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
The head of the femur articulates with which structure?
AAcetabulum
BObturator foramen
CGreater trochanter
DGlenoid cavity
2.Human anatomyHard
When inserting a chest drain or needle into the pleural space, why is it placed just above the upper border of a rib?
AThe lung does not reach the area above each rib
BThe pleura is thickest just above each rib
CThe intercostal muscles are absent above each rib
DThe intercostal vein, artery and nerve run along the lower border of each rib
3.Human physiologyMedium
How is the ejection fraction calculated?
AEnd-systolic volume ÷ stroke volume
BStroke volume ÷ end-diastolic volume
CEnd-diastolic volume ÷ end-systolic volume
DCardiac output ÷ heart rate
4.Human physiologyHard
Which change shifts the oxygen–haemoglobin dissociation curve to the right?
AHigher pH
BLower 2,3-BPG
CLower CO₂
DHigher temperature
5.MicrobiologyMedium
How is DNA transferred between bacteria during conjugation?
ABy fusion of two bacterial cells into one
BBy uptake of free DNA from the environment
CThrough direct contact using a sex pilus
DBy a bacteriophage carrying bacterial DNA
6.MicrobiologyHard
A child develops bloody diarrhoea after eating an undercooked burger, then a week later has kidney failure, anaemia and a low platelet count. Which organism is most likely?
ARotavirus
BShiga toxin-producing E. coli such as O157:H7
CStaphylococcus aureus
DClostridium perfringens
7.PharmacologyMedium
Patients on older MAO inhibitors are warned about foods rich in tyramine. What can happen?
AAn allergic rash
BA hypertensive crisis
CSevere hypoglycaemia
DSudden kidney failure
8.PharmacologyHard
A patient who takes clonidine stopped it suddenly two days ago. He is now agitated, has a headache and his blood pressure is very high. What is the most likely explanation?
ANormal variation unrelated to the medication
BAn allergic reaction to clonidine
CRebound hypertension from abrupt clonidine withdrawal
DClonidine toxicity due to slow excretion by the kidneys
9.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.51, PaCO₂ 39 mmHg (5.2 kPa) and HCO₃⁻ 30 mmol/L. The primary disorder is metabolic alkalosis. How far is it compensated? Click your answer.
pH 7.51PaCO₂ 39 mmHg (5.2 kPa)HCO₃⁻ 30 mmol/L
1uncompensated
2partially compensated
3fully compensated
10.Arterial blood gas interpretationHard
Practice values: Na⁺ 134, Cl⁻ 96, HCO₃⁻ 18 mmol/L and albumin 18 g/L. Find the albumin-corrected anion gap AGc = AG + 0.25·(40 − albumin).
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
12.ECG rhythm strip interpretationHard
Interpret this lead II strip. Which interpretation can you defend from this single 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
1first-degree AV block
2second-degree AV block type II
3second-degree AV block type I (Wenckebach)
42:1 AV block (type undetermined)
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). A new test was compared with the reference standard in a study (see the 2 × 2 table). Find the positive likelihood ratio LR+. Round to two decimal places.
a = 198b = 285c = 22d = 485
LR+ =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 18% in relative terms (RRR = 18%). In a patient group the risk without treatment (CER) is 15% over five years. Find the number needed to treat NNT for five years to prevent one event. Round up to a whole number of patients.
RRR = 18%CER = 15%
NNT =
3Solutions
1.Human anatomy
Answer: A · Acetabulum
Why this is the answer
The acetabulum is the cup-shaped socket of the hip bone, formed by the ilium, ischium and pubis. The glenoid cavity is the shoulder's socket.
Answer
A · Acetabulum
2.Human anatomy
Answer: D · The intercostal vein, artery and nerve run along the lower border of each rib
Why this is the answer
The intercostal neurovascular bundle (vein, artery, nerve from above down) lies in the costal groove under each rib. Going just above the rib below avoids it.
Answer
D · The intercostal vein, artery and nerve run along the lower border of each rib
3.Human physiology
Answer: B · 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
B · Stroke volume ÷ end-diastolic volume
4.Human physiology
Answer: D · Higher temperature
Why this is the answer
Higher temperature, CO₂, H⁺ (lower pH) and 2,3-BPG lower haemoglobin's affinity and shift the curve right, releasing more oxygen to active tissues.
Answer
D · Higher temperature
5.Microbiology
Answer: C · Through direct contact using a sex pilus
Why this is the answer
In conjugation a donor cell forms a pilus to a recipient and transfers DNA, often a plasmid. It is an important way for antibiotic resistance genes to spread.
Answer
C · Through direct contact using a sex pilus
6.Microbiology
Answer: B · Shiga toxin-producing E. coli such as O157:H7
Why this is the answer
Shiga toxin-producing E. coli damages the bowel and the small blood vessels of the kidney, causing haemolytic uraemic syndrome. Cattle are a reservoir, so undercooked minced beef is a classic source.
Answer
B · Shiga toxin-producing E. coli such as O157:H7
7.Pharmacology
Answer: B · A hypertensive crisis
Why this is the answer
MAO in the gut normally breaks down dietary tyramine. With MAO inhibited, tyramine from foods such as aged cheese reaches the circulation and releases noradrenaline, causing a dangerous rise in blood pressure.
Answer
B · A hypertensive crisis
8.Pharmacology
Answer: C · Rebound hypertension from abrupt clonidine withdrawal
Why this is the answer
Clonidine is a central alpha-2 agonist. Stopping it suddenly lets sympathetic activity surge, causing rebound hypertension, so it is tapered gradually.
Answer
C · Rebound hypertension from abrupt clonidine withdrawal
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 7.51: alkalaemia (above 7.45)
PaCO₂ 39 mmHg: normal (35–45)
HCO₃⁻ 30 mmol/L: high (22–26)
Compensation
Clinically, also check the expected compensation: for a metabolic alkalosis PaCO₂ ≈ 0.7·HCO₃⁻ + 21 = 40–44 mmHg, so the measured 39 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: AGc = 25.5 mmol/L
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.
Anion gap
AG = 134 − (96 + 18) = 20
Correction for albumin
20 + 0.25·(40 − 18) = 25.5 mmol/L
Answer
AGc = 25.5 mmol/L
Above about 12 after correction: a high anion gap.
11.ECG rhythm strip interpretation
Answer: QT = 0.46 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
11.5 small boxes from the start of the QRS to the end of the T wave
Convert
QT = 11.5 × 0.04 s = 0.46 s
Answer
QT = 0.46 s
A rough rule: at normal rates the QT should be less than half the R–R interval (here R–R = 0.84 s).
12.ECG rhythm strip interpretation
Answer: 2:1 AV block (type undetermined)
2:1 AV block: with two P waves for every QRS, no two conducted P waves follow each other, so you cannot see whether the PR would lengthen (type I) or stay constant (type II) before a non-conducted P wave. From one strip it is called a 2:1 AV block; a longer strip or a change in the conduction ratio decides the type. A narrow QRS makes type I more likely and a wide QRS type II, but neither proves it.
Five steps
1. Rate: R–R = 44 small boxes, 1500 ÷ 44 ≈ 34 per minute.
2. Rhythm: regular, every R–R is 1.76 s.
3. P waves: upright and regular (P–P 0.88 s, 68 per minute), two for every QRS.
4. PR interval: constant at 0.18 s in the conducted beats.
5. QRS: 0.08 s, narrow (below 0.12 s).
Why the type stays open
Every second P wave is not conducted, so no two conducted beats follow each other and a lengthening PR could never show.
2:1 AV block (type undetermined)
Answer
2:1 AV block (type undetermined)
A narrow QRS makes type I more likely and a wide QRS type II, but only a longer strip or a change in the conduction ratio decides.
13.Evidence-based practice statistics
Answer: LR+ = 2.43
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.
Sensitivity and specificity
Se = 198/220 = 0.9
Sp = 485/770 = 0.6299
Positive likelihood ratio
LR+ = 0.9/(1 − 0.6299) = 2.432 ≈ 2.43
Answer
LR+ = 2.43
14.Evidence-based practice statistics
Answer: NNT = 38
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.
Absolute risk reduction
ARR = 0.15·0.18 = 0.027
Number needed to treat
NNT = 1/0.027 = 37.04 → 38
Answer
NNT = 38
The same relative effect gives a smaller NNT when the baseline risk is higher.