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 part of the adrenal gland secretes adrenaline?
AAdrenal medulla
BZona glomerulosa
CZona reticularis
DZona fasciculata
2.Human anatomyHard
Which structure is found in the posterior mediastinum?
AThymus
BOesophagus
CAscending aorta
DHeart
3.Human physiologyMedium
What is the normal pH range of arterial blood?
A7.00 to 7.20
B7.35 to 7.45
C7.50 to 7.65
D6.80 to 7.00
4.Human physiologyHard
Why is severe hyperkalaemia dangerous for the heart?
AIt speeds up the SA node until tachycardia causes fainting
BIt hyperpolarises cardiac cells, so they can never fire
CIt depolarises cardiac cells, inactivating sodium channels and causing arrhythmias
DIt blocks the binding of calcium to troponin
5.MicrobiologyMedium
In the chain of infection, what is the "reservoir"?
AThe place where the pathogen normally lives and multiplies
BThe route by which the pathogen leaves the infected host's body
CThe way the pathogen enters a new host
DThe person who is most at risk of infection
6.MicrobiologyHard
Staphylococcus aureus food poisoning starts only 1–6 hours after eating. Why so fast?
AThe bacteria made a heat-stable toxin in the food before it was eaten
BThe bacteria multiply very fast inside the gut and invade the intestinal wall
CThe immune system reacts to the bacteria within minutes
DThe bacteria invade the bloodstream immediately
7.PharmacologyMedium
Which sign is typical of opioid toxicity?
AHigh body temperature
BFast, deep breathing
CWidely dilated pupils
DPinpoint pupils
8.PharmacologyHard
Why do infants absorb more of a topical corticosteroid than adults?
ATheir skin contains no blood vessels
BInfants do not absorb topical drugs at all
CTheir liver changes the cream into a much stronger form
DThey have more skin area per kilogram and thinner skin
9.Arterial blood gas interpretationMedium
A practice gas shows a metabolic acidosis with HCO₃⁻ 12 mmol/L. Use Winter’s formula to find the expected PaCO₂ (the centre of the range) in mmHg.
HCO₃⁻ 12 mmol/L
PaCO₂ = mmHg
10.Arterial blood gas interpretationHard
Practice values: Na⁺ 140, Cl⁻ 90 and HCO₃⁻ 15 mmol/L. Calculate the delta ratio (AG − 12)/(24 − HCO₃⁻), using AG = Na⁺ − (Cl⁻ + HCO₃⁻). Give two decimals.
Na⁺ 140 mmol/LCl⁻ 90 mmol/LHCO₃⁻ 15 mmol/L
delta ratio =
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
2accelerated junctional rhythm
3supraventricular tachycardia (SVT)
4ventricular tachycardia
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
1first-degree AV block
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, 350 patients received a new treatment and 710 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: 109/350control: 277/710
NNT =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 50% in relative terms (RRR = 50%). 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 = 50%CER = 29%
NNT =
3Solutions
1.Human anatomy
Answer: A · Adrenal medulla
Why this is the answer
The adrenal medulla releases adrenaline and noradrenaline in response to sympathetic stimulation. The three zones of the cortex make steroid hormones.
Answer
A · Adrenal medulla
2.Human anatomy
Answer: B · Oesophagus
Why this is the answer
The posterior mediastinum contains the oesophagus, descending aorta, azygos veins and thoracic duct. The heart and ascending aorta are in the middle mediastinum; the thymus is in the superior and anterior mediastinum.
Answer
B · Oesophagus
3.Human physiology
Answer: B · 7.35 to 7.45
Why this is the answer
Arterial pH is normally 7.35–7.45. Below 7.35 is acidaemia and above 7.45 alkalaemia.
Answer
B · 7.35 to 7.45
4.Human physiology
Answer: C · 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
C · It depolarises cardiac cells, inactivating sodium channels and causing arrhythmias
5.Microbiology
Answer: A · The place where the pathogen normally lives and multiplies
Why this is the answer
The chain of infection links the agent, reservoir, portal of exit, mode of transmission, portal of entry and susceptible host. Breaking any link stops transmission.
Answer
A · The place where the pathogen normally lives and multiplies
6.Microbiology
Answer: A · The bacteria made a heat-stable toxin in the food before it was eaten
Why this is the answer
S. aureus produces enterotoxins in food left at warm temperatures. The toxin is already there when the food is eaten and survives reheating, so symptoms start quickly.
Answer
A · The bacteria made a heat-stable toxin in the food before it was eaten
7.Pharmacology
Answer: D · Pinpoint pupils
Why this is the answer
Opioid toxicity classically causes pinpoint pupils, slow or shallow breathing and reduced consciousness. Naloxone reverses these effects.
Answer
D · Pinpoint pupils
8.Pharmacology
Answer: D · They have more skin area per kilogram and thinner skin
Why this is the answer
Infants have a high surface area-to-weight ratio and a thinner skin barrier, so topical drugs reach higher systemic levels. Nappies can act as an occlusive dressing and increase absorption further.
Answer
D · They have more skin area per kilogram and thinner skin
9.Arterial blood gas interpretation
Answer: PaCO₂ = 26 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 = 1.5·12 + 8 = 26 mmHg
range: 24–28 mmHg
Answer
PaCO₂ = 26 mmHg
10.Arterial blood gas interpretation
Answer: delta ratio = 2.556
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 = 140 − (90 + 15) = 35
Delta ratio
(35 − 12)/(24 − 15) = 23/9 = 2.556
Answer
delta ratio = 2.556
Above 2: an added metabolic alkalosis.
11.ECG rhythm strip interpretation
Answer: supraventricular tachycardia (SVT)
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. Narrow-complex tachycardia and pauses: supraventricular tachycardia (SVT) is a regular tachycardia at about 150–250/min with narrow QRS complexes; the P waves are usually not visible because they are hidden in the QRS complex or the T wave. Sinus tachycardia in an adult at rest is usually below about 150/min and has a visible P wave before every QRS; atrial flutter with 2:1 conduction often runs at about 150/min and shows sawtooth waves. In sinus arrest the sinus node stops firing for a moment: a pause with no P wave and no QRS whose length is not a whole number of P–P cycles (a pause of exactly two or three cycles points to sinus exit block instead).
Five steps
1. Rate: R–R = 6 small boxes, 1500 ÷ 6 ≈ 250 per minute.
2. Rhythm: regular, every R–R is 0.24 s.
3. P waves: none visible; at this rate they are hidden in the QRS complexes or the T waves.
4. PR interval: not measurable.
5. QRS: 0.06 s, narrow (below 0.12 s), so the rhythm starts above the ventricles.
Interpretation
supraventricular tachycardia (SVT)
Answer
supraventricular tachycardia (SVT)
12.ECG rhythm strip interpretation
Answer: second-degree AV block type II
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: 6 QRS complexes in 6 s, × 10 = 60 per minute.
2. Rhythm: the P waves are regular; R–R is regular except for a long pause where a QRS is missing.
3. P waves: upright and regular (P–P 0.84 s, 71 per minute); one P in every 4 is not followed by a QRS.
4. PR interval: constant at 0.16 s in the conducted beats.
5. QRS: 0.14 s, wide (0.12 s or more).
Interpretation
second-degree AV block type II
Answer
second-degree AV block type II
Type II and third-degree blocks are the ones that can suddenly progress, so textbooks rank them as more serious than type I.
13.Evidence-based practice statistics
Answer: NNT = 13
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 = 109/350 = 0.3114
CER = 277/710 = 0.3901
Number needed to treat
ARR = 0.3901 − 0.3114 = 0.07871
NNT = 1/0.07871 = 12.7 → 13
Answer
NNT = 13
Treating 13 patients prevents about one event.
14.Evidence-based practice statistics
Answer: NNT = 7
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 = 0.29·0.5 = 0.145
Number needed to treat
NNT = 1/0.145 = 6.897 → 7
Answer
NNT = 7
The same relative effect gives a smaller NNT when the baseline risk is higher.