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 type of epithelium lines the urinary bladder?
ATransitional epithelium (urothelium)
BSimple columnar epithelium
CSimple squamous epithelium
DPseudostratified columnar epithelium
2.Human anatomyHard
In portal hypertension, oesophageal varices form at which portosystemic anastomosis?
ABetween the paraumbilical veins and the epigastric veins
BBetween the left gastric vein and the oesophageal tributaries of the azygos vein
CBetween the splenic vein and the left renal vein
DBetween the superior rectal and middle rectal veins
3.Human physiologyMedium
Which is a classic example of positive feedback in the body?
AInsulin release after a meal
BADH release when plasma osmolality rises
CSweating when body temperature rises
DOxytocin release during labour
4.Human physiologyHard
After several days of persistent vomiting of stomach contents, which acid–base disturbance is most likely?
ARespiratory alkalosis
BRespiratory acidosis
CMetabolic acidosis
DMetabolic alkalosis
5.MicrobiologyMedium
Which organism is a common cause of infection in burns and in people with cystic fibrosis, and can produce a green pigment?
APseudomonas aeruginosa
BStreptococcus pyogenes
CClostridium tetani
DHelicobacter pylori
6.MicrobiologyHard
What did Pasteur's swan-neck flask experiment demonstrate?
AMicroorganisms need light to grow
BMicroorganisms come from other microorganisms, not from spontaneous generation
CAir itself produces microorganisms
DBoiling cannot kill microorganisms
7.PharmacologyMedium
A patient starting sulfasalazine should be told about which harmless effect?
AOrange-yellow colouring of the urine
BWhite spots on the nails
CGreen colouring of the sweat
DBlue colouring of the tongue and gums
8.PharmacologyHard
A patient with severe chronic hyponatraemia receives hypertonic saline. Why is the rate of correction strictly limited?
ARaising sodium too fast causes severe hypokalaemia and cardiac arrest
BHypertonic saline is toxic to the liver at high rates
CRaising sodium too fast can cause osmotic demyelination in the brain
DFast correction causes immediate kidney failure
9.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.57, PaCO₂ 43 mmHg (5.7 kPa) and HCO₃⁻ 38 mmol/L. The primary disorder is metabolic alkalosis. How far is it compensated? Click your answer.
pH 7.57PaCO₂ 43 mmHg (5.7 kPa)HCO₃⁻ 38 mmol/L
1uncompensated
2partially compensated
3fully compensated
10.Arterial blood gas interpretationHard
A practice gas shows a metabolic alkalosis with HCO₃⁻ 39 mmol/L. Use expected PaCO₂ = 0.7·HCO₃⁻ + 21 to find the expected PaCO₂.
HCO₃⁻ 39 mmol/L
PaCO₂ = mmHg
11.ECG rhythm strip interpretationMedium
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
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
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a cohort, groups of people were followed for different lengths of time (see the table: number of people and years of follow-up for each). During follow-up, 21 new cases occurred. Find the incidence rate IR per 1000 person-years. Round to one decimal place.
Practice exercise (made-up values, not clinical guidance). A patient has a pre-test probability of 5% for a disease. A test with sensitivity 62% and specificity 91% comes back positive. Find the post-test probability p′ of the disease as a percentage. Round to one decimal place.
pre-test 5%Se = 62%Sp = 91%positive
p′ = %
3Solutions
1.Human anatomy
Answer: A · Transitional epithelium (urothelium)
Why this is the answer
Transitional epithelium can change shape from rounded to flattened cells as the bladder stretches, while staying watertight.
Answer
A · Transitional epithelium (urothelium)
2.Human anatomy
Answer: B · Between the left gastric vein and the oesophageal tributaries of the azygos vein
Why this is the answer
At the lower oesophagus, portal blood in the left gastric vein meets systemic blood in the oesophageal veins draining to the azygos system. High portal pressure dilates these veins.
Answer
B · Between the left gastric vein and the oesophageal tributaries of the azygos vein
3.Human physiology
Answer: D · Oxytocin release during labour
Why this is the answer
Stretch of the cervix releases oxytocin, which strengthens contractions and causes more stretch. The loop ends with delivery. The other examples are negative feedback.
Answer
D · Oxytocin release during labour
4.Human physiology
Answer: D · Metabolic alkalosis
Why this is the answer
Vomiting removes HCl, and the stomach's acid secretion leaves bicarbonate in the blood. Volume loss and low potassium help keep the alkalosis going.
Answer
D · Metabolic alkalosis
5.Microbiology
Answer: A · Pseudomonas aeruginosa
Why this is the answer
Pseudomonas aeruginosa is an opportunistic pathogen of burns, ventilated patients and cystic fibrosis lungs. It is naturally resistant to many antibiotics.
Answer
A · Pseudomonas aeruginosa
6.Microbiology
Answer: B · Microorganisms come from other microorganisms, not from spontaneous generation
Why this is the answer
Boiled broth in a flask with a curved neck stayed sterile because dust and microbes were trapped in the bend. When the neck was broken, microbes entered and grew, disproving spontaneous generation.
Answer
B · Microorganisms come from other microorganisms, not from spontaneous generation
7.Pharmacology
Answer: A · Orange-yellow colouring of the urine
Why this is the answer
Sulfasalazine can colour urine and sometimes skin and contact lenses orange-yellow. More serious adverse effects include blood disorders and rash, so blood counts are monitored.
Answer
A · Orange-yellow colouring of the urine
8.Pharmacology
Answer: C · Raising sodium too fast can cause osmotic demyelination in the brain
Why this is the answer
In chronic hyponatraemia the brain adapts to low osmolality. Correcting sodium too rapidly shrinks brain cells and can cause osmotic demyelination (central pontine myelinolysis), so sodium is checked often.
Answer
C · Raising sodium too fast can cause osmotic demyelination in the brain
9.Arterial blood gas interpretation
Answer: uncompensated
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 7.57: alkalaemia (above 7.45)
PaCO₂ 43 mmHg: normal (35–45)
HCO₃⁻ 38 mmol/L: high (22–26)
Compensation
Clinically, also check the expected compensation: for a metabolic alkalosis PaCO₂ ≈ 0.7·HCO₃⁻ + 21 = 45.6–49.6 mmHg, so the measured 43 mmHg points to an additional respiratory alkalosis (a mixed disorder). The answer above is the ROME exam classification.
uncompensated
Answer
uncompensated
10.Arterial blood gas interpretation
Answer: PaCO₂ = 48.3 mmHg
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
0.7·39 + 21 = 48.3 mmHg
Answer
PaCO₂ = 48.3 mmHg
The rise has a limit: PaCO₂ rarely goes above about 55 mmHg from compensation alone.
11.ECG rhythm strip interpretation
Answer: QTc = 0.449 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 = 11 × 0.04 = 0.44 s
RR = 24 × 0.04 = 0.96 s
Bazett’s formula
QTc = 0.44/√0.96 = 0.44/0.9798 = 0.449 s
Answer
QTc = 0.449 s
Within the usual limits (about 0.45 s for men, 0.46 s for women). (449 ms)
12.ECG rhythm strip interpretation
Answer: atrial rate = 63 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 = 24 small boxes = 0.96 s → 1500 ÷ 24 = 62.5 ≈ 63 per minute
Ventricular rate
5 QRS complexes in 6 s × 10 = 50 per minute
Answer
atrial rate = 63 bpm
Answer
atrial rate = 63 bpm
Second-degree type I (Wenckebach): 3 P waves for every 2 QRS complexes.
13.Evidence-based practice statistics
Answer: IR = 28.6 per 1000 person-years
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.
Person-years
PY = 170·2 + 80·1.5 + 110·2.5 = 735
Incidence rate
IR = 21/735·1000 = 28.6 per 1000 person-years
Answer
IR = 28.6 per 1000 person-years
14.Evidence-based practice statistics
Answer: p′ = 26.6 %
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.