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 part of the stomach connects to the duodenum?
ABody
BPylorus
CCardia
DFundus
2.Human anatomyHard
A lesion of the left optic tract causes which visual field defect?
ARight homonymous hemianopia
BBitemporal hemianopia
CBlindness in the left eye only
DLeft homonymous hemianopia
3.Human physiologyMedium
How does glucagon raise blood glucose?
AIt stimulates glycogen breakdown and gluconeogenesis in the liver
BIt blocks insulin receptors on fat cells
CIt stimulates glucose uptake by muscle
DIt slows glucose absorption from the gut
4.Human physiologyHard
At the start of a fever, a patient shivers and feels cold even though body temperature is rising. Why?
APyrogens raise the hypothalamic set point above the current body temperature
BThe body has lost its ability to produce heat
CThe thermoreceptors in the skin are destroyed
DSweating has already cooled the body too much
5.MicrobiologyMedium
Which infection is a typical complication of broad-spectrum antibiotic use, causing severe diarrhoea?
ASalmonella infection
BHelicobacter pylori infection
CHepatitis A
DClostridioides difficile infection
6.MicrobiologyHard
An older adult develops a painful blistering rash in a band on one side of the chest. What explains this pattern?
AA new chickenpox infection spreads through the blood to one side
BVaricella-zoster virus reactivates from one sensory ganglion and travels along its nerve
CBacteria spread along the lymphatic vessels of the chest
DAn allergic reaction to clothing affects one dermatome
7.PharmacologyMedium
An Rh-negative mother gives birth to an Rh-positive baby. When is anti-D immunoglobulin usually given?
AWithin 72 hours after the birth
BAt least two weeks after the birth
CIt is not needed if the baby is Rh-positive
DOnly at the start of the next pregnancy
8.PharmacologyHard
In aspirin overdose the urine may be made alkaline with sodium bicarbonate. Why does this increase salicylate excretion?
ABicarbonate speeds up the liver metabolism of salicylate to inactive forms
BMore salicylate is ionised in the tubule, so less is reabsorbed
CAlkaline urine increases glomerular filtration
DBicarbonate binds salicylate and makes it inactive
9.Arterial blood gas interpretationMedium
Practice values: Na⁺ 133 mmol/L, glucose 4.6 mmol/L and urea 4.7 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 133 mmol/Lglucose 4.6 mmol/Lurea 4.7 mmol/L
Osm = mOsm/kg
10.Arterial blood gas interpretationHard
A practice gas shows a metabolic alkalosis with HCO₃⁻ 32 mmol/L. Use expected PaCO₂ = 0.7·HCO₃⁻ + 21 to find the expected PaCO₂.
HCO₃⁻ 32 mmol/L
PaCO₂ = mmHg
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
2junctional rhythm
3normal sinus rhythm
4atrial fibrillation
12.ECG rhythm strip interpretationHard
This lead II strip shows second-degree AV block type I (Wenckebach). How many P waves per group are there, counting the non-conducted one (the n of an n:(n − 1) block)?
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
P waves per group =
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). A case-control study compares 340 patients with a disease (cases) and 290 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 = 222b = 157c = 118d = 133
OR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A screening test has a sensitivity of 77% and a specificity of 94%. It is used in a population where the prevalence is 30%. Find the positive predictive value PPV as a percentage. Round to one decimal place.
Se = 77%Sp = 94%p = 30%
PPV = %
3Solutions
1.Human anatomy
Answer: B · Pylorus
Why this is the answer
The pylorus is the distal stomach. Its pyloric sphincter regulates the passage of chyme into the duodenum.
Answer
B · Pylorus
2.Human anatomy
Answer: A · Right homonymous hemianopia
Why this is the answer
The left optic tract carries fibres from the left half of each retina, which see the right visual field. Its lesion causes loss of the right half of the field in both eyes.
Answer
A · Right homonymous hemianopia
3.Human physiology
Answer: A · It stimulates glycogen breakdown and gluconeogenesis in the liver
Why this is the answer
Glucagon from alpha cells acts on the liver to break down glycogen and make new glucose from non-carbohydrate sources.
Answer
A · It stimulates glycogen breakdown and gluconeogenesis in the liver
4.Human physiology
Answer: A · Pyrogens raise the hypothalamic set point above the current body temperature
Why this is the answer
Pyrogens trigger prostaglandin E₂ in the hypothalamus, raising the set point. Until body temperature reaches the new set point, heat-gain responses such as shivering occur.
Answer
A · Pyrogens raise the hypothalamic set point above the current body temperature
5.Microbiology
Answer: D · Clostridioides difficile infection
Why this is the answer
Antibiotics kill normal gut bacteria, letting toxin-producing C. difficile overgrow. Its spores are not killed by alcohol gel, so soap and water handwashing is needed.
Answer
D · Clostridioides difficile infection
6.Microbiology
Answer: B · Varicella-zoster virus reactivates from one sensory ganglion and travels along its nerve
Why this is the answer
After chickenpox, VZV stays latent in a dorsal root or cranial nerve ganglion. When immunity weakens it can reactivate, travelling down that nerve to cause shingles in its dermatome.
Answer
B · Varicella-zoster virus reactivates from one sensory ganglion and travels along its nerve
7.Pharmacology
Answer: A · Within 72 hours after the birth
Why this is the answer
Anti-D is given within 72 hours of birth to an Rh-negative mother of an Rh-positive baby. It is also given after events such as miscarriage or abdominal trauma, and often routinely during pregnancy.
Answer
A · Within 72 hours after the birth
8.Pharmacology
Answer: B · More salicylate is ionised in the tubule, so less is reabsorbed
Why this is the answer
Salicylate is a weak acid. In alkaline urine more of it is in the ionised form, which is "trapped" in the tubule and excreted. This is called ion trapping.
Answer
B · More salicylate is ionised in the tubule, so less is reabsorbed
9.Arterial blood gas interpretation
Answer: Osm = 275 mOsm/kg
Derived values: calculated serum osmolality is 2·Na⁺ + glucose + urea with all values in mmol/L (2·Na⁺ + glucose/18 + BUN/2.8 with glucose and BUN in mg/dL); normal is about 275–295 mOsm/kg. The P/F ratio is PaO₂ divided by FiO₂ as a fraction. The alveolar oxygen pressure is PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 mmHg at sea level, and the A–a gradient is PAO₂ − PaO₂.
Calculated osmolality
Osm = 2·133 + 4.6 + 4.7 = 275.3 ≈ 275 mOsm/kg
Answer
Osm = 275 mOsm/kg
10.Arterial blood gas interpretation
Answer: PaCO₂ = 43.4 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
0.7·32 + 21 = 43.4 mmHg
Answer
PaCO₂ = 43.4 mmHg
The rise has a limit: PaCO₂ rarely goes above about 55 mmHg from compensation alone.
11.ECG rhythm strip interpretation
Answer: junctional rhythm
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 = 34 small boxes, 1500 ÷ 34 ≈ 44 per minute.
2. Rhythm: regular, every R–R is 1.36 s.
3. P waves: none visible.
4. PR interval: not measurable.
5. QRS: 0.10 s, narrow (below 0.12 s).
Interpretation
junctional rhythm
Answer
junctional rhythm
12.ECG rhythm strip interpretation
Answer: P waves per group = 3
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.
March out the P waves
P–P = 0.88 s, regular; one P wave in every 3 is not conducted
PR in each group
PR₁ = 0.18 s, PR₂ = 0.30 s
Answer
P waves per group = 3
A 3:2 Wenckebach block.
13.Evidence-based practice statistics
Answer: OR = 1.59
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: 222/118 = 1.881
controls: 157/133 = 1.18
Odds ratio
OR = (222·133)/(157·118) = 1.594 ≈ 1.59
Answer
OR = 1.59
OR above 1: exposure is more common among the cases.
14.Evidence-based practice statistics
Answer: PPV = 84.6 %
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.
True and false positives (per person)
Se·p = 0.77·0.3 = 0.231
(1 − Sp)·(1 − p) = 0.06·0.7 = 0.042
Positive predictive value
PPV = 0.231/(0.231 + 0.042) = 0.8462 ≈ 84.6%
Answer
PPV = 84.6%
The lower the prevalence, the lower the PPV of the same test.