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)
BTrochlear nerve (IV)
COculomotor nerve (III)
DAbducens nerve (VI)
2.Human anatomyHard
Early appendicitis is often felt as vague pain around the umbilicus. Why?
AInflammation spreads first to the abdominal wall at the navel region
BIts visceral nerves enter the cord at T10, like the umbilical skin
CThe phrenic nerve carries the pain to the umbilicus
DThe appendix lies directly behind the umbilicus
3.Human physiologyMedium
What is the main stimulus for breathing in a healthy person?
AA rise in blood glucose
BA fall in arterial O₂
CA rise in arterial CO₂
DA fall in blood pH caused by lactate only
4.Human physiologyHard
Why is severe hyperkalaemia dangerous for the heart?
AIt hyperpolarises cardiac cells, so they can never fire
BIt speeds up the SA node until tachycardia causes fainting
CIt blocks the binding of calcium to troponin
DIt depolarises cardiac cells, inactivating sodium channels and causing arrhythmias
5.MicrobiologyMedium
Pneumocystis jirovecii pneumonia in a person with advanced HIV is an example of:
AA food-borne infection
BA zoonotic infection
CAn opportunistic infection
DA vector-borne infection
6.MicrobiologyHard
A patient taking high-dose immunosuppressive drugs after a transplant is due for vaccinations. Which vaccine type is generally avoided?
ALive attenuated vaccines
BInactivated whole-virus vaccines
CSubunit vaccines
DToxoid vaccines
7.PharmacologyMedium
In which patient is bupropion contraindicated?
AA patient who wants to stop smoking
BA patient with high blood cholesterol and a family history of it
CA patient with seasonal allergies
DA patient with a seizure disorder or an eating disorder
8.PharmacologyHard
Why may omeprazole be avoided in a patient taking clopidogrel?
AIt increases clopidogrel absorption from the gut and causes bleeding
BIt binds clopidogrel in the stomach and makes it insoluble
CIt inhibits CYP2C19, so less clopidogrel is converted to its active form
DIt speeds up kidney excretion of clopidogrel
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 acidosis: pH 7.31, PaCO₂ 27 mmHg (3.6 kPa), HCO₃⁻ 13 mmol/L. Compare the measured PaCO₂ with Winter’s formula. Click your interpretation.
pH 7.31PaCO₂ 27 mmHg (3.6 kPa)HCO₃⁻ 13 mmol/L
1appropriate compensation
2added respiratory acidosis
3added respiratory alkalosis
11.ECG rhythm strip interpretationMedium
This lead II strip shows sinus rhythm with one pause. Measure the pause, the R–R interval across it, in seconds.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
pause = 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
1first-degree AV block
2second-degree AV block type I (Wenckebach)
3third-degree (complete) AV block
4second-degree AV block type II
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). A new test was compared with the reference standard in a study (see the 2 × 2 table). Find the positive likelihood ratio LR+. Round to two decimal places.
a = 186b = 703c = 104d = 1147
LR+ =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A screening test has a sensitivity of 93% and a specificity of 90%. It is used in a population where the prevalence is 20%. Find the positive predictive value PPV as a percentage. Round to one decimal place.
Se = 93%Sp = 90%p = 20%
PPV = %
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: B · Its visceral nerves enter the cord at T10, like the umbilical skin
Why this is the answer
Visceral pain from midgut structures is referred to the T10 dermatome around the umbilicus. When the parietal peritoneum becomes inflamed, pain localises to the right iliac fossa.
Answer
B · Its visceral nerves enter the cord at T10, like the umbilical skin
3.Human physiology
Answer: C · A rise in arterial CO₂
Why this is the answer
Central chemoreceptors respond to CO₂ (through the pH of the cerebrospinal fluid). Low O₂ adds strongly to the drive only when PaO₂ falls well below normal, for example at high altitude.
Answer
C · A rise in arterial CO₂
4.Human physiology
Answer: D · 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
D · It depolarises cardiac cells, inactivating sodium channels and causing arrhythmias
5.Microbiology
Answer: C · An opportunistic infection
Why this is the answer
Opportunistic infections take advantage of a weakened immune system. In HIV they become common when the CD4 count falls to low levels.
Answer
C · An opportunistic infection
6.Microbiology
Answer: A · Live attenuated vaccines
Why this is the answer
Live vaccines contain weakened organisms that can cause disease when the immune system is severely suppressed. Inactivated, toxoid and subunit vaccines are safe, although the response may be weaker.
Answer
A · Live attenuated vaccines
7.Pharmacology
Answer: D · A patient with a seizure disorder or an eating disorder
Why this is the answer
Bupropion increases the risk of seizures, especially with bulimia or anorexia, where electrolyte disturbances are common. It is used for depression and as a smoking cessation aid.
Answer
D · A patient with a seizure disorder or an eating disorder
8.Pharmacology
Answer: C · It inhibits CYP2C19, so less clopidogrel is converted to its active form
Why this is the answer
Clopidogrel must be activated by liver enzymes, mainly CYP2C19. Omeprazole inhibits this enzyme and can reduce its antiplatelet effect, so another acid suppressant may be chosen.
Answer
C · It inhibits CYP2C19, so less clopidogrel is converted to its active form
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: appropriate compensation
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
1.5·13 + 8 = 27.5 → 25.5–29.5 mmHg
Compare
measured: 27 mmHg → appropriate compensation
Answer
appropriate compensation
11.ECG rhythm strip interpretation
Answer: pause = 1.96 s
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).
Measure
49 small boxes from R to R across the pause
Convert
pause = 49 × 0.04 s = 1.96 s
Compare with P–P
1.96 ÷ 0.80 ≈ 2.4 P–P cycles, not a whole number
Answer
pause = 1.96 s
No P wave appears in the pause, and the pause is not a whole number of P–P cycles: sinus arrest. A pause of exactly two or three cycles would point to sinus exit block.
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 3 is not followed by a QRS.
4. PR interval: lengthens 0.16 → 0.30 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: LR+ = 1.69
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.
Sensitivity and specificity
Se = 186/290 = 0.6414
Sp = 1147/1850 = 0.62
Positive likelihood ratio
LR+ = 0.6414/(1 − 0.62) = 1.688 ≈ 1.69
Answer
LR+ = 1.69
14.Evidence-based practice statistics
Answer: PPV = 69.9 %
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.
True and false positives (per person)
Se·p = 0.93·0.2 = 0.186
(1 − Sp)·(1 − p) = 0.1·0.8 = 0.08
Positive predictive value
PPV = 0.186/(0.186 + 0.08) = 0.6992 ≈ 69.9%
Answer
PPV = 69.9%
The lower the prevalence, the lower the PPV of the same test.