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 sympathetic nervous system leaves the central nervous system from which spinal segments?
ASacral only (S2–S4)
BThoracic and upper lumbar (T1–L2)
CCranial and sacral
DCervical only
2.Human anatomyHard
A pituitary tumour grows upwards and causes loss of both outer (temporal) halves of the visual field. Where is the visual pathway compressed?
ARight optic nerve
BOccipital cortex
COptic chiasm
DLeft optic tract
3.Human physiologyMedium
Which hormonal event triggers ovulation?
AA fall in oestrogen
BA surge of prolactin
CA rise in progesterone from the corpus luteum
DA surge of luteinising hormone (LH)
4.Human physiologyHard
In Otto Loewi's classic experiment, fluid from a vagus-stimulated frog heart slowed a second heart. What did this show?
AThe vagus nerve speeds up the heart
BNerves pass signals to their targets by releasing a chemical
CThe heart beats without any nerve supply
DNerve impulses spread through the fluid directly as electricity
5.MicrobiologyMedium
Which type of immunity is given by breast milk antibodies?
ANatural passive immunity
BNatural active immunity
CArtificial passive immunity
DArtificial active immunity
6.MicrobiologyHard
A patient with suspected infectious pulmonary tuberculosis is admitted. Which precautions are needed?
AContact precautions only, with gloves and an apron
BNo special precautions, because TB spreads only by blood
CAirborne precautions, ideally a negative-pressure room and respirator masks for staff
DDroplet precautions with a surgical mask, and only within one metre
7.PharmacologyMedium
A patient receiving chemotherapy has a very low neutrophil count and is going home. Which teaching is most important?
AEat plenty of raw shellfish to boost protein
BReport a temperature of 38 °C (100.4 °F) or higher straight away
CExercise in crowded gyms to build strength
DTake paracetamol regularly so that fevers do not show
8.PharmacologyHard
A drug has a volume of distribution of 40 L and complete bioavailability. Which loading dose gives a target plasma concentration of 2 mg/L?
A80 mg
B20 mg
C42 mg
D160 mg
9.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.60, PaCO₂ 37 mmHg (4.9 kPa) and HCO₃⁻ 35 mmol/L. The primary disorder is metabolic alkalosis. How far is it compensated? Click your answer.
pH 7.60PaCO₂ 37 mmHg (4.9 kPa)HCO₃⁻ 35 mmol/L
1uncompensated
2partially compensated
3fully compensated
10.Arterial blood gas interpretationHard
Practice values: Na⁺ 143, Cl⁻ 115, HCO₃⁻ 10 mmol/L and albumin 32 g/L. Find the albumin-corrected anion gap AGc = AG + 0.25·(40 − albumin).
This lead II strip shows atrial flutter with a fixed conduction ratio. Find the ventricular rate in beats per minute.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
ventricular rate = bpm
12.ECG rhythm strip interpretationHard
This patient has a pacemaker. Interpret the lead II 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
1ventricular paced rhythm
2paced rhythm, failure to capture
3sinus rhythm with wide QRS
4third-degree (complete) AV block
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a randomised trial, 560 patients received a new treatment and 900 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: 121/560control: 324/900
NNT =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A patient has a pre-test probability of 40% for a disease. A test with sensitivity 89% and specificity 74% comes back positive. Find the post-test probability p′ of the disease as a percentage. Round to one decimal place.
pre-test 40%Se = 89%Sp = 74%positive
p′ = %
3Solutions
1.Human anatomy
Answer: B · Thoracic and upper lumbar (T1–L2)
Why this is the answer
Sympathetic preganglionic neurons lie in the lateral horn from T1 to L2. The parasympathetic system has a craniosacral outflow.
Answer
B · Thoracic and upper lumbar (T1–L2)
2.Human anatomy
Answer: C · Optic chiasm
Why this is the answer
Bitemporal hemianopia results from compression of the crossing nasal fibres at the optic chiasm, which lies directly above the pituitary gland.
Answer
C · Optic chiasm
3.Human physiology
Answer: D · A surge of luteinising hormone (LH)
Why this is the answer
Rising oestrogen from the dominant follicle triggers an LH surge from the anterior pituitary, which causes ovulation about a day later.
Answer
D · A surge of luteinising hormone (LH)
4.Human physiology
Answer: B · Nerves pass signals to their targets by releasing a chemical
Why this is the answer
The fluid carried "Vagusstoff", later identified as acetylcholine. This was the first clear evidence for chemical neurotransmission.
Answer
B · Nerves pass signals to their targets by releasing a chemical
5.Microbiology
Answer: A · Natural passive immunity
Why this is the answer
Antibodies (mainly IgA) passed in breast milk give passive immunity acquired naturally. It is short-lived because no memory cells form.
Answer
A · Natural passive immunity
6.Microbiology
Answer: C · Airborne precautions, ideally a negative-pressure room and respirator masks for staff
Why this is the answer
M. tuberculosis spreads in airborne droplet nuclei that can stay suspended for long periods. Patients need a single room with negative pressure where possible, and staff wear FFP2/FFP3 or N95 respirators.
Answer
C · Airborne precautions, ideally a negative-pressure room and respirator masks for staff
7.Pharmacology
Answer: B · Report a temperature of 38 °C (100.4 °F) or higher straight away
Why this is the answer
Neutropenic patients may have few signs of infection other than fever, and sepsis can develop quickly. A fever needs urgent assessment; good hand hygiene and avoiding sick contacts are also taught.
Answer
B · Report a temperature of 38 °C (100.4 °F) or higher straight away
8.Pharmacology
Answer: A · 80 mg
Why this is the answer
Loading dose = 40 L × 2 mg/L = 80 mg. The loading dose depends on Vd, while the maintenance dose depends on clearance.
Answer
A · 80 mg
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 7.60: alkalaemia (above 7.45)
PaCO₂ 37 mmHg: normal (35–45)
HCO₃⁻ 35 mmol/L: high (22–26)
Compensation
Clinically, also check the expected compensation: for a metabolic alkalosis PaCO₂ ≈ 0.7·HCO₃⁻ + 21 = 43.5–47.5 mmHg, so the measured 37 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 = 20 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 = 143 − (115 + 10) = 18
Correction for albumin
18 + 0.25·(40 − 32) = 20 mmol/L
Answer
AGc = 20 mmol/L
Above about 12 after correction: a high anion gap.
11.ECG rhythm strip interpretation
Answer: ventricular rate = 150 bpm
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.
Flutter waves
F–F = 5 small boxes = 0.20 s → 1500 ÷ 5 = 300 per minute
QRS complexes
R–R = 10 small boxes = 0.40 s → 1500 ÷ 10 = 150 per minute
conduction ratio = 300 ÷ 150 = 2:1
Answer
ventricular rate = 150 bpm
12.ECG rhythm strip interpretation
Answer: ventricular paced rhythm
Paced rhythms: a pacemaker impulse shows as a narrow vertical pacing spike. After a ventricular spike the QRS is wide (0.12 s or more), because the impulse spreads from the lead through the muscle instead of the conduction system; with right-ventricular pacing it is often mostly negative in lead II. A spike before each P wave means atrial pacing, and a dual-chamber (AV) pacemaker shows an atrial spike and a P wave, then, after the programmed AV delay, a ventricular spike and a wide QRS. Capture means every spike is followed by a depolarisation. Failure to capture is a spike without a P wave or QRS after it; failure to sense (spikes that ignore the patient’s own beats) and failure to pace (no spike when one is due) are the other classic malfunctions.
Five steps
1. Rate: R–R = 22 small boxes, 1500 ÷ 22 ≈ 68 per minute.
2. Rhythm: regular, every R–R is 0.88 s.
3. P waves: none visible; a sharp pacing spike comes just before every QRS.
4. PR interval: not measurable.
5. QRS: 0.14 s, wide, and every spike is followed by a QRS (capture).
Interpretation
ventricular paced rhythm
Answer
ventricular paced rhythm
The paced QRS is wide by nature; do not call it a bundle branch block or a ventricular rhythm.
13.Evidence-based practice statistics
Answer: NNT = 7
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 = 121/560 = 0.2161
CER = 324/900 = 0.36
Number needed to treat
ARR = 0.36 − 0.2161 = 0.1439
NNT = 1/0.1439 = 6.948 → 7
Answer
NNT = 7
Treating 7 patients prevents about one event.
14.Evidence-based practice statistics
Answer: p′ = 69.5 %
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.