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 part of the brain is chiefly responsible for coordination and balance?
AThalamus
BCerebellum
CHypothalamus
DMedulla oblongata
2.Human anatomyHard
In which part of the small intestine are vitamin B12 and bile salts mainly absorbed?
ATerminal ileum
BJejunum
CDuodenum
DCaecum
3.Human physiologyMedium
During which phase of the cardiac cycle are all four heart valves closed while ventricular pressure rises?
ARapid filling
BIsovolumetric contraction
CAtrial systole
DRapid ejection
4.Human physiologyHard
Which combination of signs is expected in a patient in early hypovolaemic shock after major bleeding?
ABradycardia, cool pale skin and high urine output
BTachycardia, cool pale skin and reduced urine output
CTachycardia, warm flushed skin and high urine output
DBradycardia, warm flushed skin and high urine output
5.MicrobiologyMedium
Which type of immunity is given by breast milk antibodies?
AArtificial passive immunity
BArtificial active immunity
CNatural active immunity
DNatural passive immunity
6.MicrobiologyHard
After eating home-canned vegetables, a patient develops double vision, difficulty swallowing and a descending flaccid paralysis. What is the most likely cause?
ABotulinum toxin, which blocks acetylcholine release
BA viral infection of the motor neurons
CEndotoxin from Gram-negative bacteria
DTetanus toxin, which blocks inhibitory neurotransmitters
7.PharmacologyMedium
A partial agonist is given together with a high dose of a full agonist. What happens to the response?
AIt stays exactly the same as with the high dose of full agonist alone
BIt falls, because the partial agonist competes for the receptors
CIt doubles, because both drugs activate the receptor
DIt rises above the maximum of the full agonist
8.PharmacologyHard
A patient taking a GLP-1 receptor agonist such as semaglutide has severe, constant upper abdominal pain spreading to the back, with vomiting. What should the nurse suspect?
AAcute pancreatitis; the drug is stopped and urgent assessment arranged
BExpected nausea; the patient should take the next dose with food
CConstipation; a laxative will solve the problem
DHypoglycaemia; the patient should eat sweets and continue as usual
9.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.24, PaCO₂ 24 mmHg (3.2 kPa) and HCO₃⁻ 10 mmol/L. The primary disorder is metabolic acidosis. How far is it compensated? Click your answer.
pH 7.24PaCO₂ 24 mmHg (3.2 kPa)HCO₃⁻ 10 mmol/L
1uncompensated
2partially compensated
3fully compensated
10.Arterial blood gas interpretationHard
At sea level, a practice gas taken on 30% oxygen shows PaO₂ 84 mmHg (11.2 kPa) and PaCO₂ 44 mmHg (5.9 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.
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
1junctional rhythm
2sinus bradycardia
3normal sinus rhythm
4first-degree AV block
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
1paced rhythm, failure to capture
2dual-chamber (AV) paced rhythm
3third-degree (complete) AV block
4sinus rhythm with wide QRS
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a randomised trial, 920 patients received a new treatment and 730 received the control treatment. The table shows how many in each group had the outcome (an unwanted event such as a stroke). Find the relative risk RR of the event with treatment. Round to two decimal places.
treatment: 175/920control: 277/730
RR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A patient has a pre-test probability of 60% for a disease. A test with sensitivity 93% and specificity 79% comes back negative. Find the post-test probability p′ of the disease as a percentage. Round to one decimal place.
pre-test 60%Se = 93%Sp = 79%negative
p′ = %
3Solutions
1.Human anatomy
Answer: B · Cerebellum
Why this is the answer
The cerebellum fine-tunes movement, posture and balance. Damage causes ataxia and an unsteady gait.
Answer
B · Cerebellum
2.Human anatomy
Answer: A · Terminal ileum
Why this is the answer
The terminal ileum absorbs vitamin B12 (bound to intrinsic factor) and reabsorbs bile salts. The duodenum mainly absorbs iron.
Answer
A · Terminal ileum
3.Human physiology
Answer: B · Isovolumetric contraction
Why this is the answer
After the AV valves close, the ventricles contract against closed semilunar valves. Pressure rises with no volume change until the aortic and pulmonary valves open.
Answer
B · Isovolumetric contraction
4.Human physiology
Answer: B · Tachycardia, cool pale skin and reduced urine output
Why this is the answer
Baroreceptor-driven sympathetic activation raises heart rate and constricts skin vessels. Reduced renal perfusion, ADH and aldosterone lower urine output.
Answer
B · Tachycardia, cool pale skin and reduced urine output
5.Microbiology
Answer: D · Natural passive immunity
Why this is the answer
Antibodies (mainly IgA) passed in breast milk give passive immunity acquired naturally. It is short-lived because no memory cells form.
Answer
D · Natural passive immunity
6.Microbiology
Answer: A · Botulinum toxin, which blocks acetylcholine release
Why this is the answer
Clostridium botulinum can grow in poorly canned, low-acid foods and produce botulinum toxin. The toxin blocks acetylcholine release at nerve endings, causing descending flaccid paralysis.
Answer
A · Botulinum toxin, which blocks acetylcholine release
7.Pharmacology
Answer: B · It falls, because the partial agonist competes for the receptors
Why this is the answer
A partial agonist has lower efficacy. When it displaces a full agonist from receptors, the total response drops. Buprenorphine can precipitate withdrawal in people using full opioid agonists for this reason.
Answer
B · It falls, because the partial agonist competes for the receptors
8.Pharmacology
Answer: A · Acute pancreatitis; the drug is stopped and urgent assessment arranged
Why this is the answer
Mild nausea is common with GLP-1 agonists, but severe persistent abdominal pain radiating to the back suggests pancreatitis. The drug is stopped and the patient needs urgent assessment, including serum lipase or amylase.
Answer
A · Acute pancreatitis; the drug is stopped and urgent assessment arranged
9.Arterial blood gas interpretation
Answer: partially compensated
Reading a blood gas (ROME): first the pH: below 7.35 is acidaemia, above 7.45 alkalaemia. Then find the value that explains it. PaCO₂ (normal 35–45 mmHg, 4.7–6.0 kPa) is the respiratory part and moves Opposite to the pH in a respiratory disorder; HCO₃⁻ (normal 22–26 mmol/L) is the metabolic part and moves in the same direction as the pH (Respiratory Opposite, Metabolic Equal). The other value shows the compensation: normal means uncompensated; abnormal in the direction that pulls the pH back while the pH is still abnormal means partially compensated; a normal pH with both values abnormal means fully compensated. Ranges differ slightly between laboratories; these are practice values.
ROME, value by value
pH 7.24: acidaemia (below 7.35)
PaCO₂ 24 mmHg: low (35–45)
HCO₃⁻ 10 mmol/L: low (22–26)
Compensation
partially compensated
Answer
partially compensated
10.Arterial blood gas interpretation
Answer: A–a gradient = 74.9 mmHg
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.
Alveolar oxygen
PAO₂ = 0.3·713 − 44/0.8 = 213.9 − 55 = 158.9 mmHg
Gradient
A–a = 158.9 − 84 = 74.9 mmHg
Answer
A–a gradient = 74.9 mmHg
11.ECG rhythm strip interpretation
Answer: first-degree AV block
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 = 17 small boxes, 1500 ÷ 17 ≈ 88 per minute.
2. Rhythm: regular, every R–R is 0.68 s.
3. P waves: upright, one before every QRS, all alike.
4. PR interval: 0.26 s, constant, longer than 0.20 s.
5. QRS: 0.10 s, narrow (below 0.12 s).
Interpretation
first-degree AV block
Answer
first-degree AV block
12.ECG rhythm strip interpretation
Answer: dual-chamber (AV) 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 = 25 small boxes, 1500 ÷ 25 ≈ 60 per minute.
2. Rhythm: regular, every R–R is 1.00 s.
3. P waves: each P wave follows its own pacing spike (atrial pacing).
4. AV interval: the second spike comes 0.18 s after the first one (the programmed AV delay).
5. QRS: 0.16 s, wide, right after the second spike (ventricular pacing).
Interpretation
dual-chamber (AV) paced rhythm
Answer
dual-chamber (AV) 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: RR = 0.5
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 = 175/920 = 0.1902
CER = 277/730 = 0.3795
Relative risk
RR = 0.1902/0.3795 = 0.5013 ≈ 0.5
Answer
RR = 0.5
14.Evidence-based practice statistics
Answer: p′ = 11.7 %
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.