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 valve lies between the left atrium and the left ventricle?
AAortic valve
BPulmonary valve
CTricuspid valve
DMitral (bicuspid) valve
2.Human anatomyHard
When inserting a chest drain or needle into the pleural space, why is it placed just above the upper border of a rib?
AThe lung does not reach the area above each rib
BThe pleura is thickest just above each rib
CThe intercostal vein, artery and nerve run along the lower border of each rib
DThe intercostal muscles are absent above each rib
3.Human physiologyMedium
What causes the first heart sound (S1)?
AClosure of the mitral and tricuspid valves
BBlood filling the atria
COpening of the aortic valve
DClosure of the aortic and pulmonary valves
4.Human physiologyHard
Why is severe hyperkalaemia dangerous for the heart?
AIt speeds up the SA node until tachycardia causes fainting
BIt depolarises cardiac cells, inactivating sodium channels and causing arrhythmias
CIt blocks the binding of calcium to troponin
DIt hyperpolarises cardiac cells, so they can never fire
5.MicrobiologyMedium
How does an mRNA vaccine produce immunity?
AThe mRNA kills the virus directly in the blood
BBody cells use the mRNA to make a viral protein that the immune system learns to recognise
CThe mRNA is inserted permanently into human DNA
DThe mRNA contains weakened live virus
6.MicrobiologyHard
Why are penicillins ineffective against Mycoplasma pneumoniae?
AIt lives only inside red blood cells
BIt produces large amounts of beta-lactamase
CIt has a thick waxy mycolic acid layer that blocks entry
DIt has no cell wall for penicillin to act on
7.PharmacologyMedium
A patient takes isoniazid. Which advice is most important?
AAvoid alcohol and report jaundice, dark urine or poor appetite
BStop the drug when the cough improves
CTake it with antacids to protect the stomach from irritation
DExpect the urine to turn orange
8.PharmacologyHard
Why is pyridoxine (vitamin B6) often given together with isoniazid for tuberculosis?
ATo prevent peripheral neuropathy
BTo reduce bacterial resistance
CTo prevent liver damage from isoniazid
DTo increase the absorption of isoniazid
9.Arterial blood gas interpretationMedium
Practice values: Na⁺ 144 mmol/L, glucose 24.7 mmol/L and urea 24.4 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 144 mmol/Lglucose 24.7 mmol/Lurea 24.4 mmol/L
Osm = mOsm/kg
10.Arterial blood gas interpretationHard
A practice gas shows a metabolic acidosis: pH 7.05, PaCO₂ 41 mmHg (5.5 kPa), HCO₃⁻ 11 mmol/L. Compare the measured PaCO₂ with Winter’s formula. Click your interpretation.
pH 7.05PaCO₂ 41 mmHg (5.5 kPa)HCO₃⁻ 11 mmol/L
1appropriate compensation
2added respiratory acidosis
3added respiratory alkalosis
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 Bazett’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
Identify the rhythm on this 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
1atrial flutter, 2:1 conduction
2ventricular tachycardia
3sinus tachycardia
4ventricular fibrillation
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a randomised trial, 240 patients received a new treatment and 280 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: 64/240control: 106/280
NNT =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 32% in relative terms (RRR = 32%). In a patient group the risk without treatment (CER) is 29% 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 = 32%CER = 29%
NNT =
3Solutions
1.Human anatomy
Answer: D · Mitral (bicuspid) valve
Why this is the answer
The mitral (bicuspid) valve separates the left atrium and left ventricle; the tricuspid valve does the same on the right side.
Answer
D · Mitral (bicuspid) valve
2.Human anatomy
Answer: C · The intercostal vein, artery and nerve run along the lower border of each rib
Why this is the answer
The intercostal neurovascular bundle (vein, artery, nerve from above down) lies in the costal groove under each rib. Going just above the rib below avoids it.
Answer
C · The intercostal vein, artery and nerve run along the lower border of each rib
3.Human physiology
Answer: A · Closure of the mitral and tricuspid valves
Why this is the answer
S1 ("lub") is caused by closure of the atrioventricular valves at the start of systole. S2 ("dub") is closure of the semilunar valves.
Answer
A · Closure of the mitral and tricuspid valves
4.Human physiology
Answer: B · It depolarises cardiac cells, inactivating sodium channels and causing arrhythmias
Why this is the answer
High extracellular potassium makes the resting potential less negative. Many fast sodium channels then become inactivated, slowing conduction and risking fatal arrhythmias.
Answer
B · It depolarises cardiac cells, inactivating sodium channels and causing arrhythmias
5.Microbiology
Answer: B · Body cells use the mRNA to make a viral protein that the immune system learns to recognise
Why this is the answer
The mRNA is translated by cells into an antigen, such as a coronavirus spike protein. The immune system responds to it. The mRNA is broken down and does not enter the nucleus.
Answer
B · Body cells use the mRNA to make a viral protein that the immune system learns to recognise
6.Microbiology
Answer: D · It has no cell wall for penicillin to act on
Why this is the answer
Mycoplasma lacks a cell wall, so beta-lactams have no target. Macrolides or tetracyclines, which act on the ribosome, are used instead.
Answer
D · It has no cell wall for penicillin to act on
7.Pharmacology
Answer: A · Avoid alcohol and report jaundice, dark urine or poor appetite
Why this is the answer
Isoniazid can cause hepatitis, and alcohol increases the risk. Signs of liver injury are reported at once, and liver tests may be checked.
Answer
A · Avoid alcohol and report jaundice, dark urine or poor appetite
8.Pharmacology
Answer: A · To prevent peripheral neuropathy
Why this is the answer
Isoniazid causes pyridoxine deficiency, which can lead to peripheral neuropathy. The risk is greater in slow acetylators, malnourished patients and people with diabetes or alcohol dependence.
Answer
A · To prevent peripheral neuropathy
9.Arterial blood gas interpretation
Answer: Osm = 337 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 = 2·144 + 24.7 + 24.4 = 337.1 ≈ 337 mOsm/kg
Answer
Osm = 337 mOsm/kg
10.Arterial blood gas interpretation
Answer: added respiratory acidosis
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
1.5·11 + 8 = 24.5 → 22.5–26.5 mmHg
Compare
measured: 41 mmHg → added respiratory acidosis
Answer
added respiratory acidosis
11.ECG rhythm strip interpretation
Answer: QTc = 0.426 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 × 0.04 = 0.40 s
RR = 22 × 0.04 = 0.88 s
Bazett’s formula
QTc = 0.40/√0.88 = 0.40/0.9381 = 0.426 s
Answer
QTc = 0.426 s
Within the usual limits (about 0.45 s for men, 0.46 s for women). (426 ms)
12.ECG rhythm strip interpretation
Answer: ventricular tachycardia
Ventricular rhythms: ventricular tachycardia is a run of three or more wide (0.12 s or more) complexes in a row at more than 100/min, usually 150–250/min, regular, without related P waves. Ventricular fibrillation is a chaotic, irregular waveform with no identifiable P waves, QRS complexes or T waves and no countable rate. Asystole is an almost flat line with no ventricular activity; a flat line is always confirmed by checking the patient, the leads and the gain and a second lead. In resuscitation guidelines VF and pulseless VT are the shockable rhythms and asystole is not.
Five steps
1. Rate: R–R = 10 small boxes, 1500 ÷ 10 ≈ 150 per minute.
2. Rhythm: regular, every R–R is 0.40 s.
3. P waves: none identifiable.
4. PR interval: not measurable.
5. QRS: 0.14 s, wide (0.12 s or more).
Interpretation
ventricular tachycardia
Answer
ventricular tachycardia
In resuscitation guidelines VF and pulseless VT are the shockable rhythms. This is identification practice only.
13.Evidence-based practice statistics
Answer: NNT = 9
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 = 64/240 = 0.2667
CER = 106/280 = 0.3786
Number needed to treat
ARR = 0.3786 − 0.2667 = 0.1119
NNT = 1/0.1119 = 8.936 → 9
Answer
NNT = 9
Treating 9 patients prevents about one event.
14.Evidence-based practice statistics
Answer: NNT = 11
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 = 0.29·0.32 = 0.0928
Number needed to treat
NNT = 1/0.0928 = 10.78 → 11
Answer
NNT = 11
The same relative effect gives a smaller NNT when the baseline risk is higher.