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?
AFacial nerve (VII)
BGlossopharyngeal nerve (IX)
CHypoglossal nerve (XII)
DVagus nerve (X)
2.Human anatomyHard
Why does an inhaled foreign body more often lodge in the right main bronchus than the left?
AThe right lung has stronger airflow due to the heart's position
BThe left bronchus is closed by the epiglottis
CThe right bronchus is narrower and more horizontal
DThe right bronchus is wider, shorter and more vertical
3.Human physiologyMedium
Which hormone makes the gallbladder contract after a fatty meal?
ACholecystokinin (CCK)
BGastrin
CInsulin
DSecretin
4.Human physiologyHard
A pulmonary embolus blocks blood flow to a well-ventilated region of lung. How does the ventilation–perfusion (V/Q) ratio of that region change?
AIt becomes zero, acting as a shunt
BIt becomes low, like in pneumonia
CIt becomes very high, acting as dead space
DIt stays normal because ventilation falls too
5.MicrobiologyMedium
Which organism causes typhoid fever?
AVibrio cholerae
BCampylobacter jejuni
CShigella dysenteriae
DSalmonella Typhi
6.MicrobiologyHard
What did Pasteur's swan-neck flask experiment demonstrate?
AMicroorganisms need light to grow
BAir itself produces microorganisms
CMicroorganisms come from other microorganisms, not from spontaneous generation
DBoiling cannot kill microorganisms
7.PharmacologyMedium
Which assessment is most important before giving a dose of IV morphine?
APupil size and hearing
BBowel sounds and urine colour
CRespiratory rate and level of sedation
DBlood glucose and temperature, checked before each dose
8.PharmacologyHard
A woman using the combined oral contraceptive pill starts carbamazepine for epilepsy. What is the main concern?
AContraceptive failure because the hormones are metabolised faster
BNo concern, because the drugs do not interact
CCarbamazepine toxicity because the pill strongly blocks its metabolism
DSevere hypoglycaemia from the combination
9.Arterial blood gas interpretationMedium
A practice gas shows a metabolic acidosis with HCO₃⁻ 16 mmol/L. Use Winter’s formula to find the expected PaCO₂ (the centre of the range) in mmHg.
HCO₃⁻ 16 mmol/L
PaCO₂ = mmHg
10.Arterial blood gas interpretationHard
A practice gas shows a respiratory acidosis: pH 7.33, PaCO₂ 65 mmHg (8.7 kPa), HCO₃⁻ 33 mmol/L. Use the rules ‘HCO₃⁻ rises 1 mmol/L per 10 mmHg (acute) or 3.5 mmol/L per 10 mmHg (chronic)’ to classify it. Click your answer.
pH 7.33PaCO₂ 65 mmHg (8.7 kPa)HCO₃⁻ 33 mmol/L
1acute
2chronic
3acute on chronic
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
1atrial fibrillation
2atrial flutter, 2:1 conduction
3junctional rhythm
4atrial flutter, 4:1 conduction
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
1atrial fibrillation
2third-degree (complete) AV block
3second-degree AV block type I (Wenckebach)
4second-degree AV block type II
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a randomised trial, 100 patients received a new treatment and 740 received the control treatment. The table shows how many in each group had a side effect. Find the number needed to harm NNH: how many patients must be treated for one extra patient to have the side effect? Round up to a whole number of patients.
treatment: 54/100control: 200/740
NNH =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A screening test has a sensitivity of 71% 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 = 71%Sp = 94%p = 30%
PPV = %
3Solutions
1.Human anatomy
Answer: C · 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
C · Hypoglossal nerve (XII)
2.Human anatomy
Answer: D · The right bronchus is wider, shorter and more vertical
Why this is the answer
The right main bronchus is wider, shorter and continues more directly in line with the trachea, so aspirated objects tend to fall into it.
Answer
D · The right bronchus is wider, shorter and more vertical
3.Human physiology
Answer: A · Cholecystokinin (CCK)
Why this is the answer
Fat and protein in the duodenum release CCK, which contracts the gallbladder and stimulates pancreatic enzyme secretion.
Answer
A · Cholecystokinin (CCK)
4.Human physiology
Answer: C · It becomes very high, acting as dead space
Why this is the answer
Ventilation without perfusion gives a very high V/Q ratio (dead space). A shunt, such as in pneumonia, is the opposite: perfusion without ventilation.
Answer
C · It becomes very high, acting as dead space
5.Microbiology
Answer: D · Salmonella Typhi
Why this is the answer
Salmonella enterica serotype Typhi causes typhoid fever, a systemic illness with prolonged fever. Some people become chronic carriers, often in the gallbladder.
Answer
D · Salmonella Typhi
6.Microbiology
Answer: C · Microorganisms come from other microorganisms, not from spontaneous generation
Why this is the answer
Boiled broth in a flask with a curved neck stayed sterile because dust and microbes were trapped in the bend. When the neck was broken, microbes entered and grew, disproving spontaneous generation.
Answer
C · Microorganisms come from other microorganisms, not from spontaneous generation
7.Pharmacology
Answer: C · Respiratory rate and level of sedation
Why this is the answer
Opioids can cause respiratory depression, which is usually preceded by increasing sedation. Both are checked before and after each dose.
Answer
C · Respiratory rate and level of sedation
8.Pharmacology
Answer: A · Contraceptive failure because the hormones are metabolised faster
Why this is the answer
Carbamazepine induces CYP3A4, speeding up the breakdown of oestrogen and progestogen. Contraceptive levels fall and pregnancy can occur, so another method is advised.
Answer
A · Contraceptive failure because the hormones are metabolised faster
9.Arterial blood gas interpretation
Answer: PaCO₂ = 32 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.
Winter’s formula
PaCO₂ = 1.5·16 + 8 = 32 mmHg
range: 30–34 mmHg
Answer
PaCO₂ = 32 mmHg
10.Arterial blood gas interpretation
Answer: chronic
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 HCO₃⁻
acute: 24 + 1·2.5 = 26.5 mmol/L
chronic: 24 + 3.5·2.5 = 32.75 mmol/L
Compare
measured: 33 mmol/L → chronic
Answer
chronic
11.ECG rhythm strip interpretation
Answer: atrial fibrillation
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: 8 QRS complexes in 6 s, × 10 = 80 per minute.
2. Rhythm: irregularly irregular, R–R from 0.48 to 0.92 s with no pattern.
3. P waves: none; the baseline shows small irregular fibrillatory waves.
4. PR interval: not measurable.
5. QRS: 0.10 s, narrow (below 0.12 s).
Interpretation
atrial fibrillation
Answer
atrial fibrillation
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: 5 QRS complexes in 6 s, × 10 = 50 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 0.92 s, 65 per minute); one P in every 3 is not followed by a QRS.
4. PR interval: lengthens 0.20 → 0.34 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: NNH = 4
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 = 54/100 = 0.54
CER = 200/740 = 0.2703
Number needed to harm
ARI = 0.54 − 0.2703 = 0.2697
NNH = 1/0.2697 = 3.707 → 4
Answer
NNH = 4
About one in every 4 treated patients has the side effect because of the treatment.
14.Evidence-based practice statistics
Answer: PPV = 83.5 %
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.
True and false positives (per person)
Se·p = 0.71·0.3 = 0.213
(1 − Sp)·(1 − p) = 0.06·0.7 = 0.042
Positive predictive value
PPV = 0.213/(0.213 + 0.042) = 0.8353 ≈ 83.5%
Answer
PPV = 83.5%
The lower the prevalence, the lower the PPV of the same test.