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 muscle is the main extensor of the elbow?
ATriceps brachii
BBrachialis
CCoracobrachialis
DBiceps brachii
2.Human anatomyHard
After an injury at the medial epicondyle of the elbow, a patient has numbness of the little finger and weakness of finger abduction. Which nerve is affected?
AUlnar nerve
BMedian nerve
CMusculocutaneous nerve
DRadial nerve
3.Human physiologyMedium
Cardiac output equals heart rate multiplied by:
ABlood pressure
BPeripheral resistance
CEjection fraction
DStroke volume
4.Human physiologyHard
Which adaptation develops over days to weeks at high altitude?
ALower ventilation to save oxygen
BIncreased renal bicarbonate retention
CHigher erythropoietin levels and more red blood cells
DA fall in haematocrit
5.MicrobiologyMedium
Which laboratory test separates Staphylococcus aureus from most other staphylococci?
AThe coagulase test
BThe catalase test
CThe acid-fast stain
DThe Gram stain
6.MicrobiologyHard
In the 1840s Ignaz Semmelweis lowered deaths from puerperal fever in a Vienna maternity clinic. What did he introduce?
AHandwashing with chlorinated lime before examining women
BVaccination of all mothers before birth
CKeeping mothers in separate isolation rooms after delivery
DTreating mothers with penicillin
7.PharmacologyMedium
A person who has never been vaccinated against tetanus has a deep, dirty wound. What is usually given?
ATetanus vaccine only, after the wound heals
BNo treatment unless symptoms appear
CTetanus vaccine and tetanus immunoglobulin
DOral antibiotics only
8.PharmacologyHard
A patient on lisinopril for hypertension starts taking ibuprofen every day for knee pain. What is the main concern?
AWeaker blood pressure control, kidney injury and high potassium
BFaster liver metabolism of lisinopril, so it stops working
CA large fall in blood pressure, with fainting and falls on standing up
DA higher risk of a dry cough than with lisinopril alone
9.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.42, PaCO₂ 24 mmHg (3.2 kPa) and HCO₃⁻ 15 mmol/L. The primary disorder is respiratory alkalosis. How far is it compensated? Click your answer.
pH 7.42PaCO₂ 24 mmHg (3.2 kPa)HCO₃⁻ 15 mmol/L
1uncompensated
2partially compensated
3fully compensated
10.Arterial blood gas interpretationHard
At sea level, a practice gas taken on 40% oxygen shows PaO₂ 78 mmHg (10.4 kPa) and PaCO₂ 36 mmHg (4.8 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.
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, 4:1 conduction
3atrial flutter, 2:1 conduction
4normal sinus rhythm
12.ECG rhythm strip interpretationHard
Interpret this lead II strip with the five steps in order: rate, rhythm, P waves, PR interval, QRS duration. At which step do you find the first abnormal finding? Click it.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
1rate
2rhythm
3P waves
4PR interval
5QRS duration
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). A case-control study compares 310 patients with a disease (cases) and 270 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 = 31b = 59c = 279d = 211
OR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 21% in relative terms (RRR = 21%). In a patient group the risk without treatment (CER) is 10% 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 = 21%CER = 10%
NNT =
3Solutions
1.Human anatomy
Answer: A · Triceps brachii
Why this is the answer
Triceps brachii inserts on the olecranon of the ulna and extends the elbow. Biceps and brachialis are elbow flexors.
Answer
A · Triceps brachii
2.Human anatomy
Answer: A · Ulnar nerve
Why this is the answer
The ulnar nerve passes behind the medial epicondyle. It supplies most intrinsic hand muscles, including the interossei, and the skin of the little finger.
Answer
A · Ulnar nerve
3.Human physiology
Answer: D · Stroke volume
Why this is the answer
CO = HR × SV. At rest about 70 beats/min × 70 mL ≈ 5 L/min.
Answer
D · Stroke volume
4.Human physiology
Answer: C · Higher erythropoietin levels and more red blood cells
Why this is the answer
Hypoxia stimulates EPO and red cell production, raising oxygen-carrying capacity. Hyperventilation causes alkalosis, which the kidneys compensate by excreting bicarbonate.
Answer
C · Higher erythropoietin levels and more red blood cells
5.Microbiology
Answer: A · The coagulase test
Why this is the answer
S. aureus produces coagulase, which converts fibrinogen to fibrin. All staphylococci are Gram-positive and catalase-positive, so those tests do not separate them.
Answer
A · The coagulase test
6.Microbiology
Answer: A · Handwashing with chlorinated lime before examining women
Why this is the answer
Semmelweis noticed higher mortality in the ward run by doctors who also did autopsies. Hand disinfection with chlorinated lime sharply reduced deaths, long before germ theory was accepted.
Answer
A · Handwashing with chlorinated lime before examining women
7.Pharmacology
Answer: C · Tetanus vaccine and tetanus immunoglobulin
Why this is the answer
Tetanus immunoglobulin gives immediate passive protection, while the vaccine starts active immunity. Thorough wound cleaning is also essential.
Answer
C · Tetanus vaccine and tetanus immunoglobulin
8.Pharmacology
Answer: A · Weaker blood pressure control, kidney injury and high potassium
Why this is the answer
NSAIDs block prostaglandins that dilate the afferent arteriole and promote sodium loss. With an ACE inhibitor this can reduce blood pressure control and cause kidney injury and hyperkalaemia, especially if a diuretic is also used.
Answer
A · Weaker blood pressure control, kidney injury and high potassium
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 7.42: within 7.35–7.45
PaCO₂ 24 mmHg: low (35–45)
HCO₃⁻ 15 mmol/L: low (22–26)
Compensation
fully compensated
Answer
fully compensated
10.Arterial blood gas interpretation
Answer: A–a gradient = 162.2 mmHg
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.
Alveolar oxygen
PAO₂ = 0.4·713 − 36/0.8 = 285.2 − 45 = 240.2 mmHg
Gradient
A–a = 240.2 − 78 = 162.2 mmHg
Answer
A–a gradient = 162.2 mmHg
11.ECG rhythm strip interpretation
Answer: atrial flutter, 2:1 conduction
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 = 10 small boxes, 1500 ÷ 10 ≈ 150 per minute.
2. Rhythm: regular, every R–R is 0.40 s.
3. P waves: none; regular sawtooth flutter waves every 0.20 s (300 per minute), 2 for each QRS.
4. PR interval: not measurable.
5. QRS: 0.08 s, narrow (below 0.12 s).
Interpretation
atrial flutter, 2:1 conduction
Answer
atrial flutter, 2:1 conduction
12.ECG rhythm strip interpretation
Answer: rhythm
Five steps for every strip: 1. rate (300 or 1500 method, or the 6-second method if irregular); 2. rhythm: are the R–R intervals equal; 3. P waves: present, upright in lead II, one before every QRS and all alike; 4. PR interval: 0.12–0.20 s and constant; 5. QRS duration: below 0.12 s. Only after the five steps name the rhythm. Normal sinus rhythm passes all five: 60–100/min, regular, upright P before every QRS, PR 0.12–0.20 s, QRS below 0.12 s.
Five steps
1. Rate: 8 QRS complexes in 6 s, × 10 = 80 per minute.
2. Rhythm: irregularly irregular, R–R from 0.60 to 0.94 s with no pattern.
3. P waves: none; the baseline shows small irregular fibrillatory waves.
4. PR interval: not measurable.
5. QRS: 0.08 s, narrow (below 0.12 s).
First abnormal step
rhythm → atrial fibrillation
Answer
rhythm
13.Evidence-based practice statistics
Answer: OR = 0.4
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.
Odds of exposure
cases: 31/279 = 0.1111
controls: 59/211 = 0.2796
Odds ratio
OR = (31·211)/(59·279) = 0.3974 ≈ 0.4
Answer
OR = 0.4
OR below 1: exposure is less common among the cases, which suggests a protective effect.
14.Evidence-based practice statistics
Answer: NNT = 48
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 = 0.1·0.21 = 0.021
Number needed to treat
NNT = 1/0.021 = 47.62 → 48
Answer
NNT = 48
The same relative effect gives a smaller NNT when the baseline risk is higher.