO₂ 67 mmHg (8.9 kPa) and PaCO₂ 40 mmHg (5.3 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.
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
12.ECG rhythm strip interpretationHard
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
1atrial fibrillation
2ventricular tachycardia
3torsades de pointes
4supraventricular tachycardia (SVT)
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a randomised trial, 700 patients received a new treatment and 780 received the control treatment. The table shows how many in each group had the outcome (an unwanted event such as a stroke). Find the number needed to treat NNT to prevent one event. Round up to a whole number of patients.
treatment: 84/700control: 117/780
NNT =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). Without a drug, 6% of patients have a certain side effect during a year. The drug raises this risk with a relative risk of RR = 2.1. Find the number needed to harm NNH: how many patients must take the drug for a year for one extra patient to have the side effect? Round up to a whole number of patients.
control 6%RR = 2.1
NNH =
3Solutions
1.Human anatomy
Answer: D · Median nerve
Why this is the answer
The median nerve passes through the carpal tunnel under the flexor retinaculum; compression causes tingling in the thumb, index and middle fingers.
Answer
D · Median nerve
2.Human anatomy
Answer: D · Junction of the left subclavian and left internal jugular veins
Why this is the answer
The thoracic duct drains lymph from both legs, the abdomen and the left side of the upper body into the left venous angle. The right lymphatic duct drains the rest into the right venous angle.
Answer
D · Junction of the left subclavian and left internal jugular veins
3.Human physiology
Answer: D · Stroke volume ÷ end-diastolic volume
Why this is the answer
Ejection fraction is the proportion of blood in the full ventricle that is ejected per beat. A normal value is about 55–70%.
Answer
D · Stroke volume ÷ end-diastolic volume
4.Human physiology
Answer: B · Nitrogen dissolved at high pressure forms bubbles as the pressure falls
Why this is the answer
At depth more nitrogen dissolves in the tissues. If pressure falls too fast it comes out of solution as bubbles in tissues and blood, causing "the bends".
Answer
B · Nitrogen dissolved at high pressure forms bubbles as the pressure falls
5.Microbiology
Answer: D · Salmonella Typhi
Why this is the answer
Salmonella enterica serotype Typhi causes typhoid fever, a systemic illness with prolonged fever. Some people become chronic carriers, often in the gallbladder.
Answer
D · Salmonella Typhi
6.Microbiology
Answer: C · When it carries a bacteriophage with the toxin gene
Why this is the answer
The tox gene is carried by a bacteriophage. When the phage is integrated as a prophage, the bacterium can produce toxin. Strains without the phage are non-toxigenic.
Answer
C · When it carries a bacteriophage with the toxin gene
7.Pharmacology
Answer: D · As soon as possible, and within 72 hours at the latest
Why this is the answer
Post-exposure prophylaxis aims to stop infection becoming established. It works best when started within hours, is not started after 72 hours, and is usually taken for 28 days.
Answer
D · As soon as possible, and within 72 hours at the latest
8.Pharmacology
Answer: B · Withdrawal seizures
Why this is the answer
Abrupt withdrawal after long-term benzodiazepine use causes anxiety, insomnia, tremor and sometimes seizures. Long-term users should have the dose tapered gradually.
Answer
B · Withdrawal seizures
9.Arterial blood gas interpretation
Answer: partially 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 7.53: alkalaemia (above 7.45)
PaCO₂ 51 mmHg: high (35–45)
HCO₃⁻ 41 mmol/L: high (22–26)
Compensation
partially compensated
Answer
partially compensated
10.Arterial blood gas interpretation
Answer: A–a gradient = 96.9 mmHg
Derived values: calculated serum osmolality is 2·Na⁺ + glucose + urea with all values in mmol/L (2·Na⁺ + glucose/18 + BUN/2.8 with glucose and BUN in mg/dL); normal is about 275–295 mOsm/kg. The P/F ratio is PaO₂ divided by FiO₂ as a fraction. The alveolar oxygen pressure is PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 mmHg at sea level, and the A–a gradient is PAO₂ − PaO₂.
Alveolar oxygen
PAO₂ = 0.3·713 − 40/0.8 = 213.9 − 50 = 163.9 mmHg
Gradient
A–a = 163.9 − 67 = 96.9 mmHg
Answer
A–a gradient = 96.9 mmHg
11.ECG rhythm strip interpretation
Answer: QTc = 0.481 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
QT = 12.5 × 0.04 = 0.50 s
RR = 27 × 0.04 = 1.08 s
Bazett’s formula
QTc = 0.50/√1.08 = 0.50/1.039 = 0.481 s
Answer
QTc = 0.481 s
Above the usual limits for both men (about 0.45 s) and women (about 0.46 s): prolonged. (481 ms)
12.ECG rhythm strip interpretation
Answer: ventricular tachycardia
Ventricular rhythms: ventricular tachycardia is a run of three or more wide (0.12 s or more) complexes in a row at more than 100/min, usually 150–250/min, regular, without related P waves. Ventricular fibrillation is a chaotic, irregular waveform with no identifiable P waves, QRS complexes or T waves and no countable rate. Asystole is an almost flat line with no ventricular activity; a flat line is always confirmed by checking the patient, the leads and the gain and a second lead. In resuscitation guidelines VF and pulseless VT are the shockable rhythms and asystole is not.
Five steps
1. Rate: R–R = 10 small boxes, 1500 ÷ 10 ≈ 150 per minute.
2. Rhythm: regular, every R–R is 0.40 s.
3. P waves: none identifiable.
4. PR interval: not measurable.
5. QRS: 0.14 s, wide (0.12 s or more).
Interpretation
ventricular tachycardia
Answer
ventricular tachycardia
In resuscitation guidelines VF and pulseless VT are the shockable rhythms. This is identification practice only.
13.Evidence-based practice statistics
Answer: NNT = 34
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 = 84/700 = 0.12
CER = 117/780 = 0.15
Number needed to treat
ARR = 0.15 − 0.12 = 0.03
NNT = 1/0.03 = 33.33 → 34
Answer
NNT = 34
Treating 34 patients prevents about one event.
14.Evidence-based practice statistics
Answer: NNH = 16
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.