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 supplies the diaphragm?
AAccessory nerve
BIntercostal nerves
CVagus nerve
DPhrenic nerve
2.Human anatomyHard
Which coronary artery most often supplies the sinoatrial (SA) node?
ACircumflex artery
BRight coronary artery
CLeft main coronary artery
DLeft anterior descending artery
3.Human physiologyMedium
Since action potentials are all-or-none, how does a neuron signal a stronger stimulus?
ABy changing the neurotransmitter it releases
BBy making each action potential larger
CBy making each action potential longer
DBy firing action potentials more often
4.Human physiologyHard
Which spirometry finding is typical of chronic obstructive pulmonary disease (COPD)?
AFEV1/FVC ratio above 0.9 with low FVC
BFEV1/FVC ratio below 0.7
CIncreased peak expiratory flow
DNormal FEV1 with low FVC
5.MicrobiologyMedium
How is DNA transferred between bacteria during conjugation?
ABy uptake of free DNA from the environment
BBy a bacteriophage carrying bacterial DNA
CThrough direct contact using a sex pilus
DBy fusion of two bacterial cells into one
6.MicrobiologyHard
In Griffith's 1928 experiment, mice died after injection of heat-killed smooth (S) pneumococci mixed with live rough (R) pneumococci. What did this show?
ARough pneumococci are always more virulent than smooth ones
BHeat does not kill pneumococci
CViruses carry genes between bacteria
DA "transforming principle" from dead bacteria could change harmless bacteria into virulent ones
7.PharmacologyMedium
How long does it take, approximately, for a drug given at regular intervals to reach a steady state?
A1 half-life
B4 to 5 half-lives
C10 half-lives
DIt depends only on the dose size
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
BPotassium has no link with insulin, so the infusion is started as planned
CInsulin moves potassium into cells, so the level can fall dangerously low
DInsulin raises potassium further and can cause cardiac arrest, so the infusion is given faster
9.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 70 mmHg (9.3 kPa). Use the rule that in acute respiratory acidosis HCO₃⁻ rises by 1 mmol/L for every 10 mmHg that PaCO₂ is above 40, starting from HCO₃⁻ 24 mmol/L, to find the expected HCO₃⁻.
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
This lead II strip shows a conduction problem between the atria and the ventricles. 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
1sinus bradycardia
2second-degree AV block type II
3second-degree AV block type I (Wenckebach)
4first-degree AV block
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 negative likelihood ratio LR−. Round to two decimal places.
a = 114b = 101c = 16d = 429
LR− =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). 10 000 people are screened with a test of sensitivity 72% and specificity 83%. The prevalence of the disease is 21%. How many people with the disease are missed by the test (false negatives, FN)?
N = 10 000p = 21%Se = 72%Sp = 83%
FN =
3Solutions
1.Human anatomy
Answer: D · Phrenic nerve
Why this is the answer
The phrenic nerve (roots C3–C5) is the motor supply to the diaphragm. High cervical cord injury above C3 therefore stops spontaneous breathing.
Answer
D · Phrenic nerve
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: D · By firing action potentials more often
Why this is the answer
Stimulus strength is coded by the frequency of action potentials and by how many neurons are recruited, not by their amplitude.
Answer
D · By firing action potentials more often
4.Human physiology
Answer: B · FEV1/FVC ratio below 0.7
Why this is the answer
Narrowed airways slow expiration, so FEV1 falls more than FVC and the ratio drops below 0.7. In COPD this is not fully reversed by bronchodilators.
Answer
B · FEV1/FVC ratio below 0.7
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: D · A "transforming principle" from dead bacteria could change harmless bacteria into virulent ones
Why this is the answer
Something from the dead S strain turned live R bacteria into capsule-forming, virulent S bacteria. Avery, MacLeod and McCarty later showed that this transforming principle was DNA.
Answer
D · A "transforming principle" from dead bacteria could change harmless bacteria into virulent ones
7.Pharmacology
Answer: B · 4 to 5 half-lives
Why this is the answer
With repeated dosing, the level approaches steady state after about 4–5 half-lives, whatever the dose. A loading dose can be used to reach it faster.
Answer
B · 4 to 5 half-lives
8.Pharmacology
Answer: C · 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
C · Insulin moves potassium into cells, so the level can fall dangerously low
9.Arterial blood gas interpretation
Answer: HCO₃⁻ = 27 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.
Change in PaCO₂
|70 − 40| = 30 mmHg → 3 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 + 1·3 = 27 mmol/L
Answer
HCO₃⁻ = 27 mmol/L
10.Arterial blood gas interpretation
Answer: AGc = 13.25 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 = 139 − (109 + 18) = 12
Correction for albumin
12 + 0.25·(40 − 35) = 13.25 mmol/L
Answer
AGc = 13.25 mmol/L
Above about 12 after correction: a high anion gap.
11.ECG rhythm strip interpretation
Answer: QTc = 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
QT = 10.5 × 0.04 = 0.42 s
RR = 19 × 0.04 = 0.76 s
Fridericia’s formula
QTc = 0.42/∛0.76 = 0.42/0.9126 = 0.46 s
Answer
QTc = 0.46 s
Between the usual limits for men (about 0.45 s) and women (about 0.46 s): borderline, depending on the reference used. (460 ms)
12.ECG rhythm strip interpretation
Answer: second-degree AV block type II
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.
Five steps
1. Rate: 5 QRS complexes in 6 s, × 10 = 50 per minute.
2. Rhythm: the P waves are regular; R–R is regular except for a long pause where a QRS is missing.
3. P waves: upright and regular (P–P 0.96 s, 63 per minute); one P in every 4 is not followed by a QRS.
4. PR interval: constant at 0.24 s in the conducted beats.
5. QRS: 0.14 s, wide (0.12 s or more).
Interpretation
second-degree AV block type II
Answer
second-degree AV block type II
Type II and third-degree blocks are the ones that can suddenly progress, so textbooks rank them as more serious than type I.
13.Evidence-based practice statistics
Answer: LR− = 0.15
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 = 114/130 = 0.8769
Sp = 429/530 = 0.8094
Negative likelihood ratio
LR− = (1 − 0.8769)/0.8094 = 0.1521 ≈ 0.15
Answer
LR− = 0.15
14.Evidence-based practice statistics
Answer: FN = 588
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.