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 cranial nerve supplies the lateral rectus muscle of the eye?
AOptic nerve (II)
BOculomotor nerve (III)
CTrochlear nerve (IV)
DAbducens nerve (VI)
2.Human anatomyHard
Why can an inflamed gallbladder that irritates the diaphragm cause pain at the tip of the right shoulder?
AThe phrenic nerve shares spinal segments C3–C5 with the skin of the shoulder
BThe vagus nerve supplies the skin of the shoulder
CBile travels up the lymphatics to the shoulder
DThe gallbladder is attached to the right clavicle
3.Human physiologyMedium
What is the main effect of vagus nerve stimulation on the heart?
AIt slows the heart rate
BIt increases ventricular contractility strongly
CIt constricts the coronary arteries
DIt increases the heart rate
4.Human physiologyHard
In long-standing aortic stenosis, how does the left ventricle adapt?
AConcentric hypertrophy with a thicker wall
BNo change because the valve is not part of the ventricle
CThinning of the wall with dilation
DAtrophy of the muscle due to less work
5.MicrobiologyMedium
Which white blood cells are the main phagocytes in an acute bacterial infection and the main component of pus?
AEosinophils
BB lymphocytes
CBasophils
DNeutrophils
6.MicrobiologyHard
Why are penicillins ineffective against Mycoplasma pneumoniae?
AIt has a thick waxy mycolic acid layer that blocks entry
BIt lives only inside red blood cells
CIt has no cell wall for penicillin to act on
DIt produces large amounts of beta-lactamase
7.PharmacologyMedium
Which factor increases the risk of lithium toxicity?
ATaking the dose with food
BA high-salt diet
CDrinking extra water
DDehydration
8.PharmacologyHard
A nurse gave a patient ten times the prescribed dose of hydromorphone. The patient is currently alert. What is the most appropriate sequence of actions?
ASay nothing because the patient is alert
BChange the chart so that it shows the correct dose, and continue to monitor
CGive another dose later to balance it out
DMonitor closely, have naloxone ready, inform the prescriber and report
9.Arterial blood gas interpretationMedium
A practice gas shows a metabolic acidosis with HCO₃⁻ 16 mmol/L. Use Winter’s formula to find the expected PaCO₂ (the centre of the range) in mmHg.
HCO₃⁻ 16 mmol/L
PaCO₂ = mmHg
10.Arterial blood gas interpretationHard
A practice gas shows a metabolic alkalosis with HCO₃⁻ 38 mmol/L. Use expected PaCO₂ = 0.7·HCO₃⁻ + 21 to find the expected PaCO₂.
HCO₃⁻ 38 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
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
1second-degree AV block type I (Wenckebach)
2second-degree AV block type II
3atrial fibrillation
4first-degree AV block
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, 44 new cases occurred. Find the incidence rate IR per 1000 person-years. Round to one decimal place.
290 × 1.5 years300 × 3 years44 new cases
IR = per 1000 person-years
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). Without a drug, 30% of patients have a certain side effect during a year. The drug raises this risk with a relative risk of RR = 1.28. Find the number needed to harm NNH: how many patients must take the drug for a year for one extra patient to have the side effect? Round up to a whole number of patients.
control 30%RR = 1.28
NNH =
3Solutions
1.Human anatomy
Answer: D · Abducens nerve (VI)
Why this is the answer
The abducens nerve (VI) supplies lateral rectus. A lesion stops the eye moving outwards and causes horizontal double vision.
Answer
D · Abducens nerve (VI)
2.Human anatomy
Answer: A · The phrenic nerve shares spinal segments C3–C5 with the skin of the shoulder
Why this is the answer
Diaphragmatic irritation is carried by the phrenic nerve (C3–C5). The brain refers it to skin supplied by the same segments, via the supraclavicular nerves over the shoulder.
Answer
A · The phrenic nerve shares spinal segments C3–C5 with the skin of the shoulder
3.Human physiology
Answer: A · It slows the heart rate
Why this is the answer
The vagus releases acetylcholine on muscarinic receptors in the SA and AV nodes, slowing pacemaker firing and AV conduction.
Answer
A · It slows the heart rate
4.Human physiology
Answer: A · 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
A · Concentric hypertrophy with a thicker wall
5.Microbiology
Answer: D · Neutrophils
Why this is the answer
Neutrophils arrive quickly at sites of infection and engulf bacteria. Dead neutrophils, bacteria and tissue debris form pus.
Answer
D · Neutrophils
6.Microbiology
Answer: C · 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
C · It has no cell wall for penicillin to act on
7.Pharmacology
Answer: D · Dehydration
Why this is the answer
When the body is short of water or sodium, the kidney reabsorbs more lithium, so its level rises. Thiazides, ACE inhibitors and NSAIDs can have the same effect.
Answer
D · Dehydration
8.Pharmacology
Answer: D · Monitor closely, have naloxone ready, inform the prescriber and report
Why this is the answer
The patient may become sedated and develop respiratory depression later, so close monitoring and a ready antidote are needed. The prescriber is informed, the patient is told honestly, and the error is reported to improve the system.
Answer
D · Monitor closely, have naloxone ready, inform the prescriber and report
9.Arterial blood gas interpretation
Answer: PaCO₂ = 32 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 = 1.5·16 + 8 = 32 mmHg
range: 30–34 mmHg
Answer
PaCO₂ = 32 mmHg
10.Arterial blood gas interpretation
Answer: PaCO₂ = 47.6 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
0.7·38 + 21 = 47.6 mmHg
Answer
PaCO₂ = 47.6 mmHg
The rise has a limit: PaCO₂ rarely goes above about 55 mmHg from compensation alone.
11.ECG rhythm strip interpretation
Answer: QTc = 0.403 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 = 8.5 × 0.04 = 0.34 s
RR = 15 × 0.04 = 0.60 s
Fridericia’s formula
QTc = 0.34/∛0.60 = 0.34/0.8434 = 0.403 s
Answer
QTc = 0.403 s
Within the usual limits (about 0.45 s for men, 0.46 s for women). (403 ms)
12.ECG rhythm strip interpretation
Answer: second-degree AV block type I (Wenckebach)
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: 5 QRS complexes in 6 s, × 10 = 50 per minute.
2. Rhythm: regularly irregular, the QRS complexes come in groups separated by a pause (group beating).
3. P waves: upright and regular (P–P 0.88 s, 68 per minute); one P in every 4 is not followed by a QRS.
4. PR interval: lengthens 0.18 → 0.30 → 0.38 s, then a P wave is not conducted.
5. QRS: 0.08 s, narrow (below 0.12 s).
Interpretation
second-degree AV block type I (Wenckebach)
Answer
second-degree AV block type I (Wenckebach)
13.Evidence-based practice statistics
Answer: IR = 33 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 = 290·1.5 + 300·3 = 1335
Incidence rate
IR = 44/1335·1000 = 33 per 1000 person-years
Answer
IR = 33 per 1000 person-years
14.Evidence-based practice statistics
Answer: NNH = 12
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.