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 structure connects the middle ear to the nasopharynx?
AAuditory (Eustachian) tube
BNasolacrimal duct
CCochlear duct
DExternal acoustic meatus
2.Human anatomyHard
When a patient stands on the right leg only, the left side of the pelvis drops (positive Trendelenburg sign). Which nerve is most likely damaged?
ARight inferior gluteal nerve
BRight femoral nerve
CLeft superior gluteal nerve
DRight superior gluteal nerve
3.Human physiologyMedium
The hormone secretin is released when acid enters the duodenum. What is its main effect?
AAcid secretion from the stomach
BContraction of the gallbladder
CBicarbonate secretion from the pancreas
DInsulin release from the endocrine pancreas
4.Human physiologyHard
A patient with small-cell lung cancer has low plasma sodium and low plasma osmolality but inappropriately concentrated urine. What is the most likely explanation?
APrimary hyperaldosteronism
BSyndrome of inappropriate ADH secretion
CCranial diabetes insipidus
DAddison's disease with high aldosterone
5.MicrobiologyMedium
MRSA is resistant to methicillin because it has:
AAn efflux pump for every antibiotic
BAn altered penicillin-binding protein (PBP2a)
CNo cell wall at all
DA polysaccharide capsule that blocks the entry of all beta-lactams
6.MicrobiologyHard
A deep, dirty wound becomes swollen and very painful, and gas can be felt under the skin. Which organism is most likely?
AClostridium perfringens
BStreptococcus pneumoniae
CNeisseria gonorrhoeae
DBordetella pertussis
7.PharmacologyMedium
How does metformin mainly lower blood glucose?
AIt blocks glucose absorption in the kidney
BIt reduces glucose production by the liver
CIt stimulates insulin release from the pancreas
DIt replaces missing insulin
8.PharmacologyHard
A patient is switched from oral to intravenous morphine. Why is the IV dose lower than the oral dose?
AThe oral route causes tolerance faster than the IV route
BIV morphine is a stronger chemical form of the drug
COral morphine has a large first-pass effect, so less reaches the blood
DIV morphine is excreted much more slowly by the kidney than oral morphine
9.Arterial blood gas interpretationMedium
Practice values: Na⁺ 137 mmol/L, glucose 15.5 mmol/L and urea 10.6 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 137 mmol/Lglucose 15.5 mmol/Lurea 10.6 mmol/L
Osm = mOsm/kg
10.Arterial blood gas interpretationHard
A practice gas shows a metabolic acidosis: pH 7.19, PaCO₂ 19 mmHg (2.5 kPa), HCO₃⁻ 7 mmol/L. Compare the measured PaCO₂ with Winter’s formula. Click your interpretation.
pH 7.19PaCO₂ 19 mmHg (2.5 kPa)HCO₃⁻ 7 mmol/L
1appropriate compensation
2added respiratory acidosis
3added respiratory alkalosis
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 flutter, 3:1 conduction
2atrial flutter, 2:1 conduction
3atrial flutter, 4:1 conduction
4normal sinus rhythm
12.ECG rhythm strip interpretationHard
This patient has a pacemaker. Interpret the lead II strip. 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
1third-degree (complete) AV block
2second-degree AV block type II
3idioventricular rhythm
4paced rhythm, failure to capture
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a randomised trial, 1000 patients received a new treatment and 440 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: 175/1000control: 110/440
NNT =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). 10 000 people are screened with a test of sensitivity 86% and specificity 91%. The prevalence of the disease is 18%. How many positive results n₊ are expected in total?
N = 10 000p = 18%Se = 86%Sp = 91%
n₊ =
3Solutions
1.Human anatomy
Answer: A · Auditory (Eustachian) tube
Why this is the answer
The auditory tube equalises air pressure on both sides of the eardrum. In children it is shorter and more horizontal, which favours middle ear infections.
Answer
A · Auditory (Eustachian) tube
2.Human anatomy
Answer: D · Right superior gluteal nerve
Why this is the answer
The superior gluteal nerve supplies gluteus medius and minimus. If those of the standing (right) leg are weak, the pelvis sags on the opposite, unsupported side.
Answer
D · Right superior gluteal nerve
3.Human physiology
Answer: C · Bicarbonate secretion from the pancreas
Why this is the answer
Secretin stimulates pancreatic duct cells to secrete bicarbonate-rich fluid, which neutralises acidic chyme so intestinal enzymes can work.
Answer
C · Bicarbonate secretion from the pancreas
4.Human physiology
Answer: B · Syndrome of inappropriate ADH secretion
Why this is the answer
In SIADH excess ADH makes the kidneys retain water, diluting the plasma while urine stays concentrated. Small-cell lung cancer can secrete ADH.
Answer
B · Syndrome of inappropriate ADH secretion
5.Microbiology
Answer: B · An altered penicillin-binding protein (PBP2a)
Why this is the answer
The mecA gene codes for PBP2a, which has low affinity for beta-lactams, so the bacterium can keep building its cell wall despite methicillin.
Answer
B · An altered penicillin-binding protein (PBP2a)
6.Microbiology
Answer: A · Clostridium perfringens
Why this is the answer
Clostridium perfringens causes gas gangrene in poorly oxygenated wounds. Its toxins destroy muscle and its metabolism produces gas, felt as crepitus. Urgent surgery and antibiotics are needed.
Answer
A · Clostridium perfringens
7.Pharmacology
Answer: B · It reduces glucose production by the liver
Why this is the answer
Metformin mainly reduces hepatic gluconeogenesis and improves insulin sensitivity. Because it does not stimulate insulin release, hypoglycaemia is rare when used alone.
Answer
B · It reduces glucose production by the liver
8.Pharmacology
Answer: C · Oral morphine has a large first-pass effect, so less reaches the blood
Why this is the answer
Much of an oral morphine dose is metabolised by the liver before reaching the systemic circulation. Equianalgesic tables are used, so the IV dose is a fraction of the oral dose.
Answer
C · Oral morphine has a large first-pass effect, so less reaches the blood
9.Arterial blood gas interpretation
Answer: Osm = 300 mOsm/kg
Derived values: calculated serum osmolality is 2·Na⁺ + glucose + urea with all values in mmol/L (2·Na⁺ + glucose/18 + BUN/2.8 with glucose and BUN in mg/dL); normal is about 275–295 mOsm/kg. The P/F ratio is PaO₂ divided by FiO₂ as a fraction. The alveolar oxygen pressure is PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 mmHg at sea level, and the A–a gradient is PAO₂ − PaO₂.
Calculated osmolality
Osm = 2·137 + 15.5 + 10.6 = 300.1 ≈ 300 mOsm/kg
Answer
Osm = 300 mOsm/kg
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
1.5·7 + 8 = 18.5 → 16.5–20.5 mmHg
Compare
measured: 19 mmHg → appropriate compensation
Answer
appropriate compensation
11.ECG rhythm strip interpretation
Answer: atrial flutter, 3:1 conduction
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.
Five steps
1. Rate: R–R = 15 small boxes, 1500 ÷ 15 ≈ 100 per minute.
2. Rhythm: regular, every R–R is 0.60 s.
3. P waves: none; regular sawtooth flutter waves every 0.20 s (300 per minute), 3 for each QRS.
4. PR interval: not measurable.
5. QRS: 0.08 s, narrow (below 0.12 s).
Interpretation
atrial flutter, 3:1 conduction
Answer
atrial flutter, 3:1 conduction
12.ECG rhythm strip interpretation
Answer: paced rhythm, failure to capture
Paced rhythms: a pacemaker impulse shows as a narrow vertical pacing spike. After a ventricular spike the QRS is wide (0.12 s or more), because the impulse spreads from the lead through the muscle instead of the conduction system; with right-ventricular pacing it is often mostly negative in lead II. A spike before each P wave means atrial pacing, and a dual-chamber (AV) pacemaker shows an atrial spike and a P wave, then, after the programmed AV delay, a ventricular spike and a wide QRS. Capture means every spike is followed by a depolarisation. Failure to capture is a spike without a P wave or QRS after it; failure to sense (spikes that ignore the patient’s own beats) and failure to pace (no spike when one is due) are the other classic malfunctions.
Five steps
1. Rate: 6 QRS complexes in 6 s, × 10 = 60 per minute.
2. Rhythm: the pacing spikes are regular (every 0.92 s), but the R–R is not, because a QRS is missing after some spikes.
3. P waves: none visible.
4. PR interval: not measurable.
5. QRS: 0.14 s and wide after the captured spikes; one spike is not followed by a QRS (no capture).
Interpretation
paced rhythm, failure to capture
Answer
paced rhythm, failure to capture
Failure to capture: the pacemaker fires but the heart does not respond. Textbooks treat it as urgent to report, because the patient may depend on the pacemaker.
13.Evidence-based practice statistics
Answer: NNT = 14
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 = 175/1000 = 0.175
CER = 110/440 = 0.25
Number needed to treat
ARR = 0.25 − 0.175 = 0.075
NNT = 1/0.075 = 13.33 → 14
Answer
NNT = 14
Treating 14 patients prevents about one event.
14.Evidence-based practice statistics
Answer: n₊ = 2286
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.
With and without the disease
10 000·0.18 = 1800
10 000 − 1800 = 8200
All positive results
TP = 1800·0.86 = 1548
FP = 8200·0.09 = 738
n₊ = 1548 + 738 = 2286
Answer
n₊ = 2286
Of the 2286 positives only 1548 have the disease (PPV 67.7%).