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 vessels carry oxygenated blood from the lungs to the heart?
AVenae cavae
BBronchial arteries
CPulmonary arteries
DPulmonary veins
2.Human anatomyHard
Which ligament of the knee prevents the tibia from sliding forward relative to the femur?
AAnterior cruciate ligament
BPosterior cruciate ligament
CMedial collateral ligament
DPatellar ligament
3.Human physiologyMedium
Why can cardiac muscle not be tetanised like skeletal muscle?
AIt is controlled only by hormones
BIts long refractory period lasts almost as long as the contraction
CIt has no calcium in the sarcoplasmic reticulum
DIt is not striated
4.Human physiologyHard
In long-standing aortic stenosis, how does the left ventricle adapt?
AAtrophy of the muscle due to less work
BConcentric hypertrophy with a thicker wall
CNo change because the valve is not part of the ventricle
DThinning of the wall with dilation
5.MicrobiologyMedium
How does an autoclave sterilise instruments?
AWith ultraviolet light only
BWith alcohol vapour at room temperature
CWith dry heat at about 60 °C
DWith steam under pressure at about 121 °C
6.MicrobiologyHard
Why is Chlamydia trachomatis usually diagnosed with nucleic acid amplification tests instead of routine culture on agar?
AIt is too large to grow on agar plates
BIt dies immediately in the presence of oxygen
CIt is an obligate intracellular bacterium that cannot grow on ordinary media
DIt is a virus and has no DNA
7.PharmacologyMedium
Which drug is used for torsades de pointes?
AOral digoxin
BIntravenous magnesium sulfate
CIntravenous adenosine (rapid push)
DIntravenous verapamil
8.PharmacologyHard
Which antiepileptic drug carries the highest well-known risk of neural tube defects and developmental problems when taken in pregnancy?
ASodium valproate
BLevetiracetam
CGabapentin
DLamotrigine
9.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.44, PaCO₂ 29 mmHg (3.9 kPa) and HCO₃⁻ 19 mmol/L. The primary disorder is respiratory alkalosis. How far is it compensated? Click your answer.
pH 7.44PaCO₂ 29 mmHg (3.9 kPa)HCO₃⁻ 19 mmol/L
1uncompensated
2partially compensated
3fully compensated
10.Arterial blood gas interpretationHard
A practice gas shows a metabolic alkalosis with HCO₃⁻ 37 mmol/L. Use expected PaCO₂ = 0.7·HCO₃⁻ + 21 to find the expected PaCO₂.
HCO₃⁻ 37 mmol/L
PaCO₂ = mmHg
11.ECG rhythm strip interpretationMedium
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
A patient without a palpable pulse has this organised rhythm on the monitor (pulseless electrical activity). Which treatment do adult resuscitation guidelines name for 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
1CPR (non-shockable rhythm)
2transcutaneous pacing
3immediate defibrillation
4synchronised cardioversion
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a cohort, groups of people were followed for different lengths of time (see the table: number of people and years of follow-up for each). During follow-up, 29 new cases occurred. Find the incidence rate IR per 1000 person-years. Round to one decimal place.
Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 16% in relative terms (RRR = 16%). In a patient group the risk without treatment (CER) is 12% over five years. Find the number needed to treat NNT for five years to prevent one event. Round up to a whole number of patients.
RRR = 16%CER = 12%
NNT =
3Solutions
1.Human anatomy
Answer: D · Pulmonary veins
Why this is the answer
Four pulmonary veins drain oxygenated blood into the left atrium. They are the main exception to the rule that veins carry deoxygenated blood.
Answer
D · Pulmonary veins
2.Human anatomy
Answer: A · Anterior cruciate ligament
Why this is the answer
The anterior cruciate ligament limits anterior translation of the tibia; a positive anterior drawer or Lachman test suggests an ACL tear.
Answer
A · Anterior cruciate ligament
3.Human physiology
Answer: B · Its long refractory period lasts almost as long as the contraction
Why this is the answer
The plateau of the cardiac action potential gives a long refractory period, so the muscle relaxes before it can be excited again. This allows the heart to refill.
Answer
B · Its long refractory period lasts almost as long as the contraction
4.Human physiology
Answer: B · Concentric hypertrophy with a thicker wall
Why this is the answer
Pressure overload makes the ventricular wall thicken (concentric hypertrophy) to keep wall stress normal. Volume overload, as in regurgitation, causes dilation instead.
Answer
B · Concentric hypertrophy with a thicker wall
5.Microbiology
Answer: D · With steam under pressure at about 121 °C
Why this is the answer
Pressurised steam reaches about 121 °C and kills all microorganisms, including spores, typically in about 15 minutes or more depending on the load.
Answer
D · With steam under pressure at about 121 °C
6.Microbiology
Answer: C · It is an obligate intracellular bacterium that cannot grow on ordinary media
Why this is the answer
Chlamydia replicates only inside host cells, so it cannot be grown on agar. NAATs such as PCR on urine or swabs are sensitive and are the standard diagnostic test.
Answer
C · It is an obligate intracellular bacterium that cannot grow on ordinary media
7.Pharmacology
Answer: B · Intravenous magnesium sulfate
Why this is the answer
IV magnesium is first-line for torsades de pointes, even when serum magnesium is normal. QT-prolonging drugs are stopped and potassium is corrected.
Answer
B · Intravenous magnesium sulfate
8.Pharmacology
Answer: A · Sodium valproate
Why this is the answer
Valproate is strongly teratogenic, causing neural tube defects and other malformations, and it affects neurodevelopment. It should be avoided in pregnancy unless no alternative works.
Answer
A · Sodium valproate
9.Arterial blood gas interpretation
Answer: fully compensated
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 7.44: within 7.35–7.45
PaCO₂ 29 mmHg: low (35–45)
HCO₃⁻ 19 mmol/L: low (22–26)
Compensation
fully compensated
Answer
fully compensated
10.Arterial blood gas interpretation
Answer: PaCO₂ = 46.9 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
0.7·37 + 21 = 46.9 mmHg
Answer
PaCO₂ = 46.9 mmHg
The rise has a limit: PaCO₂ rarely goes above about 55 mmHg from compensation alone.
11.ECG rhythm strip interpretation
Answer: QTc = 0.409 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 = 27 × 0.04 = 1.08 s
Fridericia’s formula
QTc = 0.42/∛1.08 = 0.42/1.026 = 0.409 s
Answer
QTc = 0.409 s
Within the usual limits (about 0.45 s for men, 0.46 s for women). (409 ms)
12.ECG rhythm strip interpretation
Answer: CPR (non-shockable rhythm)
Pulseless electrical activity (PEA): PEA cannot be seen on a strip. The monitor shows organised electrical activity (a rhythm other than VF, VT or asystole) but the patient has no palpable pulse, so it is recognised by checking the patient. In adult resuscitation guidelines PEA and asystole are the non-shockable cardiac arrest rhythms: high-quality CPR, epinephrine as soon as possible and a search for reversible causes (the Hs and Ts, such as hypovolaemia, hypoxia, tension pneumothorax and cardiac tamponade). VF and pulseless VT are the shockable rhythms.
What the strip shows
1. Rate: R–R = 19 small boxes, 1500 ÷ 19 ≈ 79 per minute.
2. Rhythm: regular, every R–R is 0.76 s.
3. P waves: upright, one before every QRS, all alike.
4. PR interval: 0.16 s, constant, within 0.12–0.20 s.
5. QRS: 0.06 s, narrow (below 0.12 s).
normal sinus rhythm
What the patient shows
no pulse despite organised electrical activity: pulseless electrical activity (PEA), a non-shockable cardiac arrest rhythm
Answer
CPR (non-shockable rhythm)
Answer
CPR (non-shockable rhythm)
PEA is recognised at the bedside, never from the strip alone. This is exam practice, not clinical guidance.
13.Evidence-based practice statistics
Answer: IR = 15.3 per 1000 person-years
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.
Person-years
PY = 270·1 + 440·2.5 + 530·1 = 1900
Incidence rate
IR = 29/1900·1000 = 15.3 per 1000 person-years
Answer
IR = 15.3 per 1000 person-years
14.Evidence-based practice statistics
Answer: NNT = 53
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.
Absolute risk reduction
ARR = 0.12·0.16 = 0.0192
Number needed to treat
NNT = 1/0.0192 = 52.08 → 53
Answer
NNT = 53
The same relative effect gives a smaller NNT when the baseline risk is higher.