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 cranial nerve controls movement of the tongue?
AHypoglossal nerve (XII)
BVagus nerve (X)
CGlossopharyngeal nerve (IX)
DFacial nerve (VII)
2.Human anatomyHard
Which coronary artery most often supplies the sinoatrial (SA) node?
ACircumflex artery
BRight coronary artery
CLeft anterior descending artery
DLeft main coronary artery
3.Human physiologyMedium
Besides lowering glucose, insulin moves which ion into cells, an effect used to treat hyperkalaemia?
APotassium
BSodium
CChloride
DCalcium
4.Human physiologyHard
Which change shifts the oxygen–haemoglobin dissociation curve to the right?
ALower 2,3-BPG
BHigher temperature
CLower CO₂
DHigher pH
5.MicrobiologyMedium
Which white blood cells are the main phagocytes in an acute bacterial infection and the main component of pus?
AB lymphocytes
BBasophils
CEosinophils
DNeutrophils
6.MicrobiologyHard
Why can an HIV antibody test be negative in the first weeks after infection?
AThe body has not yet produced enough antibodies to detect
BThe antibodies are destroyed by the test reagents
CHIV does not trigger antibody production at all
DThe virus hides only in red blood cells during the first weeks
7.PharmacologyMedium
Aciclovir is mainly used to treat infections caused by:
AInfluenza A virus
BHerpes viruses such as herpes simplex and varicella-zoster
CBacteria that produce beta-lactamase
DHIV
8.PharmacologyHard
A patient with diabetic ketoacidosis is about to start an IV insulin infusion. Serum potassium is 3.1 mmol/L (mEq/L). Why is this a concern?
ALow potassium makes insulin ineffective, so a higher dose is needed
BInsulin moves potassium into cells, so the level can fall dangerously low
CPotassium has no link with insulin, so the infusion is started as planned
DInsulin raises potassium further and can cause cardiac arrest, so the infusion is given faster
9.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.24, PaCO₂ 80 mmHg (10.7 kPa) and HCO₃⁻ 33 mmol/L. The primary disorder is respiratory acidosis. How far is it compensated? Click your answer.
pH 7.24PaCO₂ 80 mmHg (10.7 kPa)HCO₃⁻ 33 mmol/L
1uncompensated
2partially compensated
3fully compensated
10.Arterial blood gas interpretationHard
Practice values: Na⁺ 133, Cl⁻ 107 and HCO₃⁻ 9 mmol/L. Calculate the delta ratio (AG − 12)/(24 − HCO₃⁻), using AG = Na⁺ − (Cl⁻ + HCO₃⁻). Give two decimals.
Na⁺ 133 mmol/LCl⁻ 107 mmol/LHCO₃⁻ 9 mmol/L
delta ratio =
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 Fridericia’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
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
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a randomised trial, 830 patients received a new treatment and 690 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: 224/830control: 248/690
NNT =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A screening test has a sensitivity of 94% and a specificity of 95%. It is used in a population where the prevalence is 3%. Find the positive predictive value PPV as a percentage. Round to one decimal place.
Se = 94%Sp = 95%p = 3%
PPV = %
3Solutions
1.Human anatomy
Answer: A · Hypoglossal nerve (XII)
Why this is the answer
The hypoglossal nerve supplies the intrinsic and most extrinsic tongue muscles. A lesion makes the protruded tongue deviate towards the affected side.
Answer
A · Hypoglossal nerve (XII)
2.Human anatomy
Answer: B · Right coronary artery
Why this is the answer
The SA nodal artery arises from the right coronary artery in roughly 60% of people and from the circumflex artery in the rest.
Answer
B · Right coronary artery
3.Human physiology
Answer: A · Potassium
Why this is the answer
Insulin stimulates the Na⁺/K⁺-ATPase and shifts potassium into cells. Insulin with glucose is therefore used to lower dangerously high plasma potassium.
Answer
A · Potassium
4.Human physiology
Answer: B · 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
B · Higher temperature
5.Microbiology
Answer: D · Neutrophils
Why this is the answer
Neutrophils arrive quickly at sites of infection and engulf bacteria. Dead neutrophils, bacteria and tissue debris form pus.
Answer
D · Neutrophils
6.Microbiology
Answer: A · The body has not yet produced enough antibodies to detect
Why this is the answer
Antibodies take several weeks to appear. Combined tests that also detect the p24 antigen, or tests for viral RNA, shorten this window period.
Answer
A · The body has not yet produced enough antibodies to detect
7.Pharmacology
Answer: B · Herpes viruses such as herpes simplex and varicella-zoster
Why this is the answer
Aciclovir is phosphorylated by viral thymidine kinase in infected cells and then blocks viral DNA polymerase. It is used for cold sores, genital herpes, chickenpox and shingles.
Answer
B · Herpes viruses such as herpes simplex and varicella-zoster
8.Pharmacology
Answer: B · Insulin moves potassium into cells, so the level can fall dangerously low
Why this is the answer
Insulin drives potassium into cells, so a low starting level can become life-threatening hypokalaemia with arrhythmias. Potassium is replaced and monitored closely; the 2024 international consensus (ADA and others) delays insulin until potassium is above 3.5 mmol/L, while UK JBDS guidance adds potassium to the fluids and asks for senior review.
Answer
B · Insulin moves potassium into cells, so the level can fall dangerously low
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 7.24: acidaemia (below 7.35)
PaCO₂ 80 mmHg: high (35–45)
HCO₃⁻ 33 mmol/L: high (22–26)
Compensation
partially compensated
Answer
partially compensated
10.Arterial blood gas interpretation
Answer: delta ratio = 0.3333
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 = 133 − (107 + 9) = 17
Delta ratio
(17 − 12)/(24 − 9) = 5/15 = 0.3333
Answer
delta ratio = 0.3333
Below 1: an added normal anion gap acidosis.
11.ECG rhythm strip interpretation
Answer: QTc = 0.466 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 = 21 × 0.04 = 0.84 s
Fridericia’s formula
QTc = 0.44/∛0.84 = 0.44/0.9435 = 0.466 s
Answer
QTc = 0.466 s
Above the usual limits for both men (about 0.45 s) and women (about 0.46 s): prolonged. (466 ms)
12.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 = 13 small boxes, 1500 ÷ 13 ≈ 115 per minute.
2. Rhythm: regular, every R–R is 0.52 s.
3. P waves: upright, one before every QRS, all alike.
4. PR interval: 0.12 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
13.Evidence-based practice statistics
Answer: NNT = 12
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 = 224/830 = 0.2699
CER = 248/690 = 0.3594
Number needed to treat
ARR = 0.3594 − 0.2699 = 0.08954
NNT = 1/0.08954 = 11.17 → 12
Answer
NNT = 12
Treating 12 patients prevents about one event.
14.Evidence-based practice statistics
Answer: PPV = 36.8 %
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.
True and false positives (per person)
Se·p = 0.94·0.03 = 0.0282
(1 − Sp)·(1 − p) = 0.05·0.97 = 0.0485
Positive predictive value
PPV = 0.0282/(0.0282 + 0.0485) = 0.3677 ≈ 36.8%
Answer
PPV = 36.8%
The lower the prevalence, the lower the PPV of the same test.