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
In which structure do sperm mature and get stored after leaving the testis?
AProstate gland
BEpididymis
CSeminal vesicle
DSeminiferous tubules
2.Human anatomyHard
In the femoral triangle, what is the order of the main structures from lateral to medial?
AFemoral nerve, femoral vein, femoral artery
BFemoral nerve, femoral artery, femoral vein
CFemoral artery, femoral nerve, femoral vein
DFemoral vein, femoral artery, femoral nerve
3.Human physiologyMedium
Which hormone makes the gallbladder contract after a fatty meal?
AGastrin
BInsulin
CCholecystokinin (CCK)
DSecretin
4.Human physiologyHard
Why does ketoacidosis typically develop in untreated type 1 diabetes but rarely in type 2?
AType 1 patients have more glucagon receptors
BType 2 patients cannot make ketones at all
CEven low insulin levels suppress lipolysis and ketogenesis
DKetones are only made in children
5.MicrobiologyMedium
Which organism causes tuberculosis, and what makes it hard to stain with the Gram method?
ALegionella pneumophila, which lives inside amoebae
BMycoplasma pneumoniae, which has no cell wall to take up the stain
CMycobacterium tuberculosis, which has a waxy mycolic acid wall
DStreptococcus pneumoniae, which has a capsule
6.MicrobiologyHard
Why can a person get influenza many times but measles usually only once?
AMeasles is caused by a bacterium
BMeasles virus has no surface antigens
CInfluenza attacks and destroys the memory cells formed after infection
DInfluenza viruses change their surface antigens often
7.PharmacologyMedium
An adult patient is due to receive digoxin. The apical pulse, counted for a full minute, is 52 beats per minute. What should the nurse do?
AGive the dose with food to slow absorption
BGive the dose and recheck the pulse in one hour
CGive half the dose and document the pulse
DWithhold the dose and notify the prescriber
8.PharmacologyHard
Why can an ACE inhibitor cause acute kidney injury in a patient with bilateral renal artery stenosis?
AGFR depends on angiotensin II constricting the efferent arteriole
BThe drug crystallises in the renal tubules
CACE inhibitors constrict the afferent arteriole
DACE inhibitors are directly toxic to the proximal tubule cells in the kidney
9.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 75 mmHg (10.0 kPa). Use the rule that in chronic respiratory acidosis HCO₃⁻ rises by 3.5 mmol/L for every 10 mmHg that PaCO₂ is above 40, starting from HCO₃⁻ 24 mmol/L, to find the expected HCO₃⁻.
A practice gas shows a metabolic acidosis: pH 7.16, PaCO₂ 38 mmHg (5.1 kPa), HCO₃⁻ 13 mmol/L. Compare the measured PaCO₂ with Winter’s formula. Click your interpretation.
pH 7.16PaCO₂ 38 mmHg (5.1 kPa)HCO₃⁻ 13 mmol/L
1appropriate compensation
2added respiratory acidosis
3added respiratory alkalosis
11.ECG rhythm strip interpretationMedium
This lead II strip shows atrial flutter. Measure the flutter-wave interval and find the atrial rate in beats per minute.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
atrial rate = bpm
12.ECG rhythm strip interpretationHard
Identify the rhythm on this 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
1asystole
2third-degree (complete) AV block
3sinus bradycardia
4junctional rhythm
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). A case-control study compares 260 patients with a disease (cases) and 310 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 = 161b = 109c = 99d = 201
OR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A patient has a pre-test probability of 25% for a disease. A test with sensitivity 95% and specificity 75% comes back negative. Find the post-test probability p′ of the disease as a percentage. Round to one decimal place.
pre-test 25%Se = 95%Sp = 75%negative
p′ = %
3Solutions
1.Human anatomy
Answer: B · Epididymis
Why this is the answer
Sperm are made in the seminiferous tubules and then mature and are stored in the epididymis before passing into the ductus deferens.
Answer
B · Epididymis
2.Human anatomy
Answer: B · Femoral nerve, femoral artery, femoral vein
Why this is the answer
From lateral to medial: nerve, artery, vein, then empty space and lymphatics (NAVEL). This is why the femoral artery pulse lies between the nerve and vein.
Answer
B · Femoral nerve, femoral artery, femoral vein
3.Human physiology
Answer: C · Cholecystokinin (CCK)
Why this is the answer
Fat and protein in the duodenum release CCK, which contracts the gallbladder and stimulates pancreatic enzyme secretion.
Answer
C · Cholecystokinin (CCK)
4.Human physiology
Answer: C · Even low insulin levels suppress lipolysis and ketogenesis
Why this is the answer
Insulin strongly inhibits lipolysis and hepatic ketone production. In type 1 there is almost no insulin, so fatty acids flood the liver and ketones build up.
Answer
C · Even low insulin levels suppress lipolysis and ketogenesis
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: D · Influenza viruses change their surface antigens often
Why this is the answer
Influenza haemagglutinin and neuraminidase change through drift (small mutations) and shift (reassortment), so old antibodies may not recognise new strains. Measles is antigenically stable.
Answer
D · Influenza viruses change their surface antigens often
7.Pharmacology
Answer: D · Withhold the dose and notify the prescriber
Why this is the answer
Digoxin is usually withheld in adults if the apical pulse is below 60 per minute, and the prescriber is informed. Bradycardia may be an early sign of toxicity.
Answer
D · Withhold the dose and notify the prescriber
8.Pharmacology
Answer: A · GFR depends on angiotensin II constricting the efferent arteriole
Why this is the answer
When renal perfusion is low, angiotensin II constricts the efferent arteriole to maintain glomerular pressure. Blocking it lets the pressure and GFR fall, so kidney function can drop sharply.
Answer
A · GFR depends on angiotensin II constricting the efferent arteriole
9.Arterial blood gas interpretation
Answer: HCO₃⁻ = 36.25 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
|75 − 40| = 35 mmHg → 3.5 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 + 3.5·3.5 = 36.25 mmol/L
Answer
HCO₃⁻ = 36.25 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
1.5·13 + 8 = 27.5 → 25.5–29.5 mmHg
Compare
measured: 38 mmHg → added respiratory acidosis
Answer
added respiratory acidosis
11.ECG rhythm strip interpretation
Answer: atrial rate = 250 bpm
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.
Flutter waves
F–F = 6 small boxes = 0.24 s → 1500 ÷ 6 = 250 per minute
QRS complexes
R–R = 24 small boxes = 0.96 s → 1500 ÷ 24 = 63 per minute
conduction ratio = 250 ÷ 63 = 4:1
Answer
atrial rate = 250 bpm
12.ECG rhythm strip interpretation
Answer: asystole
Ventricular rhythms: ventricular tachycardia is a run of three or more wide (0.12 s or more) complexes in a row at more than 100/min, usually 150–250/min, regular, without related P waves. Ventricular fibrillation is a chaotic, irregular waveform with no identifiable P waves, QRS complexes or T waves and no countable rate. Asystole is an almost flat line with no ventricular activity; a flat line is always confirmed by checking the patient, the leads and the gain and a second lead. In resuscitation guidelines VF and pulseless VT are the shockable rhythms and asystole is not.
Five steps
1. Rate: no countable ventricular rate.
2. Rhythm: none; an almost flat line.
3. P waves: none identifiable.
4. PR interval: not measurable.
5. QRS: no QRS complexes.
Interpretation
asystole
Answer
asystole
A flat line is always checked: look at the patient, the leads and the gain, and view a second lead, because a loose electrode or fine VF can look like asystole.
13.Evidence-based practice statistics
Answer: OR = 3
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: 161/99 = 1.626
controls: 109/201 = 0.5423
Odds ratio
OR = (161·201)/(109·99) = 2.999 ≈ 3
Answer
OR = 3
OR above 1: exposure is more common among the cases.
14.Evidence-based practice statistics
Answer: p′ = 2.2 %
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.