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 controls most of the muscles of facial expression?
AOculomotor nerve (III)
BFacial nerve (VII)
CHypoglossal nerve (XII)
DTrigeminal nerve (V)
2.Human anatomyHard
When inserting a chest drain or needle into the pleural space, why is it placed just above the upper border of a rib?
AThe intercostal vein, artery and nerve run along the lower border of each rib
BThe intercostal muscles are absent above each rib
CThe pleura is thickest just above each rib
DThe lung does not reach the area above each rib
3.Human physiologyMedium
How does glucagon raise blood glucose?
AIt stimulates glycogen breakdown and gluconeogenesis in the liver
BIt stimulates glucose uptake by muscle
CIt blocks insulin receptors on fat cells
DIt slows glucose absorption from the gut
4.Human physiologyHard
Hodgkin and Huxley's classic studies of the ionic basis of the action potential used which preparation?
AThe cat spinal cord
BCultured human cortical neurons
CThe frog sciatic nerve–muscle preparation
DThe squid giant axon
5.MicrobiologyMedium
What is the incubation period?
AThe time during which a patient remains infectious to others
BThe time a culture needs in the laboratory
CThe time an antibiotic takes to work
DThe time between infection and the first symptoms
6.MicrobiologyHard
Gram-positive cocci are found. Which test quickly separates staphylococci from streptococci?
AThe catalase test
BThe acid-fast stain
CThe coagulase test
DGrowth on MacConkey agar
7.PharmacologyMedium
Typical antipsychotics such as haloperidol work mainly by blocking:
AAcetylcholinesterase
BGABA-A receptors
CSerotonin reuptake
DDopamine D2 receptors
8.PharmacologyHard
A patient took a doxycycline tablet with a sip of water just before lying down. Now swallowing is very painful. What is the most likely cause?
AAn allergic reaction to the tablet coating
BOral candidiasis from the antibiotic
CPill-induced oesophagitis
DGastric ulcer caused by doxycycline
9.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 26 mmHg (3.5 kPa). Use the rule that in chronic respiratory alkalosis HCO₃⁻ falls by 5 mmol/L for every 10 mmHg that PaCO₂ is below 40, starting from HCO₃⁻ 24 mmol/L, to find the expected HCO₃⁻.
A practice gas shows a metabolic acidosis: pH 7.20, PaCO₂ 37 mmHg (4.9 kPa), HCO₃⁻ 14 mmol/L. Compare the measured PaCO₂ with Winter’s formula. Click your interpretation.
pH 7.20PaCO₂ 37 mmHg (4.9 kPa)HCO₃⁻ 14 mmol/L
1appropriate compensation
2added respiratory acidosis
3added respiratory alkalosis
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 Bazett’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)
2third-degree (complete) AV block
3first-degree AV block
4atrial fibrillation
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). A test has a sensitivity of 65% and a specificity of 85%. Find the negative likelihood ratio LR−. Round to two decimal places.
Se = 65%Sp = 85%
LR− =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A patient has a pre-test probability of 15% for a disease. A test with sensitivity 80% and specificity 66% comes back positive. Find the post-test probability p′ of the disease as a percentage. Round to one decimal place.
pre-test 15%Se = 80%Sp = 66%positive
p′ = %
3Solutions
1.Human anatomy
Answer: B · Facial nerve (VII)
Why this is the answer
The facial nerve (cranial nerve VII) supplies the muscles of facial expression. Damage causes facial palsy, as in Bell's palsy.
Answer
B · Facial nerve (VII)
2.Human anatomy
Answer: A · The intercostal vein, artery and nerve run along the lower border of each rib
Why this is the answer
The intercostal neurovascular bundle (vein, artery, nerve from above down) lies in the costal groove under each rib. Going just above the rib below avoids it.
Answer
A · The intercostal vein, artery and nerve run along the lower border of each rib
3.Human physiology
Answer: A · It stimulates glycogen breakdown and gluconeogenesis in the liver
Why this is the answer
Glucagon from alpha cells acts on the liver to break down glycogen and make new glucose from non-carbohydrate sources.
Answer
A · It stimulates glycogen breakdown and gluconeogenesis in the liver
4.Human physiology
Answer: D · The squid giant axon
Why this is the answer
The squid giant axon is up to about 1 mm wide, so voltage-clamp electrodes could be inserted. This showed that sodium inflow and potassium outflow produce the action potential.
Answer
D · The squid giant axon
5.Microbiology
Answer: D · The time between infection and the first symptoms
Why this is the answer
The incubation period runs from exposure to the onset of symptoms. People can sometimes already transmit the infection during it.
Answer
D · The time between infection and the first symptoms
6.Microbiology
Answer: A · The catalase test
Why this is the answer
Staphylococci produce catalase, which breaks hydrogen peroxide into water and oxygen bubbles; streptococci do not. The coagulase test is then used to identify S. aureus among staphylococci.
Answer
A · The catalase test
7.Pharmacology
Answer: D · Dopamine D2 receptors
Why this is the answer
Haloperidol blocks D2 receptors in the mesolimbic pathway, reducing positive symptoms. Blocking D2 in other pathways causes extrapyramidal effects and raised prolactin.
Answer
D · Dopamine D2 receptors
8.Pharmacology
Answer: C · Pill-induced oesophagitis
Why this is the answer
Doxycycline can cause oesophageal ulceration if it stays in the oesophagus. It is taken with a full glass of water while upright, and the patient stays upright for at least 30 minutes.
Answer
C · Pill-induced oesophagitis
9.Arterial blood gas interpretation
Answer: HCO₃⁻ = 17 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.
Change in PaCO₂
|26 − 40| = 14 mmHg → 1.4 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 − 5·1.4 = 17 mmol/L
Answer
HCO₃⁻ = 17 mmol/L
10.Arterial blood gas interpretation
Answer: added respiratory acidosis
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
1.5·14 + 8 = 29 → 27–31 mmHg
Compare
measured: 37 mmHg → added respiratory acidosis
Answer
added respiratory acidosis
11.ECG rhythm strip interpretation
Answer: QTc = 0.385 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 = 10 × 0.04 = 0.40 s
RR = 27 × 0.04 = 1.08 s
Bazett’s formula
QTc = 0.40/√1.08 = 0.40/1.039 = 0.385 s
Answer
QTc = 0.385 s
Within the usual limits (about 0.45 s for men, 0.46 s for women). (385 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: 4 QRS complexes in 6 s, × 10 = 40 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 0.92 s, 65 per minute); one P in every 3 is not followed by a QRS.
4. PR interval: lengthens 0.16 → 0.28 s, then a P wave is not conducted.
5. QRS: 0.06 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− = 0.41
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.
Negative likelihood ratio
LR− = (1 − 0.65)/0.85 = 0.4118 ≈ 0.41
Answer
LR− = 0.41
14.Evidence-based practice statistics
Answer: p′ = 29.3 %
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.