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
Where does fertilisation most commonly take place?
AAmpulla of the uterine tube
BOvary
CCervix
DBody of the uterus
2.Human anatomyHard
After a difficult delivery with traction on the neck, a baby's arm hangs by the side, internally rotated with the forearm pronated ("waiter's tip"). Which part of the brachial plexus is injured?
AUpper trunk (C5–C6)
BMedial cord
CLower trunk (C8–T1)
DPosterior cord
3.Human physiologyMedium
Where is the heart's natural pacemaker?
AAtrioventricular node
BSinoatrial node
CBundle of His
DPurkinje fibres
4.Human physiologyHard
Why can a patient with carbon monoxide poisoning have a normal PaO₂ but severe tissue hypoxia?
ACO stops oxygen from dissolving in plasma
BCO lowers the barometric pressure in the alveoli
CCO occupies haemoglobin binding sites, so oxygen content falls
DCO blocks diffusion across the alveolar wall
5.MicrobiologyMedium
Which bacterium causes whooping cough (pertussis)?
ACorynebacterium diphtheriae
BHaemophilus influenzae
CMycoplasma pneumoniae
DBordetella pertussis
6.MicrobiologyHard
Why must a negative-sense RNA virus such as influenza carry its own RNA-dependent RNA polymerase into the host cell?
AHost ribosomes cannot read any viral genes
BThe polymerase is needed to integrate the genome into host DNA
CThe polymerase protects the virus from antibodies
DHost cells cannot make mRNA from an RNA template
7.PharmacologyMedium
How can the absorption of oral iron be improved?
ATake it with a cup of strong tea
BTake it with a glass of milk
CTake it with vitamin C, such as orange juice, on an empty stomach
DTake it at the same time as an antacid to protect the stomach
8.PharmacologyHard
An older patient who has taken omeprazole for many years has muscle cramps, tremor and an irregular heartbeat. Which lab finding is most likely linked to the drug?
AHigh serum calcium
BLow serum glucose
CHigh serum sodium
DLow serum magnesium
9.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 63 mmHg (8.4 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⁻.
Practice values: Na⁺ 142, Cl⁻ 95 and HCO₃⁻ 9 mmol/L. Calculate the delta ratio (AG − 12)/(24 − HCO₃⁻), using AG = Na⁺ − (Cl⁻ + HCO₃⁻). Give two decimals.
Na⁺ 142 mmol/LCl⁻ 95 mmol/LHCO₃⁻ 9 mmol/L
delta ratio =
11.ECG rhythm strip interpretationMedium
Interpret this lead II rhythm strip. 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
1first-degree AV block
2sinus bradycardia
3junctional rhythm
4sinus tachycardia
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
1second-degree AV block type I (Wenckebach)
2first-degree AV block
3atrial fibrillation
4third-degree (complete) AV block
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). A case-control study compares 300 patients with a disease (cases) and 300 people without it (controls). The table shows how many in each group were exposed to a risk factor. Find the odds ratio OR for the exposure. Round to two decimal places.
a = 212b = 180c = 88d = 120
OR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A patient has a pre-test probability of 50% for a disease. A test with sensitivity 98% and specificity 85% comes back positive. Find the post-test probability p′ of the disease as a percentage. Round to one decimal place.
pre-test 50%Se = 98%Sp = 85%positive
p′ = %
3Solutions
1.Human anatomy
Answer: A · Ampulla of the uterine tube
Why this is the answer
Fertilisation usually occurs in the ampulla of the uterine (fallopian) tube. The embryo then travels to the uterus to implant.
Answer
A · Ampulla of the uterine tube
2.Human anatomy
Answer: A · Upper trunk (C5–C6)
Why this is the answer
Erb's palsy follows stretching of the upper trunk (C5–C6), paralysing deltoid, the external rotators and biceps. Lower trunk injury (Klumpke's) affects the small hand muscles instead.
Answer
A · Upper trunk (C5–C6)
3.Human physiology
Answer: B · Sinoatrial node
Why this is the answer
The SA node depolarises spontaneously at the fastest rate and sets the heart rhythm. The AV node delays the impulse before it spreads to the ventricles.
Answer
B · Sinoatrial node
4.Human physiology
Answer: C · CO occupies haemoglobin binding sites, so oxygen content falls
Why this is the answer
CO binds haemoglobin far more strongly than oxygen and shifts the curve left. Dissolved O₂ (PaO₂) is normal, but total oxygen content and release to tissues drop sharply.
Answer
C · CO occupies haemoglobin binding sites, so oxygen content falls
5.Microbiology
Answer: D · Bordetella pertussis
Why this is the answer
Bordetella pertussis causes long coughing fits with a "whoop". It is most dangerous for young babies, which is why vaccination in pregnancy is offered in many countries.
Answer
D · Bordetella pertussis
6.Microbiology
Answer: D · Host cells cannot make mRNA from an RNA template
Why this is the answer
A negative-sense genome must first be copied into positive-sense mRNA. Human cells have no enzyme that copies RNA from RNA, so the virus brings its own polymerase.
Answer
D · Host cells cannot make mRNA from an RNA template
7.Pharmacology
Answer: C · Take it with vitamin C, such as orange juice, on an empty stomach
Why this is the answer
Vitamin C keeps iron in its more absorbable ferrous form. Milk, antacids, tea and some antibiotics reduce iron absorption. If the stomach becomes upset, it may be taken with food, accepting lower absorption.
Answer
C · Take it with vitamin C, such as orange juice, on an empty stomach
8.Pharmacology
Answer: D · Low serum magnesium
Why this is the answer
Long-term PPI use can cause hypomagnesaemia, which leads to cramps, tremor and arrhythmias. Other long-term risks include fractures, vitamin B12 deficiency and C. difficile infection.
Answer
D · Low serum magnesium
9.Arterial blood gas interpretation
Answer: HCO₃⁻ = 26.3 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
|63 − 40| = 23 mmHg → 2.3 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 + 1·2.3 = 26.3 mmol/L
Answer
HCO₃⁻ = 26.3 mmol/L
10.Arterial blood gas interpretation
Answer: delta ratio = 1.733
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 = 142 − (95 + 9) = 38
Delta ratio
(38 − 12)/(24 − 9) = 26/15 = 1.733
Answer
delta ratio = 1.733
Between 1 and 2: a pure high anion gap acidosis.
11.ECG rhythm strip interpretation
Answer: sinus bradycardia
Rhythm criteria (lead II): sinus rhythm has an upright P wave before every QRS, a constant PR of 0.12–0.20 s and a narrow QRS: normal sinus rhythm at 60–100/min, sinus bradycardia below 60, sinus tachycardia above 100. Atrial fibrillation: no P waves, a fibrillatory baseline and an irregularly irregular R–R. Atrial flutter: sawtooth flutter waves at about 250–350/min with a fixed conduction ratio such as 2:1 or 4:1. First-degree AV block: sinus rhythm with a constant PR above 0.20 s. Junctional rhythm: narrow QRS at 40–60/min with absent or inverted P waves (PR below 0.12 s, or P after the QRS). A PVC is an early, wide (0.12 s or more), bizarre QRS without a P wave, with the T wave in the opposite direction and usually a full compensatory pause; in bigeminy every second beat is a PVC.
Five steps
1. Rate: R–R = 30 small boxes, 1500 ÷ 30 ≈ 50 per minute.
2. Rhythm: regular, every R–R is 1.20 s.
3. P waves: upright, one before every QRS, all alike.
4. PR interval: 0.18 s, constant, within 0.12–0.20 s.
5. QRS: 0.10 s, narrow (below 0.12 s).
Interpretation
sinus bradycardia
Answer
sinus bradycardia
12.ECG rhythm strip interpretation
Answer: second-degree AV block type I (Wenckebach)
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: 4 QRS complexes in 6 s, × 10 = 40 per minute.
2. Rhythm: regularly irregular, the QRS complexes come in groups separated by a pause (group beating).
3. P waves: upright and regular (P–P 1.00 s, 60 per minute); one P in every 3 is not followed by a QRS.
4. PR interval: lengthens 0.16 → 0.30 s, then a P wave is not conducted.
5. QRS: 0.08 s, narrow (below 0.12 s).
Interpretation
second-degree AV block type I (Wenckebach)
Answer
second-degree AV block type I (Wenckebach)
13.Evidence-based practice statistics
Answer: OR = 1.61
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.
Odds of exposure
cases: 212/88 = 2.409
controls: 180/120 = 1.5
Odds ratio
OR = (212·120)/(180·88) = 1.606 ≈ 1.61
Answer
OR = 1.61
OR above 1: exposure is more common among the cases.
14.Evidence-based practice statistics
Answer: p′ = 86.7 %
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.