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
Injury to the common fibular (peroneal) nerve at the neck of the fibula typically causes:
AWeak hip abduction
BFoot drop
CLoss of sensation on the sole of the foot
DInability to flex the knee
3.Human physiologyMedium
Which ABO blood group has both anti-A and anti-B antibodies in the plasma?
AGroup O
BGroup B
CGroup A
DGroup AB
4.Human physiologyHard
Why can a patient with carbon monoxide poisoning have a normal PaO₂ but severe tissue hypoxia?
ACO occupies haemoglobin binding sites, so oxygen content falls
BCO blocks diffusion across the alveolar wall
CCO stops oxygen from dissolving in plasma
DCO lowers the barometric pressure in the alveoli
5.MicrobiologyMedium
Which bacterium causes whooping cough (pertussis)?
ABordetella pertussis
BCorynebacterium diphtheriae
CMycoplasma pneumoniae
DHaemophilus influenzae
6.MicrobiologyHard
Why are beta-lactamase inhibitors such as clavulanic acid combined with amoxicillin?
AThey protect amoxicillin from bacterial enzymes that destroy it
BThey kill bacteria by themselves through the cell membrane
CThey slow the excretion of amoxicillin by the kidneys, raising its level
DThey make amoxicillin active against viruses
7.PharmacologyMedium
Benzathine benzylpenicillin (penicillin G benzathine) is ordered. Which route is correct?
AIntravenous push
BDeep intramuscular only
CIntravenous infusion over 30 minutes
DSubcutaneous in the abdomen
8.PharmacologyHard
A toddler swallowed many of his mother's iron tablets. He has vomiting and bloody diarrhoea. Which drug may be used to bind iron?
ADeferoxamine (desferrioxamine)
BNaloxone
CActivated charcoal given by mouth
DAcetylcysteine
9.Arterial blood gas interpretationMedium
A practice gas shows a metabolic acidosis with HCO₃⁻ 10 mmol/L. Use Winter’s formula to find the expected PaCO₂ (the centre of the range) in mmHg.
HCO₃⁻ 10 mmol/L
PaCO₂ = mmHg
10.Arterial blood gas interpretationHard
Practice values: Na⁺ 137, Cl⁻ 115, HCO₃⁻ 14 mmol/L and albumin 19 g/L. Find the albumin-corrected anion gap AGc = AG + 0.25·(40 − albumin).
Measure the QT interval on this lead II strip, in seconds.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
QT = s
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 330 patients with a disease (cases) and 310 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 = 214b = 118c = 116d = 192
OR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 44% in relative terms (RRR = 44%). In a patient group the risk without treatment (CER) is 39% 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 = 44%CER = 39%
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: B · Foot drop
Why this is the answer
The common fibular nerve supplies the dorsiflexors and evertors. Damage at the fibular neck causes foot drop and a high-stepping gait. The sole is supplied by the tibial nerve.
Answer
B · Foot drop
3.Human physiology
Answer: A · Group O
Why this is the answer
Group O people have no A or B antigens and make both anti-A and anti-B. Group AB has both antigens and neither antibody.
Answer
A · Group O
4.Human physiology
Answer: A · CO occupies haemoglobin binding sites, so oxygen content falls
Why this is the answer
CO binds haemoglobin far more strongly than oxygen and shifts the curve left. Dissolved O₂ (PaO₂) is normal, but total oxygen content and release to tissues drop sharply.
Answer
A · CO occupies haemoglobin binding sites, so oxygen content falls
5.Microbiology
Answer: A · Bordetella pertussis
Why this is the answer
Bordetella pertussis causes long coughing fits with a "whoop". It is most dangerous for young babies, which is why vaccination in pregnancy is offered in many countries.
Answer
A · Bordetella pertussis
6.Microbiology
Answer: A · They protect amoxicillin from bacterial enzymes that destroy it
Why this is the answer
Beta-lactamases hydrolyse the beta-lactam ring. Clavulanic acid binds these enzymes, so amoxicillin stays active against beta-lactamase-producing bacteria.
Answer
A · They protect amoxicillin from bacterial enzymes that destroy it
7.Pharmacology
Answer: B · Deep intramuscular only
Why this is the answer
Benzathine penicillin is a depot suspension given deep IM. Accidental intravenous injection can cause cardiorespiratory arrest and death.
Answer
B · Deep intramuscular only
8.Pharmacology
Answer: A · Deferoxamine (desferrioxamine)
Why this is the answer
Iron poisoning can cause GI bleeding, shock, metabolic acidosis and liver failure. Deferoxamine chelates iron and is used in serious poisoning; activated charcoal does not bind iron.
Answer
A · Deferoxamine (desferrioxamine)
9.Arterial blood gas interpretation
Answer: PaCO₂ = 23 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 = 1.5·10 + 8 = 23 mmHg
range: 21–25 mmHg
Answer
PaCO₂ = 23 mmHg
10.Arterial blood gas interpretation
Answer: AGc = 13.25 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 = 137 − (115 + 14) = 8
Correction for albumin
8 + 0.25·(40 − 19) = 13.25 mmol/L
Answer
AGc = 13.25 mmol/L
Above about 12 after correction: a high anion gap.
11.ECG rhythm strip interpretation
Answer: QT = 0.44 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
11 small boxes from the start of the QRS to the end of the T wave
Convert
QT = 11 × 0.04 s = 0.44 s
Answer
QT = 0.44 s
A rough rule: at normal rates the QT should be less than half the R–R interval (here R–R = 0.88 s).
12.ECG rhythm strip interpretation
Answer: QRS duration
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: 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: 0.14 s, wide (0.12 s or more).
First abnormal step
QRS duration → sinus rhythm with wide QRS
Answer
QRS duration
13.Evidence-based practice statistics
Answer: OR = 3
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: 214/116 = 1.845
controls: 118/192 = 0.6146
Odds ratio
OR = (214·192)/(118·116) = 3.002 ≈ 3
Answer
OR = 3
OR above 1: exposure is more common among the cases.
14.Evidence-based practice statistics
Answer: NNT = 6
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 = 0.39·0.44 = 0.1716
Number needed to treat
NNT = 1/0.1716 = 5.828 → 6
Answer
NNT = 6
The same relative effect gives a smaller NNT when the baseline risk is higher.