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 is the main bundle of nerve fibres connecting the left and right cerebral hemispheres?
AFornix
BInternal capsule
CCerebral peduncle
DCorpus callosum
2.Human anatomyHard
After an injury at the medial epicondyle of the elbow, a patient has numbness of the little finger and weakness of finger abduction. Which nerve is affected?
ARadial nerve
BUlnar nerve
CMusculocutaneous nerve
DMedian nerve
3.Human physiologyMedium
Vital capacity is the sum of which lung volumes?
AInspiratory reserve volume and residual volume
BTidal volume, inspiratory reserve volume and expiratory reserve volume
CResidual volume and expiratory reserve volume
DTidal volume and residual volume
4.Human physiologyHard
In Graves' disease, what thyroid function results are expected and why?
ALow T4 and high TSH, because antibodies destroy the gland
BHigh T4 and high TSH, because the anterior pituitary gland is overactive
CHigh T4 and low TSH, because antibodies stimulate the TSH receptor
DLow T4 and low TSH, because the hypothalamus fails
5.MicrobiologyMedium
Which organism causes tuberculosis, and what makes it hard to stain with the Gram method?
AStreptococcus pneumoniae, which has a capsule
BMycoplasma pneumoniae, which has no cell wall to take up the stain
CMycobacterium tuberculosis, which has a waxy mycolic acid wall
DLegionella pneumophila, which lives inside amoebae
6.MicrobiologyHard
Why is Chlamydia trachomatis usually diagnosed with nucleic acid amplification tests instead of routine culture on agar?
AIt is an obligate intracellular bacterium that cannot grow on ordinary media
BIt is too large to grow on agar plates
CIt is a virus and has no DNA
DIt dies immediately in the presence of oxygen
7.PharmacologyMedium
A patient has severe kidney impairment. A drug is mainly excreted unchanged in the urine. What is the usual adjustment?
AReduce the dose or lengthen the dosing interval
BNo change is needed, because the liver takes over
CIncrease the dose to make up for poor absorption
DGive the drug orally instead of intravenously
8.PharmacologyHard
In aspirin overdose the urine may be made alkaline with sodium bicarbonate. Why does this increase salicylate excretion?
AAlkaline urine increases glomerular filtration
BMore salicylate is ionised in the tubule, so less is reabsorbed
CBicarbonate speeds up the liver metabolism of salicylate to inactive forms
DBicarbonate binds salicylate and makes it inactive
9.Arterial blood gas interpretationMedium
Practice values: Na⁺ 150 mmol/L, glucose 24 mmol/L and urea 18.2 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 150 mmol/Lglucose 24 mmol/Lurea 18.2 mmol/L
Osm = mOsm/kg
10.Arterial blood gas interpretationHard
Practice values: Na⁺ 133, Cl⁻ 103 and HCO₃⁻ 12 mmol/L. Calculate the delta ratio (AG − 12)/(24 − HCO₃⁻), using AG = Na⁺ − (Cl⁻ + HCO₃⁻). Give two decimals.
Na⁺ 133 mmol/LCl⁻ 103 mmol/LHCO₃⁻ 12 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 flutter, 4:1 conduction
2atrial flutter, 2:1 conduction
3atrial flutter, 3:1 conduction
4sinus tachycardia
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
1sinus rhythm with wide QRS
2paced rhythm, failure to capture
3dual-chamber (AV) paced rhythm
4ventricular paced rhythm
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). A case-control study compares 250 patients with a disease (cases) and 200 people without it (controls). The table shows how many in each group were exposed to a risk factor. Find the odds ratio OR for the exposure. Round to two decimal places.
a = 55b = 30c = 195d = 170
OR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A patient has a pre-test probability of 10% for a disease. A test with sensitivity 95% and specificity 63% comes back positive. Find the post-test probability p′ of the disease as a percentage. Round to one decimal place.
pre-test 10%Se = 95%Sp = 63%positive
p′ = %
3Solutions
1.Human anatomy
Answer: D · Corpus callosum
Why this is the answer
The corpus callosum is the largest commissure, linking matching areas of the two hemispheres. The internal capsule carries projection fibres up and down instead.
Answer
D · Corpus callosum
2.Human anatomy
Answer: B · Ulnar nerve
Why this is the answer
The ulnar nerve passes behind the medial epicondyle. It supplies most intrinsic hand muscles, including the interossei, and the skin of the little finger.
Answer
B · Ulnar nerve
3.Human physiology
Answer: B · Tidal volume, inspiratory reserve volume and expiratory reserve volume
Why this is the answer
Vital capacity = IRV + TV + ERV. It excludes the residual volume, which always stays in the lungs.
Answer
B · Tidal volume, inspiratory reserve volume and expiratory reserve volume
4.Human physiology
Answer: C · High T4 and low TSH, because antibodies stimulate the TSH receptor
Why this is the answer
TSH-receptor antibodies stimulate the thyroid independently of TSH. The resulting high T4 and T3 suppress pituitary TSH by negative feedback.
Answer
C · High T4 and low TSH, because antibodies stimulate the TSH receptor
5.Microbiology
Answer: C · Mycobacterium tuberculosis, which has a waxy mycolic acid wall
Why this is the answer
M. tuberculosis has a cell wall rich in mycolic acids, so it resists Gram staining and is identified with acid-fast stains.
Answer
C · Mycobacterium tuberculosis, which has a waxy mycolic acid wall
6.Microbiology
Answer: A · It is an obligate intracellular bacterium that cannot grow on ordinary media
Why this is the answer
Chlamydia replicates only inside host cells, so it cannot be grown on agar. NAATs such as PCR on urine or swabs are sensitive and are the standard diagnostic test.
Answer
A · It is an obligate intracellular bacterium that cannot grow on ordinary media
7.Pharmacology
Answer: A · Reduce the dose or lengthen the dosing interval
Why this is the answer
When renal clearance falls, drugs excreted by the kidney accumulate. The dose is lowered or given less often, guided by kidney function and, where possible, drug levels.
Answer
A · Reduce the dose or lengthen the dosing interval
8.Pharmacology
Answer: B · More salicylate is ionised in the tubule, so less is reabsorbed
Why this is the answer
Salicylate is a weak acid. In alkaline urine more of it is in the ionised form, which is "trapped" in the tubule and excreted. This is called ion trapping.
Answer
B · More salicylate is ionised in the tubule, so less is reabsorbed
9.Arterial blood gas interpretation
Answer: Osm = 342 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 = 2·150 + 24 + 18.2 = 342.2 ≈ 342 mOsm/kg
Answer
Osm = 342 mOsm/kg
10.Arterial blood gas interpretation
Answer: delta ratio = 0.5
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 = 133 − (103 + 12) = 18
Delta ratio
(18 − 12)/(24 − 12) = 6/12 = 0.5
Answer
delta ratio = 0.5
Below 1: an added normal anion gap acidosis.
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: ventricular paced rhythm
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: R–R = 23 small boxes, 1500 ÷ 23 ≈ 65 per minute.
2. Rhythm: regular, every R–R is 0.92 s.
3. P waves: none visible; a sharp pacing spike comes just before every QRS.
4. PR interval: not measurable.
5. QRS: 0.16 s, wide, and every spike is followed by a QRS (capture).
Interpretation
ventricular paced rhythm
Answer
ventricular paced rhythm
The paced QRS is wide by nature; do not call it a bundle branch block or a ventricular rhythm.
13.Evidence-based practice statistics
Answer: OR = 1.6
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.
Odds of exposure
cases: 55/195 = 0.2821
controls: 30/170 = 0.1765
Odds ratio
OR = (55·170)/(30·195) = 1.598 ≈ 1.6
Answer
OR = 1.6
OR above 1: exposure is more common among the cases.
14.Evidence-based practice statistics
Answer: p′ = 22.2 %
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.